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# CALC ANALYT GEOM I MATH 124

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This 179 page Class Notes was uploaded by Addison Beer on Wednesday September 9, 2015. The Class Notes belongs to MATH 124 at University of Washington taught by Staff in Fall. Since its upload, it has received 10 views. For similar materials see /class/192090/math-124-university-of-washington in Mathematics (M) at University of Washington.

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Announcements Today Chapter 20 Recall that the next homework from Ch 1821 is due on Friday No office hours for Dr Nichifor today Instead she Will hold extended hours on Wed and Thu 0 Wednesday 130330 in Padelford C326 0 Thu 2304 in MSC 0 Or by appointment Other sources of help 0 Michael s office hrs Tue 1011 or Thu 23 in MSC o Matt s of ce hrs Wed 10301130 in Padelford C113 or Thu 10301130 in MSC o MSC tutors instructors and TAs o CLUE Note The homework from Ch 20 is very short If you re done early it will help to start looking at the Ch 21 before Wednesday s lecture Chagter 20 Exgonenual Modelmg Smce an Expunemlal numb m standard farm 5 quhe farm m ADM m urda tu nd an andh Examglq Emmmmundu39mnd Lhegrncadure 1950 a mm the Ewmmal lnman PtPgb Wm model the gan data b m the actualpapulanan m 1950 mm wa about 150mm 5 Whatpapulanan due m deEIpVEdIEIfDVIhE amen USPapulanaW 2 Gwen data Pmle smce 197m xSLZZEI yrs a er 195m and F4IZSEI 203 P 21quot Plug thetwu datapumts mm Lhamudel m gettvm equanuns 25 r away m duthsherexs m 64de Zmequanun by the 5m cu ahmmate pquot 250 7 an m u b 250 in 7 b 203 Salve fur b by taking the appmpnate mut 25 In mu 7 b 203 b Ul 4 7l48 v y z Keep zsnlzny decimalszs ynu can at Else ymr answsi mayhehzdly nm a 7 2 m Replaceb m the 1 sq 203 71304110457148 5 PB 7 wmmm N 54 835 Sn Lhamudelmg funmunxs Ft164 8361U1U467148 b FIF164 m mum Made uveresnmated by abuut 14 m mlllmn 9 9 YearZEIEI7xs 57 years a eryezr 195m 5 F57164 8351n1n457148 zaa 41 mum a n COMPOUND INTEREST this is an application of exponential functions that we encounter in daily life so it helps to understand how it works Example You invest 1000 in a bank account o ering an annual interest percentage of 7 compounded annually Assuming no further deposits how much money will you have in that account after t years LetPlj be your balance after t years We are asked to figure out a formula for this function PM o P0 1 000 o P1prev balance interest applied to previous balance1 000 0 07 0001 070 o P2P10 07P1107000710701070 74911449 o Etc What s the pattern o P1P0007P0 1P0007P0107P0 o PdP1007P1107 P1 107 107P0 107iP0 o PdP2007P2107P2107 1072P0 107iP0 o Etc The pattern is l tl07t P0 where P0 is the initial deposit principal 107 comes from 1 plus the annual interest rate in decimal form and t is the number of years More generally if a bank offers an annual rate r compounded n times a year they actually mean rate rn applied n times a year EX 6 annual rate compounded monthly actually means 61205 interest applied to your balance each month which ends up being more that 6 a year The formula to use in such cases is a slight modi cation of the above me H in nylu 1 w Ex You deposit 5000 in a bank account o ering 10 annual interest rate P05000 r0 I How much money is in your account after 20 years the interest is compounded a Annually n1 P205000112 55336375 3 Quarterly n4 P2050001014 2 500010258 3604784 c Monthly n12 P20500010112 2 3664037 d Daily n365 P2050001013653652 3693516 we get Can we keep lmpmvmg the mum fursver m is there a max 3537 m mtelylzrge we dun39t get an m mte amuunt ufmuney hand we get c1 J4 e 7 u WheransthepnnmpalnsLheannualratem denmal farm txsthenumberufyezrs a ens aspenal mathemancal cunstant Type m mwlx calculate Whath dayau gem 22 712221222 xn muly many decxmal dlgts gels Inga andlargu Mme Salads m uxtbank xfyau39re mum pen Hrnquot mm eppxaeeheev a sa me mm mm Pnlrnquotbhaves asymptahca y m m mum campalmded sawmmzyo FZUSUUUEquot W36945 22 communwa eampmmedo p1pnequotquot pn1 1n517 Cumpznngtuthediscretefm mulawnhn FlFn1r wegetthat 1F1 1n517 su FU1U517 xe abuut ID 517 mu uwl 1 annually Altamanvemethud pine m Lhe ehange m 1 yezns 2 010517 5 abuut m 517 5207 Math 124 Lecture notes Announcements amp Reminders Work on Week 6 Hwk pbls Bring questions to quiz section Thursday Final exam Saturday June 239 130420 in KNE 210 More info link on website Deadline for petition May 11m Section 36 Implicit differentiation Preparation 1 Supposey x3 5x Then 3139 3x2 5 Note Here by y39 I mean the derivative of ywith respect to x ie Lg The latter notation would be better than the prime notation because it makes it clearer which letter is the variable However it is more cumbersome Consider the derivative of yquot7 o In terms ofx this is 31739x3 5x739 7x3 5x63x2 5 o In terms of y this is 0739 7y6y39 or in the other notation 017 7y6d y dx In both we use Chain Rule 2 Now suppose you don t know the precise expression for y in terms of x but you do know that y depends on x via some relationship You do 0 02 l Answer 2yy Can t write it in terms ofx because we don t know what either y or y are in terms of x 0 03 l Answer 3y2y39 0 l Answer Zly 0 83 Answer eyy Note in each of these since we re taking the derivative with respect to x not to y we must use Chain Rule hence multiply by y Here s a tougher one o xey39 Since we have a product of two functions of x namely x and eV we must use the Product Rule xey39 x ey xeyl 18y xeyy ey xeyy When is all this useful Suppose you have a curve given by some equation in x and y and you cannot solve for y in terms ofx or is difficult to do so but you d still like to know the rate ofchange derivative of that curve at some point What to do METHOD of Implicit Differentiation We cannot write y as a function ofx explicitly but we can think of it as an imglicit function of x via the given equation Further if two things are equal then so are their derivatives So STEP 1 Take the derivative of both sides of the equation with respect to x treating any part containing y as practiced above using Chain Rule You get an equation involving x y and 1 STEP 2 Solve the resulting equation for y in terms ofx and y If y is actually a function ofx and you can actually determine the formula for y in terms of x replace it in the result so you d get y in terms ofx only Otherwise you have to be content with an expression depending both on the variable x and on the implicit function y STEP 3 If you need to evaluate the derivative at an actual point ab then replace xa yb into the formula you got for y then compute EXAMPLES Ex 1 Let s compute the derivative y for the ellipse in the week 5 supplementary problem namely 2 x 2 1 4 y Here the variable is x and we consider y as an implicit function of x Step 1 Apply derivatives to both sides x2 2 T y 11 Then evaluate them 1 I Z 2x Zyy 0 Step 2 Solve this equation for y 2 We can express y in terms ofx from the original equation y i 1 x7 so replacing we get x Vi41 41 Example 2 Find the eq ofthe tangent line to the curve sinx y yzcasx at the point rt 0 if curious you can see a graph of a portion of this curve in the hwk problems from section 36 Here there is no way to solve for y in terms ofx or even vice versa so we have no choice but to use implicit differentiation to compute y Apply derivative with respect to x to both sides of the equation j xsinbc y dix yzcosx To evaluate these derivatives we need Chain Rule on the left handside and Product Rule on the right cosx y x 31 32 cosx y2 costEM dx dx dx cosx y 1 y 2yy cosx y2 sini x Now solve for y cosx y y2 sinx 2yy cosx cosx y y cosx y y2 sinx 2ycosx cosx yy cosx y y2 sinx 2ycosx cosx y cos n 00 sin n cos 397 20cos Toicos n0 ico Plug in xn and y0 to calculate y at the given point Using point TI 0 and slope 1 the equation of the desired tangent line is y x Tl Reading Look in text at other examples especially the end of this section at how to use Implicit Diff to compute derivatives of inverse trig fcts Math 124 1997 1998 by Neal Koblitz Send comments to kublitz math washingtun edu Printed on November 29 2000 Contents Preface 7 Advice for Students Chapter 1 Straight Lines7 Polygonal Lines7 Slopes7 Distance7 Circles 1 Distance Between Two Points Circles Chapter 2 Functions 1 Shifts and Dilations Chapter 3 Instantaneous Rate Of Change The Derivative Chapter 4 The Derivative Function 1 Corners and Cusps 2 Graphical Differentiation Chapter 5 The Tangent Line Approximation 1 The Power Rule 2 The Tangent Line Approximation Chapter 6 Rules For Finding Derivatives 1 Linearity of the Derivative 2 The Product Rule 3 The Quotient Rule 4 The Second Derivative and Acceleration 5 The Chain Rule Practice First Midterms Chapter 7 Implicit Functions and Implicit Differentiation 1 Implicit Functions 2 Implicit Differentiation Chapter 8 Trig Functions and Sinusoidal Functions 1 Graphs of Trig Functions 2 Sinusoidal Functions Chapter 9 Derivative of Trig Functions 1 The Derivative of sinX 2 Other Trig Derivatives Chapter 10 Curves Given Parametrically 1 Trajectories 2 The Velocity Vector Chapter 11 Combined Parametric Motions iii iv CONTENTS Chapter 12 Related Rates Chapter 13 Curve Sketching 1 MaximaMinima 2 The Second Derivative Test 3 Concavity and In ection 4 Asymptotes and Other Things to Look For Practice Second Midterms Chapter 14 MaximumMinimum Problems Chapter 15 Newton s Method Chapter 16 Inde nite Integrals 1 From Acceleration to Velocity to Distance Falling Bodies 2 Anti derivative of sin In and cos 5v 3 Substitution Chapter 17 De nite Integrals Area 1 Riemann Sums 2 Computing the Area Under a Curve 3 Substitution 4 Some Properties of De nite Integrals 5 Application to Degree Days 6 Velocity and Distance Traveled As Area Chapter 18 The Fundamental Theorem 1 The Fundamental Theorem together with the Chain Rule Practice Final Exams Final Exams 1992 1993 Answers 1 Answers to Homework Problems Answers to Practice Midterms 2 3 Answers to Practice Final Exams 4 Answers to Autumn 1992 7 Autumn 1993 Final Exams 119 120 122 122 125 125 127 129 130 130 131 133 134 137 145 155 155 158 160 162 Preface Advice for Students The overwhelming emphasis in this course is on applications 7 story problems To master story problems one needs a tremendous amount of practice 7 doing homework worksheet and practice exam problems For this reason this course will require much more of your time than would a high school or community college calculus course that does not have such a focus on word problems Here are some pointers for doing story problems 1 2 ugt on V A 01 V AA 1 VV A 3 V Carefully read each problem twice before writing anything Assign letters to quantities that are described only in words draw a dia gram if appropriate Decide which letters are constants and which are variables A letter stands for a constant if its value remains the same throughout the problem Using math notation write down what you know and then write down what you want to nd Decide what category of problem it is this might be obvious if the problem comes at the end of a particular chapter but will not necessarily be so obvious if it comes on an exam covering several chapters Double check each step as you go along don t wait until the end to check your work Use common sense if an answer is out of the range of practical possibilities then check your work to see where you went wrong Suggestions for Using These Notes Read the example problems carefully lling in any steps that are left out ask someone if you can t follow the solution to a worked example Later use the worked examples to study by covering the solutions and seeing if you can solve the problems on your own Give yourself the practice midterms and nals in an exam situation with out access to material other than a single sheet of notes and with the same time limit as on the exam The answers to homework and practice exam problems that are in the back of the notes should be used only as a nal check on your work not as a crutch Keep in mind that sometimes an answer could be expressed in various ways that are algebraically equivalent so don t assume that your answer is wrong just because it doesn t have exactly the same form as the answer in the back Formulas Used in Math 124 and 125 Algebra 72 172 Z tC 7b i V b2 7 4ac 2a I Completing the square X2 bX c X Quadratic formula roots of IX2 bX c are a Exponents 1quot ac ab E ab 0 abc ab llb 73 Geometry Circle circumference 2717quot area 7T7 2 4 3 2 Sphere vol 3717 surface area 4717 Cylinder vol 7Tr2h lateral area 27Trh total surface area 27T7 h 27T7 2 1 Cone vol 7T72h lateral area 7T7quot r2 hg total surface area 7T7quot V 72 h2 7T7 2 Analytic geometry Point slope formula for straight line y yo mm 7 x0 Circle centered at h k In 7 h2 y 7 k2 r2 2 2 Ellipse With sem1maJor ax1s along x ax1s and sem1m1nor ax1s along y ax1s 2 32 2 1 a Trigonometry opposite adjacent opposite s1n o hypotenuse 7 hypotenuse7 adJ acent7 1 sin cos sec csc cot a cot cos s1n tan os s1n 7T 7T Sln cos 7 at cosa s1n 7m sinv 7139 7 sin 5v cosa 7139 7 cos In Law of cosines a2 b2 c2 7 2176 cos A a b c Law of sines s1n A s1n B s1n C Sine of sum of angles sinv y sins cosy coscvsiny 17cos 2x 1cos 223 Sin2 and cos2 formulas s1n2 cos2 1 SIHZJI cos2 5v 2 7 2 CHAPTER 1 Straight Lines Polygonal Lines Slopes Distance Circles Much of the mathematics in this chapter will be review for you However the examples will be oriented toward applications story problems and so will take some thought In the x y coordinate system we normally write the x axis horizontally with positive numbers to the right of the origin and the y axis vertically with positive numbers above the origin That is unless stated otherwise we take rightward to be the positive x direction and upward to be the positive y direction In a pure math situation we normally choose the same scale for the a and y axes For example the line joining the origin to the point a 1 makes an angle of 45 with the x axis and also with the y axis But in applications often letters other than In and y are used and often dif ferent scales are chosen in the cv and y directions For example suppose you drop something from a window and you want to study how its height above the ground changes from second to second It is natural to let the letter If denote the time the number of seconds since the object was released and to let the letter 5 denote the height For each If say at one second intervals you have a corresponding height 5 This information can be tabulated and then plotted on the t s coordinate plane Here is an example 15 sec 5 meters Bo H 0 60 xix 1 751 2 604 40 x 3 359 20 x 4 16 k We use the word quadran for each of the four regions the plane is divided into the 1st quadrant is where points have both coordinates positive and the 2nd 2nd 1st 3rd 4th Suppose we have two points A2 1 and B3 3 in the 5vy plane We often want to know the change in m coordinate also called the horizontal distance in going from A to B This is often written Ax where the meaning of A is change in thus Act can be read as change in In 7 it denotes a single number and should not be read as delta times In In our example Ax 3 7 2 1 Similarly the change in y is written Ay In our example Ay 37 1 2 the difference between 3rd and 4th quadrants are counted off counterclockwise as follows 1 2 1 LINESSLOPESCIRCLES the y coordinates of the two points It is the vertical distance you have to move in going from A to B The general formulas for the change in In and the change in y between a point 1311 and a point 2312 is Arcc27m17 Ayy27y1 If we have two points Av1y1 and Bv2y2 then we can draw one and only one line through both points By the slupe of this line we mean the ratio of Ag to Am The slope is often denoted m m AyAa yg 7y1x2 7x1 For example the line joining the points A and B in the last paragraph has slope 2 Example 11 According to the 1990 US federal income tax schedules a head of household pays 15 on income up to 26050 If the taxable income is between 26050 and 134930 then in addition 28 must be paid on the amount between 26050 and 67200 and 33 paid on the amount over 67200 if any Interpret the tax bracket information 15 28 or 33 using mathematical terminology and graph the tax on the y axis against the taxable income on the m axis The percent when converted to a decimal 015 028 and 033 is the slupe of the straight line which is the graph of tax for the corresponding tax bracket The tax graph is what s called a polygonal line ie it s made up of several straight line segments of different slopes The rst line starts at the point 00 and heads upward with slope 015 ie it goes upward 15 for every increase of 100 in the m direction until it reaches the point above In 26050 Then the graph bends upward ie the slope changes to 028 As the horizontal coordinate goes from In 26050 to CU 67200 the line goes upward 28 for each 100 in the m direction At 5v 67200 the line turns upward again and continues with slope 033 See the graph below 30000 20000 10000 50000 100000 The most familiar form of the equation of a straight line is y mxb Here m is the slope of the line if you increase In by 1 the equation tells you that you have to increase y by m If you increase In by Any then y increases by Ag mAm The number 17 is called the y intercept because it is where the line crosses the y axis If you know two points on a line the formula m 32 7 y1a2 7 x1 gives you the slope Once you know a point and the slope then the y intercept can be found by substituting the coordinates of the point in the equation yl mxl b ie l7 yl 7 mxl Alternatively one can use the point slope form of the equation of a straight line which is y 7 7 x1 m This relation says that between the point 1311 and any other point 5vy on the line the change in y divided by the change in In is the slope m of the line It is possible to nd the equation of a line between two points directly from the relation y 7 7 x1 yg 7 y1x2 7 ml which says the slope measured 1 LINESSLOPESCIRCLES 3 between the point 1311 and the point 2312 is the same as the slope measured between the point 1311 and any other point x g on the line For example if we want to nd the equation of the line joining our earlier points A2 1 and B3 3 we can use this formula 7 1 3 7 1 72 2 sothat y712v72 ie 312273 The slope m of a line in the form y mm b tells us the direction in which the line is pointing If m is positive the line goes into the 1st quadrant as you go from left to right If m is large and positive it has a steep incline while if m is small and positive then the line has a small angle of inclination If m is negative the line goes into the 4th quadrant as you go from left to right If m is a large negative number large in absolute value then the line points steeply downward while if m is negative but near zero then it points only a little downward These four possibilities are illustrated in the graphs below 7713 m01 m74 m701 If m 0 then the line is horizontal its equation is simply y b There is one type of line that cannot be written in the form y mxb namely vertical lines A vertical line has an equation of the form In 1 Sometimes one says that a vertical line has an in nite slope Sometimes it is useful to nd the x intercept of a line y ma 17 This is the 5v value when y 0 Setting mmb equal to 0 and solving for In gives 5v 7b m For example the line y 2x 7 3 through the points A2 1 and B3 3 has x intercept 15 Example 12 Suppose that you are driving to Seattle at constant speed and notice that after you have been traveling for 1 hour ie t 1 you pass a sign saying 110 miles to Seattle and after driving another half hour you pass a sign saying 85 miles to Seattle Using the horizontal axis for the time t and the vertical axis for the distance y from Seattle graph and nd the equation y mt b for your distance from Seattle Find the slope y intercept and t intercept and describe the practical meaning of each The graph of y versus t is a straight line because you are traveling at constant speed The line passes through the two points 1110 and 15 85 So its slope is m 85 7 11015 7 1 750 The meaning of the slope is that you are traveling at 50 mph m is negative because you are traveling in ward Seattle ie your distance y is decreasing The word velocity is often used for m 750 when we want to indicate direction while the word speed refers to the magnitude absolute value 4 1 LINESSLOPESCIRCLES of velocity which is 50 mph To nd the equation of the line we use the point slope formula y 7 110 750 t 7 1 The meaning of the y intercept 160 is that when t 0 when you started the trip you were 160 miles from Seattle To nd the t intercept set 0 75025 160 so that t 16050 32 The meaning of the t intercept is the time when you ll be in Seattle After traveling 3 hrs 12 min your distance y from Seattle will be 0 so that y 750t 7 1 110 75025 160 1 Distance Between Two Points Circles Given two points 1311 and 232312 recall that their horizontal distance from one another is Ax x2 7 x1 and their vertical distance from one another is Ag yg 7 yl The actual distance from one point to the other is the hypotenuse of a right triangle with legs Am and Ag 9627312 Ay 9617111 A96 The Pythagorean theorem then says that the distance between the two points is the square root of the sum of the squares of the horizontal and vertical sides distance 7 Aw Aggt2 7 x2 7 5602 w 7 M For example the distance between our two points A2 1 and B3 3 is equal to x3722371 2236 As a special case of the distance formula suppose we want to know the distance of a point 5vy to the origin According to the distance formula this is equal to m 7 0gt2 y 7 0 7 aw 7 y2 A point 5vy is at a distance 7 from the origin if and only if V232 y 7 or if we square both sides x2 y r2 This is the equation of the circle of radius 7 centered at the origin The special case 7 1 is called the unit circle its equation is x2 y 1 Similarly if Ch k is any xed point then a point as y is at a distance 7 from the point C if and only if It 7 h2 y 7 k 7 ie if and only if m 7 h y 7 k 7 This is the equation of the circle of radius 7 centered at the point hk For example the circle of radius 5 centered at the point 0 5 on the y axis has equation In 7 02 y 7 52 25 If we expand y 7 52 32 7 10y 25 and cancel the 25 on both sides we can rewrite this as x2 32 7 10y 0 1 DISTANCE BETWEEN TWO POINTS CIRCLES 5 Homework 1 1 For each pair of points Ar1y1 and Bm2 yg nd Am and Ag in going from A to B ii the slope of the line joining A and B iii the equation of the line joining A and B in the form y mm b and iv the distance from A to B a A20 B4 3 b A1 71 B0 2 c A0 0 B72 72 d A72 3 B4 3 e A73 72 B2 3 f A001 7001 B7001005 2 Graph each of the following lines after changing to the form y mm b and also nd the y intercept and m intercept a y72x2 b xy6 c x2y71 d 3231 e 2x3y60 3 Find the equation of the circle of radius 3 centered at a 0 0 b 5 6 c 7576 d 03 e lt0 73 f 30 4 Graph the circles a m2y210y0 b x2710xy224 c x276xy278y0 5 Let 5v stand for temperature in degrees Celsius centigrade and let y stand for temperature in degrees Fahrenheit A temperature of 0 C freezing of water corresponds to 32 F and a temperature of 100 C boiling of water corresponds to 212 F Find the equation of the line that relates temperature Fahrenheit y to temperature Celsius 5v Graph the line and nd the y and m intercepts What is the practical meaning of the intercepts 6 A car rental rm has the following charges for a certain type of car 25 per day with 100 free miles included 015 per mile for more than 100 miles Suppose you want to rent a car for one day and you know you ll use it for more than 100 miles What is the equation relating the cost y to the number of miles In that you drive the car 7 An instructor gives a 100 point nal exam and decides that a score 90 or above will be a grade of 40 a score of 40 or below will be a grade of 00 and between 40 and 90 the grading will be linear Let 5v be the exam score and let y be the corresponding grade Find a formula of the form y mmb which applies to scores In between 40 and 90 8 A photocopy store advertises the following prices Ecper copy for the rst 20 copies 4cper copy for the 21st through 100th copy and Wper copy after the 100th copy Let 5v be the number of copies and let y be the total cost of photocopying a Graph the cost as In goes from 0 to 200 copies b Find the equation in the form y mm b that tells you the cost of making 5v copies when In is more than 100 9 In the Kingdom of Xyg the tax system works as follows Someone who earns less than 100 gold coins per month pays no tax Someone who earns between 100 and 1000 golds coins pays tax equal to 10 of the amount over 100 gold coins that he or she earns Someone who earns over 1000 gold coins must hand over to the King all of the money earned over 1000 in addition to the tax on the rst 1000 a Draw a graph of the tax paid y versus the money earned In and give formulas for y in terms of In in each of the regions 0 S 5v 3 100 100 S 5v 3 1000 and Ir 2 1000 b Suppose that the King of Xyg decides to use the second of these line segments for 100 S 5v 3 1000 for In S 100 as well Explain in practical terms what the King is doing and what the meaning is of the y intercept 1 LINESSLOPESCIRCLES 10 Market research tells you that if you set the price of an item at 1507 you will be able to sell 5000 items and for every 10 cents you lower the price below 150 you will be able to sell another 1000 items Let 5v be the number of items you can sell7 and let P be the price of an item a Express P linearly in terms of In in other words7 express P in the form P mm b b Express 5v linearly in terms of P 11 The tax for a single taxpayer is described in the gure below Use this informa tion to graph tax versus taxable income ie7 In is the amount on Form 10407 line 377 and y is the amount on Form 10407 line 38 Find the slope and y intercept of each line that makes up the polygonal graph 1990 Tax Rate Schedules ONLY CautIon use If your taxable Income form 1040 lIne 37 IS 50000 or more though you cannot use the tax rate schedules below If If less use the Tax Ta le Even le Inoome so that taxpayers can b your Laxa le Inoome Is less than 50000 we show all levels of taxab see the tax rate that applIes to eat lev Schedule X Use If your fllIng status Is SIngle Schedule Z Use If your fllIng status Is Head of hold If the amount If the amount Enter on on Form 1040 on Form 1040 Form 1040 Form 1040 Ilne 37 Is lIne 37 Is he 38 of the lIne 38 0ft 9 But not amount But not amount OVer over 7 over 7 Over 7 over 7 over 7 0 19450 15 0 0 26050 15 0 19450 47050 29175028 19450 26050 67200 39075028 26050 47050 97620 106455033 47050 67200 134930 154295033 67200 Use Worksheet Use Worksheet 97620 below to gure 134930 below to gure our tax our tax CHAPTER 2 Functions A function y is a rule for determining y when you re given a value of It For example the rule y 2x 1 is a function Any line y mx b is called a linear function The graph of a function looks like a curve above or below the m axis where for any value of It the rule y tells you how far to go above or below the x axis to reach the curve Functions can be de ned in various ways by an algebraic formula or several algebraic formulas by a graph or by an experimentally determined table of values In the latter case the table gives a bunch of points in the plane which we then interpolate with a smooth curve Given a value of Ir a function has to give you only one value of y Thus vertical lines are not functions For example the line It 1 has in nitely many values of y if It 1 whereas if It is any number not 1 there is no g which corresponds In addition to lines another familiar example of a function is the parabola y x2 You can draw the graph of this function by taking various values of Ir say at regular intervals and plotting the points 5vv2 Then connect the points with a smooth curve See the picture below The two examples y 2x 1 and y x2 are both functions which can be evaluated at any value of In from negative in nity to positive in nity For many functions however it only makes sense to take It in some interval or outside of some forbidden region The interval of x values where we re allowed to evaluate the function is called the domain of the function Graph ofy x2 Graph ofy Graph of y 123 For example the square root function y is the rule which says given an x value take the nonnegative number whose square is It This rule only makes sense if It is positive or zero We say that the domain of this function is It 2 0 Alternately we can use interval notation and write that the domain is 0 1n interval notation square brackets mean that the endpoint is included and a parenthesis means that the endpoint is not included The fact that the domain of y is It 2 0 means that in the graph of this function see above we have points as y only above x values on the right side of the m axis 7 8 2 FUNCTIONS Another basic example of a function whose domain is not the entire cv axis is y 1v the reciprocal function We cannot substitute 5v 0 in this formula The function makes sense however for any nonzero In so we take the domain to be In y 0 The graph of this function does not have any point as y with In 0 As 5v gets close to 0 from either side the graph goes off toward in nity We call the vertical line In 0 an asymptote To summarize two reasons why certain x values are excluded from the domain of a function are that we cannot divide by zero and ii we cannot take the square root of a negative Another reason why the domain of a function might be restricted is that in a given practical situation the x values outside of some range might have no practical meaning For example if y is the area of a square of side In then we can write y 2 In a purely mathematical context the domain of the function y In is all In But in the story problem context of nding areas of squares we restrict the domain to In gt 0 because a square with negative or zero side makes no sense In a pure math problem we take the domain to be all values of Ir where the formulas can be evaluated 7 where we don t have zero in the denominator or the square root ofa negative But in a story problem there might be further restrictions on the domain because only certain values of Ir are of interest or make practical sense In a story problem often letters different from In and y are used For example the volume V of a sphere is a function of the radius 7 given by the formula V f0 37775 Also letters different from f may be used For example if y is the velocity of something at time t we write y vt with the letter 1 instead of standing for the velocity function and t playing the role of The letter playing the role of In is called the independent variable and the letter playing the role of y is called the dependent variable because its value depends on the value of the independent variable In story problems when one has to translate from English into mathematics a crucial step is to determine what letters stand for variables If only words and no letters are given then you have to decide which letters to use Some letters are traditional For example almost always t stands for time Example 21 An open top box is made from an a X b rectangular piece of cardboard by cutting out a square of side In from each of the four corners and then folding the sides up Find a formula for the volume V of the box as a function of In and nd the domain of this function Here the box we get will have height In and rectangular base of dimensions a7 22 by b 7 25v Thus V xa 7 2xb 7 25v Here a and b are constants and V is the variable that depends on In ie V is playing the role of y This formula makes mathematical sense for any IE but in the story problem the domain is much less In the rst place In must be positive In the second place it must be less than half the length of either of the sides of the cardboard Thus the domain is 0 lt 5v lt minimum of a and b 2 FUNCTIONS 9 ln interval notation we write the domain is the interval 0 mina We now give more examples of the domain of a pure math function Example 22 Circle of radius 7 centered at the origin The equation for this circle is usually given in the form x2 32 r2 To write the equation in the form y we solve for y obtaining y iv r2 7 x2 But this is nut 11 function because when we have an In it does not give us a single value of y but rather twu values provided that In is between 7 and 77 To get a function we must choose one of the two signs in front of the square root If we choose the positive sign for example we get the upper semicircle y V r2 7 2 The domain of this function is the interval 77 7 ie 5v must be between 77 and 7 including the endpoints If In is outside of that interval then r2 7 x2 is negative and we cannot take the square root In terms of the graph this just means that there are no points on the curve whose m coordinate is greater than 7 or less than 77quot Example 23 Find the domain of 1 yffviw To answer this question we must rule out the x values that make 4x75v2 negative because we cannot take the square root of a negative and also the x values that make 42 7 x2 zero because if 42 7 CU 0 then when we take the square root we get 0 and we cannot divide by 0 In other words the domain consists of all In for which 42 7 x2 is strictly positive We give two different methods to nd out when 42 7 x2 gt 0 First methud Factor 42 7 x2 as 5v4 7 The product of two numbers is positive when either both are positive or both are negative ie if either In gt 0 and 4 7 In gt 0 or else In lt 0 and 4 7 CU lt 0 The latter alternative is impossible since if In is negative then 4 7 In is greater than 4 and so cannot be negative As for the rst alternative the condition 4 7 In gt 0 can be rewritten adding In to both sides as 4 gt In so we need In gt 0 and 4 gt In this is sometimes combined in the form 4 gt In gt 0 or equivalently 0 lt 5v lt 4 ln interval notation this says that the domain is the interval 04 Secund methud Write 423 7 x2 as 75172 7 45v and then complete the square obtaining 7 722 74 4 7 v 7 22 For this to be positive we need In 7 22 lt 4 which means that In 7 2 must be less than 2 and greater than 72 72 lt 5v 7 2 lt 2 10 2 FUNCTIONS Adding 2 to everything gives 0 lt 5v lt 4 Both of these methods are equally correct you may use either in a problem of this type A function does not always have to be given by a single formula For example suppose that y vt is the velocity function for a car which starts out from rest zero velocity at time t 0 then increases its speed steadily to 20 msec taking 10 seconds to do this then travels at constant speed 20 msec for 15 seconds and nally applies the brakes to decrease speed steadily to 0 taking 5 seconds to do this The formula for y vt is different in each of the three time intervals The graph of this function is shown below along with the three formulas ifogtgio yvt 20 if10 t 25 1207415 if 25 t 30 Not all functions are given by formulas at all A function can be given by an experimentally determined table of values For example the population y of the US is a function of the time t we can write y This is a perfectly good function 7 you could graph it if you had data for various t 7 but you couldn t nd an algebraic formula for it 1 Shifts and Dilations Many functions in applications are built up from simple functions by inserting constants in various places It is important to understand the effect such constants have on the appearance of the graph Horizontal shifts If yua replace In by In 7 C everywhere it uccars in the furmala fur then the graph shifts aver C t0 the right If C is negative then this means that the graph shifts over lCl to the left For example the graph of y 5v 7 22 is the xZ parabola shifted over to have its vertex at the point 2 on the m axis The graph of y 5v 12 is the same parabola shifted over to the left so as to have its vertex at 71 on the m axis Vertical shifts If yua replace y by y 7 D then the graph maves up D units If D is negative then this means that the graph moves down lDl units If the formula is written in the form y and if y is replaced by y 7 D to get y 7 D we can equivalently move D to the other side of the equation and write y D Thus this principal can be stated t0 get the graph ufy fvD take the graph 0f y and Wave it D am39ts up For example the function y 2742 72274 can be obtained from y 5v 7 22 see the last paragraph by moving the graph 4 units down The result is the xZ parabola shifted 2 units to the right and 4 units down so as to have its vertex at the point 2 74 1 SHIFTS AND DILATIONS 11 Warning Do not confuse D and D For example if is the function x2 then 2 is the function x2 2 while 2 is the function 517722 x24x4 Example 24 Circles An important example of the above two principles is the circle x2 y r2 This is the circle of radius r centered at the origin As we saw this is not a single function y but rather two functions y iv r2 7 x2 put together in any case the two shifting principles apply to equations like this one which are not in the function form y If we replace In by In 7 C and replace y by y 7 D 7 getting the equation In 7 C2 y 7 D2 r2 7 the effect on the circle is to move it C to the right and D up thereby obtaining the circle of radius r centered at the point CD This tells us how to write the equation of any circle not necessarily centered at the origin We will later want to use two more principles concerning the effects of constants on the appearance of the graph of a function Horizontal dilation fr is replaced by xA in a furmula then the effect an the graph is t0 expand it by a factur qu in the x directiun away from the y aatis This wording supposes that A gt 1 If A is between 0 and 1 then the effect on the graph is to contract by a factor of 1A towards the y axis We use the word dilate in both cases For example replacing In by In 05 22 has the effect of contracting toward the y axis by a factor of 2 If A is negative then we dilate by a factor of lAl and then ip about the y axis Thus replacing In by 75v has the effect of taking the mirror image of the graph with respect to the y axis For example the function y 7 x which has domain In S 0 is obtained by taking the graph of and ipping it around the y axis into the second quadrant Vertical dilation Ify is replaced by yB in a furmula and B gt 0 then the effect an the graph is t0 dilate it by a factur ufB in the vertical directiun Note that if we have a function y replacing y by yB is equivalent to multiplying the function on the right by B y The effect on the graph is to expand the picture away from the x axis by a factor of B if B gt 1 to contract it toward the x axis by a factor of 1B if 0 lt B lt 1 and to dilate by lBl and then ip about the x axis if B is negative Example 25 Ellipses A basic example of the two expansion principles is given by an ellipse of semimajor axis a and semiminor axis b We get such an ellipse by starting with the unit circle 7 the circle of radius 1 centered at the origin the equation of which is x2 y2 1 7 and dilating by a factor of a horizontally and by a factor of b vertically To get the equation of the resulting ellipse which crosses the x axis at ia and crosses the y axis at ib we replace In by xa and y by yb in the equation for the unit circle This gives m2 2 72 t 35 2 Finally if you want to analyze a function that involves bath shifts and dilations it is usually simplest to work with the dilations rst and then the shifts For instance if you want to dilate a function by a factor of A in the x direction and then shift C to the right you do this by replacing In rst by xA and then by v 7 CA in the formula As an example suppose that after dilating our unit circle by a in the x direction and by b in the y direction to get the ellipse in the last paragraph we 1 12 2 FUNCTIONS then wanted to shift it a distance h to the right and a distance k upward7 so as to be centered at the point h The new ellipse would have equation web yak a 2b 2139 1 SHIFTS AND DILATIONS 13 Homework 2 Find the domain of each of the following functions 1yfv2r73 7yfr r27x7h2 7 h are positive constants 1 2 x H 8yfv 171m 1 1 3 5v 9 5v y f 56271 y f 17856 1 4yfr ilx 10yfxEm71 5 y M a 11 y we V51 1 631 HQ 75 Starting With the graph of y the graph of y 15177 and the graph of y 1 7 x2 the upper unit semicircle7 sketch the graph of each of the follow ing functions 12yfxr72 18yfx717mi2 13yfx4 n2 19 yfrv 1m in 14 idem 20 yfm2 17x712 15yfr74 75v72 21 yfx2x1im32 1 7 7 7 7 2 16y7fx x1 22y 42x1 5r 5 9 1 17yfr1 1 23yfvx100725v7122 x 7 ln questions 247307 y is given by the following graph Using that graph7 sketch 24yfr71 28y2f3x721 25y1fx2 29y05f3x73 26y12fx 30yf 12 2731 2f3x CHAPTER 3 Instantaneous Rate Of Change The Derivative Suppose that y is a function of It say y It is often necessary to know how sensitive the value of y is to small changes in In about a xed value Example 31 Take for example y 625 7 x2 the upper semicircle of radius 25 centered at the origin When It 7 we nd that y 625 7 49 24 Suppose we want to know how much y changes when It increases a little say to 71 or 701 In the case of a straight line y mm b the slope m measures the change in g per unit change in It Let us look at the same ratio for our function y 625 7 x2 when It changes from 7 to 71 Here Ax 71 7 7 01 is the change in It and 4y f56 Am f0 f71 f0 V 625 7 712 7 V 625 7 72 239706 7 24 700294 Thus AyAa 70029401 70294 Geometrically this means that the chord of the circle drawn from the point 724 to the point 71 239706 has slope equal to 70294 In general if we draw the chord from the point 7 to a nearby point on the semicircle 7 Ax f7 Ax the slope of this chord is the so called difference quotient f7 Am 7 f7 7 625 7 7 A502 7 24 Am 7 Am I For example if It changes only from 7 to 701 then the difference quotient slope of the chord is equal to 23997081 7 24001 702919 As our second It value 7 7 Am moves in towards 7 the chord joining 7 to 7 Ax f7 Ax shifts slightly As can be seen in the picture below as An gets smaller and smaller the chord joining 7 24 to 7Acv f7Avx gets closer and closer to the tangent line to the circle at the point 724 Recall that the tangent line is the line that just grazes the circle at that point ie it doesn t meet the circle at any second point Thus as Art gets smaller and smaller the slope AyAa of the chord gets closer and closer to the slope of the tangent line slope of chord 16 3 RATE OF CHANGE tangent tangent tangent 25 11ne June June I II II II II chord chord 15 1 0rd II n II n 5 I II n n 5 10 1 20 25 30 5 10 15 20 25 30 5 10 15 20 25 30 Am 8 Am Ax 05 The slope of the tangent line to the circle at 7 24 is called the derivative of our function 625 7 x2 at 7 It is denoted f 7 we say f prime of 7 or equivalently the derivative off at 7 The slope of the chord joining 7 to 7 Ax f7 Ax namely AyAa Am 7 f7Av gets closer and closer to this value f 7 We write I f7A7f7 f 7 T A9120 Am and we say that f 7 is the limiting value or simply the limit as Art approaches zero of the difference quotient Am 7 1n the particular case of a circle there s a simple way to nd the derivative Namely the tangent to a circle at a point is perpendicular to the radius drawn to the point of contact and so its slope is the negative reciprocal of the slope of the radius The radius joining 0 0 to 7 24 has slope 247 Hence the tangent line has slope 7724 7029166 We write f 7 7029166 Notice that when A2 is small such as 001 the slope of the chord joining 7 to 7 Any f7 Ax is a good approximation to the value f 7 For example when Ax 001 we saw that this difference quotient is 702919 which is close to f 7 Now suppose that we choose a different 0 say CU 15 and we want to know how fast y is changing as CU increases a little from 15 As an approximation we could take the chord joining 15 f15 to 15 Ax f15 Ax when Ax 001 Ag 7 f1501 7 f15 7 19992476 7 20 7 E 001 001 The limiting value of this difference quotient as Art becomes smaller and smaller is what we mean by f 15 The value f 15 is the slope of the tangent to the circle at the point 15 20 As before we can nd this slope exactly since the tangent to a circle is perpendicular to the radius The answer is f 15 71520 7075 Again we see that if ACE is small like 001 then the slope of the chord joining as to 93 Ax Ax is very close to this value here CU 15 707504 3 RATE OF CHANGE 17 U For different It we generally get different values of the derivative In our example of the circle we had f 7 7724 and f 15 734 Warning The geometrical method used in this example for nding the slope of the tangent line works only for circles The only time has the same derivative for all different It is when we have a straight line y mm 17 Then the difference quotient AyAa will always be m and the derivative at any point on the line will also be m The tangent line to the line y mmb is simply the line y mmb itself and this has slope m But except for straight lines a function y has different slopes of the tangent line for different It Example 32 Suppose we are on an airless planet where the gravitational constant is 2 msec2 the moon has gravitational constant close to this Then it turns out that an object dropped from rest falls through a distance of exactly t2 meters after t seconds have elapsed Falling body problems will be taken up more systematically later That is if y denotes the distance fallen in meters we have y t2 Suppose that we are interested in the moment t 1 sec and we want to know how much the object falls as t increases from 1 to 1 At seconds The distance fallen in that short time interval is Ay 1 At2 7 12 which simpli es to 2At At2 For example in the time from 1 sec to 101 sec the object falls a distance Ay 00201 m The difference quotient for this 001 sec time interval is AyAt 00201001 201 msec This difference quotient is the average velocity during the time interval ie distance traveled divided by time More generally the average velocity between 1 and 1 At sec is Ag 7 1At2 7 12 7 12AtAt27 1 7 2AtAt2 E At At At 2 At39 Now the derivative of y t2 at the point t 1 is the limiting value of this difference quotient as At approaches zero I All f 1 Alt oE Alt 02At 2 because as At gets closer and closer to 0 the number 2 At gets closer and closer to 2 1n the case of a distance function like our distance fallen the derivative f 1 has the practical meaning of the instantaneous velucity at time t 1 The instantaneous velocity at time t is the limiting value of the average velocity during a shorter and shorter time interval from t to t At To understand the distinction between average and instantaneous velocity think of a car the average velocity during a time interval means the distance traveled divided by time while the instantaneous velocity at a particular instant means the speed indicated on the speedometer at that moment 18 3RATE OF CHANGE The graph below shows distance fallen as a function of time in this example y 252 The diagram next to it is an enlargement near the point t 1 sec where y 1 meter ie the point 11 on the curve The slope of the line joining 1 1 to 11 121 on the curve y t2 is the average velocity of the falling object during the time interval from 1 to 11 sec The slope of the line which just grazes the parabola at the point 1 1 is the instantaneous velocity of the object at time 1 sec Thus we have two interpretations of the derivative as the slope of the tangent line in the graph or as the instantaneous velocity H H o m H H H H m Example 33 As a nal example we look at the reciprocal function y 1x Suppose we want to nd the derivative when 5v 2 One way is to approximate We can join the point 205 to the point 2 Ax 12 Ax for small values of Am say 01 001 00001 or 00000001 and then compute the slope of the chord We get the following table In 2 2 2 2 CU Ax 21 201 20001 20000001 AyAm 0238095 0248756 0249988 024999999875 At this point we might guess that the slope of these chords is approaching 7025 so that the derivative is 714 But to be more precise we need an algebraic procedure to determine for sure what the limit is That is as in the example y 252 we write out the difference quotient which represents the slope of the chord joining the points 2 and 2Am f2Aa and then we try to simplify f2 Am 7 f2 7 12Av 7 12 7 22Am 12Av 7 12 Am 7 Am 7 22 Am I Am 7272Ax7 7A2 7 1 7 22 AmAm 7 22 AmAm 7 22 Ax As Art approaches zero the denominator 22 Ax approaches 2 2 4 and so we get i f2Av7f2 1 Hl f2 All o Ax Al o 22Ax 439 3 RATE OF CHANGE 19 Homework 3 1 Draw the graph of the function y V 169 7 x2 between 5v 0 and Ir 13 Find the slope AyAa of the chord between the points of the circle lying over a In 12 and Ir 13 b In 12 and Ir 121 c 5v 12 and Ir 1201 d In 12 and Ir 12001 Now use the geometry of tangent lines on a circle to nd e the exact value of the derivative f 12 Your answers to a7d should be getting closer and closer to your answer to e 2 Use geometry to nd the derivative of the function 625 7 x2 in the text for each of the following In a 20 b 24 c 77 d 715 Draw a graph of the upper semicircle and draw the tangent line at each of these four points 3 Let y t2 where t is the time in seconds and y is the distance in meters that an object falls on a certain airless planet Draw a graph of this function between t 0 and t 3 Make a table of the average velocity of the falling object between a 2 sec and 3 sec b 2 sec and 21 sec c 2 sec and 201 sec d 2 sec and 2001 sec Then use algebra to nd a simple formula for the average velocity between time 2 and time 2 At If you substitute At 1 01 001 0001 in this formula you should again get the answers to parts a7d Next in your formula for average velocity which should be in simpli ed form determine what happens as At approaches zero This is the instantaneous velocity Finally in your graph of y t2 draw the straight line through the point 2 4 whose slope is the instantaneous velocity you just computed 4 If an object is dropped from an 80 meter high window its height y above the ground at time t sec is given by the formula y 80 7 49t2 Here we are neglecting air resistance the graph of this function was shown at the beginning of the rst section Find the average velocity of the falling object between a 1 sec and 11 sec b 1 sec and 101 sec c 1 sec and 1001 sec Now use algebra to nd a simple formula for the average velocity of the falling object between 1 sec and 1 At sec Determine what happens to this average velocity as At approaches 0 That is the instantaneous velocity at time t 1 sec it ll be negative because the object is falling 5 Draw the graph of the function y 12 between 5v 12 and Ir 4 Find the slope of the chord between a In 3 and Ir 31 b In 3 and Ir 301 c 5v 3 and Ir 3001 Now use algebra to nd a simple formula for the slope of the chord between 3 and 3 Ax f3 Determine what happens when Ax approaches 0 In your graph of y 1x draw the straight line through the point 3 13 whose slope is this limiting value of the difference quotient as An approaches 0 6 Find an algebraic expression for the difference quotient Am 7 Aa when x2 7 Simplify the expression as much as possible Then deter mine what happens as Acv approaches 0 That value is f 1 7 Draw the graph of y x3 between 5v 0 and Ir 15 Find the slope of the chord between a In 1 and Ir 11 b In 1 and Ir 1001 c 5v 1 and Ir 100001 Then use algebra to nd a simple formula for the slope of the chord between 1 and 1 Ax Use the expansion A B3 A3 3A2B 314B2 B3 Determine the limit as Art approaches 0 and in your graph of y x3 draw the straight line through the point 11 whose slope is equal to the limit you just found CHAPTER 4 The Derivative Function In the last section we saw how given a function y and some particular value of It we can speak of the derivative of the function at CU denoted ln geometrical terms is the slope of the tangent line which grazes the curve at the point It In this section instead of taking just one particular value of It or a few particular values as in Problem 2 of the last homework we let 5v vary That is we are interested in the rule which for every It in the domain of our function determines the value of the derivative That gives us a new function f called f prime or the derivative of f Example 41 Find the derivative function f for the rst function discussed in the last section V 625 7 2 The point on the upper semicircle above It has coordinates 5v 625 7mg The radius out to that point has slope AyAa V 625 7x2 7 7 0 v 625 7 Since the tangent to the circle at that point is perpendicular to this radius its slope is the negative reciprocal ie 75v 6257m2 Thus f m 75vx 6257m2 For example if you plug 5v 20 24 77 or 715 into this formula you get the answers to Problem 2 of the last homework Notation There are various other ways to denote the derivative If y we can also write 3 or else dydx or else i For example using the different types of notation in Example 1 where V 625 7 x2 we can write iridiiir7277 5 y f5 dx dx 625 m m39 Example 42 Find the derivative function for the second function in the last sec tion y t2 The derivative f t is the instantaneous velocity which we nd by taking the limit as At approaches zero of the average velocity during the time interval between t and t At That average velocity is the difference quotient distance traveled 7 t At2 7 t2 time 7 At 7 t2 2tAt A252 7 t2 i 2tAt A252 7 7 t 2t At At d The limit of this as At approaches zero is 2t Thus f t E t2 2t In the case of a distance function the derivative the instantaneous velocity is often denoted with a dot rather than a prime so we can write E t2 2t 22 4 THE DERIVATIVE FUNCTION Example 43 Find the derivative function for the third function in the last section y f 93 193 We go through the procedure at the end of the last section but with any possible In rather than with 2 The difference quotient 7 the slope of the chord joining the point x 123 on the reciprocal curve to the nearby point v Ax 12 Am 7 1s fvAx7fx 1mAm71m xvAv 1mAm71m Ax Ax 5va Am I Am 5v 7 v Ax 7A5v 1 5va AmAm 7 5va AmAm 5va Ax As Art approaches zero the denominator approaches 5va 0 x2 and so we get d 1 1 y Na g E i In Example 1 we had a special geometrical situation the tangent being perpen dicular to the radius which does not apply for any function except circles But in Examples 2 and 3 we had a more general procedure for nding the derivative func tion 1 form the difference quotient 2 simplify algebraically and 3 take the limit as Art becomes smaller and smaller In other words we de ne the derivative function fI as follows A93 7 f 93 1 f 5 A9330 Am Warning If you happen to know a bunch of derivative formulas from an earlier course for the time being you should forget them In examples like the ones above and the homework below you are required to know how to nd the derivative formula starting from basic principles 1 Corners and Cusps Sometimes one encounters a point in the domain of a function y where there is no derivative because there is no tangent line In order for the notion of the tangent line at a point to make sense the curve must be smooth at that point This means that if you imagine a particle traveling at some steady speed along the curve then the particle does not experience an abrupt change of direction There are two types of situations you should be aware of 7 corners and cusps 7 where there s a sudden change of direction and hence no derivative Example 44 Discuss the derivative of the absolute value function y This is the function that for negative In drops the minus sign and for nonnegative 5v leaves it unchanged If In is positive then this is the function y 5v whose derivative is the constant 1 Recall that when y mm b the derivative is the slope m If In is negative then we re dealing with the function y 72 whose derivative is the constant 71 If 5v 0 then the function has a corner ie there is no tangent line A tangent line would have to point in the direction of the curve 7 but there are twu directions of the curve that come together at the origin Thus 2 GRAPHICAL DIFFERENTIATION 23 d 1 if In gt 0 y glxl 71 ifxlt0 unde ned if 5v 0 y lxl y M dx 2 2 1 1 2 1 1 2 2 1 1 2 1 1 Example 45 The function y 9623 2 3 y f 93 93 pictured at the right has a derivative for every nonzero value of In which we will nd a formula for later but it 73 72 71 1 2 does not have a derivative at the ori gin which is a cusp 71 2 Graphical Differentiation Often in applications one has a function y which is not given by a formula at all but rather by drawing a smooth curve through points corresponding to a table of experimental data Whenever the function is given to you as a curve without any formula you can nd the derivative function using a graphical method Namely 1 divide up the domain of your function into regular intervals 2 for each In which is an endpoint of an interval look at the point over In on the curve 3 put a small ruler along the curve at that point In so that it just grazes the curve there ie it is tangent to the curve there 4 nd the slope of the ruler by measuring distance up divided by distance across for example if you are doing this against a piece of graph paper and if you go over to the right 5 little boxes and up 8 little boxes then the slope is 8516 5 tabulate the slopes obtained for all of the CU values 6 graph these values drawing a smooth curve between them The curve you draw is the derivative function 24 4 THE DERIVATIVE FUNCTION Example 46 Use graphical differentiation to nd the graph of for the curve y pictured below 1 l l 0 5 0 0 5 Following the 6 step procedure we make a table of values of for In at intervals of 02 from 71 to 1 In 10 08 06 04 02 00 02 04 06 08 10 f x 03 04 05 06 10 in nite 10 06 05 04 03 Note that this is a very approximate procedure so our values of can be relied upon only to 1 or 2 signi cant gures Based on this table of values we obtain an approximate graph of the derivative function Note that the derivative becomes in nite as we get closer and closer to CU 0 because the tangent to the curve there is vertical has in nite slope We say that f 0 does not exist or is in nite l 5 2 GRAPHICAL DIFFERENTIATION 25 Homework 4 1 Using the procedure used above in Example 41 nd the derivative function for y 169 7 x2 2 Using the procedure that we used in Examples 42 and 43 nd the derivative function for a y 80 7 49252 b y x2 7 1 In C y 122 bmc where a b and c are constants d y x3 Please show every step clearly 3 For each of the following two graphs y use graphical differentiation to sketch the graph of the derivative function For In at regular intervals from one end of the domain to the other estimate the derivative using a ruler as explained in the text Show clearly the points where the derivative is zero and where the derivative does not exist or is in nite 26 4 THE DERIVATIVE FUNCTION 4 The 24 graphs below are labeled by letters from a to x For eaCh of the following graphs of give the letter of the graph that looks most like it could be the graph of the derivative function f v E39137 2 C7 3 e7 4 g 5 h 6 L 717 8137 9 F7 10 S7 11 t 13117 13 W7 al b e 1d 0 9 h l Ir L CHAPTER 5 The Tangent Line Approximation 1 The Power Rule Before discussing the main theme of this section we introduce the formula for the derivative of a power function ie a function of the form y asquot Here n can be anything 7 positive or negative integer or fraction or even irrational like 7139 The formula is nxn l That is P R 1 n quot1 ower u e dm In n In We shall use this derivative formula in this section even though we have not yet explained why it is valid 7 this will be done later For some special values of n we did derive this formula in the last section namely 1 27 d 3 2 gm 725v dxcv 732 d 1 7d 17 27 1 d 1 0 d2 ltTdfvm 771 775v dxm 713 7139 Here are some examples of the n mn l formula for other 71 d 7 d 1271 247 1 d dw dxx 25 Q dm 1 w d 1 3 13 723 dwx 3x The last of these formulas tells us how to nd the slopes of the tangent lines to the graph of y 2323 in Example 45 of the last section As expected as In gets closer and closer to 0 the slopes become in nite since 0 to a negative power is in nite 2 The Tangent Line Approximation The tangent line approximation is one of the most fundamental uses of the derivative Suppose we have a function which we can easily evaluate at some particular In For example the function 12 has an obvious value when 5v 1 or when 5v 100 the function V has an obvious value when 5v 4 or when 5v 001 Now suppose that we want to evaluate our function not at the easy IE but rather at some nearby 5v Ads For example we want to nd the reciprocal not of 1 but rather of 101 or we want to nd the square root not of 4 but rather of 399 The tangent line approximation is a way of doing this quickly but not with perfect precision 7 the result will be a little off the accuracy depends on the particular function and on the size of Am 7 the smaller Am the better the accuracy 27 28 5 THE TANGENT LINE APPROXIMATION In a tangent line approximation problem we will know and want to nd fa Am The derivative f will also be relatively easy to compute at the point In The formula derived below expresses Am in terms of and To derive the formula we draw a picture of our situation 96 A967 f9c P A fac Ax 7 7 7 7 7 7 7 7 7 7 7 7 me l l 96 We want to nd the y coordinate of the point on the curve above xA5v Physically we can get to that point by starting at our known point x and following the curve until we go a horizontal distance Act to the right we go to the left if ACE is negative In doing this we move upward a distance Ay Am 7 If we can determine Ay then we know what to add to our known value to get the value we re interested in Ax Now AyAa is the slope of the chord joining the two points on the curve The idea of the tangent line approximation is to use instead the slope of the tangent line to the curve at the point x This will give us a slightly different value for the change in y 7 namely it gives us the change in y if we were to follow the tangent line rather than the actual curve when we move to the right a horizontal distance of Am That is slope of chord z slope of tangent line A A MW And so Ay z Ax This gives us the VERY IMPORTANT tangent line approximation formula m M M Ag 3 M Axgtf ltmgt In using this formula the basic steps are I decide what your function f is in the given problem and what you re taking for In and Ir Am 2 nd the derivative f m and 3 evaluate the formula Example 51 Find 213 Here our function is x3 which we can evaluate in our heads at the point In 2 namely 23 8 We want to know its value at the nearby point In Art 21 Thus Ax 01 In the tangent line approximation formula we need 2 THE TANGENT LINE APPROXIMATION 29 to know Now 233 3x2 see Problem 2d in the last homework set or else use the Power Rule above Hence f 2 3 22 12 We are now ready to use the tangent line approximation formula f2 01 x f2 01f 2 8 01 12 92 Of course we could easily nd 213 exactly by calculator or by multiplying by hand the result is 9261 Since the tangent line approximation does not give the exact answer and since we could have found 213 easily enough without it you might wonder why one bothers with the tangent line approximation There are several answers 1 1n easy examples with experience you can perform the tangent line approximation in your head thereby getting a better feel for how the function changes when you make a small change from In to CU Ads For instance in Example 51 we can say that the cube of a number near 2 is going to be near 8 and the difference between the cube and 8 is going to be about 12 times the difference between our number and 2 2 In many practical applications the tangent line approximation gives simple answers to questions about experimental error especially percent errur 3 Later on in the section on implicit functions we will encounter examples where y cannot be written in terms of In and so there is no way to evaluate the y coordinate of the point on the curve above In Art except through an approximation method such as the tangent line approximation Example 52 Find 1499 Here we re interested in evaluating the reciprocal function very near to an 5v value whose reciprocal we can evaluate in our heads namely 15 02 That is our function is 1v our In is 5 and xAm 499 ie Ax 7001 Since 7122 7125 7004 we have 1499 f499 f5 7001 f5 7001f 5 02 7 001 7004 02004 Using a calculator we can nd the exact value 1499 0200400801 Thus in this case the tangent line approximation is accurate to 5 decimal places 30 5 THE TANGENT LINE APPROXIMATION Example 53 On the circle of radius 10 centered at the origin nd a point in the rst quadrant whose m coordinate is 601 Here we know the easy point 68 on the circle and we want to know the y coordinate of the nearby point 601 The function is y 100 7 x2 our In is 6 and our 5v An 601 ie Ax 001 Using the technique in the last section eg Problem 1 on the homework we can nd the derivative 75v 100 7 x2 76 100 7 62 7075 Thus the tangent line approximation tells us that f601 z 001 f 6 8 0017075 79925 The exact value computed by calculator is 7992490225 Example 54 If 1 quart equals 09464 liter and 1 liter of water occupies exactly 1000 cc cubic centimeters ie a cubic container 10 cm on a side then what are the dimensions in cm of a cubic container that holds 1 quart of water In a story problem like this we must rst analyze the operation that is being performed The question can be reworded as follows if a cube holding 1000 cc has side 10 cm then what is the side of a cube holding 09464 as much The process of going from the volume to the side is the cube root function Since 09464 as much as 1000 cc obviously means 9464 cc we can rephrase the question once again if the cube root of 1000 is 10 then what is the cube root of 9464 Let s be the length of the side of the cube and V its volume then sfV3V 1n the tangent line approximation we take V 1000 and V AV 9464 ie AV 7536 By the Power Rule we have f V V 23 When V 1000 we have 1000 23 1100 and so f 1000 13 100 1300 So the tangent line approximation says that 3 946 f9464 z f1000 7536 f 1000 10 7 536300 9821 cm Note the exact value obtained by calculator is 981804 Alternate Methud Write V 53 Then by the tangent line approximation AV z 352As Since 5 10 and AV 7536 we have 7536 x 300As Solving for As gives As x 70179 so new 5 10 As z 10 7 0179 9821 cm The next example involves percent errur If you measure a value In then the error meaning the maximum possible error or perhaps the maximum error that is likely is the amount by which the true value could possibly differ from your measured value We shall denote the error Ads In most cases the true value could be on either side of the measured value so the error is actually iAm By the percent error we mean the percent that this error is of the measured value namely 5v percent error 100 In For example if you measure the speed ofa car to be 50 mph to within 1 mph on either side that means you have a percent error of i2 If you know the percent error say ip then to convert this to absolute error Ax use the formula Ax r Example 55 If you determine the length ofa side ofa cubic container by measuring the volume of liquid it holds then an error of i1 in measuring the volume will lead to what percent error in your value for the length of a side An error of ip will lead to what percent error for the side This is a lot like Example 54 Again our process is starting with the volume of a cube determine its side That is our function is again the cube root function 2 THE TANGENT LINE APPROXIMATION 31 But we re not given any concrete values 7 and to answer the question we don t need any So let V stand for the measured value of the volume and let 5 3V be the length of a side We want to know the error As that is caused if V is replaced by V i AV where AV is a 1 error ie AV 001V Actually what we want to know is the percent error in the side 5 ie 100Ass According to the tangent line approximation 1 1 As f VAV EVZS 001V g 00137 where the last equality comes from the rule of exponents V Z3 V V Z3 V1 V ZSJr1 Vl3 But W is precisely the value 5 So our percent error in s is 3 100 z 100 l M 1 s 3 s 3 If we take iAV in place of AV we get 713 Thus the percent error in s is i This answers our rst question If the error in V is ip instead of i1 everything would go through the same way except with a factor ofp everywhere I 72 3 100 100fVAV 10013V 001pV 1p 5 s s 3 so the error in s is ip That is a certain percent errur in measuring the volume 0f a cube leads t0 a percent errur one third as great in the value camputed fur the side Note Notice where the 13 came from in the last example the exponent 13 that comes down in the rule for the derivative of asquot In the same way it turns out that whenever your function is of the form y CCUquot where C is any constant there is a simple relationship between the percent error in your measurement of In and the percent error in y obtained using the tangent line approximation More examples of this will show up in the homework inn 32 5 THE TANGENT LINE APPROXIMATION Homework 5 1 Compute the following values using the tangent line approximation with an ap propriate and in each case see how close you are by computing the true value by calculator a 212 b 111 c 1099 d 1101 e 401 f m g Q h 38012 2 What is the y coordinate of the point on the circle x2y2 25 whose m coordinate is 308 Get an approximate answer using the tangent line approximation 3 One liter is 10567 quarts Use the tangent line approximation to answer the question One quart is how many liters 4 You want to tile a square oor which you guess to be 15 feet by 15 feet Based on your guess you buy 225 square feet of tile But after you ve used up all of the tile you still have 5 square feet not covered with tile What are the actual dimensions of the square oor Use the tangent line approximation 5 Roughly speaking an acre is about 200 feet by 200 feet More precisely an acre is 43560 square feet What are the dimensions of an acre of land in the shape of a square Use the tangent line approximation 6 If a right triangle has legs 6 and 8 then the smallest circle in which one can put this right triangle has area 257139 its diameter is the hypotenuse of the triangle which is 10 Suppose that one leg of the right triangle is known to be exactly 6 but the other leg is known to be 8 with an error of iAx What is the area of the smallest circle containing the triangle Give the error estimate for this area in terms of Am 7 You measure a cubic container and nd it to be about 10 cm on a side From this you conclude that it holds 1000 cc cubic centimeters However your measurement is accurate only to within i01 cm In other words you can only be sure that a side of the cube is between 99 and 101 cm Use the tangent line approximation to estimate the error in your value of 1000 for the volume Then give the percent ermr in your measurement of the length of a side and the percent ermr in your value for the volume 8 The distance in feet that an object falls in t seconds is given by the formula y 16152 a If an object is dropped from a height of 400 ft how long will it take to fall to the ground b If your value of 400 ft for the initial height of the object is accurate only to i1 how accurate is your answer to part a Determine both the absolute error and the percent error c If your value of 400 ft is accurate to ip then how accurate as a percent error expressed in terms of p is your answer to part a 9 Suppose that you know that a circle has area A with an error of i1 a Find a formula for the radius 7 in terms of the area A b If your value for A is off by i1 how far off as a percent error is your value for 7 obtained using the formula in part a c What if your value for A is off by ip 10 If you know that the hypotenuse of a right triangle is exactly 10 and one of the legs is 6 i 020 then what is the other leg lnclude an error estimate in your answer 11 The diameter of a sphere is measured with an error of ip and this measure ment is used to calculate the volume of the sphere What is the percent error in the value computed for the volume 2 THE TANGENT LINE APPROXIMATION 33 12 What is the permissible percent error in measuring the side of a cube if you want at most a ip error in your value for a the surface area of the cube b the volume of the cube 13 The distance around a sphere s equator is measured with an accuracy of ip This value is used to compute the sphere s a radius b volume c surface area In each case determine the percent error 14 A spherical container is lled with water and the volume of water the container holds is measured to within ip This information is then used to compute the surface area of the spherical inside surface of the container What is the percent error in the value obtained for this surface area 15 The height of a cylinder is known exactly but its diameter is measured only to within ip This information is then used to compute a the volume of the cylinder b the circumference of the top of the cylinder c the lateral surface area of the cylinder top and bottom not included In each case determine the percent error 16 Use the tangent line approximation to nd an In near 2 such that x5 x2 35 Hint Replace the left hand side by its tangent line approximation at CU 2 and solve for Am 17 A company is making solid cube paperweights The material used to make the cubes costs 23per cubic cm while the paint used to paint the 6 square faces costs OEDper square cm If the paperweight is 3cm gtlt 3cm gtlt 3cm then we calculate that the total cost of the material is 54c 27c 81c If you want the total cost to be 753rather than 813 by how much do you have to decrease the side of the paperweight Use the tangent line approximation CHAPTER 6 Rules For Finding Derivatives 1 Linearity of the Derivative This means two things 1 If you know the derivative of a function then the derivative of a constant multiple is simply cf v The derivative 0f a canstant c times a functiun is equal t0 c times the derivative 0f the function 1 d gem ewe This principle is obvious if you think of the derivative as a rate of change For example it says that if is the distance function for a bicycle and if the distance function for a car is always 5 times the distance for the bicycle then the car must be going at 5 times the velocity of the bicycle d d EGND 5am 2 If you know the derivatives and g m of two functions and gv then the derivative of the sum 923 is simply g m The derivative 0f the sum 0f tum functiuns is equal t0 the sum 0f the derivatives 0f the functiuns d d 1 gm mm We was This principle is also obvious when one thinks of rates It says that the rate at which the sum of two distances is changing is equal to the sum of the rates at which each distance is changing Principle 2 can be used repeatedly if we have a sum of more than two functions It can also be used in combination with the rst principle When combined in this way the two principles together tell us that it is easy to nd the derivative of any linear combination of functions whose derivatives we know By a linear combination we mean any function that is constructed using the process of addition and multiplying by constants For example a polynomial is a linear combination of power functions Co 612 6222 cumquot Example 61 In the section on the derivative function we found the derivative of t2 or 2 of mxb and of the reciprocal function Use that together with linearity of the derivative to do Problem 2a c of that section much more quickly 36 6 RULES FOR FINDING DERIVATIVES We have d d d 80 7 492 80 7 49 2 0 7 49 225 79825 dz dz dz d 1 d d 1 x2 7 x2 7 v 1 2a 7 7x72 2m 2 g In da da 5v d 2 7 d 2 d 7 3am bxc7adxv dmbxc72axb Example 62 Find the derivative of y 1 7 5m32 Since 1 7 5x32 1 7 10133 255v6 the derivative is d d d d 1710 3 25 6 1710 3 25 6 dcv m x dd dxx dmm 0 7 10 35v2 25 6x5 730d2 1505 2 The Product Rule Suppose that we want to nd the derivative of a function that is written as a product umvm where each of the two factors and vm is a function whose derivative we know For example max 25 7 x2 is the product of a function x3 whose derivative was already computed namely u m 3x2 and a function vm 25 7 x2 whose derivative we also know namely v m 75v 25 7 x2 We derive the product rule using the tangent line approximation formula for and vv ua Ax z A cu i va Ax z vm Ax m We now look at the difference quotient for umvm whose limit as An approaches zero is the derivative f In fa Am 7 i ua Axvx Am 7 Am 7 Am N and Amym dd Amm 7 amped N Am 7 Ax Ax2u xv v 7 Am 7 waded u dvd Adu ddd As Art approaches zero the last term Axu xv m drops out and we are left with uvv v Thus if umvm then uvv v This rule can be written in shorthand as follows Product rule uv u v 5v 3 THE QUOTIENT RULE 37 In words the derivative 0f a product is equal t0 the rst functian times the deriva tive 0f the securid functian plus the derivative 0f the rst functian times the second functian WARNING Never take the derivative of a product by multiplying the derivatives of the terms With addition you have u v u v but with multiplication uv y u v Example 63 Find the derivative of max 25 7 x2 and determine the points where the curve has a horizontal tangent line By the product rule x3 75vx 25 7 2 3x2 25 7 x2 To say that a point In has a horizontal tangent line means that the slope of the tangent line is 0 ie 0 So we must solve the equation 7x4 25 7 x2 3x2 25 7 x2 0 We do this by rst clearing denominators multiplying everything through by 25 7 2 The result is 75v4 3x225 7 x2 0 When you have an equation like this it is a good idea to try to factor the expression on the left Right away we can factor out x2 then we gather terms in what remains x27x2 325 7 x2 x274x2 75 0 This has roots 5v 0 or 4x2 75 ie 5v ix754 i25 i433 Thus there are three points with horizontal tangent lines 00 43320297 and 7433 720297 Example 64 Use the product rule to show how the nxn l formula can be obtained for n 4 5 6 ie the successive values of h after the values it 0 1 2 3 for which we already derived the nxn l formula earlier So suppose we don t yet know a formula for the derivative of 4 but we do know the derivative of x3 namely 3mg Then we can write 5v 5v v so that by the product rule 1 d d 1 3x4 xs x3xx313x2x4m3 This con rms the nxn l rule when n 4 Now that we have established the rule for the derivative of 4 we can use the name method to treat 5v d 5 i d 4 7 4 d d 4 7 4 3 7 4 EcriEQ ix Ex Ex xix14xmi5x And similarly for n 6 d 6 i d 5 7 5 d d 5 7 5 4 7 5 EcriEQ ix Ex Ex xix15xmi6x By now it should be clear that continuing in this way we can establish the validity of the nxn l rule for all positive integers n 3 The Quotient Rule Suppose that we want to nd the derivative of a function that is written as the quotient of two functions whose derivatives we know The rule Quotient Rule d M W E vm v12 or in shorthand notation E This rule is a little more complicated than the product rule the derivative 0f a quatient is equal t0 the denaminatar 38 6 RULES FOR FINDING DERIVATIVES times the derivative 0f the numerator minus the numeratar times the derivative 0f the denaminatar all aver the square 0f the denaminatar We shall not prove the quotient rule here We will derive the quotient rule later in this section when we cover the chain rule Example 65 Use the quotient rule to nd the derivative of the derivative of the upper semicircle function y 25 7 2 The derivative of the derivative is called the second derivative Since the derivative is 75vx 25 7 x2 we want the derivative of an expression of the form where 75v and vx V 25 7 x2 Since u 71 and v 75v 25 7 2 the quotient rule gives us d in 257v2717 75v 7xv2575v2 7257v27 d3 25 7 x2 i v 25 7 x22 7 25 7 x2 The last expression can be simpli ed by multiplying the top and bottom by 25 7 x2 After a little cancelation we arrive at 725 25 7 x2 32 Example 66 Find the derivative of V 100 7 mg in two ways using the quotient rule and using the product rule First using the quotient rule and the fact that 100 7 x2 has derivative m 12 V25772 and V has derivative we obtain 1 100 7 x2 i 7x 100 7 x2 7 12 100 7 x2 2E 100 7 x2 in V In Q 100 7 x2 7 7mg 7 100 7 2m32100 7 x239 Alternately we could use the product rule if we rst write our function in the form x lZ 100 7 d 712 2 7 712 95 1 732 2 dwltx 1007v7x W722 V10075v 2312 1 3 7 7 21007 2 100 7 m2 293 m which is algebraically equivalent to the answer we got using the quotient rule A bit of advice on avoiding the quotient rule Do not use the quotient rule when the numerator is a constant because it is unnecessarily cumbersome in that case For example to nd the derivative of 10v2 write this as 105172 and use the power rule the derivative is 10 72x 3 720v3 4 The Second Derivative and Acceleration As mentioned before the second derivative is simply the derivative of the deriv ative If y is our function then there are various ways to denote the second C t d d f eriva 1ve 5v da dd 12 123 d H 7 H 7 y 7 f x 7 M M 652 9 d m 4 THE SECOND DERIVATIVE AND ACCELERATION 39 Example 67 Find the second derivative of a y asquot b y lm c ax2bxc where a b c are constants a We have d d n 7 d n71 7 n72 dmltdxxidmnm 7nn Us Then b is the special case n 71 ie the second derivative of 71 is 2x73 In part c we have aaj2 bx c am b 2a Dot Notation Ifthe time t is playing the role of the m variable ie if y ft then we often use dots rather than primes for the derivatives d d2 93 tlv y ft If the y variable is denoted by a letter other than y we still use the dot notation For example if the distance function is denoted s 5t then we write for the second derivative of the distance function In the case of a distance function 5 5t both the rst derivative function and the second derivative function have a familiar practical meaning The rst derivative is as we ve seen the velocity function which is often denoted vt d vt E 5t The derivative of that 7 the instantaneous rate at which the velocity is changing 7 is what is called the acceleration often denoted at d d2 at Ev f 305 5t Earlier we saw that the velocity at a certain instant t is the slope of the tangent line to the graph of the distance function at the point t 505 Similarly the acceleration at a certain instant t is the slope of the tangent line to the graph of the velocity function at the point tvt Example 68 A bicycle is traveling along a road which we take to be the x axis Its distance in meters from the reference point 0 on the x axis at time t seconds is given by the formula x05 50 5t 252 Find the derivative function and the second derivative function 5605 graph all three functions distance velocity and acceleration and explain in words what is going on d d E60 525 252 5 225 Next at 5525 Ev f 1 E65 225 2 The distance function x05 tells us that the bicycle is at 230 50 meters from the reference point at time t 0 and is moving to the right at a certain velocity vt The formula for this velocity function vt 5 2t says that the speed of the bicycle is 5 msec at rst 15 0 but it increases steadily The rate at which the velocity increases is the acceleration function at 2 which happens to be a constant function in this example The acceleration is measured in meters First vt per second per second or meters per second squared written 2 msec2 Here are the graphs of the distance velocity and acceleration functions 40 6 RULES FOR FINDING DERIVATIVES l0511 lIonsec EI llsec2 80 8 4 60 6 3 40 4 2 20 2 l Example 69 Free fall without air resistance This is a good place to discuss the vertical motion of an object moving without air resistance under the in uence of gravity Let y denote the height say in meters of the object above ground level and let t be the time in seconds Suppose that at time t 0 an object is located at height y yo and has velocity v0 Notice that v0 positive means that the object is initially moving upward Then y and t are related by the formula 1 Falling Body Formula y yt 7 gt2 vot yo Taking derivatives gives the formula for the velocity d 1 05 E3105 915 U07 and taking one more derivative gives the acceleration at 1425 9 In particular the acceleration of a falling body is constant it is called the accel eration due to gravity On the surface of the earth 9 98msec2 Notice that the acceleration is negative because the force of gravity points down 5 The Chain Rule The chain rule is the way to nd the derivative of a function that is built up from simpler functions by the composition of two functions ie by taking a function of a function For example the function 1 x4 is built up from the square root function and the function 1 4 The function lxS 7 is built up from the reciprocal square root function ie the 712 power function and the function 5 7 We shall write y to indicate that y is obtained by taking a function gu which we sometimes call the outside function and applying it to another function which we call the inside function In the two examples just mentioned 9a and um 1x4 y f0 9a1Eandux5E gt 90493 V 1 564 90103 1V 5 7 xE 5 THE CHAIN RULE 41 The chain rule enables us to nd dgdm once we know the derivative of the outside function dgdu g u and the derivative of the inside function dudm We shall derive the chain rule using the tangent line approximation The tangent line approximation formula for the function g gu can be written in the form Ag z g uAu Similarly the tangent line approximation formula for the function can be written Au z u vAv Putting these two facts together we see that a small change Ax produces a change u a times as much in u which in turn produces a change g u times as much in g ie Ag z g uAu z g uu mAm Since we also have Ag z f vAv it follows that the two proportionality factors between A2 and Ag must be equal ie g uu v This is the chain rule It is especially easy to remember when it is written in the dgdcv notation dg dg du Ch R l am u e da du dd It is as ifthe dy du and dx can be treated individually as algebraic quantities where we cancel the two du on the right Schematically we can regard a function of a function as a 2 step procedure rst going from In to u and then from u to g The chain rule says that the derivative from In to u must be multiplied by the derivative from u to g to get the derivative from In to g Before giving some computational examples we shall give a story problem to show that the chain rule makes sense from a practical point of view Example 610 Suppose your income is in the 28 tax bracket and it is increasing at the rate of 1000 per year At what rate is your tax increasing As before let us use In to denote income and g to denote the income tax Recall that 28 tax bracket means that g as a function of Ir has slope 028 over the range of values of In your income in question In this problem g depends on In which in turn depends on the time t We could write that g 923 and Ir xt together give g Thus In is playing the role of u in the chain rule and t is playing the role of The chain rule says that d d d d Z g d j 28 1000 dollars year 028 1000 dollarsyear 280 dollarsyear That is tax is increasing at the rate of 028 for every dollar increase in income and income is increasing at the rate of 1000 for every year increase in time hence tax is increasing at the rate 028 1000 280 dollars every year Example 611 a Use the chain rule to nd the derivative of the upper semicircle function g r2 7 x2 where 7 is a constant and see that the answer agrees with our earlier formula for the derivative obtained using geometry Also nd the deriv ative of the two examples mentioned above b g 1 x4 c g 1x5 7 a We take g gu for the outside function and u r2 7 x2 for the inside function Then since dudx 722 we have 1 d d 1 iii if 72xi7 5 dx du39dx Q 39 T275627 as expected 42 6 RULES FOR FINDING DERIVATIVES b Here the outside function is y 9u and the inside function is u 1 x4 So dy dy du 1 4 3 4x3 m dd du dd Q 241 x4 the answer should be written in terms of Ir ie in g u you should substitute u uv which in this case is 1 x4 c Here the outside function is y 9u u lg and the inside function is u 5 7 So dy dy du 1 732 1 1 7u 7 dd du dd 2 2 45 7 g When the outside function 9u is the power function uquot we can write the chain rule in the form d d 3 won MW For example the derivative of the square of a function is 14502 Example 612 Find the derivative of 1 3x2 in two ways a by the chain rule b by expanding the square a In the above formula for the derivative of uv2 we take 1 35v so that u v 3 Then in 3x2 21 35v 3 61 32 b We write 1 3x2 1 6x 9x2 and then 1 6x 9x2 6 9 2x 6 18x 61 35v the same answer as in part a We next use the chain rule to show how we get the quotient rule We write the quotient in the form of a product vv 1 Then we apply the chain rule to nd the derivative of vv 1 where the outside function is the reciprocal function and the inside function is According to the chain rule vm 1 7v2v m We are now ready to apply the product rule 1 Mac iumvmii umivm4 961964 dxltvm dm dm which is the quotient rule Another comment to make about the chain rule is that it can be used to nd the derivative of a function of a function of a function or even a function of a function of a function of a function and so on where in each case we build up a function by applying a chain of simple functions Here is an example to illustrate this Example 613 Find the derivative of 1 V 1 5 THE CHAIN RULE 43 The outermost function is gu where 1 V 1 1n the chain rule we need u m which in this case we nd by using the chain rule again I 741 56 uv72 1 dm1 Finally 1 f d 1 7 1 4 1 1xE7gxE7mi limH m 4W 1 7 8 x1Eq1x1EI Homework 6 1 Find ddx of a x4 7 3x3 0522 7x 7 7T b x3 7 2x2 4 c x2 13 expand out the cube rst d x2 13 use the chain rule e m169 7 x2 f x274x5257x2 g x2x117v V 25 7 xZx x1695v 7 In use the chain rule x1695v 7 CU factor out the 12 and use the product or quotient rule k m3 7 m2 7 1v 1 100100 7 x232 m W n x2 12 1 x2 12 A AAAAAA VV 2 Find the rst derivative ddt and also the second derivative iZdt2 of the following functions oft here a b and c are constants a 749152 2025 50 b 125 17 c 125 17 d V 1252 1725 c 3 Suppose that the bicycle in Example 68 has distance function xt 50 75tt2 Find vt and at Graph xt and vt and explain in words what the bicycle is doing 4 1f 3 asquot then 2 Use the tangent line approximation to nd f1 Ax Express your answer in terms of n and Ir 5 You want a ladder to reach a window which is about 24 feet above the ground You place the bottom of the ladder exactly 7 feet away from the side of the building If the window were exactly 24 feet high you compute that you would need 25 feet of ladder to reach it Suppose that the error in your value of 24 for the height of the window is iAm Find the error in the gure of 25 feet for the length of ladder 6 1n the situation of Problem 6 of the last homework suppose that you want to t the triangle in the smallest possible sphere rather than circle In that case the triangle s hypotenuse becomes a diameter of the sphere Find the volume of that smallest sphere and include an error estimate obtained using the tangent line approximation 7 Suppose that an object falling from a height of 80 meters see Problem 4 of Chapter 3 is viewed from a height of 2 meters by a person standing 10 meters from the side of the building where the object is falling a Find a formula in terms oft 44 6 RULES FOR FINDING DERIVATIVES for the distance from the person s eye to the object b Find a formula in terms of t for the rate at Which that distance is Changing Practice First Midterms Practice First Midterm 1 50 points in all time 1 hour 1 12 points The graph ofy 25v75v2 is given at the right Find the domain of each 73 of the following functions You do not need to use any algebra in this problem a mac We 7 x2 b ex Tc 7 x2 c 3 1 d fan1 2 12 points At time t 0 an object starts from 0 moving to the right at 2 msec lts velocity changes as shown in the graph at the right This graph shows the velocity with which the object is moving to the right as a function of time Give the letters of ALL labeled points where a the object is actually moving right ward b the object is actually moving leftward c the object has positive rightward ac celeration d the object has zero acceleration e the object s speed heading to the right is maximum f the object s speed heading to the left 7 is maximum 3 13 points A box with square top and bottom and rectangular sides has surface made up of two s X 5 sides and four s X t sides Thus if your box has dimensions 5 ft X 5 ft X 4 ft then the total surface area A is 130 sq ft If you want to increase 5 by 1 inch to 5 ft 1 in how should t change so that the surface area stays the same Use the tangent line approximation and give your answer in feet 4 13 points A cubic container with side 9 cm holds 93 729 cubic cm Suppose you have a bottle containing 750 ml of liquor A milliliter ml is the same thing as a cubic cm Use the tangent line approximation to nd the side of a cubic container needed to hold the 750 ml of liquor 46 PRACTICE FIRST MIDTERMS Practice First Midterm 2 50 points in all time 1 hour 1 8 points Sketch the graph of the functions ay932247 byv255fv2 2 6 points The graph of the function 923 on the interval 0 27139 is pictured below to the left Find the domain of y 1 gv This function 923 is actually the sine function restricted to the interval 0 27139 2 Graph for Problem 2 Graph for Problem 3 3 12 points At time t 0 an object starts from 0 moving to the left at 1 msec lts velocity changes as shown in the graph above at the right This graph shows the velocity with which the object is moving to the right as a function of time Give the letters of ALL labeled points where a the object is actually moving rightward b the object is actually moving leftward c the object has positive rightward acceleration d the object has zero acceleration e the object s speed heading to the right is maximum f the object s speed heading to the left is maximum AAAA 4 12 points From a point exactly 75 m above the ground you throw a stone upward at v0 msec and you time how long it takes to hit the ground If we neglect air resistance the height of the stone at time t is given by the falling body formula 5 739252 vot 50 where g 98 msec2 a If you time 5 sec for the stone to reach the ground at what upward speed did you throw the stone b If your stopwatch is accurate only to i01 sec how accurate is your answer to part a Use the tangent line approximation 5 12 points You have a 1000 square foot apartment which you re renting for the summer to someone from Europe so you have to convert its area to square meters in order to explain its size You know that there are about 10 square feet in one square meter More precisely 1 square meter 1077 square feet Use the tangent line approximation to nd the area of your apartment in square meters You do NOT need a calculator to do this problem correctly PRACTICE FIRST MIDTERMS 47 Practice First Midterm 3 50 points in all time 1 hour 1 8 points The graph of y is pictured below to the left It is called the arcsine function and its domain is the interval 71 1 Find the domain of 1 A M R H 05 1 y 71 72 Graph for Problem 1 Graph for Problem 2 2 12 points At time t 0 an object starts from 0 moving to the right at 1 msec Its velocity changes as shown in the graph above at the right This graph shows the velocity with which the object is moving to the right as a function of time Give the letters of ALL labeled points where a the object is actually moving rightward b the object is actually moving leftward c the object has positive rightward acceleration d the object has zero acceleration e the object s speed heading to the left is maximum 3 15 points In this problem use the formula 5 59252 for the distance an object falls in t seconds with no air resistance Upon arrival on a new airless planet you decide to measure its gravitational acceleration g by dropping a stone from a height of exactly 1125 meters and measuring the time t it takes to fall to the ground a If you time 300 sec for the stone to reach the ground what is g b If you do not know the time precisely but only know that it is between 300 and 310 sec what is the range you know 9 to be in Use the tangent line approximation 4 15 points You lean a yardstick exactly 36 inches 07216 long against a wall and measure the height up the wall to where the yardstick touches to be 216 inches 36 i 02 inches Find the distance the other end of the yardstick is from the bottom of the wall see the dia gram to the right Include the error which you nd using the tangent line approximation 48 PRACTICE FIRST MIDTERMS Practice First Midterm 4 50 points in all 7 time 1 hour 1 10 points Let V 625 7 x2 Find the domain of each of the following functions a L b Mm 7 24 5fac 107 2 15 points The downward velocity of a parachutist measured in msec is given by the graph at the bottom of the page on the left a Sketch the graph of downward acceleration b At what instant is the downward velocity the greatest What in practical terms is happening at that instant At what instant does the acceleration have the greatest positive value What is happening at that instant At what instant does the acceleration have the greatest negative value 3 10 points Sketch the circle y iv 100 7 x2 and the ellipse y i2 100 7 2 The point 87 12 is on the latter ellipse Use the tangent line approximation to nd the y coordinate of the point on the ellipse in the rst quadrant whose m coordinate is 77 You must use the tangent line approximation in this problem7 showing your work clearly 4 15 points A game board is to be constructed in the shape of a rectangle of dimensions 7 X s with a half circle attached to each of the two sides of length 7 see the bottom of this page at the right a Find a formula for the area of the board rectangle plus two half circles in terms of 7 and s b You rst construct a board having dimensions 7 20 cm7 s 30 cm What is the area of the board c Now you want to increase 7 to 21 cm What must you do to s so that the total area of the board remains the same Use the tangent line approximation Be sure to show clearly what the function is to which you re applying the tangent line approximation formula meterssec velocity time sec CHAPTER 7 Implicit Functions and Implicit Differentiation 1 Implicit Functions We have seen that not all functions y are given by formulas Sometimes a function is determined by interpolating a table of experimentally determined data drawing a smooth curve between the points Another way a function can be determined is implicitly That means that we have a formula which It and y to gether must satisfy The equation may take the form Fa y constant or perhaps 706711 G03 1 Example 71 a The circle can be written x2 32 r2 which is actually the implicit equation of two functions at the same time the upper semicircle function y r2 7 x2 and the lower semicircle function y 7v r2 7 x2 b We can avoid fractional powers y mmquot by raising both sides to the m th power and writing the implicit equation 3 5v c We can avoid negative powers by writing y x lm in the form ymxl l d The points x g on an ellipse with foci at 1311 and 2312 and semimajo axis a satisfy the geometrical property distance from 5vy to 1311 distance from x y to 2312 2a which leads to the following implicit equation for the ellipse 90 12 31 y12 90 22 31 392 2a e In the case when the ellipse s axes of symmetry are parallel to the cv and y axes a simpler implicit equation can be given namely 5v7h212yik2b2 1 With many implicit equations it is possible to solve for y in terms of It For example we can do this for the circle Example 7la In that case we have the choice of whether to use the explicit form y or the implicit form Fvy constant However there are other examples where it is impossible to solve algebraically for y in terms of It and we have no choice but to work with the implicit equation 2 Implicit Differentiation Suppose we want to nd the derivative 3 at a point x y of an implicitly de ned curve What you do is 1 take ddm of both sides of the equation 2 separate the terms involving 3 to one side of the equation and the terms without 3 to the other side and 3 solve for y In step 1 it is important to remember that you are differentiating with respect to It or whatever letter is playing the role of the x variable and so the chain rule must be used for expressions like 312 In other words the letter y stands for a certain function of It and so taking ddm of 32 is just like taking ddm of uv2 312 2313 1 2y 3 49 50 7 IMPLICIT FUNCTIONS AND IMPLICIT DIFFERENTIATION Example 72 Find 3 at the point 5vy for each implicitly de ned function in Example 71 a When we take ddm of both sides of x2 32 r2 remember that 1 the derivative of 32 is nut 2y but rather 2313 and 2 the derivative of r2 is nut 27 but rather 0 because r2 is just a constant Thus taking ddm of both sides gives 22 2313 0 We next move the 2x to the right and divide both sides by 2y to solve for y y 72x2y 75vy This answer agrees with our earlier formula for the derivative of y r2 7 x2 namely 3 75v r2 7 x2 since y and r2 7 x2 are the same Notice that when you differentiate an explicit function y your answer is a formula in terms of Ir whereas implicit differentiation generally leads to a formula involving bath y and Ir Parts b and c of this example will show us how to derive the nxn l rule when the power of In is a fraction n lm or n 7lm where l and m are positive integers In the section on the product rule we already saw how to get the nxn l rule for any positive integer b We take ddm of both sides of 3 ml and then solve for 31 Z71 m71 i a my 3 lxlil so that y m ym7139 The last ratio simpli es algebraically if we use the fact that ym l ymy xlxlm That is l 3171 l l I 7 7 l 71 7 71 7 y im mlxlmimm m 77w herenim c We take ddm of both sides of ymxl 1 using the product rule and then solve for y lymml71 gm lmlil mymily ml 0 so that y mym 5v Canceling ymxl from the numerator and denominator in the last expression we obtain 1x71 l yi l 17quot 1 41 l 7 1 7 7 v m nx where n7 my mm m In m m 7 n71 d When we take ddm of both sides we must use the chain rule several times rst with the square root as the outside function and then when we nd the derivative of v 7 m12 y 7 y22 etc The result is 93 7 561 y 7 MW 93 7 562 y 7 MW xw 7 mg y 7 mg xw 7 562 y 7 y22 Solving for y is not dif cult but the expression we end up with is rather messy so we shall omit it If we had speci c values for the foci 1311 and 2312 and if we wanted to nd the slope y at a particular point x y then we could substitute all of these numerical values in place of Ir g x1 y1 x2 y2 ending up with an equation of the form 04 y 0 where comes from bringing together all of the terms with y and 04 comes from bringing together all the terms without 31 Such an equation has solution simply y 7a We will see such situations later in the story problems 0 2 IMPLICIT DIFFERENTIATION 51 e Here we get 2a 7 ha2 2y 7 ky b2 0 Solving for y we obtain b2m7h y 2 a y7k Example 73 If you know that the curve Ivy4 72 y 1 C where C is a constant passes through the point 1 2 nd the y coordinate of the point 101 on the curve near 12 Use the tangent line approximation Note The constant C is not given We could easily compute it from the fact that the point 1 2 satis es the equation However we will never need to know the value of C so we won t bother to compute it In this problem there is no way you can nd an exact expression for the y coordinate corresponding to CU 101 because when you substitute 5v 101 you get an equation for y that cannot be solved algebraically The equation for y could be solved by an approximation method that we ll learn later Newton s method In the meantime the best we can do is to use the tangent line approximation to nd the y coordinate In this problem we know that the implicitly de ned y has value 2 when 5v 1 this is what it means to say that the curve passes through the point 12 We want to know f101 According to the tangent line approximation f101 z 001f 1 2 0013 where 3 denotes the derivative at the point 12 To nd that derivative we rst take ddm of both sides of our implicit equation We use the chain rule and the product rule to nd vy4 and on the right we get 0 the derivative of any constant is 0 4Ivy3wy 1 y 4973 w gy 0 Once we ve taken ddm we can substitute the particular In and g values of the point where we want the derivative After that our equation will look a lot simpler 7 just some numbers together with the unknown y 4139 330 223917272y 0 y 32 7 025 64 7 2 0 y 7623175 71953 ie ie Finally substituting this value of y f 1 in the tangent line approximation gives the y coodinate as 2 001 71953 198047 Example 74 Suppose that the ellipse in Example 71d has foci at the points 01 and 30 and passes through the point 23 Find the y coordinate of the point 199 through which it passes Use the tangent line approximation The tangent line approximation tells us that f199 z 7 001f 2 3 7 001y where y is the derivative at the point 2 3 The implicit differentiation of the equation of this ellipse has already been carried out so it remains just to substitute the particular values we re given x1 0 yl 1 x2 3 yg 0 CU 2 y 3 22y 713y lt2 1 2 3 0 sothat 7 0 x x y 8 x ie 039088 165579y 0 Thus 3 7039088165579 7023607 and so f199 z 3 7 0013 300236 52 7 IMPLICIT FUNCTIONS AND IMPLICIT DIFFERENTIATION Example 75 You are standing at 01 and receiving light from two sources source A located at 71 0 and source B located at 10 You want to start walking in the direction which keeps the total light intensity from the two sources as constant as possible Assume that light intensity from a source is proportional to the strength of the source and inversely proportional to the square of the distance from the source a If A and B are equally strong show that you should walk horizontally ie set out along the line y 1 b If B is twice as strong as A nd the line along which you should move We are interested here in the curve of constant total light intensity passing through the point 01 That is we want an equation satis ed by all points 5vy where the light intensity from A plus the light intensity from B remains constant Since light intensity is inversely proportional to the square of the distance from the source this means that K K dist of 5vy from A2 dist of x y from B2 C Here the proportionality constant K depends on the strength of the source since A and B have the same strength in part a the constants are the same In part b we change the second K to 2K because B is twice as strong as A Next we use the distance formula in the denominators the square ofthe distance from 5vy to 71 0 is v 12 312 and the square of the distance from 5vy to 10 is v 7 12 312 For convenience we also divide through by K so that there are no unknown constants on the left The result is 1 1 7 C m 12 112 m 7 1 72 f The question asks for the direction to go in from the point 0 1 that best keeps to the curve of constant total light intensity This amounts to asking for the tangent line to the curve at 0 1 To nd 3 we take ddm of both sides and then substitute the point 5vy 0 1 where we want to nd 3 It is important to note that the fact that you happen to be walking from the point 01 is not used until after performing the implicit differentiation When we take ddm we use the chain rule writing the rst term on the left as 12 y271 and similarly for the second term on the left it is more e icient to use the rule for u l than to use the quotient rule 93 DZ 31372903 1 2313 7 93 i 02 112 293 i 1 2111 0 Notice that the constants C and K have disappeared because the derivative of any constant is 0 Substituting 5v 0 y 1 gives 71 122 21 7 1 1272 21 0 Solving this we nd that y 0 ie the curve of constant total light intensity has horizontal tangent line at the point 0 1 2 b When the source B is twice as strong as the source A the effect is to re place the second K by 2K and this means having a 2 rather than 1 always in the numerator of the second term on the left Our last equation is now 71 122 21 7 21 1272 21 0 2 IMPLICIT DIFFERENTIATION 53 We can clear denominators by multiplying through by 2 obtaining 717y 272y 0 ie y 13 Thus the tangent line has slope 13 and y intercept 1 so its equation is y Ea 1 Homework 7 1 In each case use the tangent line approximation to estimate the y coordinate of the point on the curve whose horizontal distance is 001 to the right from the given point P afv2wyy277P172 2 b 7 tp725 m COHS7 77 ltcgt y71 lt974 2 Find a formula for the derivative 3 at the point x y a 21x2 b m2x 27 C x3x2 3 x2 y 7 y y 7 y y y 3 A hyperbola passing through 8 6 consists of all points whose distance from the origin is a constant more than its distance from the point 52 Find the slope of the tangent line to the hyperbola at 8 6 4 In Example 75a suppose that you are starting from the point 1 1 rather than 0 1 Find the equation of the line from this point which is the closest approxima tion to the curve of constant total light intensity After you ve walked a horizontal distance of 01 to the right what is your y coordinate 5 In the situation of Example 75 suppose that you don t know the relative strengths of the sources A and B but you know that the line y 06 1 is the line that you should start walking along from the point 01 if you want the total light intensity to be as constant as possible Find how much stronger B is than 6 Suppose that a light source A is at the origin and a source B twice as strong is located at 30 Find the slope of the direction you move in from 12 in order for the total light intensity to remain as constant as possible 7 When crude oil ows from a well water is frequently mixed with it in an emulsion To remove the water the crude is piped to a device called a heateritreater which is simply a large tank in which the oil is warmed and the water is allowed to settle out Operating experience in a particular oil eld indicates that the concentration C of water in the treater s output can be modeled by the following equation in the neighborhood of the usual operating point of 135 F and a 2 hour holding time C 004 7 00032 h2 7 106 T2 41 10 5h T where h is the holding time in hours and T is the operating temperature in degrees F a Due to random uctuations in the well s ow rate the holding time actually varies slightly around 2 hours Suppose you are given a simple control device that can change the tank temperature proportionally to the measured change in holding time What constant of proportionality would best compensate for small holding time uctuations and keep the water concentration as constant as possible b Now 54 7 IMPLICIT FUNCTIONS AND IMPLICIT DIFFERENTIATION as eld equipment ages7 its maximum operating settings are generally decreased Find the equation of the line that best approximates the way in which the holding time would have to be increased as the maximum temperature rating falls slightly below the usual operating temperature 8 One end of a string the point A is attached to the y axis at a distance 20 meters from the origin the string is threaded through a hole the point B at the end of a pole situated along the line In 10 and the other end of the string the point C is attached to a hook that is free to slide along the x axis and is being pulled by a force to the right that is suf cient to keep the string taut See the diagram at the right At time t 07 when the top of the pole is at the point 10107 the pole starts lowering at the steady rate of 30 cmsec 030 msec The sliding end of the string reaches the point In 20 at time t 10 a Write an equation relating t and In the time is an implicit function ofthe location of the sliding end of the string b Use the tangent line approximation to nd the time when the sliding end of the string has moved another 20 cm to the left from the point In 20 CHAPTER 8 Trig Functions and Sinusoidal Functions We start by reviewing trig functions of an angle measured in radians Because this is review material we will go through it rapidly We rst draw the unit circle in the Ivy plane The standard way of measuring an angle is to start with the positive x axis and rotate counterclockwise As we go through the angle we sweep around the circumference of the unit circle The radian measure of the angle is de ned to be the distance around the circumference that we travel That is a full revolution of 360 is equal to 27139 radians half a revolution is 180 7139 radians and so on Other frequently used equivalences are 90 71392 rad 60 71393 rad 45 714 rad 30 716 rad All angles in this cuurse will be assumed t0 be measured in radians unless atherwise stated The designatiun radian will often be omitted eg we write an angle 0f 716 we mean a 30 angle By a negative angle we mean a clockwise angle For example an angle of 7712 ie 790 brings us to the same position as an angle of 371392 which is 270 In general if we have a circle of radius r and sweep out an angle of 0 radians then the distance around the circumference through which we have traveled is simply 0r Of course one full circumference is 2717 corresponding to rotation through 27139 radians Returning to the unit circle see the drawing below after rotating through an angle of 0 radians let 5vy be the point where we end up on the circumference Then the six trig functions are de ned as follows 9671 l l sin0 y cosecant0 sm 0 y 1 1 AA cos0 5v secant0 cos0 5v 39 0 l 0 tan0 cot0 2 cos0 5v tan0 sm0 y From the de nition of sin0 we see that as 0 increases from 0 to 27139 ie as we go once around the unit circle sin0 goes from 0 to l to 0 to 71 and back to 0 Then as 0 goes from 27139 to 47139 ie as we go once again around the circle the same values of sin0 are repeated We say that the function sin0 has period 27139 ie we have sin0 27139 sin 0 Here are some other relations for the trig functions that follow easily from the above de nitions cos0 27139 cos 0 sin0 7139 isinb cos0 7139 7cos0 tan0 7139 tan0 sin70 7 sin 0 cos70 cos 0 55 56 8 TRIG FUNCTIONS AND SINUSOIDAL FUNCTIONS tan70 7tan0 sin20 cos20 1 and cos0 sin 7 0 The last of these relations is familiar from high school trigonometry the cosine of an angle is equal to the sine of the complementary angle It can also be seen from our de nitions as follows the m coordinate of the point on the unit circle you arrive at when you rotate through 0 counterclockwise starting at the positive x axis is the same as the y coordinate of the point you arrive at when you rotate through 0 clockwise starting at the positive y axis Sometimes one also needs the following formulas for the trig functions of a sum of two angles sina sinacos cosasin and cosa cosacos 7 sinasin As a special case we have the double angle formulas sin2a 2sina cosa and cos2a cos2 0 7 sin2 04 Also notice the sign of the trig functions in the different quadrants sins is positive in the 1st and 2nd quadrants for angles between 0 and 7139 cosx is positive in the 1st and 4th and tanx is positive in the 1st and 3rd 1 Graphs of Trig Functions The most important graphs to be familiar with are the graphs of the functions y sin 5v y cos In and y tanx We re now using In for the angle in radians In no longer denotes a coordinate of a point on the unit circle For certain In the trig functions have easily stated exact values namely when In is a multiple of 7T6 or of 774 Here it is useful to draw the 30 7 60 7 90 triangle with hypotenuse 1 and the 45 7 45 7 90 triangle with hypotenuse 1 Using these triangles we arrive at a table of values of the trig functions from which we can sketch the graphs on the next page 2 Sinusoidal Functions A sinusoidal function is a function made up from the function y sins or y sin 25 iftime is playing the role ofthe x variable by inserting constants in various places The reason for inserting these constants is that the pure sine function y sins almost never occurs in the real world but curves with a sine like appearance this is what the word sinusoidal means arise frequently At this point you should review the material at the end of the section on func tions horizontal and vertical shifts horizontal and vertical dilations and the last batch of homework problems in that section Example 81 Write the cosine function as a shift of the sine function The graphs show that the cosine function is obtained by shifting the sine function 772 to the left Thus cosx sinv 7 sinv This relation can also be obtained using the identities listed above cosx cos7a sin 7 sinv g We now give the general form for a sinusoidal function y We start with y sin In Step 1 Dilation Vertically we dilate by a factor of A which is called the amplitude We do this by replacing sinx by Asinx This makes the function go to a maximum of A and a minimum of 7A rather than a maximum of 1 and a minimum of 71 as was the case for sin Next we dilate horizontally in such a way that the function repeats its pattern from In to CU B rather than from In to CU 27139 as with sin B is called the periud That is we want to dilate horizontally by a factor of B27T We do this by replacing In by xBQz39 223 inside the sine 2 SINUSOIDAL FUNCTIONS 57 k H to Q 5 a In cosx tanx 0 0 1 0 Z 7r 1 6 2 T T 1 Q Q 1 4 32 2 g 2 2 1 g 1 0 OO 2 1 Z Tquot T 2 3 quot 1 4 2 2 5 1 3 3 m wquot 2 425 435 7T 0 71 0 7w 1 6 2 2 3 5 quot 1 4 Kg 1 7r 3 2 2 V3 7quot 71 0 oo 57r 1 7 3 3 2 2 7 quot Q Q 1 4 2 2 11 quot 1 7 3 6 2 2 3 27139 0 1 0 1 Graph of y sinm 71 NI NI 1 Graph of y cosx G h f t rap 0 y an In To summarize7 the result of the vertical and horizontal dilation is to replace sinx by the function Asin 2 The graph of this new function is shown below Notice 58 8 TRIG FUNCTIONS AND SINUSOIDAL FUNCTIONS that as In goes from 0 to B the value of the function goes from 0 to A to 0 to 7A and back to 0 w vaB Step 2 Shifting We now want to move the whole sine wave upward a certain amount D called the vertical shift and over to the right a certain amount C called the phase shift We do this by adding D outside the function and replacing In by In 7 C inside the function The result is 27139 A 39 y s1n lt B In the graph of this function below we indicate the meaning of the four constants A B C and D 900 D 4 D Example 82 Suppose that the mean daily temperature ie the temperature in CF averaged over the 24 hour period on the t th day of the year in Seattle is tabulated over a period of many years and the results for each day are averaged This is what we call the normal mean temperature on the t th day For simplicity suppose that the year consists of 12 months of 30 days each Take t 1 to be January 1 and so on until t 360 for December 30 Further assume 1 the normal mean temperature function yt is sinusoidal 2 the coldest mean temperature is 37 on January 30 and 3 the warmest mean temperature is 65 on July 30 Find a formula for y t 2 SINUSOIDAL FUNCTIONS 59 We graph yt by rst plotting the two points If 30 y 37 and t 210 ie July 30 y 65 We then draw a sine wave in such a way that its minimum is at the rst point its maximum is at the second and its period is a year ie 360 days We also draw a horizontal dotted line half way between the coldest and warmest temperature ie at y 51 Now we read the four constants A B C D off the graph The amplitude A is 14 the period B is 360 and the vertical shift the distance up to the dotted line is 51 To nd C we want to know when the curve crosses the dotted line on its way from a minimum to a maximum ie when the curve looks like sins looks when it crosses the origin This can be found by taking the t value half way between the coldest day which is t 30 and the warmest day which is t 210 This t value is 30 2102 120 ie April 30 Thus the phase shift is C 120 and our equation is 27139 3125 14sin 36025 7 120 51 Homework 8 1 Convert to radians a 15 b 120 c a quarter revolution d 10 revolutions 2 A bicycle wheel has radius 40 cm Find the rate in radsec at which the wheel is turning if the bicycle is traveling at a 10 msec b 10 kmhr 3 A drawbridge whose two spans are each 30 meters long is opened as shown below Express in terms of 0 a the height y of the end of a span b the slope of the left span c the distance In between the ends of the spans 60 8 TRIG FUNCTIONS AND SINUSOIDAL FUNCTIONS B F 4 Using the diagram at the right show that sina sina cos cosasin 5 Sketch the graph of each of the following 1 E functions a y 1 sin7Tv b y 2sin gt 7 39 I 7 E iiiiii fg i 1 A C D 6 The mean daily temperature in Fairbanks Alaska was tabulated over a 30 year period the average was taken over the 30 values for each day and the result was found to be very close to the sinusoidal function 37 sin ii 7505 7 101 25 where t is the day of the year1 Here we are NOT assuming 30 days per month Sketch the graph of this temperature function nd the maximum and minimum mean daily temperatures and nd the warmest and coolest days of the year 7 Studies are made of the variation in population of foxes and rabbits in a certain large forest The number changes from year to year Roughly speaking when there are a lot of rabbits the foxes will start to increase in number because they can eat rabbits plentifully but then this decreases the number of rabbits causing the foxes to be short of food and this causes a decrease in the fox population and nally when there are fewer foxes the rabbits will be able to increase in number without being eaten so we are back to where we started Thus we might expect a sinusoidal oscillation in the number of foxes and in the number of rabbits This is the so called predatoriprey problem in mathematical biology Suppose that t stands for the number of years since the study began F stands for the number of foxes and R stands for the number of rabbits Suppose that F and R are each a sinusoidal function of 25 Use the following table of data which gives the maximum and minimum population of each to nd a formula for F and for R in terms of t t 2 F 150 200 150 100 150 R 10000 7500 5000 7500 10000 8 Write a formula for each of the following functions of t assuming that the function is sinusoidal also suppose that a year consists of 12 months of 30 days each a the mean low daily temperature if the coldest is 20 F on January 19 and the highest low temperature reading is 60 F on July 19 b the time of sunset each day if the sun sets at 9 pm on June 21 and at 5 pm on December 21 c the temperature in a well insulated building which varies between 25 C at 5 pm which is 1700 on the 24 hour clock and 15 C at 5 am d the velocity of a weight attached to the end of a spring which is extended and released at time t 0 if the weight reaches its maximum velocity of 10 cmsec after 01 sec e the height above the ground of a pebble that is picked up in the tread of a bicycle tire at time t 0 if the tire has radius 40 cm and the bicycle is traveling at 10 m sec 1This example is from an article by B M Lando and C A Lando in The Mathematics Teacher Sept 1977 CHAPTER 9 Derivative of Trig Functions 1 The Derivative of sinX To nd the slope of the tangent line to the curve y sin 5v we must go back to the de nition of the derivative as the limit of the difference quotient dy Am 7 sinv Am 7 sins llm llm dm A170 A23 A170 Ax Recall the procedure used to nd the derivative of the much simpler function x2 expand v A502 simplify algebraically and then take the limit as An gets closer and closer to 0 We do essentially the same thing for sin 5v but we need the following three basic facts in order to do this Fact 1 sina sinacos cosa sin Fact 2 lim 8mg 1 9H0 0 Fact 3 lim 1 7 C050 0 9H0 0 Fact 1 was proved in the last section Problem 4 of the homework Facts 2 and 3 can be seen from the diagram below where 0 is a small angle in radians By the de nition of the radian measurement of an angle 0 is the distance along the circumference from A to C Meanwhile sin0 E and 1 7 cos0 O B C Thus Fact 2 says that although the line E is smaller than the arc AC the ratio of the two lengths gets closer and closer to 1 as 0 gets smaller Fact 3 says that the ratio of the line W to the arc AC gets closer and closer to 0 as 0 gets smaller We are now ready to simplify the above difference quotient We have sinv Am 7 sins i sinxcosAm cosx sinAv 7 sins Act Act by Fact 1 with 04 It and Am i cosmsinAm 7 sin 5v1 7 cos Am 7 Am 62 9 DERIVATIVE OF TRIG FUNCTIONS putting sincvcosAx last on top and factoring out 7 sin 2 sinAv 1 7 cosAm cos x 7 sm ac Ax 7gtcosv17sinx0cosv as Ax7gt0 The last step comes from Facts 2 and 3 with 0 replaced by Any We therefore conclude the following basic derivative formula 1 s1n 5v cos 5v d2 Example 91 You know the exact value sin30 05 Find sin31 using the tangent line approximation and compare with the exact value found by calculator In the tangent line approximation we choose sincv 5v 30O 776 and ACE 1o 7T180 Here it was important that we changed As from degrees to radians By the tangent line approximation o 7 Z 77 g 77 I s1n317 f6 180 05 180f 6 7T 7T 7T 7 05 m COSE 7 05 E 39 7 7 0515115 the exact value is 0515038 When we have a derivative formula for a new function we can combine it with other rules to nd the derivative of a more complicated function that is built up from the basic function Example 92 Find the derivative of sin2E 2 The notation sin2something means sinsomething To evaluate this deriv ative we have to use the chain rule twice d 2 d d s1nE 2s1nE s1n 5v 2s1nEcosE E da da da 2sinEcosE12E W 2 Other Trig Derivatives All of the other trig functions can be expressed in terms of the sine and so their derivatives can easily be calculated The most important ones are d d Ecosx Esinw cosa sinv sinv 7139 7sin5v d d d Sinai cos mg sma 7 s1nxg cosa COS2 x Sin2 x 1 d t 2 ana sec 90 d2 dd cos 5v cos2 5v cos2 5v cos2 5v 7 d d 4 1lt 2 sins t seca coscv 7 coscv 7s1n5v anmsecx d2 dd cosgcv Homework 9 1 You know the exact value tan45 1 Find tan44 using the tangent line approximation and compare with the exact value found by calculator 2 OTHER TRIG DERIVATIVES 63 2 Find the derivative of a y cot 5v b y cscx csc is the abbreviation of cosecant 3 Use the tangent line approximation to nd a cos62 b sin29 c cot47 4 Find the derivative of each of the functions in Problem 8a7e of the last homework set In each case nd the rst positive time t for which the derivative is 0 and the rst positive t for which the derivative reaches its maximum value then explain in practical terms what is happening at each of these moments 5 Find the derivative of a y sins cos 5v b y sincosv c y vtan 5v d an In y 7 1 sin a 6 Suppose that the point 7T3 7T4 is on the implicitly de ned curve const Use the tangent line approximation to determine the y coordinate of the 3point on the curve With m coordinate g m 7 Suppose that the point 1 15 is on the curve xsin7Ty3 ysin7Tv4 const Use the tangent line approximation to determine the y coordinate of the point on the curve with m coordinate 11 8 A rope is tied at its left end and its right end is moved up and down to create a sinusoidal wave of period 6 ft The equation of the shape of the rope standing wave is then y Asin Asin where we are taking the origin to be the point where the rope is fastened and A is the amplitude Now actually the amplitude A varies with time as the rope oscillates Suppose that the maximum value of A is 2 ft and it varies sinusoidally with time t with period 2 sec A 2 sin7rt Thus the vertical coordinate y of a point on the rope depends on both its 5v coordinate and the time t y 2sin7rt sin a Sketch the appearance of the rope at quarter second intervals from t 0 to t 2 sec all on the same graph b Suppose your vision is blocked by a horizontal ledge at a height of 6 inches above the m axis so you see only the tips of the waves Write an equation that gives the x value where the wave emerges above the ledge implicitly as a function oft ie write an equation of the form Fvt const c What is the rst x value where the rope emerges above the ledge when t 16 sec d How rapidly is that rst x value moving leftward at the instant t 16 sec e Approximately where will that rst point where you see the rope be located when t 02 sec Use parts c and d and the tangent line approximation CHAPTER 10 Curves Given Parametrically So far we have had two different ways of giving a curve The rst was explicitly ie in the form y For example the equation y r2 7 x2 gives us the upper semicircle of radius 7 centered at the origin The second way was implicitly usually in the form Fa y constant For example the equation x2 312 r2 gives us the circle both upper and lower semicircles of radius 7 centered at the origin There is a third fundamental way a curve can be given paralnetrically This means that we have a third variable called the paralneter which is used to de termine where we are on the curve in the Ivy plane To give a curve parametrically means to give twu formulas one for In and one for g each in terms of the parameter This parameter might have a geometrical meaning 7 for example an angle denoted 0 7 or it may be the time t In the case of an angle parameter our two formulas take the form In 90 y h0 in the case of a time parameter the formulas take the form In gt y ht Often we don t bother to use separate letters like 9 and h instead writing simply xt for the formula for In in terms oft and yt for the formula for y in terms of t The best way to understand parameterized curves is through examples Example 101 Circles Suppose we want to describe the points on the circum ference of the circle of radius 7 centered at the origin A common way to do this is in terms of the angle 0 which the line from the point to the origin makes with the positive x axis measured counterclockwise That is as 0 goes from 0 to 27139 radians we sweep once around the circle What we want to do is express the cv and y coordinates of our points on the cir cumference in terms of 0 To do this we drop the perpendicular from the point to the m axis obtaining a triangle with hypotenuse 7 and with legs In and y So we can use the trig functions sin and cos to express In and y in terms of 0 5vv0rcos0 yy0rsin0 0303277 65 66 10 CURVES GIVEN PARAMETRICALLY This says that as 0 goes from 0 to 27139 if we determine In and y by these two equations then the corresponding point x y goes exactly once around the circumference of the circle A special case of this is the unit circle 7 this is simply the case 7 1 Thus the points on the unit circle can be described parametrically by the equations xvt9cos0 yy0sin0 0303271 The fact that the m coordinate is simply the cosine and the y coordinate is simply the sine is not surprising since we de ned the cosine and sine of an angle to be the cv and y coordinates of the point that you get by going around the unit circle through the given angle If we want to shift the circle of radius 7 so that its center is at the point h k we can do this by adding h to the x formula and k to the y formula 5vv0hrcos0 yy0krsin0 0303271 This is the circle which is given implicitly by the equation In 7 h2 y 7 k2 r2 1f instead of 0 S 0 S 27139 we had written say 0 S 0 3 7T then we would have described only half of the circle the upper semicircle If we had written 0 S 0 S 47139 then we would have described going around the circle twice Example 102 Ellipses Let us suppose that we have put our ellipse on the axes in the following convenient way its center is at the origin its longer axis of symmetry 7 of length 2a 7 is along the m axis and its shorter axis of symmetry 7 of length 217 7 is along the y axis Such an ellipse can be obtained by taking the unit circle and dilating it by a factor of a in the x direction and by a factor of b in the y direction If a or b is less than 1 the dilation is actually a contraction In terms of the parametric equations we can accomplish this dilation simply by multiplying the x equation for the unit circle by a and multiplying the y equation for the unit circle by b 5vvt9acos0 yy0bsin0 0303277 This is the parametric form for our ellipse 7 the ellipse which can be given implicitly in the form 72 72 1 Note A word of warning though about the parameter 0 for an ellipse 1t no longer has the meaning of the angle formed by the line from the point 5vy to the origin For example suppose a 2 l7 1 Then the point on the ellipse corresponding to the parameter value 0 7T4 450 is the point In 2cos7T4 y sin7T4 Q and it is easy to see that the line joining the point 1414 0707 to the origin does not make a 450 angle with the m axis 1 Trajectories It is very common to want to describe a curve in terms of the time parameter t In that case we sometimes call the curve the path or trajectory of our point which we think of as a small object or particle As If goes from an initial time a to a nal time b the equations 5vt yt tell us where the particle is at that particular time t 1 TRAJECTORIES 67 Example 103 Suppose a particle is traveling counterclockwise at 3 revsec around the circle of radius 7 centered at the origin It starts at the point 7 0 at time 25 0 We want to give a parametric description of the motion in terms of 25 To do this we nd an expression in terms of 25 for the angle 0 through which the particle has rotated and then we substitute this expression in the equations in Example 101 Because we shall be measuring angles in radians we rst translate the angular velocity of 3 revsec into units of radsec Since 1 rev 27139 rad we obtain d0d25 3 27139 67139 radsec Thus 0 starts at 0 at time 25 0 because our point starts at 7 0 on the positive x axis and it increases at the constant rate of 67139 Hence after 25 seconds it has value 0 6713925 Finally substituting this expression for 0 in the equations in Example 101 we obtain 5v 5v25 rcos6713925 y 3125 739sin6713925 0 S 25 S 13 where we chose to take 25 between 0 and 13 so that we would describe exactly one rotation of the particle around the circle Example 104 Motion at constant velocity between two points Suppose that at time 25 0 you start at the point P 2 5 and travel at constant velocity along the line from P to the point Q 6 3 arriving 5 seconds later at Find parametric formulas for this motion To do this we use a general principle for setting up parametric equations work with the x and ydirections separately In the x direction in this example we are starting at CU 2 and traveling to CU 6 in 5 seconds moving at constant speed This constant speed 7 called the horizontal component of velocity and denoted 5v d7 7 is the distance traveled divided by time 6 7 25 08 Thus the formula for 5v25 is 2 0825 ie starting x value plus speed times time Similarly in the y direction we compute the vertical component of velocity y to be 3 7 55 704 so that y25 is given by the formula 5 7 0425 This is all going on between 25 0 and 25 5 Thus our parametric description of the motion is 5va2520825 yy25570425 032535 We can check these formulas by noting that when 25 0 they give us 5v0 2 310 5 ie we re at the point P and when 25 5 they give us 735 6 315 3 ie we re at the point Example 105 Falling body in two dimensions Again we are interested in an object falling under the in uence of gravity we neglect air resistance In the past see Example 69 our object was moving only in one dimension 7 up and down For example if we drop a ball from a window 196 meters high its height y at time 25 is given by the formula y 7 9252 yo 749t2196 here 9 98 msec2 the acceleration is negative ie it is 79 because we are choosing up to be the positive direction Now suppose that instead of dropping the ball from the window we throw it horizontally out from the window at 10 msec The ball then moves in the Ivy plane where we have chosen the side of the building as the y axis and the line along the ground under the ball s path as the m axis Its path is a parabola as we shall soon see Our task in this example is to write parametric equations for this motion 68 10 CURVES GIVEN PARAMETRICALLY To do this we again consider the cv and y directions separately In the m direction there is no force affecting the velocity and so the 10 msec hor izontal component of velocity of the ball remains unchanged That is as in Example 104 in the x direction we are simply traveling at constant speed Since we start at initial loca tion In 0 the formula for In at time t is x05 1015 In the vertical direction we have a falling body problem and so can use the formula y 759252 vot yo Here it is important to understand what v0 means v0 is the initial velocity in y direction In this example v0 is 0 since the ball s initial velocity is entirely horizontal In other words in the y direction the problem is exactly as if we had simply dropped the ball from a height of 196 Thus we have 3125 749252 196 Finally we want to know the time interval where the formulas for In and y apply The motion starts at time t 0 and it ends when the ball hits the ground The ball hits the ground when y 0 and so to nd when this happens we set y 749252 196 equal to 0 and solve fort We obtain 252 4 t i2 and so If 2 since 72 does not make practical sense To summarize we have the following parametric description of the motion In mos 102 y 3125 749252 196 0 g t g 2 The actual trajectory in Example 104 see the picture above is a graph in the Ivy plane that is a function of the form y In this example we can easily nd the formula for this function by eliminating t in the formulas for x05 and That is since In 1015 we can write t 01x and then substitute this in place of t in the formula for y y 74901v2 196 7004922 196 Thus the path is a parabola more precisely part of a downward opening parabola 2 The Velocity Vector The velocity of an object moving in the plane is an example of a vector A vector is something which has magnitude and direction It can be visualized as an arrow of a certain length or magnitude pointing in a certain direction The magnitude of the velocity vector is called the speed of the object Example 106 The same as Example 105 except that at time t 0 the ball is thrown at an angle of 600 above the horizontal Here the initial velocity vector makes an angle of 600 above the horizontal and has a magnitude of 10 msec That is it is partly horizontal and partly vertical The rst step in solving this problem is to resolve the initial velocity vector into its horizontal and vertical components To resolve the velocity vector we draw the velocity triangle having the velocity vector as its hypotenuse see 2 THE VELOCITY VECTOR 69 gure This triangle has hypotenuse 10 msec mak ing an angle of 60 above the horizontal Then the horizontal leg of the triangle 7 the initial cv velocity vojmriz 7 can be found using the cosine and the vertical leg of the triangle 7 the initial y velocity vo ert 7 can be found using the sine We thus ave 10 msec vmhoriz 10 cos 60 10 05 5msec vow 10 sin 60 10 0866 866 msec We are now ready to write the equations for x05 and 3105 as we did in Exam ple 105 The horizontal component of velocity which remains constant is not 10 msec but rather only the horizontal part namely 5 msec So In 525 In the falling body formula y 759252 v0ty0 we now have a nonzero initial A vertical component of velocity v0 vo ert 866 msec Thus we obtain 5v x05 525 y 3125 749252 86625 196 The trajectory of the ball is shown in the gure at the right 10 To nd the time interval until the ball hits the ground we again set y 0 and use the quadratic for mula to nd the time t when we re on the ground There will be 2 roots of m which we want the positive one 31 7866 7 8662 7 47491962749 307 sec So the time interval is 0 S t S 307 Example 107 The same as Example 106 except that the ball is thrown at an angle of 60 beluw the horizontal Here the only change is that the tip of the velocity triangle points downward ie its vertical leg points in the negative direction Thus in the equations of motion the only thing that changes is the sign of vo ert in the falling body formula 5v ms 525 y 3125 749252 7 86625 196 Of course the ball hits the ground much sooner see gure on the following page In all three cases in Examples 1057107 the trajectory is part of a parabola as we can see by eliminating t and writing y in terms of Ir rather than in terms of 25 Suppose that we have a trajectory 05 yt as in Examples 1037107 There are three types of derivatives we might be interested in One is the horizontal component of velocity dcvdt The second is the vertical component of 70 10 CURVES GIVEN PARAMETRICALLY velocity dydt The third is not a velocity at all but rather the slope of the curve in the Ivy plane dydx If we nd dydcv at some instant we can determine the direction in which Ay the object is headed at that instant How do we nd dydcv if we do not have y written in terms of Ir ie if we do not have a formula for the function y Well according to the chain rule 15 at any instant our three derivatives are connected by the relation dy dy dd 10 dt dm 39 dt39 In the past we have used the chain rule to nd the left hand side Now we want to nd dydx So the 5 form of the chain rule we need is x dy dydt 3 W0 dd 7 dcvdt 7 d In words this says that the slupe 0f the path is equal t0 the vertical compunent 0f velocity divided by the hurizzmtdl compunent 0f velucity This makes sense if the horizontal component of velocity is small and the vertical component of velocity is large we are traveling at a steep angle while if the horizontal component of velocity is large and the vertical component of velocity is small we are traveling at a shallow angle In nding dydcv it is important always to remember that you take the ratio of velucities over EU NOT simply the ratio of y to In In other words you must take the derivatives with respect to t of the yt and xt formulas before taking their ratio Example 108 Suppose we want to know the instant when the ball in Example 106 reaches the peak of its trajectory At that instant the path has a horizontal tangent line ie dydcv 0 Since dydcv this slope is zero if and only if the numerator is 0 Thus to nd the instant when the ball reaches its highest point we set the vertical velocity equal to 0 and solve for t This is the same principle that we saw in our earlier one dimensional falling body problems the ball is at the peak when its vertical velocity is zero Warning Be sure not to confuse this procedure where we set equal to zero with setting y itself equal to 0 which we do to nd the time when the ball hits the ground Since 749t2 866t 196 79825 866 we have 79825 866 0 ie t 86698 088 sec If we also want to know the maximum height reached by the ball we take this value oft and substitute it in the formula for the height y ie ymax y088 749 0882 866 088 196 234 m Example 109 Suppose that we want to know the direction in which the ball in Example 106 is headed after 05 sec and also at the instant when it hits the ground We rst take the time derivatives of y and Ir dy i d 2 i dd 7 d i E 7 dt 49t 866t 196 7 9815 866 and dt 7 dt5t 7 5 Then we have d y 7 7 1 dm 7 7 98t 8665 2 THE VELOCITY VECTOR 71 Thus when t 05 we have dydcv 798 05 8665 0752 At that instant the ball s location is m05y05 25227 so that the tangent line to the path at the point 25 227 has slope 0752 We can also determine the angle at which the ball is traveling at time t 05 By this we mean the angle above the horizontal of the tangent line to the path To do that we observe that the slope of a line AyAa is opposite over adjacent ie the tan of the angle that the line makes with the horizontal So to determine the angle we use the inverse tan on our calculatorl Arctan0752 37 At the same time t 05 we could also construct the velocity triangle having horizontal leg 9305 5 and vertical leg y05 376 and compute the hypotenuse which is the speed speed at time 05 V 52 3762 626 msec 5 msec 372 376 J msec i 5 msec 777 t 05 sec 25 307 sec 7214 msec Next we make the same computations at the instant when the ball hits the ground ie at t 307 sec We have 5amp307 5 y307 798 307 866 7214 and so speed M52 7214 220 The angle at which the ball hits the ground is Arctan72145 777 Thus the ball hits the ground traveling at 22 msec at an angle of 770 below the horizontal 1Arctan is usually denoted tan 1 on calculators 72 10 CURVES GIVEN PARAMETRICALLY Homework 10 1 Find parametric equations 05 yt for the point P if a P travels at constant velocity from 23 to 1173 between t 0 and t 3 b P travels at constant speed 65 unitssec from 12 in the direction of 137 c P travels at constant speed 10 unitssec from the origin at an angle of 300 above the m axis 2 A bicycle has wheels of radius 04 meter Find a the angular velocity of the wheels in radsec if the bicycle is traveling at 10 msec b the angular velocity of the wheels in revsec if the bicycle is traveling at 10 msec c the speed of the bicycle in msec if the wheels are turning at 5 revsec d the speed of the bicycle in kmhr if the wheels are turning at 15 radsec 3 A truck has wheels of radius 16 ft Find a the speed of the truck in mph if the wheels are turning at 5 revsec b the angular velocity of the wheels in radsec and in revsec if the truck is traveling at 60 mph 4 Find a formula for the distance in feet that a car has traveled during the rst If sec if its 10 inch radius wheels a are turning at the constant rate of 24 radsec b are turning at the constant rate of 9 revsec c start from rest and accelerate at the rate of 18 radsec per second 5 A wheel of radius 3 is centered at the origin Find parameteric equations for the point P on the rim if a the wheel is spinning counterclockwise at 05 radsec and P is at 30 at time t 0 b the wheel is spinning clockwise at 2 revsec and P is at 03 at time t 0 c the wheel is spinning clockwise at 1 revolution each 5 seconds and P is at at time t 0 d the wheel is spinning counterclockwise at 10 revsec and P is at 0 73 at time t 0 6 A girl is playing on a atcar which is moving in the positive x direction at 10 msec At time t 0 she throws the ball straight upward as viewed from the moving train with an initial velocity 245 msec from a height 1 meter above the atcar From her point of view the ball goes straight up and down and lands on the atcar directly below the point from which she threw it However an observer on the ground sees the ball travel in an arc Set up a coordinate system which is xed to the earth ie one which is not moving along with the train with the y axis pointing up and the x axis pointing in the direction of motion of the train Choose the origin to be the point in space which is 1 meter below the point from which the ball was released Neglect air resistance and take 9 98 msec2 a Find parametric equations for the motion of the ball in this xed coordinate system b Find the instant when the ball reaches the peak of its trajectory c Find the instant when the ball hits the atcar 2 THE VELOCITY VECTOR 73 d Describe in words how you would nd the angle at which the ball hits the atcar in these coordinates 7 Suppose that an object is thrown at 70 ftsec at an angle of 45 beluw the horizontal from a window 163 ft above the ground a Express both the horizontal distance In from the window and the height y above the ground as functions of t b Find dydcv in terms of t c Find the angle below the horizontal that the object is pointed when t 05 sec and when t 1 sec d Find the velocity vector its magnitude and angle below the horizontal of the object at the instant when it hits the ground 8 Suppose that a ball is thrown from ground level at v msec at an angle of 04 above the horizontal Take the origin to be the point from which the ball is thrown a Find formulas for x05 and The constants v and 04 should appear in your formulas b Find the time t when the ball hits the ground expressing your answer in terms of v and 04 There are two ways to do this Set yt 0 and solve for 25 ii Find the time when the ball reaches its peak see Example 108 and double it because it takes an equal time to fall back to the ground Use both methods and see that your answers agree c Find the horizontal distance traveled by the time the ball hits the ground expressing your answer in terms of v and Oz CHAPTER 11 Combined Parametric Motions Sometimes we are interested in the motion of a point P which has a simple motion relative to some other point C which in turn also is moving in a particular way It is not hard to write parametric formulas 5vt yt for the point P if we proceed carefully in a step by step manner Example 111 The baton Suppose we want to describe the motion of a tip of a baton as it rotates after being thrown into the air More precisely suppose that at time t 0 a 2 foot long baton is situated vertically with the tip we are interested in at height 6 ft above ground level and the other tip at height 4 ft At that moment the baton is tossed at 20 ftsec at an angle of 45 above the horizontal and it is also set spinning at 3 revsec clockwise in the plane of its trajectory It could have been set spinning in a different plane but that would have made this into a three dimensional problem and that would be too complicated for our present purposes To set up the parametric equations for the motion of the tip of the baton the basic procedure is to divide the motion into two parts the motion of the center of the baton and the rotation of the tip around the center The rst of these is a falling body problem and the second is a circular motion problem We shall take the y axis to be the vertical line along which the baton is situated at time t 0 and we shall take the x axis to be the line on the ground under the baton s trajectory According to a principle of physics the center of the baton follows the same path as a small object say a ball that is thrown with the same initial velocity In other words the rst half of our procedure 7 the motion of the center of the baton 7 is like Example 106 of the last chapter Let CECth t ycent denote the a and y coordinates of the center of the baton at time t Repeating the procedure used in Example 106 7 with g 32 ftsec2 vmhoriz 20 cos 45 14 ftsec rounded to the nearest ftsec vow 20 sin 45 14 ftsec 7 we obtain meandt 141 yeah 716252 1425 5 Notice that yo 5 ft because the initial location of the center is half way between 4 and 6 Now let x05 denote the a and y coordinates of the tip of the baton at time t The second half of our procedure consists in determining how much xt differs from xcentt and how much yt differs from ycent t ie how far the tip is from the center in the cv and y directions Let 0 be the angle in radians through which the baton has spun at time t We have the following diagram 75 76 11 COMBINED PARAMETRIC MOTIONS mcenhyc t Ply i Since the tip is 1 ft from the center7 the hypotenuse of this triangle is 1 Hence 5v 7 CECth sin0 and y 7 ycent cos 0 ie7 9305 1425 sine 3125 716252 142 5 cost9 It remains to write 0 in terms of t We do this as in Example 103 of the last chapter7 obtaining 0 67725 We conclude that x05 1425 sin67139t7 yt 716252 1425 5 cos67rt Notice that here it is the m coordinate that involves the sine and the y coordinate that involves the cosine7 the reverse of what we got in our earlier examples of circular motion The reason is that in this problem it was convenient to measure the angle 0 clockwise from the positive y direction rather than counterclockwise from the positive m direction This different way of de ning the angle accounts for the reversal of the sine and cosine The dotted line is the path of the center of the baton xcentt 1425 The solid line is the path of the ycent t 716252 1425 5 top of the baton x05 1425 sin67rt 3125 716252 1425 5 cos67rt 1 9 20 3 x 20 2 Numbers indicate how many twientieths of a 1 second have elapsed The graph above shows the trajectory of the center of the baton dotted line and the trajectory of the tip of the baton solid line between t 0 and t 1 sec 11 COMBINED PARAMETRIC MOTIONS 77 The numbered points from 0 to 20 give the location at intervals of 005 seconds If you join the numbers on the dotted line with the corresponding numbers on the solid line you will see the spinning of the top half of the baton Even though the equations for xt and yt are not tremendously complicated the path of the tip of the baton is a rather elaborate curve In this example we cannot nd an algebraic or trig formula for y in terms of Ir because we cannot eliminate 25 So the parametric method is the only reasonable way to describe this curve Example 112 The cycloid Suppose a bicycle wheel of radius 7 is rolling along the m axis A pebble lying at the origin is picked up in the tread of the tire as it rolls over the origin We would like to describe the path of the pebble as it moves along the m axis spinning around the center of the wheel as the wheel turns This path is called a cycloid As in Example 111 the procedure is to split the problem into two parts describe the motion of the center of the wheel and then determine the location of the pebble relative to the center The rst part is easy because the center of the wheel is simply moving along the straight line y 7 parallel to the x axis at a height of 7 the second part is a circular motion problem We shall express the coordinates of the center meant ycent and the coordinates of the pebble 5v y in terms of the following parameter the angle 0 in radians through which the wheel has rotated after picking up the pebble at the origin We shall also introduce another letter 5 to denote the linear distance the wheel has traveled s is the distance along the x axis from the origin to the point of contact of the wheel and s is also the m coordinate of the center of the wheel There is a simple relation between the distance 5 the wheel has traveled and the angle 0 in radians through which it has rotated Namely the distance traveled along the x axis is also equal to the distance measured around the circumference through which the wheel has turned here we are assuming that the wheel rolls without slipping or sliding For example every time the wheel rotates through 1 revolution ie 27139 radians the wheel moves a distance equal to the circumference 2777 More generally if the wheel rotates through an angle of 0 radians the wheel moves a distance equal to 07 This is because the distance along the circumference of the unit circle of the part of the circumference corresponding to an angle of 0 radians is simply 0 by the de nition of the measure of an angle in radians In a circle of radius 7 7 ie the unit circle magni ed by a factor of 7 7 the distance along the circumference will be 7 times as great 9 Q To summarize we have the relation 57 0 78 11 COMBINED PARAMETRIC MOTIONS The center of the wheel always has the same y coordinate 7 a constant and its m coordinate is s r0 Thus we have meant 7 07 ycent0 7 The second part of our procedure is to nd the difference between the coordinates as y of the pebble and the coordinates meant ycent of the center Since the wheel is rolling from left to right and the radius from the center to the pebble is pointing down when we start out at the origin it is convenient to measure the angle 0 clockwise from the negative y direction We have the following diagram ent 937 3 If 0 is an acute angle as in the diagram then sin0 and cos0 are positive Since the pebble is to the left and beluw the center we must subtract the horizontal leg of the triangle from CECth to get In and we must subtract the vertical leg of the triangle from ycent to get y Using the trig functions of 0 to get the horizontal and vertical legs of the triangle whose hypotenuse is equal to 7 we end up with 230 Cccent0 7 7 sin0 r0 7 7 sin 0 310 ycentw 7 7 cos0 7 7 7 cos 0 This parametric curve is drawn below Each time 0 goes through 27139 radians the wheel turns around once and the pebble goes through one motion from cusp to cusp 27 7T7 27 quot 37Wquot 11 COMBINED PARAMETRIC MOTIONS 79 The gure below shows the point on the curve for several values of 0 Now suppose that we know how the bicycle is moving with time For example suppose that it is traveling at constant speed 1 and it passes the origin at time t 0 That is dsdt v To express the position of the pebble in terms of 25 all we have to do is express 0 in terms of t and then substitute in the above equations for as Since 0 has a simple relation to s 7 namely 0 s 7 it follows that dOdt has a simple relation to dsdt namely taking ddt of both sides 10 1 d5 1 1 dt 7 dt 7 Because 0 is 0 at time t 0 and it increases as the constant rate vr it follows that we have 0 vrt Thus when the bicycle is traveling at constant velocity v we obtain the following equations for the motion of the pebble x05 rvrt 7 rsinvrt vt 7 rsinvtr 3125 7 7 rcosvtr 80 11 COMBINED PARAMETRIC MOTIONS Example 113 The trochoid The same as in Example 112 except that When the Wheel rolls over the origin instead of getting stuck in the tread on the outside of the tire the pebble ies up and gets stuck in a spoke at a distance 1 directly below the center Where a lt 7 Thus the pebble again rotates around the center but at a distance a rather than 7 from the center The path of this pebble is called a trochoid In this problem everything is just like in Example 112 except that in the circular motion part the hypotenuse of the triangle which is the line from the center to the pebble after the Wheel has rotated through an angle of 0 radians is now a rather than 7 This leads to the formulas 230 Cccent0 7 asin0 r0 7 a sin 0 310 ycentw 7 acos0 7 7 acos0 Here is a picture of the trochoid path in the case When a rQ 27 7T7quot 2717quot 37 11 COMBINED PARAMETRIC MOTIONS 81 Homework 11 1 Suppose that the wheel in Example 112 has radius 1 a Graph the cycloid by plotting the points where the pebble is located after the wheel rotates through all multiples of 300 from 0 to 360C ie 0 goes through all multiples of 776 from 0 to 27139 b Use the chain rule in the form dy i dyd0 E docd0 to nd a formula for dydcv in terms of 0 c Draw the tangent lines to the curve in part a at all of the plotted points from 0 776 to 0 11776 and use part b to nd the slope of each tangent line 2 A spring is causing an object to oscillate up and down sinusoidally between y 3 and y 73 At t 0 the object is at maximum height y 3 and it makes one full motion from 3 to 73 and back to 3 twice a second Meanwhile the whole apparatus is moving at constant velocity 2 unitssec down the m axis starting at 0 at time t 0 a Find b Find Namely use the information given to graph y as a sinusoidal function of time in the ty plane and ii nd A B C and D in the formula y Asin t 7 C D c Graph the path of the object in the cry plane Find the formula for y in two ways Use the graph of the path to nd a b c and d in y asin 7 6 d where we re using small letters for the constants so as to avoid confusion with the sinusoidal function in part ii Eliminate t and use part b to write y directly in terms of In Use both methods and check that your answers agree 3 The same as Problem 2 except suppose that the whole apparatus is moving down the x axis according to the formula x05 252 Notice that the vertical motion yt as a function of time is unaffected by this change However in the horizontal direction the object is accelerating ie its velocity rather than being constant is steadily increasing according to the formula vt 2t and the acceleration at 2 is constant Here x05 t2 can be thought of as 59252 where the constant acceleration 9 here is 2 unitssec2 and of course unlike gravity which pulls in the negative y direction here the constant acceleration is pulling in the positive x direction Also notice that your graph of the path y is no longer sinusoidal the distance between peaks becomes greater as the object accelerates to the right The formula for y can be obtained by substituting t in place of t in the formula in part 4 A wheel of radius 2 ft is centered at the point 0 100 ft and is rotating clockwise at the constant rate of one revolution every 2 seconds At time t 0 a pebble on the perimeter of the wheel is located at the point 0102 At time t 0 the wheel is dropped Write down formulas for the cv and y coordinates of the pebble as functions of time and graph its trajectory for t between 0 and 25 sec plot points at intervals of 14 sec 5 A wheel rolls down the x axis from left to right without slipping at constant speed 6 msec The diameter of the wheel is 15 m As it passes over the origin at 82 11 COMBINED PARAMETRIC MOTIONS time t 0 a pebble gets stuck in a spoke right under the center at a distance 25 cm from the rim Find x05 and yt for the pebble 6 In an episode of the TV program Northern Exposure the hero Chris and a mathematician named Amy send his motorcycle careening over a cliff We want to describe the path of the point P at the top of one of the Wheels at the instant t 0 When the Wheel starts over the cliff Suppose that the motorcycle Wheel has a radius 025 H17 and until time t 0 the motorcycle is traveling at 10 msec Neglect air resistance7 and take 9 98 msec2 Suppose that the Wheel is traveling in the Ivy plane7 the motorcycle is moving horizontally along the negative x axis until time t 07 and the origin is the point at the edge of the cliff Where the Wheel is touching at the instant t 0 Find the coordinates 05 yt of P at time t While the motorcycle is falling CHAPTER 12 Related Rates Suppose we have two variables In and y in most story problems the letters will be different but for now let s use In and g which are both changing with time A related rates problem is a problem where we know one of the rates of change at a given instant 7 say dcvdt 7 and we want to nd the other rate dydt at that instant If y is written in terms of Ir ie y then this is easy to do using the chain rule dy 7 dy dd dt dcv dt39 That is nd the derivative of plug in the value of In at the instant in question and multiply by the given value of Cr dcvdt to get dydt But often In and y will be related in some other way for example In or Fvy const or perhaps Fvy Gvy where Fvy and Gvy are expressions involving both variables In all cases you can solve the related rates problem by taking of both sides plugging in all the known values namely 5v y and EU and then solving for To summarize here are the steps in doing a related rates problem 1 Decide what the two variables are 2 Find an equation relating them 3 Take ddt of both sides 4 Plug in all known values at the instant in question 5 Solve for the unknown rate Example 121 The radius of a circular puddle is 3 m and it is increasing at the rate of 1 cmmin How fast is the puddle s area increasing Here the two things that change with time are the radius 7 and the area A We know drdt 1 cmmin 001 mmin We want to nd dAdt Thus in this problem 7 is playing the role of In and A is playing the role of y The relation between the two variables is A 7772 Taking ddt of both sides we obtain dA 2 d7 7T r dt dt Notice that since we are taking derivatives with respect to t the derivative of r2 is 2rd T by the rule for the derivative of a power of a function the un rule In this it 7 case we alternately could have used the chain rule for the variables A 7 and t dA dA 17 d7 27T7quot dt dr dt dt We now plug in the information given At the instant in question 7 3 and drdt 001 Thus A 27139 3 001 01885 mZmin 83 84 12 RELATED RATES Note A crucial point to notice in this and similar problems is that even though a numerical value is given for 7 7 must be regarded as a variable not a constant The value 0f the variable at the instant in questiun cannut be plugged intu the equatiun until after taking ddt 0f the equation Example 122 You are in ating a spherical balloon at the rate of 7 cmssec How fast is its radius increasing at the instant when 7 4 Here the variables are the radius 7 and the volume V We know dVdt and we want drdt The two variables are related by means of the equation V gums Taking ddt of both sides gives dVdt 4717 2 drdt We now substitute the values we know at the instant in question 7 47142 drdt Solving for drdt we obtain drdt 76471 00348 cmsec Example 123 A plane is ying at 500 mph at an altitude of 3 miles in a direction away from where you re standing ie the point on the ground directly beneath the plane is moving away from you How fast is the plane s distance from you increasing at the moment when the plane is ying over a point on the ground 4 miles from you To see what s going on we rst draw a triangle whose hypotenuse is the line from you to the plane see gure at right The vertical leg of the triangle is a constant 7 3 mi 7 which does not change in the course of the problem The horizontal leg In is the variable whose rate we know 7 its rate EU is the speed of the plane The hypotenuse y is the variable whose rate we want The equation relating In and y is the Pythagorean theorem x2 32 312 We next take ddt of both sides getting 22 5v 2y At the instant in question we re told that In 4 and Cr 500 What about y From the Pythagorean theorem we have y 42 32 5 at the instant in question Thus we obtain 2 4 500 2 5 g and solving for gives the value 400 mph That is the plane is receding from you at 400 mph Until you think about this for a while it might seem counterintuitive that although the side y is greater than the side In its rate of increase is less Example 124 Water is poured into a conical container at the rate of 10 cmssec The cone points directly down see side view at right and it has a height of 30 cm and a base radius of 10 cm How fast is the water level rising when the water is 4 cm deep at its deepest point In this problem the water forms a conical shape within the big cone whose height and base radius 7 and hence also its volume 7 are all increasing as water is poured into the container This means that we ac tually have three things varying with time the water level h the height of the cone of water the radius r of the circular top surface of water the base radius of the cone of water and the volume of water V And we have the relation V 7Tr2h We know dVdt and we want dhdt At rst something seems to be wrong we have a third variable 7 whose rate we don t know But the dimensions of the cone of water must have the same proportions as those of the container In other words 7 and h t into the above diagram of similar 00 ac 12 RELATED RATES 85 triangles from which it follows that ie 7 h3 So we can eliminate 7 from the problem entirely V 7Th32h 2 quot7h3 We now take ddt of both sides and then plug in h 4 and dVdt 10 obtaining 10 3 42 Thus dhdt 901671 179 cmsec Example 125 A swing consists ofa board at the end ofa 10 ft long rope Think of the board as a point P at the end of the rope and let Q be the point of attachment at the other end Suppose that the swing is directly below Q at time t 0 and is being pushed by someone who walks at the speed of 6 ftsec from left to right Find a how fast the swing is rising after 1 sec b the angular speed of the rope in degsec after 1 sec Q In doing this problem we must start out by ask ing What is the geometric quantity whose rate of change we know and what is the geometric quantity whose rate of change we re being asked about Note that the person pushing the swing is moving hor izontally at a rate we know In other words the horizontal coordinate of P is increasing at 6 ftsec 1n the Ivy plane let us make the convenient choice of putting the origin at the location of P at time t 0 ie a distance 10 directly below the point of attach ment Then the rate we know is dxdt and in part a the rate we want is dydt the rate at which P is rising In part b the rate we want is dOdt where 0 stands for the angle in radians through which the swing has swung from the vertical Actually since we want our answer in degsec at the end we must convert dOdt from radsec by multiplying by 18077 a From the diagram above we see that we have a right triangle whose legs are In and 10 7 y and whose hypotenuse is 10 Hence x2 10 7 y2 100 Taking ddt of both sides 7 and recalling that the derivative of the square of a function of t is twice the function being squared times the time derivative of that function 7 we obtain 2x5r 210 7 y0 7 0 We now look at what we know after 1 sec namely 5v 6 because 5v started at 0 and has been increasing at the rate of 6 ftsec for 1 sec y 2 because we get 10 7 y 8 from the Pythagorean theorem applied to the triangle with hypotenuse 10 and leg 6 and Er 6 Putting in these values leaves us with 2 6 6 7 2 8g 0 from which we can easily solve for 3 getting 3 45 ftsec b Here our two variables are In and 0 so we want to use the same right triangle as in part a but this time to relate 0 to CU Since the hypotenuse is constant equal to 10 the best way to do this is to use the sine sin0 x10 Taking ddt of both sides we obtain cos 015 At the instant in question 15 1 sec when we have a right triangle with sides 6 8 10 the cos0 in this equation is 810 We also substitute 5v 6 on the right We get 086 06 ie 0608 075 radsec 43 degsec Sometimes there are several variables that change with time you know the rates of all but one of them and you want to know the remaining rate As in the case when there are just two variables take ddt of both sides of the equation relating all of the variables and then plug in all of the known values and solve for the unknown rate 86 12 RELATED RATES Example 126 A road going SouthiNorth crosses a road going WestiEast at the point 0 Car A is driving North along the rst road and car B is driving East along the second road At time t car A is at to the North of O and car B is 1705 to the East of 0 Suppose that at some instant you know the values of at 1705 the velocity U105 125 of car A and the velocity v2t of car B Find a formula in terms of these known values for the rate at which the distance between the two cars is increasing Let c be the distance from car A to car B By the Pythagorean Theorem c2 a2 172 We now take ddt of both sides of A 0at the Pythagorean Theorem obtaining 2cc39 2az39z 21717 Dividing by 2c gives the following formula for the unknown rate ail bli avl Mg 6 a2b239 North B W 0 South Notice how this problem differs from Example 123 In both cases we took ddt of the Pythagorean Theorem However in Example 123 one of the sides was a constant the altitude of the plane and so ddt of the square of that side was simply zero In Example 126 on the other hand all three sides of the right triangle are variables As always in a story problem it s important to read the problem carefully enough to determine at the start what are the variables and what are the constants 12 RELATED RATES 87 Homework 12 Part A Basic problems 1 A cylindrical tank has radius 20 cm How fast does the water level in the tank drop when the water is being drained at 25 cmssec 2 A ladder 13 m long rests on horizontal ground and leans against a vertical wall The foot of the ladder is pulled away from the wall at the rate of 06 msec How fast is the top sliding down the wall when the foot of the ladder is 5 m from the wall 3 A rotating beacon is located 2 miles out in the water Let A be the point on the shore that is closest to the beacon As the beacon rotates at 10 revmin7 the beam sweeps down the shore once each time it revolves Assume that the shore is straight How fast is the point where the beam hits the shore moving at an instant when the beam is lighting up a point 2 miles downshore from the point A 4 A baseball diamond is a square 90 ft on a side A player runs from rst base to second base at 15 ftsec At what rate is the player s distance from third base decreasing when she is half way from rst to second base 5 Sand is poured onto a surface at 15 cma sec7 forming a conical pile whose base radius is always half its altitude How fast is the altitude of the pile increasing when the pile is 3 cm high 6 A boat is pulled in to a dock by a rope with one end attached to the front of the boat and the other end passing through a ring attached to the dock at a point 5 ft higher than the front of the boat The rope is being pulled through the ring at the rate of 06 ftsec How fast is the boat approaching the dock when 13 ft of rope are out 7 A balloon is at a height of 50 meters7 and is rising at the constant rate of 5 msec At that instant a bicycle passes beneath it7 traveling in a straight line at the constant speed of 10 msec How fast is the distance between them increasing 2 seconds later 8 A cone like vat has square cross section rather than circular cross section7 as in the case of the usual cone The dimensions at the top are 2 m X 2 H17 and the depth is 5 m If water is owing into the vat at 3 m3min7 how fast is the water level rising when the depth of water at the deepest point is 4 m Note the volume of any conical shape is 13 X height gtlt area of base 9 The sun is setting at the rate of 14 degmin on a day when the sun passed di rectly overhead How fast is the shadow of a 25 meter wall lengthening at the moment when the shadow is 50 meters See the pic ture 10 A woman 5 ft tall walks at the rate of 35 ftsec away from a streetlight that is 12 ft above the ground At what rate is the tip of her shadow moving At what rate is her shadow lengthening Notice that in this problem these rates do not depend on the instant in question 7 both rates are constant7 provided that the woman s speed is constant 88 12 RELATED RATES 11 A police helicopter is ying at 150 mph at a constant altitude of 05 mile above a straight road The pilot uses radar to determine that an oncoming car is at a distance of exactly 1 mile from the helicopter and that this distance is decreasing at 190 mph Find the speed of the car 12 The trough shown on the right is constructed by fastening together three slabs of wood of dimensions 10 ft X 1 ft and then attaching the con struction to a wooden wall at each end The angle 0 was originally 30 but because of poor construction the sides are collapsing The trough is full of water At what rate in ftssec is the water spilling out over the top of the trough if the sides have each fallen to an angle of 45 and are collapsing at the rate of 10 per second 13 Suppose that in Example 126 the angle between the two roads is 120 instead of 900 that is the SouthiNorth road actually goes in a somewhat north westerly direction from O In this problem use the law of cosines c2 a2 b2 7 Qab cos0 a Express 6 in terms of a b 391 b b Notice that the angle 0 is constant and so cos 0 is also constant State a story problem perhaps involving a bug on a door rather than a car on a road in which not only a b c but also 0 are all variables In that case suppose you know the values of 117 0 and their rates Find a formula for c39 Part B Additional Practice do these as soon as possible certainly before the midterm 14 A light shines from the top of a pole 20 m high An object is dropped from the same height from a point 10 m away How fast is the object s shadow moving on the ground one second later Neglect air resistance and take 9 98 msec2 for the falling object 15 The two blades of a pair of scissors are fastened at the point A Let 1 denote the distance from A to the tip of the blade the point B Let denote the angle at the tip of the blade that is formed by the line E and the bottom edge of the blade see the diagram Suppose that a piece of paper is cut in such a way that the center of the scissors at A is xed and the paper is also xed As the blades are closed ie the angle 0 in the diagram is decreased the distance y between A and D increases cutting the paper 12 RELATED RATES 89 a Express y in terms of a 9 and b Suppose you know the rate dOdt at which the angle between the blades is decreas ing Express dydt in terms of a 9 and dOdt c Suppose that the distance a is 20 cm and the angle is 5 Further suppose that you are closing the scissors at the rate of 50 degsec At the instant when 9 30 nd the rate in cmsec at which the paper is being cut 16 In the previous problem a express in terms of dOdt the speed of the point on the scissors that moves fastest this point is B b if we have a 20 kilometers ie we have scissors of grand proportions and if 0001 radians nd a value of dOdt for which dydt is a little faster than the speed of light 6 300000 kmsec when the pair of scissors is almost closed c would the situation in your answer to part b be possible or would it violate the Theory of Relativity Explain 17 In the situation of Example 126 nd the rate at which the distance between the cars is increasing or decreasing if a car A is 300 meters north of 0 car B is 400 meters east of 0 both cars are going at constant speed toward 0 and the two cars will collide in 10 seconds b 8 seconds ago car A started from rest at O and has been picking up speed at the steady rate of 5 msec2 and 6 seconds after car A started car B passed 0 moving east at constant speed 60 msec c the functions at and 17t are given by the graphs below and you want to nd the rate at which the cars are separating at time t 2 sec Hint Estimate derivatives by drawing tangent lines IIIIIIIIIIIIIIIIIIIIIIIIIIIIIllllllllllllllllllhl IIIIIIIIIIIIIIIIIIIIIlllllllllllllllllllll 1 IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII 1 IIIIIIIIIIIIIIIIIIIIIIIIIII lIIIIIIIIIIII IIIIIIIIIIIIIIIIIIIIIIIIII39quot IIIIIIIIIIIIIIIIIIII m IllIlllllllIIlllllllIIllllllllllllllllllllllllllll 130 IIIIIIIIIIIIIIIIIIIIIIIIIIIIII m IIIIIIIIIIIIIIIIIIIIIIIIIIIIII no IIIIIIIIIIIIEIIIIIIIIIIIIIIIIII 1 lllllllllllllllllllllllllllllll llllllllllllllllllllllllllllllhl lllll lllllllllllllllllllllll illllllllllllllllllllllll 6 lllllllllIIIIIIIUIIIIIIIIIllllllllllllllll 5 llllllllllll l39hllllIIIIIIIIIIEIEIIIIIIIIIII at meters bt meters 0 2lIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII 10 IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII 20 IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII 30 IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII 40IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII 5 IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII time seconds time seconds 90 12 RELATED RATES 18 In Example 126 suppose that instead of car B you have a helicopter ying at speed v2 to the east of 0 Let h be the altitude of the helicopter7 and let c be the distance from car A to the helicopter Then the three dimensional Pythagorean Theorem states c2 a2 b2 W Find a formula for c39 in terms of 1177 h and the various rates if a the helicopter is ying at speed 17 at constant altitude b the helicopter is ying at horizontal speed 17 ie7 the point on the ground under the helicopter is moving at this speed7 and at the same time is gaining altitude at rate CHAPTER 13 Curve Sketching In this section we discuss how to sketch the graph of a function y without plotting many points Without having to make up a table of values for the function we can obtain a good qualitative picture of the graph using certain crucial information 7 local maxima and local minima in ection points asymptotes etc Our aim is not to draw an exact graph but rather to get an accurate overall picture of the graph and to pinpoint the points where something special happens We start by describing the steps to take in curve sketching For a particular not all of the steps below will necessarily lead to useful information The various possibilities will be illustrated later in the examples 1 MaximaMinima If is a point where reaches a maximum or a minimum then the tangent line at that paint is horizuntal ie 0 Such points can thus be found by setting the derivative equal to zero and solving for In There may be one In for which 0 there may be many or there may be none Once you nd the In for which 0 and the corresponding y coordinate of each possible maxmin point you have to determine whether it really is a maximum or minimum In simple examples it might be obvious A systematic way to tell is by the second derivative test which will be described below X a xb First we make a few remarks 1 We will use the term local maximum point instead of maximum point because it may not actually be a maximum value of the function As in the above drawing the local maximum might only be the maximum for all nearby 5v 7 the function might later turn again and go up to a higher value Similarly we will use the term local minimum instead of minimum poin Sometimes local maxima and minima are referred to as upper turning points and lower turning points 2 A maximum or minimum might occur at a point where there is no tangent line ie where does not exist For example has its minimum point at 0 0 92 13 CURVE SKETCHING 3 Suppose you want to nd the maximum of a function on the interval from a to b To do this you have to try a all points In where 0 b all points In where does not exist and c the endpoints a and b for example in the above drawing the maximum point is at the endpoint l7 4 Later we will see examples of points where even though the slope of the tangent line is 0 we have neither a maximum nor a minimum there 2 The Second Derivative Test Suppose that In is a solution of 0 The second derivative test is a method that in most cases will tell us whether the point is a local maximum or minimum The test works as follows Compute the secund derivative 0f the functiun at the same In ie f v If f v is positive that means that is an increasing function at CU ie is negative a little to the left of In and is positive a little to the right of In This is what happens near a Hall minimum On the other hand if f v is negative then is positive a little to the left of In and is negative a little to the right of In which is what happens near a Hall maximum If it so happens that f v 0 ie if bath and f v are zero at the same In then the second derivative test doesn t give you any information To summarize gt 0 gtmin pt f v lt 0 gtmax pt 0 gtcan t tell 3 Concavity and In ection We know that the sign of the derivative tells whether a function is increasing or decreasing For example any time that gt 0 that means that is increasing We now explain the geometrical meaning of the sign of the second derivative Suppose that f v gt 0 for a certain interval of In This means that as we go from left to right across the interval the slope of the tangent line increases We say that the function is concave upward over this interval The pictures below all show concave upward functions Conversely if f v lt 0 over a certain interval then the slope of the tangent line decreases and we say that the function is concave downward The pictures at the top of the next page all show concave downward functions 4 ASYMPTOTES AND OTHER THINGS TO LOOK FOR 93 There are several ways to describe the difference between upward concavity and downward concavity 1 An upward concave function looks like part of a right side up bowl holds water while a concave downward function looks like part of an upside down bowl 2 If you draw the tangent line to a concave upward function at a point then the curve sits on the tangent line whereas in the case of a concave downward function the tangent line rests on the curve 3 As you go from left to right the tangent line to a concave upward curve rotates counterclockwise while the tangent line to a concave downward curve rotates clockwise A point that separates a region of upward concavity and a region of downward concavity is called a point of in ection The geometrical meaning of a point of in ection is that as you roll the tangent line along the curve from left to right at a point of in ection it reverses its direction of rotation from clockwise to counter clockwise or vice versa That is the Sign 0f second derivative changes as gnu Tull past a paint 0f in ectiun Thus in ection points can be found by setting f v 0 and solving for In Each of the drawings for Examples 131 and 132 below shows a point of in ection at the origin separating a concave downward part of the curve on the left from a concave upward part on the right Warning Notice that as may not be a point of in ection even though f v 0 After nding the points where f v 0 you then must check that f v changes sign See Example 135 below 4 Asymptotes and Other Things to Look For A vertical asymptote is a place where the function becomes in nite because the formula for the function has a denominator that becomes zero For example the reciprocal function has a vertical asymptote at CU 0 and the function tanx has a vertical asymptote at CU 772 and also at CU 7T2 5v 3772 etc Vertical asymptotes are found by asking when the denominator is zero A horizontal asymptote is a horizontal line to which gets closer and closer as In approaches 00 or as In approaches 700 For example the reciprocal function has the x axis for a horizontal asymptote Sometimes one also encounters slanted lines for asymptotes For example the function 2x has the line y 2x as a slanted asymptote because when In is large 22 i and 22 are very close together If the domain of our function does not extend out to in nity we should also ask what happens as In approaches the boundary of the domain For example the function y 1 r2 7 x2 has domain 77 lt 5v lt 7 and y becomes in nite as In approaches either 7 or 77 If there are any points where the derivative fails to exist a cusp or corner then we should take special note of what the function does at such a point 94 13 CURVE SKETCHING Finally it is worthwhile to notice any symmetry A function that has the same value for 75v as for In ie is called an even function lts graph is symmetric with respect to the y axis Some examples of even functions are Inquot when n is an even number cos In and sin2 In On the other hand a function that satis es the property is called an odd function lts graph is symmetric with respect to the origin Some examples of odd functions are Inquot when n is an odd number sin In and tan 5v Of course most functions are neither even nor odd and do not have any particular symmetry Example 131 Sketch y x3 7 In First we set 0 y 3x2 7 1 which has so lutions 5v i33 i0577 The corresponding y coordinates are 70385 and 0385 ie the two crit ical points are 057770385 and 705770385 The second derivative test gives y 65v which is positive for the rst point and negative for the sec ond Thus the rst of the two points in the fourth quadrant is a local minimum and the second is a local maximum Since y gt 0 when In gt 0 and y lt 0 when 5v lt 0 it follows that the curve is concave up ward in the right half of the graph and concave down ward in the left half The two halves are separated by the in ection point at the origin This curve has no asymptotes It does have symmetry however because it is an odd function lts graph is shown above Example 132 Sketch y x3 In The only difference with Example 131 is the in front of the CU But this means that the derivative 3 3x2 1 is never zero and hence there are no maxima or minima In fact the function is always increasing because 3 is always positive The second derivative y 6a is the same as in the last problem and hence the concavity situation is the same In particular this curve also has an in ection point at the origin Example 133 Sketch y x2 7 cos2v for 7772 S 5v 3 772 When we set 0 y 2x 2sin2v we obtain the equality sin2x 72 However from a quick sketch of the two curves sin2x and 7x we immediately see that the only In for which they are equal is In 0 When In 0 the y coordinate is 02 7 cos2 0 71 so our critical point is 0 71 Since y 2 4cos2x which is positive when 5v 0 the second derivative test tells us that 01 is a local minimum To nd in ection points we set 0 y 2 4cos2x This gives cos2v 705 Looking at our table in the section on trig functions we see that in 4 ASYMPTOTES AND OTHER THINGS TO LOOK FOR 95 the range from In 0 to CU 7T2 the equality cos2v 705 holds when 22 7T ie 5v 773 Since cos72x cos2v the equality also holds when 5v 77139 Thus the points 7T315966 and 7T315966 are in ection points Between these two in ection points the second derivative is posi tive concave up whereas for In gt 7T3 and for In lt 7773 the second derivative is nega tive concave downward Finally note that x2 7 cos2v is an even function and so the graph is symmetrical with respect to the y ax1s Example 134 Sketch y x2 Setting 0 y 2x 7 1 we solve this by bringing 122 to the left and clearing denominators 1 2x x2 2x3 So In 07937 The corresponding y coordinate is 18899 Using the second derivative y 2 2 x73 we see that this is a local minimum To nd in ections we se 22x3 0 Clearing denominators and solv ing for In gives x3 71 and so In 71 Thus the point 71 0 is an in ection In this ex ample we have an asymptote when 5v 0 To the right of the asymptote ie for positive In the second derivative is always positive whereas for negative In the second derivative is positive when 5v lt 71 and negative when In is between 71 and 0 Thus the interval 71 lt 5v lt 0 is a region of downward concav ity the graph is concave upward outside of this interval Putting all this together leads to the graph at the right There is another way to think of this example Our function is the sum of two functions x2 and 1v The former function is by far the larger of the two when In is large positive or large negative whereas the reciprocal function is by far the more important when In is near 0 Thus the graph resembles 12 when In is near 0 and resembles x2 when In is far from 0 Roughly speaking the in ection point 71 0 and the local minimum 07937 18899 mark the transition from behaving like the graph of x2 to behaving like the graph of 1v Example 135 Sketch a y x3 and b y 4 a Setting 0 y 3x2 we see that the origin is a possible maximum or minimum However the second derivative test tells us nothing since y 62 also is zero when 5v 0 In fact even though 3 0 when 5v 0 the origin is neither a maximum nor a minimum Rather it is a point of in ection separating the concave downward region in the third quadrant from the concave upward region in the rst quadrant b Again we see that both the rst derivative and the second derivative vanish at the origin and neither derivative is zero anywhere else This time however the origin is a local minimum Even though the second derivative test doesn t tell us 96 13 CURVE SKETCHING this we can see directly that since x4 is positive for nonzero In its smallest possible value is when 5v 0 Note that the origin is nut a point of in ection even though y 0 there This is because y 1212 gt 0 both for In gt 0 and for In lt 0 so everywhere we have upward concavity It is rare for a point where y 0 not to be an in ection point this can occur only when the third derivative 3 is also zero at the same point Example 136 Sketch y x5 7 5x4 5mg First we set 0 y 5x4 7 20133 15mg To solve this we factor out what we can namely 5x2 This leaves a quadratic that can be factored either by inspection or by the quadratic formula The result is 0 y 5x2v 7 1v 7 3 Thus the critical points are 00 11 and 3727 Using y 2013 7 60132 30x 10125172 7 6x 3 we see from the second derivative test that 11 is a local maximum and 3 727 is a local minimum but we get no information about 0 0 Setting 0 y and using the quadratic formula to nd the roots of 2x2 7 6x 3 we nd the following three points of in ection 0 0 0634 0569 2366 71632 In a complicated case like this it is also worth while to see what the function is doing when In is large positive or large negative 1f In is large the x5 term in our function dominates is greater in absolute value than all the other terms Thus the function heads upward steeply into the rst quadrant as In 7 00 and it heads steeply down into the third quad rant as v 7 700 Putting this information together we obtain the graph shown above The curve is concave downward in the third quadrant and also between the two points of in ection 0634 0569 and 2366 71632 For 0 lt 5v lt 0634 and for In gt 2366 the curve is concave upward 4 ASYMPTOTES AND OTHER THINGS TO LOOK FOR 97 Homework 13 1 Suppose that n is an integer greater than 2 On the curve y asquot what sort of point is the origin Sketch the curve and indicate the concavity 2 ln parts a j below nd each maxmin point using the second derivative test to be sure what type of point it is point of in ection and asymptote vertical horizontal or slanted if there are any Also indicate where the curve is concave up and concave down Sketch the graph a yv27x b y23xia 3 c yx379m224x d yx472x23 e y 3x4 7 4x3 0 y x2 e was g y 3x2 41922 h g coslt2xgt e m i y sinx cosx y tana2 7 CU f m gt 0 for gt 2 f m lt 0 for lt 2 f v lt 0 for In lt 0 and I lf m gt 0 for In gt 0 From the given information sketch a possible graph of How could your answer vary and still be correct 4 Suppose f v lt 0 for In lt 1 and f v gt 0 for In gt 1 Suppose also that f1 1 Sketch a possible graph of assuming that i f 1 0 ii f 1gt 0 and iii f 1 lt 0 3 Suppose 5 At the right is a sketch of y x6 7 5x4 a Find the exact coordinates as y of the local 3 maxima and local minima b For what values of In is the function concave upward c Find the exact m coordinates of all points of in ection Hint Color the part of the curve that is concave up blue and color the part that is concave down red Points of in ection occur unly where the curve changes color 720 d Explain how you knuw from gnuquot calculatiuns not from the sketch you were given which of your answers to part a are minima and which are maxima Practice Second Midterms Practice Second Midterm 1 50 points in all time 1 hour 1 13 points A rectangle having di mensions 2a and 217 has semicircles on three sides as shown in the drawing at the right Suppose that at rst a 10 and b 20 but then you want to increase a to 10100 and decrease a b in such a way that the total area of the shape the rectangle plus the three semicircles remains constant Use the tangent line approximation to deter mine your new value for b 2 12 points Let y 5v 7 12m2 a Find the cv and y coordinates of any upper and lower turning points and use the second derivative test to determine if they are max or min b Find all intervals of values of Ir where the function is increasing c Find all vertical asymptotes if there are any d Find all points of in ection if there are any and determine the upward or downward concavity of each part of the curve e Sketch the curve Show in your sketch what looks like as It approaches 00 and 700 and as It approaches the vertical asymptote if there is one from the left and from the right 3 10 points Suppose that in Wanganui New Zealand the mean daily temperature averaged over many years is a sinusoidal function of time Let t be the time in months from the beginning of the year and let y be the temperature in CF Suppose that the hottest temperature is 75 F in mid January that is t 05 and the coldest is 35 F in mid July a Graph the temperature function y b Give its equation in the form y Asin t 7 C D c Suppose that the typical Wanganuian homeowner turns on the heat when the mean daily temperature drops below 65 F In an average year nd the time interval during the year when she has the heat turned on 4 15 points According to the law of cosines for the triangle ABC below we have c2 a2b2 72117 cos C Suppose that C 120 17 50 meters and you are running from D to C at 570 msec How fast is your distance c from A decreasing when you are at a distance a 160 m from C 99 100 PRACTICE SECOND MIDTERMS C Practice Second Midterm 2 50 points in all7 time 1 hour 1 13 points Let y x2 7 gr Find the cv and y coordinates of any upper and lower turning points Find all intervals of values of Ir where the function is increasing Find all points of in ection For what values of In is the function concave upward Sketch the curve Show in your sketch what does as In approaches 00 and 700 ElOU Q VVVVV e 2 12 points A bicycle has a wheel whose circumference is exactly 2 meters It is traveling at 5 msec from left to right At time t 0 it picks up a pebble in the tread of the wheel Take 07 0 to be the point where the pebble attaches itself to the tread In this problem use the parametric equations of the cycloid 5v r0 7 rsin07 y 7 7 7 cos 0 a What is the angular velocity dOdt of the wheel in radsec b Find formulas for the vertical velocity and the horizontal velocity of the pebble in terms of 0 c Find the slope of the direction of motion of the pebble when t 13 sec Draw a picture of the wheel showing the position of the pebble when t 13 sec7 and draw an arrow showing the direction of motion of the pebble when t 13 sec 3 12 points a Write the implicit equation of the ellipse whose bottom half is pictured below to the right7 and solve for In in terms of y b Suppose that the drawing is the side view of a container which has a circular top and circular cross section at any level That is7 the con tainer is obtained by rotating the bot tom half of the ellipse around the y axis Suppose the container is being lled with water Let A be the surface area of the top of the water when it reaches the level g on the y axis The units are meters Write a formula for A in term of y c If the water is 250 cm 25 m deep at its deepest point and is rising at 1 cmsec7 how fast is the surface area of the top of the water increasing PRACTICE SECOND MIDTERMS 101 4 13 points The ellipse pictured to the right has foci at the two points 750 and 0 75 The distance of any point 5vy on the ellipse from 750 plus its distance from 0 75 has the constant value 20 a Write an equation for this el lipse in the form Fx y Const b Notice that the point 0 712 is on the ellipse Find the slope of the tangent line to the ellipse at 0 712 Practice Second Midterm 3 50 points in all time 1 hour 1 12 points Let y 025 x4 7 2x3 45 2 a Find the cv and y coordinates of any upper and lower turning points b Find all points of in ection c min in ec d Give the exact intervals where the curve is concave downward Sketch the curve labeling all of the points found in parts a and b max 7 4 2 12 points In the triangle shown below the sides a and b are pieces of metal hinged at C Here a is exactly 150 m long and b is exactly 250 m long We C measure the side c to be 350 m but our measurement 1 of c has an error of up to a centimeter ie i001 m Using the law of cosines B c2 a2 b2 7 2ab cosC C we nd that the angle C is 120 Use the tangent line approximation and implicit differentiation to determine the error in degrees not radians in our value for the angle C 3 12 points You are 5 ft tall and the sun is casting a 5 ft shadow See the diagram at the right If the sun is descending toward the horizontal at the rate of 14 degree per minute how fast is your shadow lengthening First express the length of your shadow as a function of the sun s angle above the horizon tal 4 14 points An object oscillates horizontally at the end of a spring Its sinusoidal horizontal motion goes from a minimum position of Ir 0 to a maximum position of Ir 4m It goes back and forth once every 2 seconds starting at CU 0 at time t 0 Meanwhile at time t 0 the whole spring system is dropped from a height of 196 meters Neglect air resistance and take 9 98 msec2 a Write a formula for b Draw a graph of In as a function of t and nd a formula for c At the instant t 125 sec nd the horizontal and vertical components of velocity and the direction slope of the path 102 PRACTICE SECOND MIDTERMS Practice Second Midterm 4 50 points in all time 1 hour 1 15 points Let y 2x6 7 9x4 7 108x2 a Find the cv and y coordinates of any upper and lower turning points b Find all intervals of values of In Where the function is increasing c Find all points of in ection d For What values of In is the function concave downward e Sketch the curve Show in your sketch What does as In approaches 00 and 700 2 10 points A circular oil slick of uniform thickness is caused by a spill of 1 cubic meter of oil The thickness of the oil is decreasing at the rate of 01 cmhr as the oil spreads note 1 cm 001 m At What rate is the radius of the slick increasing When it is 8 meters Note the volume of a circular oil slick is equal to 7T gtlt radius2 Xthickness 3 10 points In the triangle pictured at the right let A B C be the angles at the three vertices and let 117 c be the sides opposite those angles According to the law of sines you always have A B b sinB C E 7 sinA39 Suppose that a and b are pieces of metal Which are hinged at C At rst the angle A is 45 and the angle B is 60 You then Widen A to 46 Without changing the sides a and b What happens to the angle B Use the tangent line approximation 92 9 4 15 points A conical pile of sand is exactly 10 ft high Suppose that you have an instru ment to measure the vertex angle 0 see the diagram at the right a Express the volume V of sand in terms of 0 b Suppose you measure 0 to be 600 With an error of i1 Find V and use the tangent line approximation to determine the error in your value for V Warning use radians for 0 angles CHAPTER 14 MaximumMinimum Problems A maxmin problem is a story problem where we want to uptimize something that is we want to determine the situation where something is the biggest possible such as pro t volume of a container etc or where something is the smallest possible cost amount of material needed time of travel etc Solving a maxmin problem is the two step process of 1 reformulating the story problem as a pure math problem and 2 solving the pure math problem We begin with a brief outline of the procedure then discuss how to solve the pure math problem and then present several examples illustrating the procedure The pure math problem just alluded to was described at the beginning of Chapter 13 Find the maximum 01quot minimum value Ufa functiun 0n the interval fmm a t0 7 Here is a procedure for solving this type of problem 1 Find the value of at each point In between a and b where 0 2 Find the value of at all points In between a and b where does not exist 3 Find the value of at the endpoints a and b If a 700 andor 17 00 you have to nd the values that approaches as In approaches ioo 4 The maximum minimum value of is the largest smallest of the values found in steps 173 Here are three simple examples which illustrate the procedure Example 141 Let 7mg 4x 7 3 Find the maximum value of In on the interval between 0 and 4 First note that 725v 4 0 when 5v 2 and 1 Next observe that is de ned for all In so step 2 gives no m values Finally 73 and f4 73 The largest value of on the interval between 0 and 4 is 1 Example 142 Again let 7mg 4x 7 3 Find the maximum value of In on the interval between 71 and 1 First note that 725v 4 0 when 5v 2 But 5v 2 is not in the interval so we don t use it Next observe that is de ned for all In so step 2 gives no m values Finally f71 78 and 0 So the largest value of on the interval between 71 and 1 is 0 Note The maximum value depends not only on the function but on the interval as well Later in this chapter we will give examples of story problems where this observation is important Example 143 Find the minimum value of the function 7 lx 7 2l for In between 1 and 4 104 14 MAXIMUMMINIMUM PROBLEMS In this example is never zero but is unde ned at CU 2 So we compute 7 Finally we check the end points 8 and 9 The smallest of these numbers is 7 which is therefore the minimum value of on the interval 1 S 5v 3 4 Example 144 Of all rectangles of area 100 which has the smallest perimeter The rst step toward solving this problem is to reduce it to a pure math problem If 5v denotes one of the sides of the rectangle then the adjacent side must be 1005v in order that the area be 100 So the function we want to minimize is perimeter of In by lm rectangle 2x 21mg The perimeter function is given by the formula 22 20023 where the dumam is In gt 0 because it makes no sense to talk of a rectangle with negative side We have reduced the problem to nding the minimum value of for In between 0 and 00 We next nd and set it equal to zero 0 2 7 200v2 To solve for In we clear denominators and then move the constant to the left obtaining 200 2x2 which gives us In 10 actually 5v i10 but In 710 does not have practical meaning Since is de ned for all In gt 0 so there are no corners or cusps to check Our next step is to examine the behavior of at the endpoints The sketch this function 2x 200v in the region of positive In is shown below at the right We nd that as In gets close to 0 the function becomes larger and larger and the perimeter function also becomes large when 5v gets very large More precisely the line In 0 is a ver tical asymptote and the line y 22 is a slanted asymptote that the function approaches as In gt Consequently the minimum value of occurs at CU 10 and the desired rectangle is the 10 X 10 rectangle whose perime ter is 40 Any other rectangle of area 100 has perimeter greater than 40 In the same way it is not hard to show that for any xed area A the rectan 5 10 15 20 25 30 35 40 gle of area A with smallest perimeter Length of Side is the square of side Another way to see that In 10 gives a minimum is to notice that decreases for a while as we get away from 0 but if we go too far to the right is increasing In other words lt 0 if 5v lt 10 while gt 0 if In gt 10 So it is clear that the point In 10 where 0 really is a minimum Example 145 You want to sell a certain number In of items in order to maximize your pro t Market research tells you that if you set the price at 150 you will be able to sell 5000 items and for every 10 cents you lower the price below 150 you will be able to sell another 1000 items Suppose that your xed costs start up costs total 2000 and the per item cost of production marginal cost is 050 Find the price to set per item and the number of items sold in order to maximize pro t and also determine the maximum pro t you can get 14 MAXIMUMMINIMUM PROBLEMS 105 To answer these questions we rst observe that our pro t is equal to the number of items sold times the price per item minus the total cost to us If P denotes the price then this pro t is CUP 7 2000 050 The term in parentheses is xed cost plus number of items times per item cost To nd P in terms of Ir we rst note that the information given allows us to write In in terms of P In 5000 10001507P010 ie 5000 plus 1000 times the number of multiples of 10 cents that P is below 150 Solving for P and simplifying gives P 2 7 000015v Thus the pro t function that we want to maximize is CUP 7 2000 050m 5v2 7 00001 2 7 2000 7 050m 15 7 00001 m2 7 2000 We are now ready to set the derivative equal to zero 0 15 7 000025v from which we obtain 5v 7500 Thus we plan to sell 7500 items which we accomplish by lowering the price to 125 It costs us 2000 050 7500 5750 to produce these items and we take in 7500 125 9375 leaving us a pro t of f7500 9375 7 5750 3625 This is the largest pro t we can get under the assumptions of this problem 4000 3000 2000 1000 Profit 71000 2 4 6 8 10 12 14 Number of items in thousands To be safe we should check that the value of In which made the pro t function have derivative 0 is really a maximum It is not hard to compute that is increasing ie gt 0 when 5v lt 7500 and is decreasing when In gt 7500 so that reaches a peak at CU 7500 In most of the examples that follow and when you do the homework problems it is clear enough from the practical set up that the value we nd really is the maximum point if we are looking for a maximum or the minimum point if we are looking for a minimum So it is not necessary to belabor the matter However we should be cautious as shown by Example 149 below Example 146 Find the largest rectangle that ts inside the graph of the parabola y m2 below the line y a The top side of the rectangle should be on the horizontal line y a 7 see the drawing on the following page Here the problem is to nd where along the parabola to put the lower right corner of the rectangle That corner will determine all the other corners Thus if we let 5v x2 be the point on the parabola where the lower right corner is put then the upper right corner will be 5va the lower left corner will be 75vv2 and the 106 14 MAXIMUMMINIMUM PROBLEMS upper left corner will be 72 a We can now 2 write the area of the rectangle in terms of Ir Area 25va 7 x2 2am 7 2x3 To maximize this area function we set 0f 2a76 a ie 5v m The resulting rectangle has dimensions 2 13 X 2a3 and its area is 4a332 Example 147 If you t the largest possible cone inside a sphere what fraction of the volume of the sphere is occupied by the cone Here by cone we mean a right circular cone ie a cone for which the base is perpendicular to the axis of symmetry and for which the cross section cut perpendicular to the axis of symmetry at any point is a circle Let R be the radius of the sphere and let 7 and h be the base radius and height of the cone inside the sphere What we want to maximize is the volume of the cone 7T7quot2h Here R is a xed value but 7 and h can vary Namely we could choose 7 to be as large as possible 7 equal to R 7 by taking the height equal to R or we could make the cone s height h larger at the expense of making 7 a little less than R See the side view depicted at the right where we have situated the picture in a convenient way relative to the cv and y axes namely with the center of the sphere at the origin and the vertex of the cone at the far left on the m axis Notice that the function we want to maximize 7 7T7quot2h 7 depends on m variables This happens in many maxmin problems However the two variables are related in some way 7 by a certain condition of the form Cr h constant So our next step is to nd such a condition and use it to solve for one of the variables in terms of the other so as to have a function of only one variable to maximize In this problem the condition can be read off the above drawing the upper corner of the triangle whose coordinates are h 7 R7 must be on the circle of radius R That is 7RVWR We can solve for h in terms of 7 or for 7 in terms of h Let s do the latter Then we substitute the result for 7 in 7T7quot2h obtaining 7139 R2 7 h 7 R2h Recall that R is just a constant 7 the only variable now is h Let us take R2 7 h 7R2h R2 7 h2 2hR 7 R2h 2h2R 7 ha What happened to the 7T Well we dropped this constant factor for convenience The value of h that makes 7Tfh maximum is the same as the value of h that makes maximum so we don t need to keep that constant factor when we take the derivative and set it equal to zero We re now ready to set 0 f h 4hR 7 3h2 h4R 7 3h This gives us h 4R3 In other words we should in fact choose h a little larger than R at the 14 MAXIMUMMINIMUM PROBLEMS 107 expense of having 7 somewhat less than R The exact value of 7 is determined by the condition above h 7R2 r2 R2 Solving for 7 here after setting h 4R3 gives 7 89R Finally the fraction of the sphere occupied by this cone is vol of cone m2h 8R294R3 8 vol of sphere 377133 4R3 2739 The cone takes up a little less than 30 of the sphere Example 148 You are making cylindrical containers to contain a given volume Suppose that the top and bottom are made of a material that is N times as expensive cost per unit area as the material used for the lateral side of the cylinder Find in terms of N the ratio of height to base radius of the cylinder that minimizes the cost of making the containers Let us rst choose letters for various things h for the height 7 for the base radius and V for the volume of the cylinder and c for the cost per unit area of the lateral side of the cylinder Using the area formulas 2717 2 for the top and bottom together and 27T7 h for the lateral side we nd that the total cost of the material is C27T7 h Nc27rr2 Here c and N are constants that are given to us but 7 and h are variables To eliminate one of the two variables we use the canditiun that the volume must be V ie 77r2h V We use this relation to eliminate h we could eliminate 7 but it s a little easier if we eliminate h which appears in only one place in the above formula for cost The result is fr 7T7 V7T7quot2 N711 2 Z N7TT 2 7 What happened to the c and the 2 in the above formula For convenience we can factor out the constant 2c and drop it from the expression to be minimized as in the previous example We now set 0 f r 7V7 2 2N7T7quot giving 7 3 V2N7T Next we know that h V7TT 2 and so the ratio hr is equal to h V V m 7TV2N7T QN ie we choose the height to be 2N times the base radius for the most economical container Example 149 Suppose you want to reach a point A that is located across the sand from a nearby road Suppose that the road is straight and b is the distance from A to the closest point C on the road Let U be your speed on the road and let w which is less than u be your speed on the sand Right now you are at the point D which is a distance a from C At what point B should you turn off the road and head across the sand in order to minimize your travel time to A Let 5v be the distance short of C where you turn off ie the distance from B to C We want to minimize the total travel time When you re traveling at constant velocity you have timedistancevelocity 108 14 MAXIMUMMINIMUM PROBLEMS In this problem see the diagram at the right we travel the distance DB at speed 1 and then the distance BA at speed w Since a 7 In and by the Pythagorean theorem BA V x2 172 we want to minimize the time of travel function a 7 In x2 b2 M 7 T V We set 0 and next simplify by moving 11 to the left and cross multiplying 1 CU 0f m7m so that wxx2b2vx We now square both sides and solve for In obtaining U wb 2x2 ie 5v 212 v2 7 w This is our solution Notice by the way that our answer turns out not to depend on a that is the point where you should turn off the road does not depend on how far back you started But something s not completely right here What if a is very small ie suppose a lt wb v2 7 1112 Then our solution is saying that we should go back along the road and then head across the sand This makes no sense What have we done wrong in the case of small a The problem is with the rst term a 7 5vv in our function If a lt In that gives us negative time What we really should have is la 7 In the case when a lt wbVU2 71112 the graph of our revised function x MW v is given at the right The minimum of this function cannot be obtained by setting 0 in fact the derivative of this function is never 0 Rather it occurs at a point where there is no derivative 7 namely at the corner at CU a In practical terms this says that if a lt wbx v2 7 1112 then what you should do is head out immediately across the sand in a straight line to A U CL SUMMARY STEPS IN A MAXMIN STORY PROBLEM 1 Decide what the variables are and what the constants are draw a diagram if appropriate understand clearly what it is that is to be maximized or minimized 2 Write a formula for what you are maximizing or minimizing 3 Express that formula in terms of only one variable that is in the form 4 Set 0 and solve 5 Be sure that your answer makes sense and that the maxmin value is not really somewhere else 14 MAXIMUMMINIMUM PROBLEMS 109 Homework 14 Part A Basic problems 14x7x2 form 3 1 Let 5v 5 i for In gt 3 a Find the maximum value of for In between 0 and 4 b Find the minimum value of for In between 0 and 4 c Graph to check your answers 2 Find the dimensions of the rectangle of largest area having xed perimeter P Compare with Example 144 above 3 A box with square base and no top is to hold a volume V Find in terms of V the dimensions of the box that requires the least material for the ve sides Also nd the ratio of height to side of the base This ratio will not involve V 4 You have l feet of fence to make a rectangular play area alongside the wall of your house The wall of the house bounds one side What is the largest size possible in g for the play area 5 Marketing tells you that if you set the price of an item at 10 then you will be unable to sell it but that you can sell 500 items for each dollar below 10 you set the price Suppose your xed costs total 3000 and your marginal cost is 2 per item What is the most pro t you can make 6 Find the area of the largest rectangle that ts inside a semicircle of radius 7 one side of the rectangle is along the diameter of the semicircle 7 For a cylinder with given surface area S nd the ratio of height to base radius that maximizes the volume 8 You want to make cylindrical containers of a given volume V using the least amount of construction material The lateral side is made from a rectangular piece of material and this can be done with no material wasted However the top and bottom are cut from squares of side 27 so that 227 2 87 2 of material is needed rather than 27772 which is the total area of the top and bottom Find the optimal ratio of height to radius 9 Given a right circular cone you put an upside down cone inside it so that its vertex is at the center of the base of the larger cone and its base is parallel to the base of the larger cone If you choose the upside down cone to have the largest possible volume what fraction of the volume of the larger cone does it occupy Let H and R be the height and base radius of the larger cone and let h and 7 be the height and base radius of the smaller cone Hint Use similar triangles to get an equation relating h and 7 10 In Example 149 what happens if w gt v ie your speed on sand is greater than your speed on the road 11 A container holding a xed volume is being made in the shape of a cylinder with a hemispherical bottom The hemispherical bottom has the same radius as the cylinder Find the ratio of height to radius of the cylinder which minimizes the cost of the container if a the cost per unit area of the bottom is twice as great as the cost per unit area of the side and the container is made with no top b the same as in a except that the container is made with a circular top for which the cost per unit area is 1 12 times the cost per unit area of the side 110 14 MAXIMUMMINIMUM PROBLEMS 12 a A square piece of cardboard of side a is used to make an open top box by cutting out a small square from each corner and bending up the sides How large a square should be cut from each corner in order that the box have maximum volume b What if the piece of cardboard used to make the box is a rectangle of sides a and 17 Note that the side In of the square to be cut from each corner must be less than a2 and 72 and in this way you can eliminate one of the two solutions of f x 0 In addition use the second derivative test to be sure your value of Ir really gives the maximum volume 13 A window consists of a rectangular piece of clear glass with a semicircular piece of colored glass on top Suppose that the colored glass transmits only k times as much light per unit area as the clear glass k is some fraction If the distance from top to bottom across both the rectangle and the semicircle is xed nd in terms of k the ratio of vertical side to horizontal side of the rectangle for which the window lets through the most light For what range of values of k does this question make sense 14 In the same situation as Problem 4 suppose that instead of a rectangular shape you decide to make the play area in the form of a trapezoid using three sections of fence of length l3 each see the diagram at the right Find the angle 0 that the sides should be turned outward in order for the enclosed area to be maximal and nd the maximum play area of this form Is the maximum play area greater or less than in Problem 4 Part B Additional Practice do these as soon as possible certainly before the nal exam 15 A terrorist sees a plane ying away from him at 500 mph at a height of 2 miles At time t 0 it is ying over a point on the ground which is 15 miles from him At that instant he shoots a missile at the plane but because of nervousness and perhaps a subconscious sense of morality he mistakenly aims directly at the present location of the plane rather than slightly ahead For this reason he misses the plane The purpose of this problem is to determine by how much Suppose the missile travels at 2000 mph and neglect both air resistance and gravity First nd parametric equations for both the plane and the missile Then nd the time at which the two objects are closest Hint Use the distance formula and save yourself unnecessary algebra by nding the time at which the square of the distance is minimal By how much does the missile miss the plane 16 You are designing a poster to contain a xed amount A of printing measured in square inches and have margins of 1 inches at the top and bottom and 17 inches at the sides Find the ratio of vertical dimension to horizontal dimension of the printing on the poster if you want to minimize the amount of posterboard needed 17 The strength of a rectangular beam is proportional to the product of its width times the square of its depth Find the dimensions of the strongest beam that can be cut from a cylindrical log of radius 7 18 What fraction of the volume of a sphere is taken up by the largest cylinder that can be t inside the sphere 14 MAXIMUMMINIMUM PROBLEMS 111 19 The US post of ce will accept a box for shipment only if the sum of the length and girth distance around is at most 108 in Find the dimensions of the largest acceptable box with square front and back 20 Find the dimensions of the lightest cylindrical can containing 025 liter 250 cm3 if the top and bottom are made of a material that is twice as heavy per unit area as the material used for the side 21 A conical paper cup is to hold a xed volume of water Find the ratio of height to base radius of the cone which minimizes the amount of paper needed to make the cup Use the formula 7T7quot r2 h2 for the area of the side of a cone 22 If you t the cone with the largest possible surface area lateral area plus area of base into a sphere what percent of the volume of the sphere is occupied by the cone Compare your answer with the answer to Example 147 23 Two electrical charges one a positive charge A of magnitude 1 and the other a negative charge B of magnitude 7 are located a distance c apart A positively charged particle P is situated on the line between A and B Find where P should be put so that the pull away from A towards B is minimal Here assume that the force from each charge is proportional to the strength of the source and inversely proportional to the square of the distance from the source 24 Find the fraction ofthe area of a triangle that is occupied by the largest rectangle that can be drawn in the triangle with one of its sides along a side of the triangle Show that this fraction does not depend on the dimensions of the given triangle 25 How are your answers to Problem 5 affected if the cost per item for the CU items instead of being simply 2 decreases below 2 in proportion to CU because of economy of scale and volume discounts by 1 cent for each 25 items produced 26 You are standing near the side of a large wading pool of uniform depth when you see a child in trouble You can run at a speed v1 on land and at a slower speed v2 in the water Your perpendicular distance from the side of the pool is a the child s perpendicular distance is b and the distance along the side of the pool between the closest point to you and the closest point to the child is c see the drawing Without even stopping to do any calculus you instinctively choose the quickest route shown in the diagram at the right and save the child Our purpose is to derive a rela tion between the angle 01 your path makes with the perpendicular to the side of the pool when you re on land and the angle 02 your path makes with the perpendicular when you re in the water To do this let 5v be the distance between the closest point to you at the side of the pool and the point where you enter the water Write the total time you run on land and in the water in terms of In and also the constants a 76 v1v2 Then set the derivative equal to zero The result called Snell s law or the law of refraction also governs the bending of light when it goes into water I I I l I 91 I I CHAPTER 15 Newton s Method Suppose you have a function and you want to nd as accurately as possible where it crosses the m axis in other words you want to solve 0 Suppose you know of no way to nd an exact solution by any algebraic procedure Newton s method is a way to nd a solution to the equation to as many decimal places as you want It is what is called an iterative procedure meaning that it can be repeated again and again to get an answer of greater and greater accuracy Iterative procedures like Newton s method are well suited to programming for a computer We start with a rough guess which we call we Starting at the point we on the m axis we go up to the curve hitting it at the point 230 fv0 then slide down the tangent line to the curve at 20 rm and nally take the point where this tangent line hits the x axis as our improved estimate for the desired cv value We call it x1 See the diagram below If we let Ax denote the distance from 21 to we then we have 231 x0 7 Am and also f v0slope of tangent linefx0Av Hence Arc f m f9 0 slope f 960 9607 f 960 and so x1 x0 7 Am fltm0gt rootA Ag 5v 7 l 0 f Wo a l l This equation is the basic formula Ax for Newton s method quot quot If the improved estimate x1 is not close enough to where crosses the axis then we repeat the procedure starting with 231 instead of CEO We let x2 denote the point obtained 7 still closer to where 0 The formula relating mg to 21 is the same as the one relating ml to CEO We continue the process as long as we want using the formulas Have f961 H902 H903 M mquot may 2 9 f rv1 3 2 mm M m How do we know when we are close enough Two things happen 1 when we plug our approximate value In into we get something very near zero and 2 our successive approximations mn1 and mu are very close together For example if we want three decimal places of accuracy then once we have two successive ap proximations which agree out to three decimal places we need go no further In 113 114 15 NEWTON S METHOD the homework and exam problems you will usually be told how many iterations to perform one iteration means nding 01 two iterations means nding 02 and so on If on the other hand the directions specify the desired accuracy three decimal places for example then just keep iterating Newton s method until the only changes in your value from one iteration to the next occur beyond the decimal places needed Example 151 Find m without using the square root function on a calculator Use Newton s method with 3 iterations The rst step is to set up the function we want to equal zero We set x2 7 102 since m is a root of that function Since we know m 10 we can make a rough guess of mo 10 We are now ready to apply the formula for Newton s method Here is of course 250 We obtain 2 51107 107 101 f 10 2 10 Actually x1 is the same value we would have obtained using the tangent line ap proximation But with the tangent line approximation if our approximate value isn t good enough there is nothing further we can do However with Newton s method we can improve this approximate value 101 That is we succes sively compute f101 001 101 7 101 7 1009950495 932 f 101 202 f1009950495 00000002 1009950495 7 1009950495 7 1009950494 933 20199 f 1009950495 Thus V 102 1009950494 If we computed x4 05 etc we would get the same as 03 out to 8 places past the decimal point STEPS IN A NEWTON METHOD STORY PROBLEM 1 Decide what equation you want to solve and what variable is playing the role of x 2 Simplify algebraically if you can dividing out constants clearing denomina tors 3 Bring everything to one side of the so you have 0 4 Apply Newton s method for the required number of iterations Example 152 Suppose that the material you are are using to manufacture marbles costs 10 centins and the coating that is painted on the surface costs 015 centin2 Find the radius of the marbles if you want to spend 03 cent per marble on the material and coating Use Newton s method with one iteration and take 03 inch as a rough guess for the radius We rst write out the condition on the total cost using the formulas 3717 3 for the volume and 4717 2 for the surface area of a marble 10 g rs 015 4717 2 03 In Newton s method we need our equation in the form 0 rather 0 in this problem and this means bringing everything over to the left of the equal sign In addition for convenience we ll divide through by the coef cient of TS thereby obtaining r3 0457 2 7 0 Notice that this is a third degree 15 NEWTON S METHOD 115 equation A third degree equation can be solved algebraically but the formula is much more complicated than the quadratic formula If you want an approximate answer then Newton s method is much faster We use the Newton s method formula with In replaced by 7 and with To 03 Here f 739 37 2 097 We obtain fr0 0027 045009 7 007162 r1 7 r0 7 fmm 7 03 7 7 030761nch Example 153 Recall the example of the path of a baton s tip Example 111 From the picture of the trajectory it s clear that the tip reaches its highest point at roughly 720035 sec Find to several more decimal places the time when this peak is reached and nd what the maximum height is Note In reality of course it is pointless to nd something like this to such a great accuracy since our model for the physical situation has some important sources of error 7 the neglect of air resistance for example 7 and our measurement of the constants given in the problem was probably fairly inaccurate In this example we want to nd the maximum value of We do this by setting the vertical component of velocity dydt equal to zero 0 716t2 1425 5 cos67139t 73225 14 7 67139 sin67139t We take to 035 as our rst approximate value and apply Newton s method with ft taken to be the vertical velocity function 73225 14 7 67139 sin67139t Then f t 732 7 3671392 cos67139t this function is actually the vertical component of the acceleration of the tip of the baton So Newton s method gives ft0 7 0 35 7 732 035 14 7 67139 sin67139 035 t 2 7 0341323 1 0 f t0 732 7 3671392cos67139 035 22 21 7 0341939 253 0341975 254 0341975 1 So 253 is accurate to 6 decimal places the last digit could conceivably be off by 1 due to rounding Thus the baton s tip reaches its maximum height at t 0341975 sec Finally to nd the maximum height reached we substitute this value oft into the height function y 716252 1425 5 cos67139t We obtain 890326 ft Example 154 Recall Problem 15 of the maxmin homework Suppose that the pilot was daydreaming did not notice the missile passing a half mile away and so the next day takes the plane again on the exact same route The terrorist is waiting and this time he decides to adjust the angle at which he shoots the missile aiming at a point roughly a half mile ahead of the plane Again suppose that the missile travels at 2000 mph and neglect both air resistance and gravity Find the angle 04 above the horizontal at which the terrorist should re the missile in order to hit the plane Use Newton s method with three iterations Also determine how the function in Newton s method would be affected if you took gravity into account in the motion of the missile To solve this we rst write the parametrized equations for the plane mplanea 15 l 500157 yplanet 27 and for the missile ivmissjlet 2000 cos at ymiss et 2000 sin at 116 15 NEWTON S METHOD At the time t when the missile hits the plane the coordinates of the missile must be equal to the corresponding coordinates of the plane This means that in the horizon tal direction we have 15750015 2000 cos at ie t 152000 cosa7500 and in the vertical direction 2 2000 sin at We substitute the value for t determined from the horizontal motion into the last equation clear denominators and move everything to the left of the equal sign 22000 cosa 7 500 7 2000 sina15 0 Dividing through by 1000 for convenience we arrive at the equation fa 4cosa 7 1 7 3sina 0 Since the problem mentioned that a rough guess would be to aim the missile at the point 22 a half mile in front of the plane which means an angle of 774 radians it makes sense to take 040 774 Here we have f a 74 sina 7 3cos 04 so Newton s method gives an 3 7 774 0785398 7 M 0726225 4 f 7T4 7494975 042 0725937 043 0725937 rad 4159310 Finally if we want to take gravity into account we must add the term 759252 to ymiss et where the gravitational constant 9 is converted from 980665 msec2 to 78980 mihr2 using the fact that 1 mi 16093 m and 1 hr 3600 sec Since nothing changes in the horizontal direction we still have t 152000 cosa 7 500 As before we equate the y coordinates of the plane and the missile and clear denominators which now means multiplying through by 2000 cosa 7 5002 We divide through by 1000000 for convenience and bring everything over to the left The result is 8cos2a 7 4cosa 7 6sinacosa 15sina 058885 0 This is what we use for our fa in Newton s method Homework 15 1 Find the fth root of 7 using me 15 as a rst guess Apply Newton s method with three iterations 2 A rectangular piece of cardboard of dimensions 8 X 17 is used to make an open top box by cutting out a small square of side In from each corner and bending up the sides See Problem 12 of the last homework set If 5v 2 then the volume of the box is 2 4 13 104 Use Newton s method to nd the value of In for which the box has volume 100 Do your computations using at least 3 signi cant gures of accuracy and apply Newton s method with enough iterations so that you know that your value of In is accurate to 3 signi cant gures 3 A wheel of radius 1 picks up a pebble in its tread How much farther will the wheel rotate in radians before the pebble has traveled a horizontal distance of 1 Use the cycloid formula and take 0 2 radians as a rough guess Use Newton s method with three iterations 15 NEWTON S METHOD 117 4 Consider a vertical line passing through the point 7772 71 on the curve y sin In This line is rotated clockwise until it rests against the sine curve at a point somewhere between 07 sin 07 and 08sin 08 See gure at right Use New ton s method to nd the point where the line rests with enough iterations to be sure that your answer is accurate to 4 signi cant gures 5 In the baton example notice that the tip of the baton makes a downward loop at around If 05 sec Find the time to four decimal places when the tip of the baton is at the bottom of the loop and nd its height at that moment 6 In the baton example suppose that there is a wall located at exactly 5v 10 ft Use Newton s method to nd to three decimal places when the tip of the baton hits the wall 7 You are riding a double ferris wheel at an amuse ment park The large circle has radius 40 ft and the smaller circle has radius 10 ft See the diagram be low Your seat rotates clockwise around the smaller circle once every 6 sec and the center of the small circle rotates counterclockwise around the big circle once every 20 sec At time t 0 you are at the low est possible point 50 ft below the center of the large circle Call that point the origin First write para metric equations for your position at time t Then use Newton s method to nd to at least three places past the decimal point the instant when you reach the highest point during the rst revolution of the large circle Also nd your height at that moment 8 Two mine shafts meet at an angle of 120 as shown above The horizontal shaft has a width of 7 ft while the entrance shaft is 9 ft wide What is the longest ladder that can negotiate the turn Neglect the thick ness of the ladder and assume it is not tipped as it is maneuvered around the corner At a given moment the ladder makes an angle of 0 as shown First con vince yourself that the problem can be rephrased as follows Find the angle 0 at which the length l E is minimal and then nd the corresponding value of l To do this rst express l E E in terms of 9 ft 0 notice that 04 can be expressed in terms of 0 Then set dld0 0 and use New ton s method with 00 30 to nd 0 and l to at least 3 signi cant gures of accuracy HH Position at t 0 CHAPTER 16 Inde nite Integrals By the inde nite integral of a function we mean a function whose derivative is Another term for inde nite integral is anti derivative since we re going in the opposite direction from taking the derivative For example the anti derivative of the function x2 is m3 since the function 443 has derivative F v x2 But note that 443 5 is also an anti derivative of 2 and in fact so is 443 C for any constant C This is because a constant has derivative zero and so we can add it to our function without affecting the derivative Notice that a constant C is the only thing that can be added to without affecting the derivative If it were possible to add some other function C23 then we would have to have F v G v F v ie G v 0 But any function that is not constant must either increase or decrease for certain values of In and at such In it has a nonzero derivative To put this another way the only function C23 whose graph is always pointing horizontally is a constant function In this discussion we are dealing only with functions C23 that have a derivative for every 5v Non constant step functionsifunctions that jump from one constant value to anotherialso have zero derivative at all points where the derivative is de ned But at the jump pointsicalled points of discontinuity isuch a function does not have a derivative The notation for the anti derivative of is dm where the integral sign I says take the anti derivative and the dos indicates what the variable is Thus we write 1 x2da 3x3C The term inde nite is used because of the constant C that always appears in the answer to an anti derivative problem C is called the constant ofintegration The function that is being integrated is called the integrand Another way to state an anti derivative problem is to say that we want to nd a function y that satis es the equation 3 7 Such an equation 3 in which y stands for an unknown function is called a differential equa tion In general a differential equation is an equation for an unknown function in which the derivative and maybe the second or higher derivatives occur Differential equations will be treated at greater length next quarter The above example x2 suggests how we can take the anti derivative of any power function asquot Just increase the power by 1 and divide by n 1 in order to cancel the factor of n 1 that arises when taking the derivative of xn1 In other words 1 xndv mn1 C for any n except 7 1 n 1 119 120 16 INDEFINITE INTEGRALS the formula makes no sense if n 71 the anti derivative of the function x71 is a special case and it will be dealt with next quarter This formula can be regarded as the reverse of the nxn l rule for taking derivatives As in that rule n is allowed to be negative or fractional The same formula can be used with other names for the m variable for example 1 1 tquotdt tn1C uquotdu un1C etc n1 n1 Anti derivatives like derivatives are linear This means that if we know the anti derivatives of a bunch of functions then we can compute the anti derivative of any linear combination of those functions For example using the above formula for fxndx we can nd the anti derivative of any polynomial Example 161 a fx376x23x4dx im476x33x24xC ix472x3 1592 45v C a f we 12 am f M a H dm yams2me www In practical problems often the time t is playing the role of the m variable and the function to be integrated is the rate of change of something Just as the derivative procedure takes us from the distance function 5t to the velocity function vt and from the velocity to the acceleration at the anti derivative procedure takes us from at to vt and then to 5t Below we will give examples of this Being able to nd anti derivatives is at least as important as being able to nd derivatives for the following reason Very often in the real world we do not start out knowing a formula for something but rather we know some information about how it is changing ie about its derivative For example in physics we have Newton s law of motion that says that ForceMassgtlt Acceleration Usually the basic thing we know about is the forces This immediately gives us the acceleration function at From that we want to go backwards ie nd anti derivatives to get the velocity function vt and then nally what we are really interested in which is the location 5t of the object A basic example of this is falling body problems 1 From Acceleration to Velocity to Distance Falling Bodies Earlier in the course we encountered the falling body formula 5t 759252 vot so for the height of an object above the ground after t seconds However we did not give any explanation of how that formula arises 7 in fact at rst the whole formula especially the at the beginning looks pulled out of a hat Now we are ready to derive the formula starting from a basic principle of physics near the surface 0f the earth gravity causes a cansttmt force 0f acceleratiun That gravitational acceleration g is 9806194 msec2 we ll use simply 98 msec2 or 32 ftsec2 More precisely the gravitational acceleration is 9 if we take upward to be the positive direction it is y if downward is taken to be the positive direction We shall usually take upward to be the positive direction If we are near the surface of some other planet the same principle applies but with a different value of g The basic principle can be restated as follows if no force other than gravity is affecting the object s motion in particular we neglect air resistance then the acceleration function at is the constant g at 9 Our object is to derive the formula for the height function 5t We do this in two steps 1 FROM ACCELERATION TO VELOCITY TO DISTANCE FALLING BODIES 121 Finding vt Since dvdt at it follows that vt is an anti derivative of at 79 But the anti derivative of the constant 79 is 7 fgdt 7gt C Thus vt fat dt 7fgdt 7gt C How do we nd C Here we have to suppose that we know initial information ie the value of vt when t to which we call v0 Often but not always the initial time to will be 0 In order for vt0 to equal v0 we must have v0 7g to C ie our constant of integration is v0 gto In general gnu solve fur the canstant 0f integration using initial infurmatiun Thus vt 7gt v0 gto 7gt 7 to M In this simple situation we could have obtained the same result by common sense if something starts at value v0 and decreases at canstant rate 9 then after t 7 to seconds its value is v0 7 gt 7 to If the initial time is t 0 as is often the case then u 7gt m Finding 5t Let us suppose that to 0 ie the starting time for the falling body is taken to be t 0 Since d5dt vt it follows that 5t is an anti derivative of vt 7gt v0 ie 5t vtdt 7gt v0 dt 79 v0t C 7th vot C We used C rather than C for the constant of integration because it has no relation to the C in the last paragraph To nd the constant of integration C we must again have initial information Suppose we know the height at time 0 50 50 Substituting this value for 5t whent 0 gives 50 50 7902v00C C ie C 50 We obtain our nal formula 5t 7th2 vot 50 This completes the derivation of the falling body formula Having the derivative relations d5dt vt and dvdt at together with a formula for at we can work back to nd 5t provided we also have two pieces of initial information that allow us to nd the two constants of integration that arise We assumed that the two known values are the initial velocity and the initial height Example 162 Suppose that starting at t 1 your acceleration is given by the formula at lts Your velocity is 2 unitsmin at that instant and your location is 0 Find the distance function 5t The rst step is to nd vt fat dt frsdt grhc 7 0 The initial information is that v1 2 Hence substituting t 1 gives 2 v1 712 C and so C 25 Thus vt 7 25 The second step is to nd 5tvtdt7t 225dt7 gt 125t0 25t0 Now we use the initial information that 51 0 0 51 251C Solving for C we obtain C 73 Thus 5t i 25t 7 3 We can check this answer by verifying three things it has the right second derivative 5 lts ii its value when t 1 is 0 and iii the value of its derivative when t 1 is 2 These computations are easy to do and they give us the assurance that we did not make any careless errors in the last paragraph 122 16 INDEFINITE INTEGRALS 2 Antiderivative of sin In and cos In From the derivative formulas sinx cosx and cosx 7 sins we obtain the corresponding anti derivative formulas cosxdmsinmC sinmdm7cosxC Notice that the derivative of sin is cos but its anti derivative is minus cos 3 Substitution In practical applications we rarely encounter the pure sine or cosine function As we have seen it is useful to be able to work with the more general sinusoidal function ft Asin 2amp0 7 0 D where A B C and D stand for various constants called the amplitude period phase shift and vertical shift respectively In order to nd anti derivatives of func tions of this type and many other functions we describe a basic method for nding anti derivatives called the method of substitution 1n the method of substitution we choose a new variable which will usually be denoted u and set u equal to an inside function as in the chain rule so that the function we are integrating takes on a simpler appearance Then EVERYTHING must be converted to the new variable u Here everything includes the dds or 12577 We convert dd to du by taking the derivative dudcv We illustrate by some examples In each case the reason for having to use dudcv to convert dd to du will become apparent when we check our answer Recall that the answer to an anti derivative problem can always be checked by taking the derivative and verifying that we get our original function back again Example 163 If a and b are constants nd formulas for a amb dd for any n y 71 b sinambdm c cosambdm a We substitute u as b We also nd dudm which in this case is simply the constant a We then treat du and da as if they were algebraic quantities writing 3 a gt du adds and so dd idu Thus in our integral we change as b to u and we change dd to idu thereby obtaining an integral purely in terms of the new variable u ambndv unldu l 1 un1 C ambn1 C a a n 1 an 1 I Your nal answer should be written in terms of IE not It is instructive to check this answer by taking its derivative We nd ourselves using the chain rule 7 as always happens when checking an anti derivative that was found by a u substitution i aa l7 1 1aa b a as 17 dd an 1 an 1 I Thus the term that we obtained when converting dd to du is needed to cancel the factor of a dudm that comes out from the chain rule 3 SUBSTITUTION 123 This also explains the importance of the da or dt or du depending on what the variable of integration is inside the integral Keeping track of that part of the integral in the course of a u substitution will give you the right anti derivative Do not get into the careless habit of omitting the da or dt or du in an integral b Again we take u as b so that dd idu and our integral becomes sinav 17 dd sinuldu licosu C ilcos w 17 C a a a You should check this formula by taking the derivative of ii cosam b a c This worksjust like part The result is fcosavbdv isinambC Example 164 Find a xcosx2dx b W C Sinl Sdt sin27rt 7 1 COS 1 Note that in parts b7c we have brought the dt into the numerator instead of putting it at the far right of the integral a Here we take u equal to the expression inside the cosine namely u x2 Then dudm 25v and so du 2x dm Fortunately the integral contains In as well as dm and so bringing the In and da together we can replace the product mdm by du 2 1 1 1 1 2 xcosv dm cosu du 5 cosudu s1nu C s1nv C Notice how when we check by taking sinx2 we get In cosx2 because of the chain rule b Here we can greatly simplify the denominator if we take u sin7rt Then iii 7 7T cos7Tt and so du 7T cos7rtdt which 7 fortunately for us 7 happens to be what is in the numerator except for the constant 7T which does not cause us any problems That is we can replace cos7rtdt by idu t dt ldu 1 1 1 1 cos277 W2 u72duu71cici I 0 s1n 7715 u 7T 7T 71 7m 7T s1n7rt c Here we can obtain a much simpler integrand by substituting u in place of the expression in the radical u 1 cos f O Then 1 7 0 7 sin f 3 and so du 7 sin 1 dt So the numerator in the integral is 7du and we obtain 71077 7 10 712 7 10 1 12 7 20 7715 E duii u duii lQu C777T 1cos10C 124 16 INDEFINITE INTEGRALS Homework 16 1 Find the anti derivative of each of the following a 1 WE 5 b lt1 7 tr c H DZ d 13 1 7 525 e sin3 xcos 5v f xv 100 7 mg g x2 1 7 x3 h sin t 7 C D cos 1 7 sin1t 0 cos7rt cos sin7rt 2 Suppose that the cos7rt term were absent in the numerator of Example 164b Find the anti derivative by substituting u 7725 and then using the formula for the derivative of cot 3 Find the formula for 5t for a falling body if you know v0 vt0 and so 5050 at some time to not necessarily at time 0 4 Find the distance function 5t if at 1254 and both velocity and distance are zero at time t 1 5 Find the distance function 5t ifthe acceleration function is given by the following formula7 and if the initial velocity v0 and the initial distance so at time t 0 are known a at kt where k is a constant b at 225 1 c at sin5t d at sin01tcos301t CHAPTER 17 De nite Integrals Area A basic type of problem in calculus is to nd the area under a curve More precisely we want to nd the area bounded on the top by the graph of some function y with gt 0 on the bottom by the m axis on the left by a vertical line In a and on the right by another vertical line In b We call this the area under between a and b We also call it the de nite integral of from a to b 1 Riemann Sums To understand the meaning of the de nite integral it is instructive to discuss a method for obtaining an approximation to the true area area This method is called the Riemann sum approximation more precisely the nth left endpoint Riemann sum approximation where n is some integer The procedure is 1 divide the interval from a to 17 into a certain number n of equal subintervals each of length Ax l7 7 an 2 draw a vertical line up to the curve at the endpoints of the subintervals 3 over each subinterval form a rectangle by going up to the curve at the left endpoint and drawing a horizontal line to the right from that point on the curve 4 take the sum of the areas of the n thin rectangles I I l a b x Area under the curve 5 th left endpoint Riemann sum For example ifn 5 so that Act bia5 then the heights ofthe 5 rectangles are Ha fa Ax fa 2m fa 3m and W 4m see the drawing above All of the rectangles have the same horizontal dimension Ax Hence the total area which is the 5 th left endpoint Riemann sum is equal to fa Ax fa Ax Ax fa QAx Ax fa 3Am Ax fa 4Am Ax In general the n th left endpoint Riemann sum obtained from the rectangles formed over n small subintervals is equal to fa Ax fa Ax Ax fa QAx Ax fa n 7 1Am Am 125 126 17 DEFINITE INTEGRALS AREA Using the summation symbol 2 we can write this sum more compactly as follows 7271 72 th Riemann Sum 2 f a 2 Am Ax Where Am b 7 a 20 Here the summation notation tells us to take all integer values of239 starting with 2 0 and ending with 2 72 7 1 put each value in place of239 in fa 2Am Am and take the sum Example 171 Find the 5 th left endpoint Riemann sum for the de nite integral of x2 from 0 to 1 ie for the area under the parabola x2 between 0 and 1 We have a 0 l7 1 Am 02 and the Riemann sum is 4 21022 02 02 022 042 062 082 02 024 20 Later we will see that the actual area is a little larger namely 13 Notice that rather than multipling each term by 02 and then summing it is faster to factor out the term 02 add the remaining terms and then multiply by 02 In general if is an increasing function on the interval 217 then the left endpoint Riemann sum will be a little less than the true area but if is a decreasing function on 217 then the left endpoint Riemann sum will be a little greater than the true area The larger 72 is the closer the Riemann sum will be to the true area We say that the area is the limit of the Riemann sums as 72 approaches 00 and we write it in symbols this way 7271 Z fa 2A2 Act In old books the printed form of the letter S is f and in nitely small values of Act are written dm Thus a good symbol for the limit is aquotfxdx This number is called the de nite integral of from a to b The function is called the integrand a is called the lower limit of integration and b is called the upper limit of integration The letter In is called the variable of integration In a de nite integral it makes no difference what the variable of integration is called In other words the area under the function y between 2 a and z b means exactly the same thing as the area under the function y between 5v a and Ir b That is both I and f are equal to the same thing because in a de nite integral the variable of integration only holds a place and does not affect the answer In the next section we will discover that there is a relation between the de nite integral and the inde nite integral which you studied in the last chapter 2 COMPUTING THE AREA UNDER A CURVE 127 2 Computing the Area Under a Curve We now describe how to nd the exact area under between a and b To do this we introduce an area function Au as follows for any u between a and b we set Au area under between a and u If we could nd a formula for the function Au then we would just substitute u b y f9c Au Au 7 Au a Q l a9 u u Au in order to obtain the area between a and b that we want To nd such a formula we proceed indirectly Instead of asking for Au itself we ask haw much the area functiun changes when we go from u to u Au and then we take the difference quotient Au Au 7 Au By the de nition of the area function Au Au 7 Au area under between u and u Au z area of the rectangle above the interval u u Au of height Au f This approximate equality between the area under the curve and the area of the thin rectangle becomes very close to equality when Au is small Dividing through by Au we nd that Au Au 7 Au R f u with the approximate equality becoming an equality in the limit as Au goes to 0 That is the limit of the difference quotient 7 the derivative of Au 7 is equal to f W M E H To see what this says imagine that the area under the curve to the right of u is covered over by a card The derivative of the area function Au means the rate at which area is being uncovered as you move the card to the right Clearly this depends on how wide the area is at the edge of the card 7 and this is precisely u Once we know that dAdu is this means that Au is an anti derivative of So suppose that we nd the anti derivative C using the methods in the last section Our function Au is not necessarily the same anti derivative Fu because we still have to determine the constant of integration C 128 17 DEFINITE INTEGRALS AREA for which Au C As in the case of the acceleration7velocity7distance problems in the last section we determine C using initial information For the area function Au the initial information is when u a there is no area ie Aa 0 Thus 0 Aa Fa C and hence C 7Fa We conclude that Au 7 Fa This means that the total area under the curve between a and b is Ab Fb 7 Fa In other words the area under between a and b can be camputed by nding an anti derivative 0f subtituting b and a in that anti derivative and subtracting Using the de nite integral notation we conclude that 17 Area under between a and b b Fb 7 Fa a where is an anti derivative of Here we used the notation l to mean substitute b and a into and then subtract Note You can tell whether an integral is a de nite integral or an inde nite integral by whether or not there are limits of integration on the integral sign The answer t0 a de nite integral prublem is a speci c value nut a functiun with an unknuwn canstant C as in an inde nite integral prublem Example 172 Find the area under a y x2 between 0 and 1 and b one arch of the sine curve 1 2 1 3 1 1 3 3 1 alfomdm m l0 1 0 39 b An arch of the sine curve is the region under y sins between 5v 0 and Ir 7T so its area is fowsinxda 7cos5v lg 7cos7T77cos 0 771771 2 y sinx Example 173 Find the area contained between the curve y and the line y CE The way to think of the area between two curves is the area under the top curve between a and b minus the area under the bottom curve between a and b Here a and b are the m coordinates of the intersection points In this case the square root function and the line y 5v intersect at the origin and at the point 11 Between 0 and 1 the top curve is So the area between them is 1 1 1 2 1 de7 mdx 75v32 7 v2 0 0 3 2 72 1 3 2 639 1 0 0 3 SUBSTITUTION 129 3 Substitution ln integrals evaluated using a u substitution as before you must change EV ERYTHING from the old variable In or t to the new variable u In the case of a de nite integral everything includes the limits of integration You get the new limits of integration the u limits of integration by asking the question as In goes from a to b the variable u goes from what to what To answer this question substitute a in place of In in the formula that gives u in terms of 5v this gives the lower u limit of integration Then substitute 7 in place of In in the same formula to get the upper u limit of integration Here are some examples Example 174 Find the area under a y 3 7a 7 6 between 1 and 2 and b y sin 1 cos between 0 and 1 a In If 3 7a 7 6 d2 we make the substitution u 7a 7 6 so that du 7dr and d2 du This transforms the integral to 1 1 37du7u13du What about the limits of integration When In is equal to the lower limit of inte gration 1 we have u 7 1 7 6 1 whereas when In is equal to the upper limit of integration 2 we have u 7 2 7 6 8 Thus our new limits of integration are 1 and 8 So the area is 8 8 l ulsdu l u43 7 1 7 4 3 437 43 3 7 7 17288 1 2816 1 45 28 39 b If we make the substitution u 1 cos so that du 7 sin dm we see that as In goes from 0 to 1 the variable u goes from 2 to 1 We obtain 01sin 11c0s7T 2mdx21Elt7 du 1 7 32 2 3 2 3339 lm 1 7 232 3i22 7 1 0776 7T 130 17 DEFINITE INTEGRALS AREA 4 Some Properties of De nite Integrals 1 Like inde nite integrals they are linear This means for example that any constant inside the integral can be pulled outside as we did in Example 174a with the constant 17 Also if you have the de nite integral from a to b of gv it makes no difference whether you work with the sum all at once or nd the de nite integral of and 923 separately and then add the results The same observation applies to 7 If C is intermediate between a and b then the area between a and b is equal A 3 V to the sum of the area between a and c and the area between c and b jammyImmmf xmm 1f the lower limit of integration is greater than the upper limit ie if b is on the bottom and a is on the top so you are traversing the area under the curve from right to left rather than from left to right then the effect is to put in a negative sign This is because I Fa 7 Fb 7Fb 7 Fa 1n Example 174b after making the u substitution we obtained an integral of the form I21 which was negative We also had A 03 V a negative constant in front of the integral so that our nal answer was positive 1f the function is negative then f is negative assuming that b gt a ie area below the x axis is counted as negative area For example A g V if we compute f0 7r sinxdm see Example 172b we obtain 7 cosx 0quot 0 as our answer This makes sense because ffquot sin 5v dd is an arch of the sine curve under the m axis and so has area 72 By property 2 above we have fog sinxdm f0 sinxdm L sinxda 2 72 0 ie the two equal arches one above and one below the m axis cancel 5 Application to DegreeDays The concept of degree days arises when one wants to estimate heating costs Suppose that in the winter you keep a building heated to a constant temperature of 65 F Let be the outside temperature Then the cost 0f heating the building between time t a and time t b is Taughly prupurtiunal t0 the area between the line y 65 and the temperature curve y fmm t a tu t b For example if the temperature outside has the constant value D then the area between the two lines y 65 and y D is simply 65 7 Db 7 a the units are units of temperature times time ie degree days In that case all we re saying is that the heating cost is proportional to the temperature difference times the time interval But when is not constant we need to use de nite integration to nd this area Example 175 Using the temperature function in Example 82 in the section on trig functions nd the number of degree days in January in an average year in Seattle Our temperature function is 14 sin t 7 120 51 Subtracting this from 65 and simplifying we obtain 65 7 14 sin 51 141 7 sin Our time 6 VELOCITY AND DISTANCE TRAVELED AS AREA 131 interval is 0 to 30 recall that we supposed that every month has 30 days So the number of degree days is 030 14 1 7 Sin 31 7 120 dt 14 t 7 cos1lt3t 7 120 0 7 14 30 7 Cos lt7 COS 730 14 30 0 0395ll 82139 Thus based on this information every January we can get a rough idea of whether to expect high or low heating bills by comparing the total number of degree days measured in the particular month with the norm for January which is 821 6 Velocity and Distance Traveled As Area In Chapter 16 we saw that the velocity is the anti derivative of the acceleration Another way to say this is that velucity is the area under the acceleratiuri graph More precisely the area under at between t t1 and t t2 is equal to the amount the velocity increases during this time interval In mathematical notation t2 meg 7 vt1 atdt t1 The graph at the left illustrates this in the special case when at k is a constant function In Chapter 16 we also saw that the distance is the anti derivative of the velocity In other words distance traveled is the area under the velocity graph More precisely the area under vt between t t1 and t t2 is equal to the distance traveled during this time interval t2 502 7 501 7 vtdt 1 The graph at the right illustrates this in the special case when vt kt Shaded area is change in velocity Shaded area is distance traveled during time interval during time interval Note By distance traveled we mean the net distance In other words the backward distance we travel when the velocity is negative represented by area that 132 17 DEFINITE INTEGRALS AREA is beneath the t axis counts with a negative sign ie it cancels some of the positive distance we traveled earlier Homework 17 1 In each case nd the n th left endpoint Riemann sum for the indicated value of n and compare with the true value of the area obtained by the anti derivative method In each case draw a picture showing the area in question and the rectangles in the Riemann sum from which it should be clear why the true value is greater or less than the Riemann sum 2 2 4 a 4 7 x2dv n 4 b 4 7 x2da n 8 c daj n 4 07r2 07r2 0 d cosxdx n 3 e cosxdx n 6 0 0 2 Compute a 2122 b 20 c 20 cosi 3 Find the area under a x2 In between 1 and 4 b 4 7 32 between 0 and 1 c 14 7 32 between 0 and 1 d 122 13 between 0 and 2 e cos7Tv between 712 and 12 4 Find the area contained a between the curve y 1 7 In and the coordinate axes b under both sins and cosx in the region from In 0 to CU 772 c between the curve y cosx and the curve y 5v 7 1 d between the line y 2 and the branch of the curve y sec2 5v lying between the asymptotes at i7T2 5 Evaluatesthe following de nite integralsl 1 a 1 V 4x73dx b 1m212dx c 1x3m3dm 0395 quot4 sins 5v 3 d 0 sin37rcv cos7Tvdv e 0 3 f 0 5v 9 7x2dx COS5 6 Find the normal number of degree days in Fairbanks a during an entire year b during the coldest 2 month period ie the time period extending 30 days before and 30 days after the coldest day and c during the warmest 2 month period Assume that the inside temperature is maintained at 65 F and use the formula for the normal mean daily temperature that is given in Problem 6 of the homework set on trig functions 7 In each case nd formulas and draw graphs for at and vt and nd the net distance traveled during the given time interval a The initial velocity is zero The acceleration starts at 0 increases linearly to 4 unitssec2 during the rst two seconds remains constant at 4 unitssec2 for 2 S t S 4 and drops off linearly to 0 during the last 4 second period 4 S t S 8 b The initial velocity is zero The accleration is the minimum of sint and 1 7 t for 0 S t S 1 Use Newton s method to nd the point of intersection of these two functions maintaining four signi cant gures of accuracy throughout the problem CHAPTER 18 The Fundamental Theorem The so called fundamental theorem of calculus says that taking the derivative and taking the integral are inverse processes More precisely suppose you de ne a function using a de nite integral as follows Fltugt mobs where u is the variable and In is only a dummy variable variable of integration This is simply the area function we de ned before except that we called it Au rather than in the last section In the last section we showed that the derivative of this area function is Thus taking the derivative of F returns us to the original function f that we integrated to get F In words the fundamental theorem says that a functiun is de ned by taking the de nite integral frum a xed luwer limit 0f integration up t0 a variable upper limit 0 f integratiun then its derivative is ubtained by simply drupping the integral sign and taking the integrand with the dummy variable replaced by the variable in the upper limit 0f integratiun ln symbols 1 d 7 Emma fltxgtdxfltugt Example 181 Find ddu of the function de ned as u u u a x2 dm b x2 dm c V l tanxdm 0 l 0 ln parts a7b check your answer by nding a formula for and computing the derivative of this formula a By the fundamental theorem f5 dex u2 Let us check this by nding F u udea 1x3 1us Since A lus u2 the answer checks 0 3 0 3 tn 3 b By the fundamental theorem 1 udea u2 the same answer as in part du 1 a But our function is not quite the same Namely u u Fu mgdx Ems 1 2 1 l 3 l T 3 339 Of course this still has derivative u2 This example illustrates the fact that the derivative of a function of the form does not depend on what constant value we choose for the lower limit of integration a c By the fundamental theorem f5 l tanxdm l tan u Here we cannot nd a formula for Fu since we do not know how to nd the anti derivative of a complicated function like l tan 5v But thanks to the fundamental theorem if all we want is the derivative of and not itself we never have to worry about anti derivatives The derivative of can be determined right away without any work 134 18 THE FUNDAMENTAL THEOREM 1 The Fundalnental Theorem together with the Chain Rule Suppose that u is a function of some other variable like time and we want to nd Fut i If This can be done using the chain rule dF dF du E E 39 E K140 Example 182 Suppose that in Example 1c we have u t2 and we want to nd the time derivative of the integral Then d 2 d u d E 1tanxdx 1tanxdv 1tanu2t2t 1tant2 0 u 0 If the function is de ned by an integral in which ut occurs in the luwer limit of integration then simply reverse the upper and lower limits and put in a minus sign ie see property 3 of de nite integrals What if a function appears in bath limits of integration ie what if you want to take ddt of fig fxda where u and v are each functions of t Then choose a constant c and use properties 2 and 3 of de nite integrals with 1 replaced by u and 7 replaced by v mommmf mxmdwfmdm We already know how to take the derivative of each of the two terms on the right Thus d W dv du a um frvdrv fvtg 7 futg Example 183 Suppose that the curve 1 sins forms the top of a shape that is bounded below by the CU axis and on the left and right by vertical lines whose position changes with time Suppose that the vertical line on the left is moving steadily to the right at speed starting at 71 at time t 0 y m Meanwhile the vertical line on the right is oscillating between 1 and 3 according to the formula 2 sin0257rt 7 Find a formula in terms oft for the rate of change of the area of the shape between 1 um 1 2 1t the two vertical lines We have 1 CHAIN RULE 135 1sinxdv d 2sin0257rt76 d 101t 1 1 sin2 sin0257rt 7 sin0257rt 7 7 1 sin71 012 71 0125 0257139 cos0257rt 7 6 1 sin2 sin0257rt 7 7 01 1 sin71 012 Homework 18 1 Find ddu of the following functions of u u u 7r a sin2 mdm b 1 t t2 dt c 1 1 sin2 ydy On 7711 u d sin2 5v dm e sin3 5v dcr 71 71 2 Find ddt of the following functions of t 1 2 abt 7 2gt vot50 a 1 93 0 sing y dy 1 3 33 0 t Asinat7C c cosv2dx d 1 ugdu t2 abt 3 The stage of a puppet show has its upper edge de ned by y 20 10 cos2017139r the units are cm lts left side is the y axis and its lower edge is the cr axis see the drawing After the show the curtain closes from left to right at 2 cmsec a Write a de nite integral for the curtain s area as perceived by the audience after t seconds b What is the rate of change of the curtain s area at time t 4 The same as Problem 3 except that now the upper edge is de ned by y 40 1 0012 7 12 5 The opening to a stage has its highest point 48 feet above the stage it is symmet rical about the center line which we ll take as the y axis and it is bounded by the at 39 I stage and the curve In i418 cos7Ty96 see the drawing A curtain with at bottom is lowered by attaching two heavy weights one at each end and letting them fall with accel eration 32 ftsec2 bringing down the curtain a Write a de nite integral for the amount of the stage blocked by the curtain after t seconds b At time 25 what is the rate of change of the area blocked by the curtain 136 18 THE FUNDAMENTAL THEOREM 6 A at vertical plate is moving from left to right back and forth across a water pipe 1 cm in radius a Write a de nite integral for the area blocked off by the plate When it is u cm to the right of center the area shaded in the draw ing b Suppose that u is a sinusoidal func tion of time If u starts at 71 When t 0 and sweeps back and forth once every 2 seconds7 nd a formula for c Find a formula for the rate of increase of the area blocked in cmZsec at time t Practice Final Exams Practice Final Exaln 1 100 points in all time 3 hours checking your work carefully 1 15 points Evaluate a fog 31 7 35vd5v b fsin t 7 1 cos t 7 1 dt 2 15 points A light is at the top of a pole 14 meters high A ball is dropped from the same height from a point 7 meters from the light How fast is the shadow of the ball moving along the ground 08 sec later Assume that the ball falls a distance 5 49252 meters in t seconds 3 15 points At time t 0 an object is at rest It then accelerates accord ing to the graph at the right where the acceleration at is measured in msec2 a Sketch the graph of the velocity function vt during these 5 seconds b Find a formula for vt for t be tween 0 and 1 for t between 1 and 2 and 705 2 3 A 5 for t between 2 and 3 71 c How far does the object travel dur ing the rst 3 seconds 4 15 points Cylindrical cans with circular tops and bottoms are to be manufac tured to contain a given volume There is no waste involved in cutting the tin that goes into the vertical side of the can But the two circular end pieces are cut from a sheet as shown at the bottom of the page If the circles for the tops and bottoms of the cans are t as snugly as possible each uses up a hexagonal area of tin as shown in the picture above First nd the area of one of the hexagons in terms of 7 Hint It consists of 6 equal equilateral triangles Now nd the ratio of height to radius for the most economical cans In other words minimize the total area of tin needed to make a can 5 20 points A 4 centimeter rod is attached at one end A to a point on a wheel of radius 2 cm The other end B is free to move back and forth along a horizontal bar that goes through the center of the wheel At time t 0 the rod is situated as in the diagram at the left below The wheel rotates counterclockwise at 3 12 revsec Thus when t 121 sec the rod is situated as in the diagram at the right below 137 138 PRACTICE FINAL EXAMS a How far is the right end of the rod the point B from the center of the wheel at time t 121 sec b Express the position of the right end of the rod as a function of t c Express the speed of the right end of the rod as a function of t d Use parts a7 c and the tangent line approximation to nd the distance from B to the center of the wheel at time t 120 sec 7 7i 7 42 sec t7 21 sec a l l o m a o a 6 20 points A wheel 20 cm in radius rolls along a at surface without slipping at a speed of 5 cmsec A particle is caught exactly half way up one of the spokes At time t 0 the particle is at its lowest point a height of 10 cm above the surface The drawing below shows the wheel at time t 0 a Find an expression for the horizontal position ofthe particle its x coordinate at time t b What is the m coordinate of the particle when t 3 sec In this part and the next7 carry out your computations to ve places beyond the decimal point c Suppose that you want to nd the time when the particle has moved a horizontal distance of 10 cm7 ie7 when its m coordinate is 10 Write a function ft that equals zero at this instant Then use Newton s method with to 3 to nd successive approximations 2517 252 Also compute the horizontal position of the particle at time 252 Please show your work clearly PRACTICE FINAL EXAMS 139 Practice Final Exaln 2 100 points time 3 hours7 checking your work carefully 1 15 points Evaluate a f g b Off6 cos3v sin3x dcv 2 10 points A wheel 9 inches in radius is rolling along a pavement without slipping lts angular velocity at time t seconds is 60 20 sin77t revsec a How many revolutions does the wheel go through between t 0 and t sec b How many radians does the wheel rotate during this time c How far does the wheel roll during this time Give your answer in feet 3 10 points A girl ies a kite at a height of 240 ft7 the wind carrying the kite horizontally away from her at a rate of 20 ftsec How fast must she let out the string when the kite is 400 ft away from her 4 10 points You are making a rectangular box with square bottom If the bottom has side 2 ft and the height is 4 ft7 then the total area of all six sides is 40 ft2 If you decrease the height to 39 ft7 what should the dimensions of the square bottom be in order for the surface area to remain the same Use implicit differentiation and the tangent line approximation 5 10 points You are building a trough whose sides are 45 745 7 90C triangles see the dia gram at the right Those triangular sides are three times more expensive in cost per unit area than the two rectangular pieces that run the length of the trough You want the trough e to hold 6 ft3 What dimensions 5 and 6 will minimize the cost 6 25 points The purpose of this problem is to sketch the function y x2 sin2x between 5v 7772 and Ir 772 a On the same graph7 sketch both the curve y x2 and the curve y sin2x between 5v 7772 and Ir 772 b What equation must be satis ed in order for y x2 sin2x to have a maximum or minimum at In C Find the a and y coordinates of all points of in ection of y x2 sin2x between 5v 7772 and Ir 772 d Give a value of Ir near a minimum of y x2 sin2v Do this by nding a minimum of sin2x Call this value me e Use Newton s method with two iterations to get a more accurate value x2 for the m coordinate of the minimum point of y x2 sin2v f Do you think the function y x2 sin2x has any other maxima or minima besides the minimum you found in part e You are not expected to do any computations or take a long time on this part g Sketch the graph of y x2 sin2x Continue to Problem 7 on the next page 140 PRACTICE FINAL EXAMS 7 20 points You are standing at B7 a dis tance of 75 ft from the bottom of a ferris wheel which is 20 ft in radius Your arm is at the same level as the bottom of the ferris wheel At time t 0 your friend is at the point A rotating counterclockwise at the rate of 1 revolution every 12 sec onds At that instant you throw a ball to her You throw the ball at 60 ftsec at an angle of 60 above the horizontal Take 9 32 ftsec27 and neglect air resistance a What is your friend s angular velocity in radsec b Using the cv and y coordinates in the diagram7 write parametric equations for your friend s position c Find the horizontal and vertical components of the velocity of the ball at time t 0 Notice that the horizontal motion is in the negative x direction Round off your answers to the nearest ftsec d Write parametric equations for the cv and y coordinates of the position of the ball after t sec e At time t 2 sec7 how far is the ball from your friend f Express the distance from the ball to your friend at time t in terms of t g Suppose you want to nd the closest distance the ball gets to your friend Find a function oft that must be equal to zero at the moment of this minimum distance This is the function that would be put into the Newton s method formula with initial guess to 2 sec to nd the time when the ball is closest to your friend PRACTICE FINAL EXAMS 141 Practice Final Exaln 3 100 points in all time 3 hours checking your work carefully 1 15 points Evaluate a b 07r sinv4 cos2a4 dm 2 15 points At time t 0 an object is moving to the right at 2 msec It then decelerates according to the graph below which shows rightward acceleration as a 1 function of 25 Here at is measured in msec2 0395 a Draw the graph of the velocity i function vt during these 4 seconds I 4 b Find formulas for vt for t in the 705 39 intervals 01 12 23 and 34 c How far does the object travel be 71 7 tween t 0 and t 4 second 3 10 points A baseball diamond is a square 90 ft on a side A player runs from Thir First rst base to second base at 16 ftsec At what rate is the player s distance from third base decreasing when the player is 20 ft from rst b 7 ase Home 4 15 points Two legs of a right triangle are 7 and 24 so the hypotenuse is 25 If you want to shorten the shorter leg to 6980 and keep the perimeter of the right triangle the sum of the three sides constant how long should you make the longer leg Use the tangent line approximation and implicit differentiation 5 20 points In this problem you want to nd the best ratio of height to radius 5v hr so that a cone whose side has xed area holds maximum volume Use the formulas 7T7quot2h for the volume of a cone and 7T7quot r2 h2 for the side of a cone a Express both the volume and area of side in terms of 7 and Ir without h h b Suppose the side area must be a constant C Solve for 7 in terms of In The constants 7T and C should also appear in your formula c Find the ratio of height to radius for which the cone of side area C holds the most volume Continue to Problems 6 and 7 on the next page 142 PRACTICE FINAL EXAMS 6 10 points Use Newton s method with two iterations to nd a number that is equal to the sum of its square root and its cube root 3 is a good initial guess7 since 33 31743 7 15 points At time t 0 an object P is dropped from a height of4 m on the y axis At the same instant an object Q at the origin is released so that it can oscillate back and forth on a horizontal spring The point Q moves sinusoidally7 returning to the origin each 1 sec When the spring is most compressed7 Q is at 1 m on the m axis Your purpose in this problem is to nd the closest distance between P and a Write a formula for Q s position as a function of time in the form 27139 In 7 As1n 7C D b Write a formula for the square of the distance from P to Q as a function of time c Find a function of t that is zero at the instant when P and Q are closest This is the function that would be put into Newton s method to nd the time when P and Q are closest P 04 7 498 Range of oscillation PRACTICE FINAL EXAMS 143 Practice Final Exaln 4 100 points in all7 time 3 hours7 checking your work carefully I 20 points Evaluate in part b express your answer in terms of B 3 dt 36 27w 27w a04mb0 cosltBs1nltBda 2 15 points At time t 0 an object is mov 3 ing to the right at 3 msec It then acceler 2 ates according to the graph at the right7 which shows rightward acceleration as a function of t measured in msec a Find a formula for vt when t is be tween 0 and 2 and when t is between 2 and 1 3 b How far does the object travel in the 3 seconds 73 3 15 points The trough shown at the right is to be constructed from a piece of metal 50 inches long and 22 inches wide The trough is to be made by turning up strips of width 4 inches to make equal angles 0 with the vertical See the diagraln at the right a Express the volume of the trough in 14 terms of the angle 0 b Let X sin 0 Find an expression of the form 1X2 bX c which must equal 0 if 0 is chosen so as to maximize the volume of the c Find the maximum possible volume ofthe H trough In this problem you will need to use the 4 50 quadratic formula 717 i V b2 7 4ac2a to solve MN for X sin 0 4 10 points Use Newton s method to nd the side 5 of a cube for which the sum of its volume and its surface area is 1500 Use enough iterations so as to be sure that your answer is accurate to four signi cant gures Continue to Problems 5 and 6 on the next page 144 PRACTICE FINAL EXAMS 5 15 points The point A revolves counterclockwise around the unit circle at the rate of 1 revolution ev ery 9 seconds At t 0 it starts at the point 17 0 The point B starts at the point 20 and travels in a straight line at constant velocity7 reaching the point 02 after 2 seconds a Find parametric equations for both points A and B b Find a function which must be equal to zero at the instant when A and B are closest This is the function that you could put into Newton s method to nd the instant when A and B are closest 6 25 points At time t 0 a ball is thrown out of a 50 meter high window at a velocity of 10 msec at an angle of Oz radians above the horizontal Take the ground to be the x axis and the side of the building from which the ball was thrown to be the y axis Neglect air resistance7 and take 9 98 msec2 a Find parametric equations for the 5v and y coordinates of the ball at time 25 Some trig functions of 04 should appear in your equa tions b Let T be the time when the ball hits y the ground Express T as a function of 04 you will need the quadratic formula 50 quot 10 msec I I 717i 72 74ac2a 0 where the sign in front of the square root should be chosen so as to give positive T c Let X be the distance of the ball from the building when it hits the ground Express X as a function of 04 d Suppose that 04 is measured to be 60 with an error of is degrees 04 60 i 5 Use the tangent line approximation to deter mine how sensitive X is to the error in 04 In other words7 nd the constant of proportion ality which relates the error in X in meters to the error e in Oz in degrees X70 Final Exams 19921993 Math 124 Final Exalnination Autumn 1992 1 15 points total Evaluate a 3 points 561 1 b 6 points0 Wdt c 6 points cosv sinm 1dr NOTE Give exact answers 2 15 points Find the CU coordinate of the point on the parabola y x2 which is closest to the point 17 71 Use Newton s method with enough iterations to be sure that your answer is accurate to 4 decimal places Pick your own initial guess 560 NOTE Use yuar calculatur t0 answer this problem T0 get credit yua must clearly indicate the functiun t0 which yua applied Newtun s methud haw yua found that functiun and the intermediate steps 0f Newtun s methud D0 nut simply give the nal result 0f applying Newtun s methud 3 15 points total An object is traveling along the m axis At time t 0 sec it is located at the origin and it is moving along the x axis to the left at 3 msec lt accelerates according to the graph at the top left of the next page7 which shows rightward acceleration as a function of t for the time interval 0 S t S 3 Here the acceleration at is measured in msec a 5 points Find formulas for vt for t between 0 sec and 3 sec b 5 points In the gure at the top righthand corner of the next page7 draw the graph of the velocity function vt for t between 0 sec and 3 sec c 5 points What is the the m coordinate of the object when t 3 sec 146 FINAL EXAMS 19921993 4 10 points total The graph of the function y 5v 2sin5v for 76 lt 5v lt 6 is shown below a 4 points Find the exact Ly coordinates of each local minimum of y whose m coordinate is between 76 and 6 b 3 points Find the exact Ly coordinates of each point of in ection of y whose m coordinate is between 76 and 6 c 3 points For what values of Ir be tween 76 and 6 is the graph of y concave down Again give exact values NOTE In this prublem yuu must give exact values ie in terms 0f such things as 7139 3 etc 7 decimal answers will nut receive full credit 5 15 points total Assume that the height y of the river bed above its lowest point is given by a formula of the form PM Asin gaim D where In is the number of feet mea sured horizontally from the lowest point of the river bed The river bed is 20 ft deep at its center and mea sures 40 ft across See the gure at the right a 5 points Suppose that the river bed is partially lled with water Let h be the depth of the water at the deepest point of the river measured in feet and let w be the width of the river see gure Find a formula relating h and w Be sure to evaluate all constants in your formula b 10 points Suppose that on a certain day w 30 ft and that h is increasing at a rate of 1 inch a day 1 ft 12 inches At what rate in ftday is w increasing FINAL EXAMS 19921993 147 NOTE In bath parts a and 2 0f this problem yau must give exact values ie in terms 0f such things as 7139 etc 6 15 points total A metal rod 1 meter in length is placed horizontally on the m axis with its ends located at the points P 0 0 and Q 1 0 A second long metal rod is attached to the rst one at the point Q 1 0 and makes an angle of 45 with the rst see the gure on the next page A third long metal rod is attached to the rst rod at the point P 0 0 and is free to rotate about P Thus the angle 0 made by the rst and third bars is free to change For 0 between 0 and 90 the second and third bars cross at a point R 5vy see gure a 5 points Find a relationship between 5v y and 0 then eliminate y by expressing y in terms of In This gives you a relationship between In and 0 b 10 points Notice that In 05 meter when 0 45 By approximately how much should you increase 0 if you want the CU coor dinate of the point R to decrease to CU 045 meters Use the tangent line approximation NOTE Use yaur calculatar in part b and give yaur answer in decimal farm accurate t0 4 decimal places Hint In part 17 use implicit di erentiatian t0 campute the derivative 7 15 points total A bicycle wheel of diameter 1 meter rolls up an inclined plane at constant speed The x axis is horizontal and the y axis is vertical At time t 0 seconds the center of the wheel is located at the point 8 05 exactly After 20 seconds the center of the wheel crosses the y axis at the point 0 65 exactly NOTE Give exact answers d0 nut use a calculator Simplify yaur answers as much as yau can a 5 points Denote the coordinates of the center ofthe wheel at time t by 230 t and ye Position of center Find the horizontal and vertical velocities of y and pebble at time the center and then nd formulas for mct and 7 t 20 sec yct b 5 points Find the speed of the center of Position of center the wheel Then nd the speed at which the and pebble at time wheel is rotating that is the wheel s angular t 0 sec velocity about its center c 5 points At time t 0 a pebble is I located on the rim of the wheel at the point 8505 Let 5vtyt denote the position of the pebble at time t Find formulas for xt and 148 FINAL EXAMS 19921993 Math 124 Final Exalnination Winter 1993 1 15 points total Evaluate a 5 points sin45m b 5 points 011 7 dt c 5 points sinv cosm 1dr Note Give exact answers 2 13 points A publishing company prints books with 1 inch margins at the top and the bottom of a page and 12 inch margin on the left and right of a page Find the dimensions of a rectangular page which will hold 32 square inches of text and use as little paper as possible Hint Let the independent variable In be the width 0f the printed area 3 15 points total An object is traveling to the right along the m axis It starts at the origin with zero velocity speeds up slows to zero velocity again and again speeds up Its velocity in this time interval is given by the following formulas sint for0lttlt7139 vt 7 7 t77139 for71 t 2711 See gure below Units of time are seconds and units of length are meters a 3 points Find formulas for the acceleration of the object for 0 S t S 27139 b 8 points Find formulas for the position of the object for 0 S t S 27139 c 4 points In the grid below give a rough sketch of the position of the object for 0 S t S 27139 FINAL EXAMS 19921993 149 4 15 points total The graph of the function msinv is sketched below for In in the interval 73 lt 5v lt 3 In the questions below we are only concerned with values of In between 73 and 3 a 3 points By inspection that is by looking at the graph but without com puting anything give the best estimate you can for the values of the endpoints of the interval on which the graph of is concave up b 6 points What equation is satis ed by the endpoints of the interval on which the graph is concave up Be as speci c as you can c 6 points Use Newton s method with 1 2 y iteration to nd a better approximation to the endpoints ofthe interval you found in part a Be sure to make clear the function to which you are applying Newton s method and yuui 3 initial guess Clearly show all steps leading to rms your nal answer 1 Note In part c give yuur nal answer in decimal farm 5 10 points total The two ends of a rubber band are attached to two nails in a at board The nails are 8 cm apart You stretch the rubber band by placing your nger at its midpoint and moving your nger along the board while keeping your nger equidistant from the two nails see gure Your nger moves at a constant speed of 2 cmsec Find the rate at which the length of the rubber band is increasing at the instant its length is 12 cm Note Give exact answers nger 6 13 points total One end of a 100 ft long straight metal beam rests on a oor at the point A The other end point B in the gure on the following page is supported by another straight metal beam which is attached to the rst by a hinge The other end of the second beam touches the oor at the point C Let 0 be the angle above the horizontal of the rst beam and let 5v be the distance between the points A and C Notice that the angle 0 can be decreased by increasing the distance In a 3 points Let L be the length of the second beam Using the law of cosines1 nd an equation relating In L and 0 1The law of cosines states that if a b and c are the sides of a triangle and 39y is the angle opposite the side 0 then c a 1 7 Zab cos y 150 FINAL EXAMS 19921993 b 10 points When the distance between B A and C is 120 ft the angle of the rst beam 100 ft is found to be 30 What Will be the approx L ft imate value of the CU coordinate of the point C if the angle 0 is decreased by 2 Use the 9c tangent line approximation Note Give exact answers 96 ft 7 19 points total A ferris Wheel With a radius of 10 ft has center 14 feet off the ground The cars people ride in are mounted at the edge of the Wheel and a child carrying an ice cream cone enters a car at the bottom of the Wheel and sits With the ice cream cone two feet directly above the point Where the car is attached to the Wheel The ferris Wheel starts rotating clockwise at a constant rate of 2 revolutions per minute Let the ground be the x axis and let the center of the Wheel be on the y axis with In and y measured in feet see gure a 7 points Let t denote the time in seconds from the time the ferris Wheel begins rotating Find parametric equations for the position of the ice cream cone as a function of t b 5 points What are the position and velocity of the ice cream cone after 23 of a revolution Be sure to give both the horizontal and vertical velocities c 7 points At the instant the ferris Wheel completes 23 of a revolution the ferris Wheel suddenly stops and the ice cream cone ies out of the child s hand starting at the velocity you computed in part b and falling under the in uence of gravity In this part of the problem let t denote the time in seconds after the cone ies out of the child s hand Find parametric equations in terms of t for the trajectory of the ice cream cone lgnore air resistance use 9 32 ftsec2 Note Give exact answers 0 0714 FINAL EXAMS 19921993 151 Math 124 Final Exalnination Spring 1993 1 15 points total a 5 points Find dydcv if x5 2313 32 constant b 5 points few4 cos2x sin2x d2 c 5 points fcva2 132 d2 2 15 points A commercial cattle ranch currently allows 20 steers per acre of grazing land on the average its steers weigh 2000 lb at market Estimates by the Department of Agriculture indicate that the average market weight per steer will be reduced by 50 lb for each additional steer added per acre of grazing land How many steers per acre should the ranch have in order for the ranch to get the largest possible total market weight per acre for its cattle 3 16 points The graph of a function is given below Suppose that is an antiderivative ofthis function so that the graph below gives the DERIVATIVE F v of the function The following questions ask you to gure out the properties of the graph of the function from this graph of Note Must 0f the credit will be far the reaszms justifying yuur answers a 4 points Is increasing or decreasing from B to C How do you know b 4 points Is the graph of F concave up or concave down from A to B How do you know c 4 points List all labeled values of In for which F has a local maximum How do you know that these are maxima d 4 points List all labeled values of In for which F has a point of in ection How do you know that these are points of in ection 4 16 points A box for candy is made in the the shape of a cylinder with a hemisphere of the same radius on top The TOTAL height y of the box is 6 inches and the radius 7 is 2 inches The total volume of the box is V1247T7 337T7 2y77 1 Then the candy manufacturer is informed by its ship ping company that it must decrease the height y of the box by 14 inch but the candy manufaturer wants to keep the volume the same Use the tangent line ap proximation to estimate the new radius 5 18 points total A small valley 12 feet wide from crest to crest and 6 feet deep has the shape of a sinusoidal curve A long stick reaches from the left side of the valley near the crest to the right near the bottom At its top end it is tangent to the valley oor at the point which is 1 foot horizontally from the crest of the 152 FINAL EXAMS 19921993 valley see picture The goal of this problem is to nd where the bottom end of the stick rests a 4 points Find the equation of the valley if the origin is placed at the lowest point of the valley b 6 points Find the equation of the stick7 that is of the tangent line to the sinusoidal curve at the point which is 1 foot horizontally from the crest of the valley c 3 points Write down an equation whose solution is the m coordinate of the point where the bottom end of the stick intersects the valley oor d 5 points Make a reasonable guess for the solution of your equation in part c7 and use one iteration of Newton s method to improve your es timate Note T0 get full credit fur part a gun must clearly indicate the functiun t0 which gnu applied Newtun s methud and the intermediate steps leading t0 yuur nal answer In part a use yuur calculatur t0 give the nal answer t0 at least 5 decimal places 6 20 points total A four foot rod is attached at one end A to a point on a wheel of radius two feet7 centered at the origin The other end B is free to move along the m axis The point A is at 2 0 at t 07 and the wheel rotates counterclockwise at constant speed with a period of 3 seconds a 4 points Find the coordinates of point A as functions of the time t b 6 points What is the m coordinate of point B when t 1 sec c 10 points At t 1 sec7 what is the horizontal velocity of B FINAL EXAMS 19921993 153 Math 124 Final Exalnination Autumn 1993 1 15 points total a 5 points Let x2 1cos7rv Find DO NOT SIM PLTFY b 5 points Evaluate x3 224 1dm 7r c 5 points Evaluate sinv2 dm 0 2 15 points total A forest re is burning along a valley Officials estimate that if the re burns for h hours the cost in lost timber is 1000 h dollars They believe that if they send 5v re ghters to cut a re break the re can be stopped in 700 hours On the other hand each re ghter is paid 18 per hour and costs 20 to transport round trip to and from the re a 5 points Let C be the cost from lost timber and ghting the re This depends on how many hours h the re burns and how many re ghters are used to ght it Give a formula for C in terms of h and Ir b 3 points Give a formula that relates In and h c 7 points How many re ghters should be used to ght the re if the cost C is to be minimized Justify why your solution gives the minimum cost 3 12 points You are standing on the ground directly under a balloon Your friend is standing on the ground exactly 500 feet away from you and she measures the angle from the horizon to the balloon to be 774 radians so she calculates that the height of the balloon is 500 feet The measurement of the angle from the horizon is correct to within 002 radians Use the tangent line approximation to estimate how much the height of the balloon can differ from 500 feet 4 18 points total A bird ies at a constant speed of 10 msec along a straight line from a windowsill that is 30 meters above the ground to a point on the ground that is 40 meters away from the point directly beneath the windowsill A child is in another window 20 meters above the ground directly beneath the windowsill Two secunds AFTER the bird leaves the windowsill the child throws a ball horizontally with initial speed 15 msec Take the acceleration due to gravity to be 9 10 msec2 and neglect air resistance a 10 points Taking time t 0 to be the instant when the child throws the ball nd parametric equations for the motion of the ball and for the motion of the bird b 3 points Where is the bird at the instant when the ball hits the ground c 5 points Find a function oft which must be equal to zero at the instant when the ball is closest to the bird Nate Newtzm s methud cauld be applied t0 this functiun t0 appruzimate the instant 0f clusest approach 7 but DO NOT APPLY NEWTON S METHOD HERE 154 FINAL EXAMS 19921993 5 14 points total The graph of the function y sins cosx for 7 S 5v 3 g is shown below Your answers to all parts must be EXACT ie in terms of such thin s as 7T etc a 7 points Find the exact as y coordinates of each local maximum of y whose m coordinate satis es 1 7 S 5v 3 Explain why each is a local maximum b 4 points Find the exact as y coordinates of each point of in ection of y whose m coordinate is be 96 tween 7 and Explain why each is a point of in ec tion 77r2 7r2 c 3 points For what values of In between 7 and g is the graph of y concave up 6 14 points A train is backing away from a vertical wall The headlight of the train is pointed at the wall and the beam from the headlight is in the shape of a right circular cone The headlight illuminates a circular region on the wall which is the base of the cone of illumination The lateral surface of the cone of illumination makes an angle of 450 with the horizontal line joining the headlight and the center of the circle on the wall When the radius of this circle is 4 feet the area of the circle is increasing at a rate of 247139 square feet per second How fast is the train moving away at that time 7 12 points total Suppose a particle is mov ing along a line so that its position 5 in feet at time tin seconds is given by s du where the 0 function is given by the graph y4 a 3 points What is the velocity of the particle at time t 3 Justify your answer b 3 points Is the acceleration of the particle at time t 4 positive or negative Justify your answer c 3 points Find the position 5 of the particle at time t 2 d 3 points At what time in the rst 9 seconds does 5 have its largest value Justify your answer Answers 1 Answers to Homework Problems Answers to Homework 1 1 m b and distance are a 15 73 xE b 73 2 m c 1 0 22 d 0 3 6 e 1 1 52 f 73 002 002m 2 y and xintercepts are a 2 71 b 639 12 71 d 32 none e 72 73 3 a 962 322 9 b x 7 5 2 y 7 62 7 9 c x52y62 9 d x2y732 9 or x2y2 76y0 e x2y26y f x2 7 6x y2 0 4 radius and center are a 5 0 75 b 7 5 0 c 5 34 5 y 1896 32 6 y 01596 10 7 y 00890 7 32 81 3 3x 120 9 a y 0 for 0 S no S 100 y 0190 7 10 for 100 S no S 1000 y no 7 910 for no 2 1000 b he s introducing a negative income tax as a rather cheap way of getting good PR 7 so that for example a penniless person receives 10 gold coins from the King since the yintercept is 710 10a P 700001x2 b no 710000P20000 11 y 01595 for 0 S no S 19450 y 02895 7 25285 for 19450 S no S 47050 y 03395 7 4881 for 47050 S no S 97620 Answers to Homework 2 1 15oo 2 ac 7E 71 3 ac 7E i1 4 7000 5 all ac 6 000 7 h 7Thr 8 no negative or in 100 9 7 10 ac nonnegative but no 7E 1 11 ac nonnegative but no 7E 1 Answers to Homework 3 1 a 75 b 7247106 c 7240679 d 7240068 e 724 2 a 743 b 7247 c 724 d 34 3 a 5 b 41 c 401 d 4001 limit4 4 a 71029 b 79849 c 798049 798 7 49At 798 msec 5 a 70107527 b 70110742 c 70111074 7133Ax f 3 719 6 2Ax11Axf 1 3 7 a 331 b 3003001 c 30000300001 3 3Ax Ax2 f 1 Answers to Homework 4 rst two problems 1 f ac 7x 169 7 x2 2 a f t 79815 b f ac 2961x2 c f ac 2axb d f ac 3962 Answers to Homework 5 1a 44 b 09 c 101 d 00099 e 20025 f 51 g 49 h 2001 2 394 3 09433 4 15 ft 2 in 5 2089 ft 6 257139 i4739rAac 7 i30 i1 and i3 8 a 5 sec b i0025 sec i05 c i1 p 9 a r CAlZ Where C 1 b 405 c 4171927 10 3401511 i3p 12 aip b ip 13 a 419 b i3p c i2p 14 ip 15 a i2p b ip c ip 16 18384 17 Am 7112 cm Answers to Homework 6 1 a 4x3 7 9x2 x 7 b 3962 7 4x 2E cd 695w 2x2 1 e 169 7 x2 7 xZx169 7 962 f 296 7 4N2 7 x2 7 we 7 4x 5 25 7 962 g 7x22x217x2 h 725 x2 25 7 x2 ij 7E169 7 x271 169 7 9029032 k 3x2 7 2x 1x2 x3 7 x2 7 1x 1 300x100 7 x252 m x2xx3 23 n 962 12 7 1 x2 12 12xx2121 7 x2 12quot2 155 156 ANSWERS 2 a 7981520 798 b 010 c 1 aatb 127a2a15b 32 d 4011257100122 b15 c 12 0101152 b15 c 12 72a15 bZ01152 b15 c 32 which simpli es to 4ac 7 b2a152 b15 c732 3 7115 75 215 0115 2 the bicycle is moving leftward initially at 5 msec but its speed is decreasing after 25 sec it has slowed to 0 msec at which point it reverses direction and starts moving to the right 4 2 Aac 5 i096Aac 6 77 i 4077Ax 7 a f15 x 100 78 7 491522 b 7981578 7 49152 100 78 7 49222 Answers to Homework 7 1 a 1992 b 50129 c 400133 2 a g 7 xy b y 7 72x yx 7 2y c y 3902y2 72acgy3y2 902 72907 3 y 1 4 y 7113x 09923 5 B is 4 times as strong as A 6 y 319 7 a dTdh 73864 Fhour b h 700259T 5493 8 a 100 10 03152 7 102 10 7 03152 const b 12492 sec Answers to Homework 8 1 a 7712 b 2773 c 772 d 2077 2 a 25 radsec b 6944 radsec 3 a 30sin6 b tan c 601 7cos6 4 sina W W W Esino r cosa cos sina sin3cos Oz 6 warmest 620 on July 12 coolest 7120 on January 10 7 F 7 50 sin gt 7 150 R 7 2500 sin 40 7 3 7500 8 a 20sin3115 7 109 40 b 2sin327315 7 81 7 c 5 sin 22 7 7 11 20 d 10sin57715 c 04sin 2515 7 04 or equivalently 041 7 cos2515 Answers to Homework 9 1 096509 exact value096569 2 a 7csc2x b 7cotaccsc ac 3 a 046977 b 0484885 c 093018 4 a g cos 37 7 109 b cos 37 7 81 c g cos 22 7 7 11 d 5077cos57715 e 10 sin2515 5 a coszx 7 sin2 ac b 7sinaccoscos ac c xseczx tanx 2x tanx d 1sinx sec2 x7sinx1sinx2 6 07632 7 1241 8 b 05 2 sin7715 sin77x3 c x 7 05 d dxdt 7 73 c x 7 05 02 7 1 73 7 04 Answers to Homework 10 1 a no 2315 y 37215 b no 1615 y 22515 C no 5x315 y 515 2 a 25 radsec b revsec c 477 msec d 216 kmhr 3 a 343 mph b 55 radsec 875 revsec 4 a 2015 b 157715 c 75152 5 a no 3cos 0515 y 3sin0515 b no 3sin477 y 3cos47715 C no 73sin047715 y 73cos047715 d ac 3sin20739rt y 73cos207715 6 a no 1015 y 749152 24515 1 b 25 sec c 504 sec 7 a no 49515 y 716152 749515163 b 732157495495 c 5290 5870 d 124 ftsec at 6640 below horizontal hits ground at 15 2 8 a no 7 cos 115 y 79152 7 sin 115 b 271sinag c 2712 sina cosa 9 Answers to Homework 11 1 b sin61 7 cos6 2 a no 215 b y 3sin47715 7 g c y 3sin277ac 7 3 y 3sin477 7 4 ac 2sin7715 y 100 7 16152 2cos7715 5 ac 615 7 05sin815 y 075 7 05cos815 6 ac 1015 025 sin4015 y 025 7 49152 025 cos4015 Answers to Homework 12 1 00199 cmsec 2 025 msec 3 8077 mimin 42 misec 4 671 ftsec 5 212 cmsec 6 065 ftsec 7 791 msec 8 1172 mmin 9 05454 mmin 10 6 ftsec 25 ftsec 11 694 mph 12 01234 ft3sec 13 a so 1213 05111 0555 a2 b2 ab b 01131 bl 7 011 cost 7 bizcosf 01b lsin6OL2 172 7 2abcos6 14 816 msec 15 a y acos6 7 asin6 cot 6 or else using the law of sines an equivalent answer is y asin sin6 b dydt 7asin61 7 csc2 6 7 acos6 cot 6d6d15 or else dydt 7asin cos6 Sin26 d6d15 c 379 cmsec 16 a ad6d15 1 ANSWERS TO HOMEWORK PROBLEMS 157 b closing at a little over 15 radsec c no violation of relativity because no part of the scissorsis moving faster than 300 kmsec 17 a 50 msec b 68 msec 18 a mi bba2 b2 h b ail bb 7770711 b2 hi Answers to Homework 13 1 Both f ac 0 and f ac 0 when no 0 If n is even the origin is a minimum and the curve is always concave up If n is odd the origin is a point of in ection and the curve is concave up for no gt 0 and concave down for ac lt 2 a min at 05 7025 always concave up no in ection b max at 14 min at 710 in ection at 02 conc up for ac lt 0 conc down for no gt 0 c max at 220 min at 416 in ection at 318 conc up for no gt 3 conc down for ac lt 3 d max at 03 min at i12 in ections at i0577 2444 conc down for 70577 lt ac lt 0577 otherwise conc up symmetric with respect to yaxis e min at 1 71 in ections at 00 and 23 705926 conc down for 0 lt ac lt 23 otherwise conc up f no max min or in ections yaxis is asymptote conc up for ac lt 0 conc down for no gt 0 the line y no is a slanted asymptote g no max or min in ections at i12 yaxis is asymptote conc up for ac lt 71 and no gt 1 otherwise conc down symmetric with respect to yaxis h min when no 77712 and ac 77712iany multiple of 77 max when no 117712 again can add or subtract any multiple of 77 in ections when no 774 3774 i any multiple of 77 conc up for 774 lt ac lt 3774 general appearance is oscillation around the slanted line y 7x i max at all points i 2k77 Where k is any integer min at 5177r i 2k77 7x2 in ections at 87quot i k770 you get a sinusoidal function asymptotes at odd multiples of 77 max when no 77239 i2k77 min when no i2k77 in ections at even multiples of 77 in each section between asymptotes the left half is conc down the right half is conc up 5 a max at 00 min at i 103 750027 b no gt and ac lt 72 c ix2 712 d use 2nd derivative test Answers to Homework 14 1 a max value f2 5 b min value f0 1 2 P4 by P4 3 Wby by ratio12 4 l28 5 ac 2000 P 6 pro t5000 6 72 by 7x22 area72 7 hr 2 8 hr 877 9 427 10 Take ac a Where has a corner if 71 gt 7 then f ac is never zero 11 hr 2 hr 35 12 a 016 on a side b a b 7 xm6 on a side 13 05 7 k provided that k lt 277 otherwise the optimal Window consists of only the semicircular piece 14 6 30 max area l2312 m 0144l2 bigger than in Problem 3 15 000139 hours m 5 sec misses by 05726 mi 16 ratioab 17 by am 18 1N8 M58976 19 18 in X 18 in X 36 in 20 7 271 h 1084 21 hr 2 22 h 11798R 7 09837R 2854 23 c1 W from A 24 12 25 ac 2500 P 5 pro t7000 26 Snell s law L 292 Answers to Homework 15 1 x1 14765432 x2 14757740 x3 14757732 2 x1 225 exactly 903 x2 219 to three signi cant gures x2 219369 x3 219192 3 61 19359512 62 19345639 63 19345632 4 Hint Set the slope of the line joining 7772 71 to acsin ac equal to the derivative of sin ac then multiply through by the denominator to get the in Newton s method ac cosx 7 sinx 7 1 then 900 07 901 07633 903 x2 07603 5 t 05062 sec height69938 ft 6 take to 07 then 151 068674 158 ANSWERS t2 068726 7 ac 405in739rt10 7 10 sin739rt37 y 50 7 40 COS7Tt10 7 1OCos739rt3 92655 sec 9356 ft 3 l 9sin 7 6 7sin6 049254 rad 7 2320 3139 ft Answers to Homework 16 1 a 7x525x0 b 70117t100 c x5x3x0 d 703175t23C e 175m mo f 71007x232C g 7 1 7 x3C h 7cos 2amp6 7 0 Dt C 723 0 1 7 sin 32 0 1sinsin739rt 0 2 7cot739rt 0 3 717915 7t02 00t7t0so 4 1 1 71 161525 a 16kt3votsm b 11 52t152vo 7tso 711 5 0 721 5 sin5t 110 t 30 d 50 tan01t 110 7 5t 30 Answers to Homework 17 1 a 625 vs 1635333 b 58125 vs 5333 c 4146 vs 5333 d 1239 vs 1 e 1125 vs 1 2 a 55 b 2512 C 3 a 285 b 149 c 23 d 024 e 2 4 a 23 b 2 7 2 c 4 4 d 7 7 2 5 a 133 b 5615 0 0 d 14 e 34 f 9 6 a 14600 degreedays b 4523 degreedays c 277 degreedays 7 a 215 0532 252 032532 at 4 2 t 4vt 42574 2 t 4s873088bat 3775 4953 325715252712 43533 392 0lttlt05110 17 t 0lttlt05110 5111 i i r 7 COS i i 17s001234 1775 051103531 7t71 t2702527 051103531 3 Answers to Homework 18 1 a sin2 u b 1 uu2 c 71 sin2 u d 2sin2 u e 0 2 a ba b bt 7gtvo sin2 7ggt2 vat 30 c 3152 cost672tcost4 d Aacosat7C 1 A sinat 7 C27 b 1 a bt2 3 a foz 20 10 0052017Tx dz b 2 20 10 0052027Tt 4 a 13 tm v b Wing 5 a 2 4887th 43 cos ggdy b 30722 cos g3 7 252 6 a 2 f 1 7 952 dz b 1415 sin739rt 7 c 27139 cos739rt 7 1 7 sin2 77t 7 2 Answers to Practice Midterms Answers to Practice First Midterm 1 1 a0 x 2 b 72 9630 c allacexcept2and4 d 71ltaclt 1 2 a ABlJ b DEFG c GHl d R e J f F 3 t 7 fs 7 7 153 f511 2 m f5 11 2f 5 7 4 117713 7 335 ft 4 987 1 7 90364 Answers to Practice First Midterm 2 1 a like the xZparabola but shifted to have its vertex at 72 74 b bottom half of ellipse passes through i1 on the xaxis and 75 on the yaxis 2 7772 lt ac lt 772 3 a CDJ b AFGH c ABHl d CGJ e C f G 4 a 110 95 msec b 50 7 ft 7 492 7 7 5 5 401 m 95 i 01 79 7 95 4079 msec 5 y fx 100096 f1077 m f10077f 10 100 077710 923 1112 Answers to Practice First Midterm 3 1 05 lt no 3 2 2 a AFG b CD1 0 DE d CFl e C 3 g ft 23739s a 250 b m 01 f 3 233 so 9 is between 233 and 250 msecz 4 y 7 fx 7 362 7 952 f216 i 02 m 233 in i 015 in Answers to Practice First Midterm 4 1 a 735 lt ac lt 15 b 77 S no 3 7 2 b when t 12 parachute opens when t 0 drop from plane right after parachute opens 3 128 4 a A T2 rs b 1007T600 c s 7 fr 710 quot 300 7 r f21 m f20 1 f 20 7 2693 cm 2 ANSWERS TO PRACTICE MIDTERMS 159 Answers to Practice Second Midterm 1 1 19861 2 a 71 715 is max b no lt 71 and no gt 0 c yaxis d none e see bottom of page 3 b y 20 sin 7 95 55 c midMarch to midNovember 4 b CcosC are all constants a and c are variables take ddt of everything you know dadt 7570 you get dcdt 7555 msec Answers to Practice Second Midterm 2 1 a 00 315 7 15 b 700 lt x lt 7 and 0 lt x lt E c 156 7156 d 71 lt ac lt 1 e see below 2 a 57139 radsec b dc 51 7 cos 6 g 5sin6 0 dc 250 msec g 7433 msec dydac 7173 3 a x 100 7 4y2 b A 77100 7 4 c 063 mZsec 4 a x52 y2 x2 y5 20 b 3 15 Answers to Practice Second Midterm 3 1 a 00 b 1275 and 3 675 c see bottom of page d 1 lt ac lt 3 2 C fc 11 are const i06175 3 y 5cot 6 00436 ftmin 4 a y 196 749152 b no 2 sin739rt 7 2 c dc 74443 msec g 71225 msec dydx 2757 160 ANSWERS Answers to Practice Second Midterm 4 1 a max 007 min i67 745 b 76lt ac lt 0and 6 lt ac lt 00 c i37 72925 d 73 lt ac lt e see bottom of page 2 WRZT 1 1113 take ddt of both sides 0804 m3hr 3 B 4 fA ab are const 617320 4 a V 4 130113112076 b 349 113 i 14 ft 3 3 11e 2 Zle 2 1 1 43 42 41 1 2 3 43 42 41 1 2 3 41 41 concavel down I 3 6 4 2 3 Answers to Practice Final Exalns Answers to Practice Final Exam 1 1 a 154 b gsin2 t 7 1 C7 or else 70052 t 7 1 0 2 78125 msec 3 bv1 t2 032531 1115717 132532 1132 23t33c83m 4 area of hexagon 237 27 optimal ratio hr is 43739r 22053 5 a 460555 cm b no 2 cos7739rt 416 7 4sin27739rt c dacdt 71477 sin7739rt 7 287139 sin7739rt cos7739rt 16 7 4sin27739rt d f 121 4 443654 f120 m 44897 cm 6 a no 5t 7 10 sint4 b 818361 cm 0 t1 357285 sec7 t2 355148 sec7 90152 1000011 Cm 712 Answers to Practice Final Exam 2 1 a 1 4 2072 C b 29 2 a 20 g rev b 407T 20 rad C 10925 ft 3 16 ftsec 4 2033 ft 5 s 159 ft7 l 476 ft 6 b 290 2cos2x 0 c 77127772144 12 and 577122577214412 d 474 e 901 4 405236 902 705150 f no other maxmin g graph at right 7 a 776 radsec b no 2OCos739rt67 y 20 20 sin739rt6 vmmrlz 730 ftsec7 vowert 52 ftsec x 4 75 4 302 y 4 716152 527 e 56737 ft f dist 75 7 3015 7 20cos739rt62 716152 5215 7 20 7 205in739rt62 g let ft 4 011512 4 275 4 30 4 20cos7Tt6730 207T6sin7rt6 2716t2 5215 7 20 7 20 sin739rt6732t 52 7 20776 cos739rt6 min o P A L in ec pts 3 ANSWERS TO PRACTICE FINAL EXAMS 161 Answers to Practice Final Exam 3 1 a 771 7 7z 7 C b 4 7 3 7 08619 2 b27tfor0 32 g 1 050527t2for1 gt g 205for2 gt g 327052 for 3 S t S 4 c 292 m 3 79823 ftsec 4 differentiate x y V962 y const get 3 73249 answer is 24013 5 a vol 1 7Tr3x area 7T7 21 902 b r xC739r1 x2 C no 6 Choose 9612 x13 7 no in Newton s method so that f ac x 12 x 23 7 1 then 900 3 x1 33163 902 33094 7 a no 1isin27 rt 7 15 b 4 7 491522 1 Sin2739rt 7 1 2 c ft 7196t4 7 49252 sin2739rt 7 1739rcos2739rt 7 Answers to Practice Final Exam 4 714 1 a 2815 b 3B16739r 2 a 33t 7 t2 0 S t S 2 9 7 3t 2 S t S 3 b 95 m 3 a Vol 504cos 144sin6 b dVold6 74004X27X 72 Where X sin c 29041113 4 A good initial guess is so 10 since vol1000 surface area600 Choose fs s3 632 7 1500 then 31 7 9762 33 7 32 7 9757 5 a 90A cos2739rt9 yA sin2739rt9 x3 2 7t yB t b 22 725 7cos2739rt971 2779 sin27Tt9 722 7sin27Tt91 7 279 cos2739rt9 6 a no 10cos 125 y 749252 10 sinat 50 b T 10204sina sin2 04 98 da 0 X 10cos aT Write T in terms of 04 using b d nd and then take 04 60 705781 the minus sign means that a positive error in 04 leads to a negative error 11 1 ANSWERS 4 Answers to Autumn 1992 Autumn 1993 Final Exalns Math 124 Autumn 1992 Final Exam Solutions 1 a 7 bf5z vd 5 4 43 0 c fcosamsinx 1dac sinx 132 C 2 The square of the distance from the point ac 902 on the parabola and the point 1 71 is given by the formula ac 7 12 x2 12 We need only nd the point 90902 at which the square of the distance to the point 1 71 is minimum For very large is large so the minimum distance occurs at a point at which f ac 0 Thus we want to nd the solutions of the equation f ac 2x714xx21 7496376967270 Notice that f ac 12902 6 gt 0 so f ac is an increasing function and can only have one real root Since for ac lt 0 f ac is negative and for no gt 1 it is positive there is exactly one and it must be between in the interval 0 lt ac lt 1 To nd the root we apply Newton s method As a rst guess at a root try 900 05 Then an x0 7 f ac0f ac0 03333 902 x1 7 f ac1f ac1 03131 903 x2 7 f ac2f ac2 03129 904 x3 7 f ac3f ac3 03129 So the nearest point to 1 71 on the parabola y x2 is ac y 03129 00979 215 for 0 S t S 1 47215 for1 t 3 t2 7 3 for 0 g t g 1 425725275 for1 gtg3 d sln7 7 21cos7 72 sln7 lt1 a 71 f 317 a Q a Integration of the function at gives 1115 c lntegrate of the formula for 1115 to get velocity t 2533732 for0 t 1 ac 222 7t3375t23 for 1 gt g3 Hence 7163 m tme 4 a Since f ac 1 2cosx the critical points of are the points for which cosx 712 For 76 lt ac lt 6 this equation implies that no i2739r3 i4739r3 At a local minimum f ac 72 sinx gt 0 Thus the local minim a occur at the points acy 47T34739r3 7 3 and ac y 72773 72773 7 b Since f ac 72 sinac the second derivative changes sign as no crosses each zero of f ac Thus the in ection points for 76 lt ac lt 6 are acy 00 7 m and 9w 7 7777 7n c The graph is concave down for f ac 72 sinx lt 0 For 76 lt ac lt 6 this implies that 76 lt ac lt 77139 and 0 lt ac lt 7139 a h 7 10sin i gm 7 1010 7 mm long 7 2 b 7 7f When 11 7 30 ft 7 sinw 7 ftft and 1 inday 7 112 ftday So when w 7 30 ft 1 7 7 w 7 Ag ftday as a tan61f or xtan617ac b By the tangent line approximation we should decrease 6 by 7005 radi ans or by 7005 degrees To compute d6dac use implicit differentiation xsec26 tan6 71 When 6 774 and ac 05 we nd 72 radmeter So we should increase 6 by approximately 2005 01 radians or 57296O degrees round 5729580 to 4 decimal places 01 0 10 or h 10 1 7 cos w 10 7 4 ANSWERS TO AUTUMN 1992 7 AUTUMN 1993 FINAL EXAMS 163 7 a Horizontal velocity 7820 725 msec Vertical velocity 620 310 msec ace 725 8 yct t 05 b Speed ofcenter v 7252 3102 25100 12 msec Angular velocity 0 iii 16 1 radsec c 9025 xct coswt 725 8 05 cost and yt yct sinwt t 05 05 sint Math 124 Winter 1993 Final Exam Solutions a Esin45x 20Esin35x cos5x 524 01095 30559 sin5x 1 bf011 7tt dt g 7 t52 0 11 0 c fsinac cosx 1dac 7cosx 13 2 0 Let ac by width of printed area y the height of printed area and total page area Then y ac 32 so y 3290 Hence x 1y 2 x 1 32x 2 or 3296 2x 34 To nd the minimum we solve f ac 0 for ac f ac 732902 2 or 7 32962 2 0 Hence no 4 Since f ac 64903 we have f 4 gt 0 Therefore no 4 is a minimum The optimal dimensions of the rectangular page of 5 inches by 10 inches ie ac 4 inches and y 8 inches cost for 0 S t lt 7T 3 a at v t unde ned for t 7T a corner 1 for 7T lt t 3 7T T3 b 9025 x0 f 11u du 0 fo vmmu 7 7 cost cos0 for t 3 7T 7 7 7cos739r cos0 7 7T dt for 7T lt t S 27139 7 7 cost 1 for t 3 7T 2 7277Tt for7rltt 27r C E a The graph appears to be concave up on the interval 71 lt ac lt 1 b At the endpoints f ac 0 Since xsinac the endpoints satisfy the equation 2cosac 7 xsin ac 0 c Let 990 2 cosac 7x sinac We apply Newton s method nd an approximate solution of 990 0 with initial guess you 1 Then x1 x0 7 gs 07 1 W1 m 107803 Let ac be the distance from the midpoint of the line joining the nails to your nger and let 2 be the length of the rubber band Then by the Pythagorean Theorem 2 2V 42 902 Compute as follows 2 12 ill f When 2 12 cm x2 16 1222 36 or no Hence when 2 12 cm 243E 2 cmsec o1 443E cmsec 6 a L2 1002 x2 7 200m cos6 1 erent1ate a 1mp1c1t yw1t respect to was 7 ac cos 7xsm bD39lf 39 39lquotl39h 6002 200 6 396 Solve for 90 90 501 A6 7 so 906 A6 N no x 6A6 5m1 20 120 0 2000057r672120 120 3g1 a e angu ar ve oc1ty o t e erris w ee a out 1ts center IS 0 47139 7 Th 1 l 39 f h f 39 h l b 39 39 60 7715 sec l Hence the position of the point of attachement of the car to the T3 E 01 ANSWERS ferris wheel is given by the formulas you 710 sin yam 14 7 10 cos The cone is 2 ft above the point of attachment so its position is given by the formulae ac 710 sin t y 16 7 10 cos b The wheel completes 23 revolution when t 27 r or when t 20 sec At that time no 5V3 and y 21 The horizontal and vertical components of velocity are given by dc 723 quot cos and g 2quot sin When 25 20 sec the horizontal and vertical components of the velocity of the cone are 5020 g and 920 7 c x53tandy217t716t2 Math 124 Spring 1993 Final Exam Solutions a Use implicit differentiation 965 x2y3 y2 const gt 5904 2xy3 3x2y2y A 3 2yy 0 Solve for y 39022 2yy 75904 2903f gt 7 b l sin22x4 1 c 1 x2 152 C a Let ac be the number of cattle per acre let M be the weight of each steer and let y be the total market weight per acre b Then 11 2000 7 5090 7 20 3000 7 5090 so y you 903000 7 50m 300095 7 50962 c Maximize over no gt 0 Since y 3000 7 10090 solving y ac 0 for ac gives no 30 d Since y 30 acre a F is increasing from B to C because F is gt 0 there b F is concave up because F is rising there so F ddacF is gt 0 there c F has a local max at C only because it s the only point with F 0 and F changing from to 7 OR say F lt 0 there d Need F 0 and F changes sign local max or min of F only happens at 7100 lt 0 ac 30 gives the largest total market weight per D G a r y so differentiate with respect to y 0 772r27quot 2rr y r2 7 3727 Cancel739rand evaluate atr 2 y6 024T 22r 64734r 824712r 4 20r 4 Sowheny6r 71 5 b New radius r6 7 14 W 2inchesr 6Ay 2 7 15714 205 inches a y 33sin 7 33sin b y gcos 3 7774 Also y75 33sin 75 7 33sin 74 So the equation of the tangent line is y 7x5 3 geos gt 7 c Tangent line and curve intersect 7x5 3 33 sin 7 d Guess x0 1 and for Newton s method use 3sin 7 x 1 73 2 3 273 2 g2 x11 f7 434 a 9025 2 cos yt 2 sin b Let B h0 At time t 1 sec A ac1y1 271223 71 Since the distance between A and B is 4 71 7 h2 16 gt h12 13 gt h 71 c Because the distance between A and B is always 4 h 7 x2 y2 16 Differentiate with respect to t 2h 7 7 2y 0 Cancel h 7 7 4 ANSWERS TO AUTUMN 1992 i AUTUMN 1993 FINAL EXAMS 165 ac yy0 Whent 1 sec7 9390 74sin2Tquot 74 andy 4COSZTquot g SouE71471 h i 0 Henceh gw 39 47 ANSWERS Math 124 Autumn 1993 Final Exam Solutions 1 a f ac 7711962 lsin739rac wcoswx b Set u 2x4 l du 8x3 dac so the integral becomes 1 fu12du u32 C 11 22ac4 l32 C 7r c fa sinac2dac 72 cosac2 720 7 l 2 a O 1000h 18am 2095 b h 720096 c O 7200000x 18 7200 2096 0 C 77200000962 20 so 7200000 20902 and ac 600 re ghters Since C 14400000903 gt 0 this value really is a local minimum it s also clear that at the extremes ac near zero and ac very large C is much larger Thus no 600 gives the absolute minimum possible C Note that h 12 hours for this value of no but you were not asked to determine Let h height of balloon 6 angle Then h f6 500 tan 6 You know 500 f739r4 and you want to know f739r4 i 002 Here f 739r4 500 sec2739r4 500 cos2739r4 5002 1000 So f7r4 i002 m f7r4i002 1000 500i20 eet a xban 15157 yball 5152 20 xburd 16 815 yblrd 18 i 6t b yba 0 at t 2 when we have xblrd 32 yblrd 6 c 72 7 162 75252 625 2 1472 7 16 275252 625 2710t 6 a 0 f ac cosac 7 x3sinx so cosac sinx cotac which gives the 30 angle in the 30 7 60 7 90 triangle ie ac 776 The ycoordinate of this point is sin739r6 x3cos739r6 12 32 2 Notice that f 739r6 7 sin739r6 7x3cos 776 72 lt 0 so this point 7T6 2 really is a local maximum b 0 f ac 7sinx 7 x3cosx so 7tanx which gives no 7773 The ycoordinate of this point is 0 c Concave up for 7772 3 ac lt 7773 Let h and r be the height of the cone which is the distance of the train from the wall and its base radius Let A 7T7 2 be the area of the circle on the wall We know that dAdt 247T and we want dhdt Because of the 45 angle we have h r Then 247T dAdt 717 2 27Trdrdt 27139 4drdt from which drdt 247787139 3 Then dhdt also 3 ftsec a 3 ftsec because the velocity is g ft and 3 b negative because the acceleration is the slope of the graph of ft and this slope is negative at 42 c 2 ft because 15 fudu is the area of the triangle with vertices at 00 20 and 22 d t 5 sec because after that the area under the uaxis starts canceling the area above and the little portion that is positive after t 9 sec is not enough to make up for the negative area T3 E 01 3

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