ELEM CALC & ITS APPLICS
ELEM CALC & ITS APPLICS MA 123
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MA123 Chapter 10 Formulas for integrals integrals antiderivatives and Date the Fundamental Theorem of Calculus pp 207233 chapter Goals 0 Understand the statement of the Fundamental Theorem of Calculus 0 Learn how to compute the antiderivative of some basic functions 0 Learn how to use the substitution method to compute the antiderivative of more complex functions 0 Learn how to solve area and distance traveled problems by means of antiderivatives Assignment 21 Due date Dec 4 Assignment 22 Due date Dec 8 Assignment 23 review Due date Dec 11 So far we have learned about the idea of the integral and what is meant by computing the de nite integral of a function fz over the interval 11 As in the case of derivatives we study now procedures for computing the de nite integral of a function fz over the interval 11 that are easier than computing limits of Riemann sums As with derivatives however the de nition is important because it is only through the de nition that we can understand why the integral gives the answers to particular problems J gt The main ideas of the Fu Theorem of Calculus The easiest procedure for computing de nite integrals is not by computing a limit of a Riemann sum but by relating integrals to antiderivatives This relationship is so important in Calculus that the theorem that describes the relationships is called the Fundamental Theorem of Calculus We introduce the theorem by rst analyzing the following example 1 Example 1 Find a formula for Am 4t 2dt that is evaluate the 1 de nite integral of the function ft 4t 2 over the interval 1 m inside 1 10 Hint think of this de nite integral as an area Find the values A5 A10 A1 What is the derivative of Am with respect to z There are two important things to notice about the function Am analyzed in Example 1 A1114t2dt0 A zlt4t2dtgt4z2 AW 1 Suppose that for any function ft it were true that the area function Am ft dt satis es I M fflttgtdto Aw from m Moreover suppose that Bm is any function whose derivative also equals m so that B z WE 1490 Any such function Bm is called an antiderivative of By a consequence of the Mean Value Theorem that we studied in Chapter 6 we have that there is a constant value c such that Bm Am c b All these facts put together help us easily evaluate ft dt Indeed I bftdt Ab 14170 The above speculations are actually true for any W function on the interval over which we are integrating These results are stated in the following theorem which is divided into two parts The Fundamental Theorem of Calculus PART I Let ft be a continuous function on the interval 11 Then the function Az de ned by the formula Am x mm for all z in the interval 11 is an antiderivative of fm that is A ltzgtjfflttgtdt m for all z in the interval 11 PART II Let be any antiderivative of x on 11 so that for all z in the interval 11 Then 17 31 dz Fb 7 F1 Special notations The above theorem tells us that evaluating a de nite integral is a two step process nd any antiderivative of the function x and then compute the difference Fb 7 F1 A notation has 17 been devised to separate the two steps of this process stands for the difference Fb 7 F1 Thus a b bmdz Fltzgt Fb 7 F1 a 92 gt Some properties of de nite integrals 1 f gngw 2 Abkmmggmabmm b 3 Q xigxdx ab xdxgt i abgxdxgt 4 ab xdxbc xdxAc x ix 5 ab xdx7ba xdx 6 If mg x M 011w then mb7a b xdx Mbia Geometric illustration of some of the above properties Property 3 says that if f and g are positive valued Property 4 says that if x is a positive valued functions With f greater than 9 Then function then the area underneath the graph of x I y y between a and 1 plus 1 7 my the area underneath the graph of x be tween b and 0 equals the area underneath the graph of x C m between a and 0 gives the area between the graphs of f and g m ab xdxiabgxdx 0 a b Property 5 follows from Properties 4 and 1 by letting c a Alternatively7 the Fundamental Theorem of a b b Calculus gives us x dx Fb7Fa 7Fa7Fb 7Fx iba xdx Property 6 is illustrated in the picture that accompanies the proof of the Fundamental Theorem of Calculus gt An idea of the proof of the Fundamental Theorem of Calculus y We already gave an explanation of why the second part of the Fundamental Theorem of Calculus follows from the rst one To prove the rst part we need to use the de nition of the derivative More precisely7 we must show m that Ax h 7 Ax 7 A z i 5310 h M For convenience7 let us assume that f is a positive valued function Given 1 that Ax is de ned by t cit7 the numerator of the above difference I quotient is Axh7Axmh tdtiw tdt 0 25 35 xh Using properties 4 and 5 of de nite integrals7 the above difference equals t dt As the function f 12 is continuous over the interval xx h the Extreme Value Theorem says that there are values 01 and 02 in x7 xh where f attains the minimum and maximum values7 say m and M7 respectively Thus m S t g M on x7 x h As the length of the interval x7 x h is h by property 6 of de nite integrals we have that AM its dt xh 01h mh t dt Mh f62h or7 equivalently7 f01 h f62 Finally7 as f is continuous we have that Linij cl x Linij cZ This concludes the proof 93 gt Computing some antiderivatives and de nite integrals Based on Part ll of the Fundamental Theorem of Calculus we can see that antiderivatives play an important role in Calculus They are helpful in evaluating x dm This leads to the following de nition 1 De nition of the inde nite integral The Lnde m39te integral of m denoted by 35 dz without limits of integration is the general antiderivative of For example it is easy to check that 3t2 dt t3 c where c is any constant Thus if we then want to 12 compute 3252 dt the Fundamental Theorem of Calculus tells us that 0 m z 3t2dtt3c 3C 03Cz3 0 0 The formulas below can be veri ed by differentiating the righthand side of each expression Some basic inde nite integrals 1 1 mndm7mn1c ny il 2 idmlnlmlc n1 m 3 emdmemc A cfzdz c fzdz B fxigmdx mwgg i ltgmdmgt 1 Example 2 Evaluate the integrals t3 3252 4t 9 dt and t3 3252 4t 9 dt 0 6 m 6 Exam le 3 Evaluate the inte rals 7 dt and 7 dt g y y 3 Warning We do not have simple derivative rules for products and quotients7 so we should not expect simple integral rules for products and quotients 1 Example 4 Evaluate the integrals t3t 2 dt and t3t 2 dt 0 2 g T 2 9 Example 5 Evaluate the integrals m dz and dm m 1 m 1 Example 6 Compute the derivative of if t4 t3 t 9 dt 2 S 8 Exam le 7 Com ute the derivative of 3 if s du p g g 5 uz u 2 1 Example 8 Suppose x V752 7 7t 1225 dt For which positive value of z does f z equal 0 1 Example 9 1 Find the value of x at which t8 t6 t4 t2 1 dt takes its minimum on the interval 37 100 3 The value of x that gives a minimum of is Example 10 1 Find the value of x at which Cm 2 dt takes its maximum 0n the interval 757 100 75 The value of x that gives a maximum of Cm is 5 Example 11 Evaluate the integral t2 1dt 0 2 Example 12 Evaluate the integral t5 t4 t3 t2 t 1 dt 72 75 2 1 Example 13 Evaluate the integral dt 77 12 Example 14 Evaluate the integral ltl dt 76 2 Example 15 Evaluate the integral 51 dm 0 Example 16 Geometric meaning of the natural logarithmic function 12 Show that idt lnz for all z 2 1 1 In other words lnm7 with z 2 17 gives the area of the region in the rst quadrant underneath the graph of the function y E and between 1 and z 98 79192 yn The average of nitely many numbers y17y27 yn is gave 7 What if we are dealing with in nitely many values More generally7 how can we compute the average of a function 1 de ned on an interval Average of a function The average ofa function f on an interval 11 y equals the integral of 1 over the interval divided by the length of the interval 1 195 b 7 a fave a a l Geometric meaning If f is a positive valued function7 fave is that number such that the rectangle With base 11 and height fave has the same area as the region underneath the graph of 1 from a to 1 Example 17 What is the average of x 2 over the interval 07 6 gt The substitution rule for integrals If u gt is a differentiable function Whose range is a subinterval I and f is continuous on I7 then fgtg t dt fudu b 95 In case of de nite integrals the substitution rule becomes fgt gt dt u du a 9 a 1 Example 18 Evaluate the integral t 92 dt 0 12 Example 19 Evaluate the integral xt 7 dt 0 12 1 Exam le 20 Evaluate the inte ral 7 dt g t 42 5 Example 21 Evaluate the integral xt 1dt 0 1 Example 22 Evaluate the integral 72m 5 2 dm 0 3 2 Example 23 Evaluate the integral 2 m dm 0 m 1 100 m2 Example 24 Compute the derivative of if 2t dt 0 1 Hint Write gfm where fz 2 and gz 2t dt and use the chain rule to nd 0 f1 w General formula lf Htdt then F z Htdt fz I it I gt Word problems If the velocity vt of an object at time t is always positive7 then the area underneath the graph of the velocity function and lying above the t axis represents the total distance traveled by the object from t a to t 1 Example 25 A train travels along a track and its speed in miles per hour is given by 3t 762 for the rst half hour of travel lts speed is constant and equal to 3t 762 after the rst half hour Here time t is measured in hours How far in miles does the train travel in the rst hour of travel Example 26 A train travels along a track and its speed in miles per hour is given by 3t 762 for the rst half hour of travel lts speed is constant and equal to 3t 762 after the rst half hour Here time t is measured in hours How far in miles does the train travel in the second hour of travel 101 Example 27 A rock is dropped from a height of 21 feet lts speed in feet per second at time t after it is dropped is given by 3t 7322f where time t is measured in seconds How far is the rock from the ground one second after it is dropped Example 28 Suppose an object is dropped from a cliff and its speed in ftsec after 25 seconds is given by vt 102 7 5 If the object lands after 7 seconds7 how high in ft is the cliff Hint how far did the object travel Example 29 A distance traveled problem revisited A car is traveling due east lts velocity in miles per hour at time 25 hours is given by 1125 725252 10t 50 How far did the car travel during the rst ve hours of the trip 102 MA123 Chapter 2 Change and the idea of the derivative pp 1745 Date Chapter s Goal Understand average rates of change understand the ideas leading to instantaneous rates of change understand the connection between instantaneous rates of change and the derivative know the de nition of the derivative at a point use the de nition of the derivative to calculate derivatives understand the connection between a position function7 a velocity function7 and the derivative 0 understand the connection between the derivative and the slope of a tangent line Assignment 02 Due date Sept 11 Assignment 03 Due date Sept 15 Roughly speaking7 Calculus describes how quantities change7 and uses this description of change to give us extra information about the quantities themselves gt Average rates of change We are all familiar with the concept of velocity speed If you drive a distance of 120 miles in two hours7 then your average velocity7 or rate of travel7 is 1202 60 miles per hour In other words7 the average velocity is equal to the ratio of the distance traveled over the time elapsed distance traveled 7 As average velocity time elapsed 7 7 A In general7 the quantity M 7y m2 7 m1 Ax Often7 a change in a quantity q is expressed by the symbol Aq you should not think of this as A times q7 but rather as one quantityl is called the average rate of change of y with respect to m Finding average rates of change is important in many contexts For instance7 we may be interested in knowing how quickly the air temperature is dropping as a storm approaches7 or how fast revenues are increasing from the sale of a new product In this course we use the terms speed and velocity for the same concept This is not the case in some other courses Thus instantaneous speed and instantaneous velocity have the same meaning7 and average speed and average velocity have the same meaning Example 1 A train travels from city A to city B It leaves A at 1000 am and arrives at B at 230 pm The distance between the cities is 150 miles What was the average velocity of the train in miles per hour mph Do you think the train was always traveling at the same speed A train leaves station A at 800 am and arrives at station B at 1000 am The train stops at station B for 1 hour and then continues to station C It arrives at station C at 300 pm The average velocity from A to B was 40 mph and the average velocity from B to C was 50 mph What was the average velocity from A to C including stopping time Generally7 in computing average rates of change of a quantity y with respect to a quantity m there is a function that shows how the values of z and y are related gt Average rates of change of a function The average rate of change of the function y x between z ml and z 2 is h 7 average rate of change w M HM HM change in z 2 7 1 The average rate of change is the slope of the secant line between z 1 and z 2 on the graph of 1 that is7 the line that passes through ml u and 2 fz2 Example 3 Find the average rate of change of gz 2 4m 7 1 with respect to x as x changes from 72 to 5 Could you have predicted your answer using your knowledge of linear equations Example 4 Find the average rate of change of kt x3t 1 with respect to t as if changes from 1 to 5 Example 5 A particle is traveling along a straight line lts position at time t seconds is given by 3t 2t23 Find the average velocity of the particle as t changes from 0 seconds to 4 seconds Example 6 1 1 Let 9a E Find a value for x such that the average rate of change of 9a from 1 to z equals 7 Example 7 Find the average rate of change of kt t3 7 5 with respect to t as t changes from 1 to 1 h gt Instantaneous rates of change The phrase instantaneous rate of change seems like an oxymoron7 a contradiction in terms like the phrases thunderous silence or sweet sorrow However7 because of your experience with traveling and looking at speedometers7 both the concept of average velocity and the concept of velocity at an instant have an intuitive meaning to you The connection between the two concepts is that if you compute the average velocity over smaller and smaller time periods you should get numbers that are closer and closer to the speedometer reading at the instant you look at it The instantaneous rate of change is de ned to be the result of computing the average rate of change over smaller and smaller intervals The following algebraic approach makes this idea more precise Algebraic approach Let 3t denote7 for sake of simplicity7 the position of an object at time 25 Our goal is to nd the instantaneous velocity at a xed time t i7 say va Let the rst value be 251 i7 and the second time value 252 a h The corresponding positions of the object are 31 3t1 M 32 3t2 8a h7 respectively Thus the average velocity between times 251 a and t2 a h is 32 7 31 sa h 7 3a t2 7 251 h 39 To see what happens to this average velocity over smaller and smaller time intervals we let h get closer and closer to O This latter process is called nding a limit Symbolically 8a h 7 3a lim Ma h70 h We can discuss the instantaneous rate of change of any function using the method above When we discuss the instantaneous rate of change of the position of an object7 then we call this change the instantaneous velocity of the object or the velocity at an instant We often shorten this phrase and speak simply of the velocity of the object Thus7 the velocity of an object is obtained by computing the average velocity of the object over smaller and smaller time intervals Example 8 A particle is traveling along a straight line lts position at time t is given by 3t t2 3 Find the velocity of the particle when t 4 seconds Example 9 A particle is traveling along a straight line lts position at time t is given by 3t t2 3 Find the velocity of the particle when t 2 seconds The approach we will see now has the tremendous advantage that it yields a formula for the instantaneous velocity of this object as a function of time 25 Example 10 A particle is traveling along a straight line lts position at time t is given by 3t t2 3 Find the velocity of the particle as a function of 25 Even if you have a formula for a quantity knowing how the quantity is changing can give you extra information that is not obvious from the formula For example if 3t denotes the position of a ball being thrown up into the air how high does the ball go Observe that when the ball is going up it has positive velocity because its height is increasing in time whereas when the ball is going down it has negative velocity because its height is decreasing in time Thus the instant at which the ball reaches its highest point is exactly the one when its velocity is 0 Example 11 A particle is traveling along a straight line lts position at time t is given by 3t t2 3 1 Find the velocity of the particle when t 4 seconds 2 Find the velocity of the particle when t 2 seconds 3 Find the velocity of the particle when t 7 seconds 4 When is the velocity of the particle equal to 5 feet per second 5 When is the velocity of the particle equal to 0 feet per second Example 12 Find the instantaneous rate of change of gk 2k2 k 7 1 at k 3 Example 13 Find the instantaneous rate of change of gk 2k2 k 7 1 as a function of k gt The derivative of f with respect to x at z m1 is the instantaneous rate of change of f with respect to m at z m1 and is thus given by the formula 3 901 712 HM Now just drop the subscript 1 from the z in the above formula and you obtain the instantaneous rate of change of f with respect to x at a general point a This is called the derivative of f at z and is denoted with M f 21 f m 7 m 1 W 53 As we remarked earlier a change in a quantity q is often expressed by the symbol Aq you should not think of this as A times q but rather as one quantityl Thus the above formulas are often rewritten as W 7 1 Aigo dm Given a general function fm it is often common to think in terms of y x so that the above formulas are often rewritten as 11 Ly A170 Am dm Often the information you have about a quantity is not about the quantity itself but about its rate of change This means that you know the derivative of a function and want to nd the function This 39 occurs frequently in Physics Newton s formula gives information about the acceleration of an object that is about the rate of change of velocity with respect to time From this one can often get information about the velocity the rate of change of position with respect to time and then information about the position itself of the object Example 14 Let x mm b be an arbitrary linear function here m and b are constants Prove that fm m Example 15 Let x azz bx c be an arbitrary quadratic function here 017 b and c are constants Prove that f z 2am 1 Example 16 Let gz 2x2 z 7 1 Find a value 0 between 1 and 4 such that the average rate of change of gz from x 1 to z 4 is equal to the instantaneous rate of change of gz at z c Example 17 Let gz 2x2 z 7 1 Find a value mo such that g m0 4 gt The geometric meaning of derivatives Let x be a function and consider its graph y x in the Cartesian coordinate system Consider the following two points Pz17 u and Qm1 h fz1 71 on the graph of f It should be clear from our discussion that the slope of the straight secant line through P and Q fz1h f1i f1hf1 Ay m1h7m1 h E is nothing but the average rate of change of f with respect to m as the variable changes from 1 to 1 h As the value of h changes7 you get a succession or pencil of different lines7 all passing through Pz17 fz1 As h gets closer and closer to 07 the lines get closer and closer to what you probably intuitively think of as the tangent line to the graph of f at the point Pz17 fz1 y yf tangent line 351 O The tangent line to the graph of a function f at a point Pz17 fm1 on the graph is the line passing through the point Pz17 u and with slope equal to f z1 Thus7 the equation of the tangent line at the point Pm1fz1 is y 901 f190 901 Example 18 Let x x2 x 14 What is the value of x for which the slope of the tangent line to the graph of y x is equal to 5 Example 19 Let x x2 x 14 What is the value of x for which the tangent line to the graph of y x is parallel to the x axis EXPONENTIAL AND LOGARITHMIC FUNCTIONS HANDOUT 2 The de nitions and basic properties of exponential functions and logarithmic functions are given in Appendix B of the text7 pages 3157 321 This material is a continuation of Appendix B Many problems involve studying how a quantity changes with respect to time Let Qt denote the amount of a quantity as a function of time We say that Qt grows exponentially as a function of time if Qt Q06 where Q0 and k are constants that depend on the speci c problem and t denotes time When It 07 we see that Q0 Qoek390 Q0391 Q0 Thus Q0 denotes the amount of the quantity at t 0 In other words7 Q0 is the initial amount of the quantity when t 0 Taking the derivative and using the chain rule7 we see that Q t Q0 k ekt M205 kQt Since Q t kQt7 it follows that if a quantity grows exponentially7 then its rate of growth is proportional to the quantity present7 and the proportionality constant is given by k The constant k satis es k gt 0 because Q t gt 0 Some quantities decrease exponentially In this case we have Qt 064 where k is a positive constant Note that we have Q0 Q0 and Q t Q0 7k 5 7kQOe kt ich We see that Q t lt 0 because 7k lt 0 Thus the rate of increase of Qt is proportional to the quantity present7 and the proportionality constant is given by 7k Typical examples in this course of quantities that grow exponentially are the number of bacteria in a culture and the amount of money in a bank account where the interest is compounded continuously A typical example of a quantity that decreases exponentially or decays exponentially is a radioactive substance The half life of a radioactive substance which decays exponentially is the length of time it takes for the substance to lose half of its mass In other words7 the half life would be the time to such that Qt0 12Q0 This leads to the equation 12Q0 Q0571 1 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 12 e k O 1n12 7m kto to 1n2k MA123 Chapter 1 Equations functions and graphs pp 115 Date Chapter Goals 0 Identify solutions to an equation 0 Solve an equation for one variable in terms of another 0 What is a function 0 Understand function notation c Find inverse functions 0 What is a graph 0 Sketch the graph of an equation in two variables 0 Find the equation of a line and a parabola c Find the intersection points of two graphs Assignment 00 Due date Jan 20 Assignment 01 Due date Jan 22 gt Equations and solutions to equations One way in which humanity increases its understanding of the universe is by discovering relationships between various objects concepts quantities and so on Our understanding of a relationship between two quantities is sharpest when this relationship can be completely quanti ed and expressed in an equation Roughly speaking an equation is a statement that two mathematical expressions are equal For instance 3 7 2mg y2 5 is an equation relating z and y A set of numbers that can be substituted for the variables in an equation so that the equality is true is a solution for the equation A solution is said to satisfy the equation Example 1 ls z 1 and y 2 a solution for the equation 2 y 2mg What about z 1 and y 1 Many problems in the sciences economics nance medicine and numerous other elds can be formulated into algebraic terms by identifying variables expressing unknown quantities and by setting up appropriate equations relating these variables Example 2 Suppose a fuel mixture is 4 ethanol and 96 gasoline How much ethanol in gallons must you add to one gallon of fuel so that the new fuel mixture is 10 ethanol gt Equations into functions An equation in two or more variables can sometimes be solved in terms of one of the variables This type of equation is closely related to the notion of a function Example 3 Solve the equation x3 2xy 5y 7 for y in terms of x i i 7 7 3 Observe that in the equation y m i i 295 f 5 i associates to any given value of x precisely one corresponding value for y De nition of function A function f is a rule that assigns to each element x in a set A exactly one element called x in a set B The set A is called the domain of 1 whereas the set B is called the codomain of 1 x is called the value of f at x or the image of x under 1 the expression on the right hand side can be viewed as a recipe that Arrow diagram of f The range of f is the set of all possible values of x as x varies throughout the domain range of f lx E A f input output Machine diagram of 1 Evaluating a function The symbol that represents an arbitrary number in the domain of a function f is called an independent variable The symbol that represents a number in the range of f is called a dependent variable In the de nition of a function the independent variable plays the role of a placeholder For example the function x 2x2 7 3x 1 can be thought of as D2D273D1 To evaluate f at a number expression we substitute the number expression for the placeholder If f is a function of x then y x is a special kind of equation in which the variable y appears alone on the left side of the equal sign and the expression on the right side of the equal sign involves only the other variable x Conversely when we have this special kind of equation such as y e1 x3 7 3x 5 it is common to think of the right hand side as de ning a function x and of the equation as being simply yf Example 4 Find the domain of the following functions If 1035 M635 4 write an expression for f1 h 7 101 f1 h 161 lf Pm 3x3 2x2 z 11 and we rewrite Pm in the form Px A Bx 71 Cm 71m 7 2 Dm 71m 7 2m 7 3 what are the values of A and B Example 7 If we rewrite the function x in the form 13 13 7 13 7 what are the values of A B and C gt Recall that two functions x and gz are said to be inverse of each other if m and m Example 8 If ht 3t 7 nd a function gt such that hgt t gt Cartesian plane and the graph of a functionzl Points in a plane can be identi ed with ordered pairs of numbers to form the coordinate plane To do this7 we draw two perpendicular oriented lines one horizontal and the other vertical that intersect at 0 on each line The horizontal line with positive direction to the right is called the z axis the other line with positive direction upward is called the y axis The point of intersection of the two axes is the origin 0 The two axes divide the plane into four quadrants7 labeled l7 ll7 lll7 and IV The coordinate plane is also called Cartesian plane in honor of the French mathematicianphilosopher Rene Descartes 1596 1650 Any point P in the coordinate plane can be located by a unique ordered pair of numbers a7 1 as shown in the picture The rst number a is called the m coordinate of P the second number I is called the y coordinate of P Graphing functions If f is a function with domain A7 then the graph of f is the set of ordered pairs graph of f la 6 A In other words7 the graph of f is the set of all points my such that y m that is7 the graph of f is the graph of the equation y Obtaining information from the graph of a function The values of a function are represented by the height of its graph above the z axis So7 we can read off the values of a function from its graph ln addition7 the graph of a function helps us picture the domain and range of the function on the m axis and y axis as shown in the picture The graph of a function is a curve in the xy plane But the question arises Which curves in the zy plane are graphs of functions The vertical line test A curve in the coordinate plane is the graph of a function if and only if no vertical II 34 I b i130 b i I I I I I I O a 111 IV z y I I I I I line intersects the curve more than once Graph of a function I I I z I I Not a graph of a function gt Lines and parabolas The simplest types of functions are linear and quadratic functions A linear function is a function f of the form x mx b7 where m and b are real numbers The graph of the equation y mx b is a non vertical line in the xy plane The numbers m and b are called the slope and yintercept7 respectively A quadratic function is a function f of the form axz bx 37 where 11 and c are real numbers and a 7 O The graph of any quadratic function is a parabola it can be obtained from the graph of x x2 by using shifting7 re ecting and stretching trans formations lndeed7 by pleting qw a quadratic function x ax2 bx c can be expressed in the standard form ax 7 h2 k The graph of f is a parabola with vertex 17 k the parabola opens upward if a gt 07 or downward if a lt 0 Vertex 17 k Minimum m7ae7hrh agt0 Maximum Vertex 17 k m7ae7hrh alt0 Example 9 If the equation of the line through the points 37 4 and 717 6 is written as yAB17 what are the values of A and B Example 10 The parabola y x2 7 15x 54 intersects the x axis at the two points P and Q What is the distance from P to Q If we rewrite the inequality x2 7 15x 54 lt 0 in the form A lt x lt B7 what are the values of A and B gt Intersection points The graphs of two equations intersect at a point if and only if that point is a solution for both equations Example 11 Find the points of intersection between the graph of the equation 4x2 9342 36 and the line with equation Example 12 Find all points where the graph of y 7 12 7 2 z crosses o the y axis o the z axis Example 13 The area of a right triangle is 7 The sum of the lengths of the two sides adjacent to the right angle of the triangle is 11 What is the length of the hypotenuse of the triangle Example 14 What is the smallest root of the polynomial Qm 3 7 12x2 44x 7 48 MA123 Handout 1 Higher Derivatives Let y f be a differentiable function and f its derivative lf f is again differentiable7 we write y f and call it the second derivative of f x In Leibniz notation d2 2 W or w Similarly7 we can de ne higher derivatives of f if they exist For example7 the third derivative f z of f is the derivative of f 7 etc The higher derivatives are denoted by f4z7 f5z7 and so on Example 1 Let Hs s5 252 35 2 We can nd the second derivative by using the power rule twice H s 554 45 3 and H s 2053 4 Example 2 Let fx Let us nd f x Rewrite f as f xi Then 1 1 7 5 f z 2x Taking the derivative of this7 we get 1 1 1 we 5 7253 7r Finally7 taking the derivative one more time gives 1 3 5 3 5 3 m 77777 77777 f 95 7 4 2V 2 895 2 825W Example 3 Suppose Ft ftgt Then Ft ft9t ft9 t and FW f t9t 2ft9t ft9quott Here we used the product rule twice for F t and collected similar terms Example 4 Suppose 5t equals the distance or height of an object from a given point at time t Remember that the derivative of 5t is it s t the velocity of the object The derivative of the velocity function gives the acceleration of the object7 usually denoted by at Thus7 at t Exercise 1 If f x 237 nd f 2 If fgz7 nd F z 3 If f 4 nd f57 the 5th derivative of Can you make a guess about the n 1st derivative of f z 4 Suppose the height of an object above ground at time t in seconds is given by ht 716t2 12t 200 Find the acceleration of the object after 3 seconds