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# Calculus With Analytic Geometry I MATH 140

Penn State

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This 0 page Class Notes was uploaded by Elaina Osinski on Sunday November 1, 2015. The Class Notes belongs to MATH 140 at Pennsylvania State University taught by Staff in Fall. Since its upload, it has received 21 views. For similar materials see /class/233010/math-140-pennsylvania-state-university in Mathematics (M) at Pennsylvania State University.

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Date Created: 11/01/15

Math 140 Calculus with Analytic Geometry I Spring 2007 Penn State University Section 47 10 amp ll A Brlef Rev1ew of Algebra and I rlgonometry The following is a list of questions from the realms of Algebra and Trigonometry that tend to come up in introductory Calculus courses Several of these questions are in fact taken from the Readiness Quiz77 given on the rst day of class Other questions are a compilation of mistakes seen time and time again on homework and exams All of them7 however7 are questions regarding algebra andor trigonometry that the successful calculus student should be capable of answering 1 A Quick Quiz Determine whether each of the following statements is True or False Be careful Try to think of speci c examples before answering with your rst instinct li 5i1y212y2 9ilfazltaythenzltyi 13 700 Z 6 1 Jr 10ilfa1aythenzyi 141 3 13 7 3r 3z 11 z z 4 g 8 wsinz 12 314159 2 Simplify H 01 i Simplify h where 312 4 H on i Simplify w where Fz 13 i Simplify where 7i 5 142 3 i Simplify 64 7 t2 7 t2 64 7 t2 12 by getting a common denominator i Simplify 2amp2 for n 2 l where N read N factorial represents the product of the rst N positive integers For examp1e112x1223112364I12 3 4 HH 0071 H to 3 Inequalities 20 For what values of z is 11 16 positive negative 21 For what values of z is 12 31 03521 7 1 positive to to i Find all values of I such that 31 7 2 2 5 to CA3 i Find all values of I such that 2 5 lt ll to bu i Find all values of t such that x 64 7 t2 7 t2 64 7 t2 12 is negative 25 Suppose 7l S I S ll Find upper and lower bounds for 12 and 13 4 Completing the Square and the Quadratic Formula We can solve any quadratic equation of the form ax2 121 c 0 by completing the square The result is known as the Quadratic Formula One application of the quadratic formula is to factor quadratic polynomials In particular a12bzca 20 712xb274ac 7b7xb274ac 1 2a 1 26 Find values A and B such that 2 7 81 7 l fl A2 B by completing the square 27 Find all values of I such that x2 7 51 7 36 0 28 Factor 213 512 7 31 29 Factor 2 cos3 9 5 cos2 9 7 3cos 9i 5 The Pythagorean Theorem and the Equation of a Circle Hm fo ML ln pictures7 the Pythagorean Theorem says that the area of the green square below is equal to the sum of the areas of the red and blue squaresi If the points P and Q are thought of as the end points of the hypotenuse of a right triangle7 then the distance between these points can be determined using the Pythagorean Theoremi A circle can be described as the set of points zy that are a xed distance 7 from a central point abi The equation for a circle can thus be derived using the distance formula from above CA3 O 1 Determine the value of 1 in the triangle below CA3 H 1 Determine the value of z in the triangle below 1 2 1 2 32 Find the distance between the points 714 and 5 2 33 Determine the radius and the center of the circle described by the equation 12101y2712y 3 Hint Complete the square twice 6 Degrees vs Radians There are two common units for measuring angles degrees and radiansi Most people are familiar with the fact that there are 360 degrees in a circle But why 360 Why not 100 degrees The answer can be traced back to several ancient civilizations The Babylonians used a base60 number system as opposed to our base10 number systemi While the exact reason for the base60 number system is a mystery it does have it s advantages over our seemingly more natural number system of today Any multiple of 60 can be divided by 1 2 3 4 5 6 10 12 15 20 and 30 No number smaller than 60 has as many proper divisorsi In particular a circle divided into 360 equal parts can easily be divided into halves thirds fourths fths sixths eighths ninths tenths etc Try to do that with a circle divided into 100 equal parts Another civilization the Sumerians used a 360 day calendar to chart the progress of the Suns circular path across the sky While our base10 number system is prevalent today the base60 system does in fact still live on Positions on the globe are still measured in degrees of longitude and latitude Each hour is divided into 60 minutes each minute is divided into 60 seconds Even though the use of both number systems can be easily justi ed they will always have a sense of arbitrariness to themi While the concept of a radian is less widely known it is in fact a much more useful and more logical choice for measuring anglesi Many identities involving trigonometric functions would lose their luster if they were expressed in degrees or any unit other than radiansi The de nition of a radian is based on the following fundamental fact The ratio of the circumference of a circle to its diameter is constant In other words if you take my circle and calculate the circumference of that circle divided by its diameter you always get the same number no matter what circle you started with This constant is denoted by the greek letter 7r for perimeter and yields the familiar formula for the circumference C of a circle of radius 7 2 C 27ml The use of the symbol Tr to represent this number began in the 18th century however the number itself was known to the Babylonians and Egyptians over four thousand years ago In particular a circle of radius 1 has circumference of length 27f and if we mark off one unit length along the perimeter of this circle the angle formed is de ned to be a radiani This construction of a radian is illustrated belowi Thus by de nition there are 27f radians in a circle and 27T radians 360 degrees 34 Complete the following table converting degrees into radiansi degrees l 0 radians l H 01 w o ugt U1 OJ 0 O o H M o H w 01 H 01 o H 00 o M H o M M 01 M ugt o M w o w o o w H 01 w w o w OJ 0 35 Determine the value of each angle in radians of the triangles shown in Problems 30 and 31 7 The Unit Circle and Trigonometric Functions Consider the unit circle de ned by the equation 12 y2 1 In other words the unit circle is the set of points that is exactly 1 unit away from the origin Now draw a ray that originates from the origin 0 Let 9 be the angle in radians that the ray makes with the positive zaxis measured in the counterclockwise direction Let P represent the point of intersection of the ray and the unit circle 5 7w 2 777 571 W 2 Figure 1 Graph of y cost9 A 1 W Jr 73 7 g g W77 71 Figure 2 Graph of y sint9 An alternate de nition for the cosine and sine functions is based on a right triangle For 0 lt 9 lt 7r2 consider the triangle shown below A l a where a is the length of the leg adjacen 77 to 9 0 is the length of the leg opposite t9 and h is the length of the hypotenuse The remaining trigonometric functions secant7 cosecant7 tangent and cotangent are de ned in terms of cosine and sine as follows sec E sint9 0 9 30319 Z tame cos9 E csc 7 7 7 cost9 7 a 6 sint9 0 COth 7 sint9 7 g 36 Use your answers from Questions 307 31 and 35 to compute the following values exactly a cos7r3 c cos7r4 e cos7r6 b sin7r3 d sin7r4 f sin7r6 37 Use your answers from the previous problem to label the points on the unit circle below by its 1 and y coordinates Each line is labeled by the angle in radians it makes with the positive zaxisl 1021 38 Compute cosx given that tanz 2 and 7r lt E lt 37r2l 39 Compute the area of the triangle with two sides of length 4 cm and 9 cm resp and an included angle of 7r6l 40 Compute sec 77r6l 41 Draw a graph of y cosz3l 42 Find all the values of 9 such that 1 7 2 cos2 39 0 8 Trigonometric Identities The following is a list of several trigonometric identities that calculus students should be at least familiar with if they have not already committed them to memory This list is by no means complete7 however7 all other trigonometric identities can be derived from this collection For a complete list of identities7 see Appendix D page A28 in Stewart7s Calculus 43 Show cost9 7r2 7 sint9 45 Show sina 7 sina cos 7 cosa sin 44 Show sint9 7r4 sint9 cos t92 46 Show cos 29 1 7 2sin2 t9 47 Compute sin7rl2i 49 Show sin 29 2 sin9cos9 48 Find all angles 9 where cos 29 sin9 50 Find all angles 9 where sin 29 cos9 9 Solutions Each statement presented in Questions 1 14 is false ll An easy way to see that this is false is to convert each fraction into it s decimal formi The left hand side becomes 05 03 08g while the right hand side becomes 02 Clearly these two values are not the same What Is True 2 This is simply a generalization of the previous question While there are speci c cases where this identity is true eg I 0 this identity in general is false A more simple example of why this is false is obtained by setting I y 2 ll What Is True Remember in order to add two fractions you must rst nd a common denominator and then and only then may you add numerators That is z I I2 my 7 J 7 i i y Z 92 92 7 12 my yZ 7 12 y 7 9 This mistake can be explained by confusing multiplication with addition In particular any common factors that appear in the numerator and denominator of a fraction can be cancelled out Nothing can done with common terms that appear in numerator and denominatori What Is True 4 We can easily verify that this is false by considering the case when I ll Note that the I l in the numerator is a factor of only one of the terms in the numerator Therefore it cannot be canceled out by the I l in the denominator What Is True zlz76 7 12z76 z lz 72 7 1 lz 7 2 7 I 3 I 7 2 7 I lz 7 2 7 z 3 7 z l 5 This common mistake can be chalked up to wishful thinking Exponentiation does not have a distributive property like multiplicationi It certainly would be nice if it did but we can easily see that it does not Again try a simple example like I y l to see that this is not the case 533 7 9 50 D What Is True Using the distributive property of multiplication we have Iy2IyIy IIyyry 12zyyzy2 1221yy2 Another example of wishful thinking If we write the statement as I y12 112 y12 we see that this is exactly the same type of error as in the previous question Again exponentiation does not have a distributive property like multiplication A simple example like I y 1 should convince you that the above statement is false What Is True Like in the previous problem there is a way to express xz y as a sumi However in this case it would require an in nite number of terms to do so Exactly how this is done is one of the main topics covered in Math 141 Remember anything in parentheses must be evaluated rst In other words the quantity 31 is being raised to the nth power on the left hand side Again letting n 0 and z 1 should convince you that the statement is false What Is True 31 31 x ntimes 333zzz qP ntimes 371171 n times Rememeber sin21 denotes the sine of 21 The sine function must be applied to the quantity 21 before dividing by 2 Consider the case when I 7r2 to see that this statement is indeed false What Is True The only simpli cation that can be done here would be to apply a trigonometric identity For example 2 sinz cosz sin21 7 2 7 sinz cosz This one is a bit tricky The tendency is to simply divide both sides by a But remember if a lt 0 then you must switch the inequality What Is True The following statements are true ifagt0andazltaythenzltyi ifalt0andazltaythenzgtyi Another tricky one This statement is true a is nonzero but if a 0 then I and y need not be the same Again this common error arises when you try to divide both sides of am ay by a Remember division by zero is not de ned Whenever you divide both sides of an equations by something you must make sure that you are not dividing by zero What Is True The following statement is true ifaa Oandazaythenzyi H to H CA3 H H H 01 One more tricky one This statement is certainly true if z is nonnegative that is greater than or equal to zero However it is not true if z is negative Try 1 71 What Is True For all real numbers I xaT m where denotes the absolute value of I We also have that z with the understanding that the left hand side is only de ned when I 2 0 since you cannot take the square root of a negative number i This is simply an approximation of 7r which is an irrational number This means that 7r cannot be written as a ratlo of integers What Is True 7T7T This may seem like a trivial statement but the point is that 7r is a number Leaving an answer with 7r in it is perfectly acceptable In fact its preferred For example if the answer to a problem is 7r37 then leave your answer as 7r37 since this is the exact answer Do not replace this answer with the approximation 00849079095565 or worse yet 3 1415937 0 084907Wi The ratio 10 is not de ned since division by zero is not allowed The 00 symbol is just that a symboli In nity is not a number it is a concept What Is True The following limits are true lim 7 00 xao I If the above line looks foreign to you then you are in the right class The concept of a limit is fundamental to the study of Calculus and will be covered in this course Again the ratio 00 is not de ned since division by zero is not allowed regardless of the value of the numerator What Is True There are certain limits that we will eventually refer to as indeterminate There are several types of indeterminate forms 00 being one of themi There are certainly examples of indeterminate forms of type 00 that are equal to 1 But not all of them are Some are equal to zero Some are equal to 00 In fact any number you choose is the value of some indeterminate form of type 00 This topic will be investigated in Math 141 i The most common mistake here is to assume that h h or h fh both of which are very rarely the case Remember that the z in can be replaced by anything In this case we want to replace it by z it That means we must also replace 1 by z h in 312 4 That is fzh3zh24 31221hh24 3z26zh3h24 Fzh7Fz 7 1h3713 7 7 13312h31h2h3713 Hf 7 312h31h2h3 h 312 31h h2 17 First7 note that 735n171 f 71 1 W 735n4 427171 Therefore fn1 7 735n4 421173 n 7 42n71 35n71 735n475n7142n7 s72n71 lt73gt544 35 7 t2 M647 t2 7 t2 647 t2 12 7 M647 t2 7 7 647 7 M64 7 464 7 t2 t2 7 M647 t2 M64 7 t2 7 647 t2 7 mim 7 6472 7m 19 To get used to the notation7 it may help to try a few examples For n 17 we have 5 712345 i 2x 7 2 3 4 5 7 2 3 4 5 Forn27 we have 7 1234567 4x 4567 4 567 Forn37 we have 9 123456789 a 6 6789 6 789 Now for the general case7 2n 3 7 2n 2n 12n 22n 3 2n 7 2n 271 12n 22n 3 10 to O to H to to to CAD to g i The quantity 11 16 is positive when I gt 0 and z 16 gt 0 ie I gt 0 or when I lt 0 and z 16 lt 0 ie I lt 716 The quantity 11 16 is negative when I gt 0 and z 16 lt 0 notice that there are no values of I such thatzgt0andz16lt0 orwhenzlt0andz16gt0 iiei 716ltzlt0i 1 Notice that the quantity 1231 1332z 7 1 is positive when 12 f 0 ie I f 0 and a 31 13 gt 0 and 321 71gt 0 or b 31 13 lt 0 and 321 71 lt 0 31 13 gt 0 when 311gt 0 which occurs for z gt 7131 321 71gt 0 when 21 71gt 0 which occurs for z gt 121 Therefore both of these conditions are met when I gt 12 31 13 lt 0 when 311lt 0 which occurs for z lt 713 321 71 lt 0 when 21 71 lt 0 which occurs for z lt 12 Therefore both of these conditions are met when I lt 713 Thus 61 13x32z 71 is positive when I gt 12 or z lt 7131 1 31 7 2 2 5 when 31 7 2 2 5 or 31 7 2 S 751 The rst situation occurs when I 2 73 and the second situation occurs when I S 711 1 12753751 lt 1 when 71 lt 213 5 lt 1 Multiplying each side by 3 yields 73 lt 21 7 5 lt 3 Adding 5 to each side yields 2 lt 21 lt 8 And nally dividing by 2 yields 1 lt z lt 4 1 Recall that in Problem 18 we determined that 64 7 272 M64 7 t2 7 t2 64 7 t2 12 7 l x64 7 t2 Notice that the above function is de ned as long as 64 7 t2 gt 0 ie 78 lt t lt 8 since you cannot take the square root of a negative number Also the ratio 5472 is negative whenever 64 7 2t2 lt 0 ie t2 gt 32 since the x647t2 denominator is always positive for 78 lt t lt 8 Therefore 34 is negative when 78 lt t lt 7S2 and x 32 lt t lt 8 i From the graphs below we can easily see that 0 S 12 S 1 and 71 S 13 S 1 if 71 S I S 1 A common mistake in this situation is to say the following if 71 S I S 1 then 17712g12g121 The reason that this is incorrect is that 12 is decreasing on 710 and increasing on the interval 011 On the other hand it is true that 717713gz3g131 11 since 13 is nondecreasing on the interval 711 In general if is nondecreasing for a S I S b then M fz M if is nonincreasing for a S I S b then M fz W 26 A 74 and B 717 That is 1278171z742717 to N i Using the Quadratic Formula we have that 12 7 5x 7 36 0 when 5 i x25 144 I 2 7 5 i V169 7 f 75i13 2 Therefore 12 7131 36 0 when I 9 and z 74 28 213 5x2 7 31 1212 5x 7 3 211 712z 3 121 7 lz 3 29 If we let 213 5x2 7 31 then fcos t9 2 cos3 9 5 cos2 9 7 3cos 9 Using our answer to the previous problem we have 2cos3 9 5cos2 t9 7 30086 00862 3086 7 lcost9 3 30 Using the Pythagorean Theorem we have 12 12 1 212 1 12 12 I 2 2 CAD H i Using the Pythagorean Theorem we have 122z2 1 121714 1234 z2 CAD to i Using the distance formula we have that the distance between 71 4 and 5 2 is x5122742 364 m N 331 By completing the square7 we have 12101y2712y 152725y762736 Therefore 152725y7627363 or equivalently 152y76232536 Which is the equation of a circle of radius 8 centered at the point 757 61 341 degrees 0 l 15 l 30 l 45 l 60 l 90 l 120 l 135 l 150 l 180 l 210 l 225 l 240 l 270 l 300 l 315 l 330 l 360 radiansl0lllll ll5l lll ll lll quotl27r 351 4 A I A I A 361 a cos7r3 12 c cos7r4 Q e cos7r6 Q b sin7r3 Q cl sin7r4 Q f sin7r6 12 371 To help remember these values7 notice that all of these numbers are of the form i T for i 0717273741 1 071 lt1 27 27 2 38 Since 2 tanz sin 1 cos I we must have sinz 2 cos 1 Using the identity cos2 zsin2 z 1 see Trigonometric Identities7 we have 1 cos2 I sin2 I cos2 I 4cos2 z 5 cos2 1 Therefore COSQI 15 which implies that cosz i1gi Since 7r lt z lt 37r27 we must conclude that cosz 1 N3 39 The given triangle is drawn below Therefore the area of the triangle is given by Base gtlt Height 7 9 X 4sin7r6 2 7 9 X 4 gtlt12 2 9 40 7 6 7 sec 7r 7 cos77r6 7 1 JP 41 76w 73w 1 N W 6w 2 2 2 2 71 42 Notice that 1 7 2cos2 39 0 when cos2 39 121 In turn7 this happens when cos39 i12 Therefore 39 2n 17r4 where n is any integer In other words7 9 must be of the form 2n 17r12 for some integer n 43 0086 7r2 cost9 cos7r2 7 81116 sin7r2 cost9 0 7 81116 1 7 sint9 44 sint9 7r4 81116 cos7r4 sin7r4 cost9 81116 22 22 cost9 2sint9 cost92 14 45 sinlta 7 7 sinlta 75gt sina cos7 sin7 cosa sina cos 7 sin cosa 46 00826 0086 9 cost9 cost9 7 sint9 sint9 cos2 9 7 sin2 9 l 7 sin2 7 sin2 9 1 7 2 sin26 47 Using the identity from Problem 46 With 9 7rl2 we have cos7r6 17 2 sin27rl2 Solving for sin27rl2 yields sin2 7r 12 7 17 2 7 f 242 7 4 1 7 cos7r6 2 Taking the square root of both sides produces Sin7r12 But since 7rl2 is in the rst quadrant sin7rl2 must be positive and therefore sin7rl2 g 00 i Using the identity from Problem 46 we have 17 2sin2t9 sint9 or equivalently 02sin2 9sin 971 2 sint9 71sint9 1 Therefore either sint9 71 ie 9 37r2 27m Where n is any integer or sint9 12 ie 9 7r6 27m or 9 57r6 27m Where n is any integer 49 sin 29 sint9 9 sint900st9 sint900st9 2sint900st9 50 Using the identity from the previous problem7 we have 2sint900st9 cost or equivalently7 2sint900st9 7 3086 2 sint9 71COSO 0 Therefore either cost 0 ie 9 7r2 27m or 9 37r2 27m Where n is any integer or sint9 12 ie 9 7r6 27m or 9 57r6 27m Where n is any integer Working with Fractions Misunderstanding the rules for working with fractions is by far the most common reason for mistakes in Calculus problems That is more than enough to justify repeating these rules 0 The best thing about fractions is being able to reduce them 3 3 lFba Oandxa Oi 121 12 Notice that all bets are off when the denominator is zero since we7re fun damentally unable to deal with that situation Here s an example how many piles of quarters can you make with 1077 has an answer but how many piles of nothing can you make with 1077 has no comprehensi ble answeri Thus we avoid the whole problem by de ning a fraction as g a b with I f 0 Remember division by zero never makes sense even in limits 0 To multiply fractions we simply multiply tops and bottoms77 or numer ators and denominators a C ac a a a Notice that I f 0 and d f 0 since we wrote the rst two fractions and this means 12d 0 by the Zero Factor Law so the resulting fraction also makes sense 0 To divide fractions we ip the one we7re dividing by and then multiply those d d a c a a a T 2 a a In this case we have to assume 5 f 0 and then be 0 again by the Zero Factor Lawi However 5 0 would mean we were secretly trying to divide by 0 which we already know is not allowed 0 Unfortunately addition of fractions is harder basically because addition and division donlt go together very well Up to reducing fractions the rule is still pretty simple though a 5 ad be b d bd We should expect to reduce this answer in most cases The right way to think of this rule is that we know how to add fractions with the same denominator a c ac 33 T and we can always unreduce two fractions to get their product as a common denominator a 5 ad be ad be 7 7 i 7 b d bd bd bd o The rule for subtracting fractions is basically the same a c adibc b d bd although we should expect to reduce this answer in most cases o It is often important to nd out which of two fractions is larger However you can always unreduce to get either the numerators or denominators to be the same whichever is easier in your specific case If the numerators are the same the fraction with the larger denominator is smaller because you are diViding the same numerator into more pieces If the denominators are the same the fraction with the larger numerator is larger since you are diViding a bigger numerator into the same number of pieces o If there is a plus or minus sign in the denominator of a fraction there is no rule telling us how to break the fraction into two reasonable piecesi Let7s look at some examples ltgtltzgtltgtlt2gt2 lt2g2gtlt215gt lt2gtltgt lt2gtltgt 9 Some examples of comparing two fractions lt ltL5 5 851116640 40 Lgt gt 223 560 mm 446 560 Now some examples to show that plusminus signs in the denominator are not something we can just break up into two pieces l l l 1 32 5 72 gigwhile 7776 76 5 5 5 35740 75 78775while 77 56 i56 Laws of Logarithms There are very few laws of logarithms that let us work with them very effectively despite the fact that logarithms are very hard to evaluate in genera i Assume a gt 0 is a positive real number 0 logaz y means I ay for real numbers I gt 0 and y So 0 logaax z for every real number 0 034 z for every I gt 0 The combination of the last two statements says that logs and exponential functions are inverse functions However you should be careful to keep track of when I must be strictly positive since no logarithm can be de ned at zero and the logarithm of a negative number is a complex number which we also won7t deal with The laws of exponents lead to the following laws of logarithmsi Here we assume I and y are positive real numbers 1 loamy 10ml 10gay 2 loga 10ml 7 10gay 3 log 17 Tlogaz for any real number Ti We will generally only care about the case when a e in which case loga lnz is the natural logarithmi One excuse for only working with this most important logarithm is the following change of base formula77 for a f l lnz lna logam For the natural log the laws become 1 lnzy lnz lny 2 ln lnz 7 lny 3i lnzT Tlnz for any real number Ti Also the earlier statements become 0 lnz y means I ey for real numbers I gt 0 and y o lne z for every real number 0 elm z for every I gt 0 One way we can use the laws is to collapse some large expression involving several logs Example log3 z 5 log3 z 7 5 7 4log32 log312 7 25 7 log316 12 7 25 log3 16 Another that Will be very important later is really the opposite idea We take the natural log of a complicated expression and break it down into a number of manageable piecesi Example 156I2 47 in1 5 nI27 77nI37 8 lnlt IMP 71lt5gt1lt 4 1lt s 6lnz 5 7lnz2 7 4 7 Slnz3 7 5

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