SAMPLE EXAM III SOLUTIONS: Section I Part A: Non calculator 3 1) B. This is straight forward 3  x 2 x 3  x  3 2) C. lim  3x 2  2 x dx  34 1  lim x 3 3  x 3  x   lim  3  x   0 x3 x 3 3) C. The problem states that C(s) has units gallons/mile. The problem also states that s has unites miles per hour…therefore ds (a very small part of s) also has units miles per hour. The integral merely sums up all the products of C, and the very small ds. gallons miles gallons   mile hour hour dr 4  2 , r  10 and the volume of a sphere is V   r 3 . We want to find the rate of dt 3 dV change in volume over time…in other words we want to find . We take the derivative of Volume dt with respect to Time, but it is important to remember that “r” is a function of time, and so a chain rule must be used. 4) D. We know dV dr dr dV  4 r 2  2 and r  10   800 Now we simply plug in dt dt dt dt 5) B. We know position is the integral of velocity so we merely integrate. The integration itself is straightforward. 3 26 2 1 t dt  3 6) D. Multiplying the original two terms would yield a coefficient that is unpleasantly large to deal with in the absence of a calculator. Arguably the best way to do this is with product and chain rules. Then we’ll see what we have. d 4 x 3  2 x 6  34 x 2 4 2 x 2  62 x 5 24 x 3  ...... Convince yourself of this product rule with chain dx rules sprinkled in before you continue! Looking at many of the solutions, it seems factoring out a 4 x 2 2 x 5 may be the correct course of action.    4 x  2 x  3  4  2 x  6  2  4 x  4 x  2 x  72 x 2 5 2 5 7) D. First consider choice I: Pick any spot that satisfies y(0)  2 meaning any spot on the y-axis higher than 2. For instance pick (0,4). That is our “initial condition”. Start there and follow the slope lines to the right. We will see that the slope lines lead us in a path that approaches the line y = 2, which is the claim of the choice. We can see that no matter what choice we make above 2, the result will be the same. This makes choice a good one. Next consider choice II: Pick any spot that satisfies 0  y(0)  2 for instance (0,1). Starting here and following the slope lines (up this time) to the right, we see this also approaches the line y = 2. This matches the claim of the choice. We can see that any choice between 0 and 2 will have the same result. This is also a good one. Lastly consider choice III Normally we’d pick any spot that satisfies y(0)  2 but notice the graph. Any spot chosen below 0, heads down…in other words the limit goes toward negative infinity. We cannot say this is a good choice because not all choices for y(0)  2 lead us to the same result. The valid choices are I and II only. 8) D. Make a u-substitution of u  x 2  1  du  2 x dx  dx  limits of integration. x  1  u  2 and 3 10 x  3  u  10 1 dx Be sure also to change your 2x 10 x x 1 1 1 1 1 1 10 1 x 2  1 dx  2 u 2 x du  2 2 u  2 ln( x)2  2 ln(10)  ln(2)  2 ln(5) 1 1 1  2x  2x  x 2  1   2x  2 2 2 2 2 1 x 1 x 1 x 1 2 x 1 10) C. This problem wants us to integrate 2 tan( x) . The first part is just the messy bit to handle the problems that come with the domain of the tangent function. The best way to do this is to use a quotient rule and u-substitution of u = cos(x). Also toward the end of the problem, some choice uses of properties of logs must be used to achieve a form that is an answer choice. 1 u  cos( x)  du  sin( x)dx  dx  du sin( x) 9) B. This is a 2-link chain rule. sin( x)  2 tan(x) dx  2 cos( x) dx  2   d  1  ln   dx  x 2  1       sin( x)  1 1  du  2 du  2 ln(u )  ln u 2   .... u sin( x) u  ... ln cos( x) 2  ln sec 2 ( x)   11) A. Being mindful to change our limits of integration, it is fairly straight forward to use the substitution they give you to achieve… 2 5 x3 x2 2 dx  0 x 2  1 1 2u du but this doesn’t get rid of x !!! But looking at our original substitution it is easy to find out what x 2 is in terms of u. if 2 so….  0 u  x 2  1 then x2  u 1 x3 x2 u 1 dx  du   du 2  2u 2u x 1 1 1 5 5 12) D. Because the line y = x + 4 is tangent to the curve at x = -2…this obviously means they meet at this point. Look at the line y = x + 4. This must go through (-2,2). The curve must also go through this point. So…. kx  8 k (2)  8 and multiplying both sides by the denominator 2 x  4  2 x  8 which y 2 kx k  (2) leads quickly to x = 3 13) E. For this problem a picture works best. Remember we have 4 regions, therefore 4 rectangles. The width of each rectangle is determined by the interval. The height is determined by the right hand side meaning the first rectangle’s height is f(6)=7 and so on 2x12 3x11 1x8 1x7 5 6 9 7 + 33 + 24 + 8 = 72 11 12 14) C. This can only really be done by making a picture. Clearly a vertical asymptote exists at x = 2. Since the slope is positive everywhere, the graph must look something like the following (keep in mind there are a few variations, but all the key elements are the same. Notice that the limit as x goes to 2 from the left is positive infinity. While the limit as x goes to 2 from the right hand side is negative infinity. This makes choice II false, and choice III true. In addition, the limit of a function does not exist unless the left hand, and right hand limits agree…which is not the case here. So choice I is also false. Choice III only. 15) B. This problem is straightforward as long as you remember the often forgot exponential rule: d x A  A x ln( A) Just apply a chain rule to the given expression and you’ll find dx 3 dy  5 x 2  3x 2 ln(5) dx 16) B. This problem is tricky. Firstly, the fact that you see a sum of an infinite number of terms makes this reek of some sort of integral. Notice also, that if n is infinity, the first term in the series is sin(0), and the last term is sin(). Giving us a hint to the limits of integration. Now take a step back. We are adding “n” number of terms of sin(x), and then dividing by n. When we add up many things and divide by the number of terms, that’s an average! This problem is asking “what is the average value of sin(x) from 0 to .  av value  1 1 sin( x) dx   2   0 0  17) D. This is asking us to relate f’ to f. The relation should be instinctual. When f’>0, f is increasing. When f’<0, f is decreasing. When f’=0, f is flat. This last part is the quickest way to do this specific example. We see the derivative is 0 at x = – 2 and x = 1. Simply see that the only graph that is “flat” at these x values is D. You could also use the fact that f should increase for x > -2, and f should decrease for x < -2 18) B. This is a simple application of one part of the Fundamental Theorem of Calculus (F.T.O.C). 0 x d du d du 1    Any time you see a derivative of an integral, you are not intended 2 2   dx x 1  u dx 0 1  u 1 x2 to either find the derivative, or the integral, but merely apply the F.T.O.C. 19) C. The segment connecting (-2,26) and (10,2) is parallel to some tangent line for which we are intended to find. To find the equation of this (or any) mystery tangent line we will need it’s slope and a point it goes through. Since it’s parallel to the mentioned segment, we quickly see it’s slope must be –2. Now we need the point it goes through. Notice in the table that the slope of our curve is –2 at x = 4 and y = 23. This must be our point of tangency, and therefore our mystery line goes through (4,23) y –23 = -2(x-4) The y-intercept is easily found to be (0,31) 20) C. We are given dy  4 xy . This is most easily done by separation of variables. dx 1 dy 1 dy 1  4x   dx   4 xdx   dy   4 xdx  ln( y)  2 x 2  C  either y dx y dx y or y  Ce Looking at the possible answers, the second form has more promise. Plug in the “initial condition” (0,4) to easily find C = 4…showing us the answer to the question is choice “C” (not to be confused with our constant C. y  e2x 2 C 2 x2 21) A. Inflection points happen when the second derivative changes sign. y'  3x 2  12 x so y' '  6 x  12 This has a root at x = 2. The sign of y’’ changes over this root (remember if something is a root once, three times, five times, etc it will change sign. If something is a root an even number of times it will not change sign). Anyway, the point at the graph when x = 2 is (2,-16) by substituting into y. The slope of the tangent at this point is –12 by substituting into y’. This easily leads to the equation of the line y +16 = -12(x –2) which can be manipulated into the proper form. 1 1 sin(2k )  or more simply 2 2  5 9  5 9 ,... So 2k  , , ,.. sin(2k )  1 . The sine function equals 1 at , , 2 2 2 2 2 2 22) B. Actually performing the integration leads us to the conclusion or k   5 9 4 , 4 , 4 and we see the only value of k in the allotted range is  4 23) C. Firstly, there is a typo in the question. Choice I is missing a dx, and not intentionally. Choice I is obviously true…we often take constant out of an integral. Choice II is ridiculous. Choice III, much like Choice I, is a basic property of integration. Choice I and III only. 24) C. Use a few chain rules: f ' ( x)  1 1 1 2  2e 2 x Plug in zero to find f ' (0)   2 x 2 e 1 2 2 25) E. You must use implicit differentiation twice. Firstly you can simplify what is given immediately to 2 x 2 y  6 or x 2 y  3 Differentiating on both sides with respect to x we get  2 xy  2 y  2 y' x  2 y Differentiating again … y ' '  now substituting  2 xy  y' x 2  0 so y '  2 x x x2 the known expression for y’   2y   2   2y x   Simplify if you wish, or just plug in (1,3) to find y’’=18 y' '  x2 26) A. This problem is poorly worded. First consider the “base” of our solid. First imagine a series of very thin, half-discs protruding out from the page. This will make a half of a cone….fat on the left, getting thinner as it goes to the right. The diameter of any given disc is the function value at that point. Therefore the radius of each disc is half the value of the function. There are two ways to go about this. 1 2  r h . In this case the radius of the base 3 of the cone is 1 (half the diameter on the left hand side). The height is 4. This would give the cone a volume of 4/3, but since this is only half of a cone, we get only 2/3. 1. Geometrically. The volume of a normal cone is 2 1 1  2. Using an integral. The volume of each half disc is   x  1 dx . The first half is because 2 4  it’s only half a disc. The ¼ x+1 is the radius (the line itself is the diameter and y = ½ x+2). dx is the very thin thickness of the disc. Integrating this from 0 to 4 will also yield the correct answer, but this is far more difficult. 27) D. An increasing velocity means a positive acceleration, which means a positive 2nd derivative. 3 2 3 2 2 v(t )  1  t  3  3t  3 t  1  t  3  t  3 3t  3 now for ease, we factor out t  3 v(t )  t  3 t  3  3t  3  t  3 4t  and taking the derivative again we get 2 2 a(t )  2t  34t  4t  3  8t t  3  4t  3 again, factoring out a t  3 2 2 a(t )  t  38t  4t  3  t  312t  12 This has roots at t = 3 and t = 1. by picking test points we see that a(t) is positive when t < 1 and t > 3 28) D. The size of the initial population has no bearing on growth rates. If a population quadruples in 2 days, that logically means that it doubles each day. So if something doubles 3 times (once each day) that is a growth factor of 8.