Solutions to the 2005 AP Calculus AB Exam Free Response Questions

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1 Solutions to the 00 AP Calculus AB Exam Free Response Questions Louis A. Talman Department of Mathematical & Computer Sciences Metropolitan State College of Denver Problem. FindRootA ÅÅÅÅ 4 + Sin@p xd ã 4-x, 8x, 0<E 8x Ø < a = x ê. % So the required area is i j4 0a -x - ÅÅÅÅ k 4 - Sin@p xdy z x {

2 FRQ00AB.nb i j ÅÅÅÅ a k 4 + Sin@p xd - y 4-x z x { p a i z j i j ÅÅÅÅ kk 4 + Sin@p xd + y { H4 -x + L y z x { Problem. R@t_D = + Sin@4 p t ê D + SinA 4 p t ÅÅÅÅÅÅÅÅÅÅÅ E S@t_D = t ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ + 3 t t ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ + 3 t

3 FRQ00AB.nb 3 6 R@tD t 0 + ÅÅÅÅÅÅÅÅÅÅ 4 p - Cos@ 4 p ÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ D 4 p N@%D The tide will remove 3.86 cubic yards of sand during this 6-hour period. YHtL = 00 - Ÿ 0 t H + sin@4 p t ê DL t + Ÿ 0 t t ÅÅÅÅÅÅÅÅÅÅÅÅ +3 t t. When t = 4 the total amount of sand is changing at the rate S@4D - R@4D 34 ÅÅÅÅÅÅÅ 3 - SinA 6 p ÅÅÅÅÅÅÅÅÅÅ E N@%D d) We examine a plot of the rate at which sand accumulates on the beach:

4 4 FRQ00AB.nb - 8t, 0, 6<D Ü Graphics Ü As the plot shows, the rate of accumulation is negative when t < t 0 and positive when t > t 0, for a certain value t 0 which is approximately. We solve for t 0, which must be the time when the amount of sand on the beach is minimal: FindRoot@S@tD - R@tD ã 0, 8t, <D 8t Ø.7863< The amount of sand on the beach is minimal at about t =.8 hours. The minimal amount is about ` 00 - H + Sin@4 p t ê DL t ` t ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ t t

5 FRQ00AB.nb Problem ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅ TH8L - TH6L Thus T ' H7L is approximately ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ degrees C per centimeter, or -7/ degrees C. per centimeter. 8-6 Average temperature of the wire is ÅÅÅÅ 8 Ÿ 0 8 THxL x. This is ÅÅÅÅ 8 i j ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ k ÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ H - L ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ H8-6L y z { Answer: ê 6 degrees Celsius. We are given that T is twice differentiable (although we are not told where). This means that T ' is continuous--on the interval [0,8], we hope, so that the problem is meaningful. By the Fundamental Theorem of Calculus, Ÿ 0 8 T ' HxL x = TH8L - TH0L = -4 degrees Celsius. The integrand T ' HxL is the (instantaneous) rate at which THxL changes per unit length at each point of the 8D, and the integral gives net temperature change over the 8D. d) By the Mean Value Theorem, there is a point x D where T ' HxL - THLD ê H - L = -3 ê 4. By the Mean Value Theorem again, there is a point h 6D where T ' HhL = TH6L - THL = -8. Note that, necessarily, 0 < x < h < 8. A third application of the Mean Value Theorem---this time to T '---assures us that there is a point z hd Œ H0, 8L such that T '' HzL ' HhL - T ' HxLD ê Hh - xl = H ê 4L ê Hh - xl = -9 Hh - xld < 0. Consequently, these data are not consistent with the assertion that T '' HxL > 0 for every x in H0, 8L.

6 6 FRQ00AB.nb Problem 4. The derivative of f must be zero or non-existent at any point of H0, 4L where f has a relative extremum. Thus, x = and x = are the only values we need to consider. We find that f ' HxL > 0 for all x in the interval H0, L. Consequently, f increases throughout that interval and can't have a relative extremum at x =. At x =, we find that f HL is meaningful, and that f ' HxL is positive on the interval H, L but negative on the interval H, 3L. Consequently, f is increasing on H, L but decreasing on H, 3L. This means that f has a relative maximum at x = x If ghxl = Ÿ f HtL t on H0, 4L, then g' HxL = f HxL by the Fundamental Theorem of calculus. Thus, g can have relative extrema only at points where f HxL = 0. At x =, f HxL undergoes a sign change from negative to positive, and so ghxl passes from a region where it is decreasing into a region where it is increasing as x passes through. Consequently, g has a relative minimum at x =. Similar reasoning shows that g has a relative maximum at x = 3. d) The points of inflection for g are to be found where g' has relative extrema. But g' is f, and, according to part a) of this problem, f has a relative extremum only at x =. We conclude that g has just one inflection point, at x =.

7 FRQ00AB.nb 7 Problem. 4 Ÿ 0 vhtl t = ÅÅÅÅ ÿ H4-0L ÿ 0 + H6-4L ÿ 0 + ÅÅÅÅ ÿ H4-6L ÿ 0 = = 360 meters. 4 Ÿ 0 vhtl t gives the distance, in meters, traveled by the car during the time period 0 t 4. v' H4L = lim + hl - vh4ld ê h is the definition of v' H4L. We note that for 0 < t 4, we have vhtl = t, so that vh4 + hl - vh4l = H4 + hl - ÿ 4 = h when h < 0 but» h» is small. + hl - vh4ld ê h = for such values of h. But when 4 t 6, we have vhtl = 0, so + hl - vh4ld = 0-0 = 0 when h > 0 and» h» is small. For such h we therefore + hl - vh4ld ê h = 0. Consequently, lim + hl - vh4ld ê h = 0, but lim + hl - vh4ld ê h =. The two one-sided limits have different values, so the two-sided limit does not exist. this means that v' H4L does not exist. Using vhtl = 60 - ÅÅÅÅ t when 6 t < 4, we find that lim + hl - vh0ld ê h = lim - ÅÅÅÅ t - ÅÅÅÅ hl - H60 - ÅÅÅÅ tld ê h. This is lim hø0 H- hl ê H hl = - ÅÅÅÅ. It follows that v' H0L = - ÅÅÅÅ. Acceleration is given by ahtl = when 0 < t < 4, by ahtl = 0 when 4 < t < 6, and by ahtl = - ÅÅÅÅ when 6 < t < 4. d) The average rate of change of v over 8 t 0 - vh8ld ê H0-8L = H0-0L ê H0-8L = - ê 6. The hypotheses of the Mean Value Theorem require that v' HtL exist at every point of the interior of an interval on which we wish to apply the theorem. The Mean Value Theorem is therefore inapplicable to v 0D because v' H6L does not exist.

8 8 FRQ00AB.nb Problem At H, -L, we have y' = - HL ê H-L =, so the equation of the line tangent to the solution for which yhl = - is y = - + Hx - L, or y = x - 3. This gives an approximate value for yh.l of y = ÿ H.L - 3 = x x If y' HxL = - x ê yhxl, then yhxl y' HxL = - x. Thus Ÿ yhtl y' HtL t = - Ÿ t t, or - = -x +. Replacing yhl with - and simplifying, we = 3 - x. This leads to yhxl = è!!!!!!!!!!!!!!! 3 - x. We resolve the ambiguity of sign by noting that we must have yhl = -, and this gives our solution: yhxl = - è!!!!!!!!!!!!!!! 3 - x.