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AB2009 Solution

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AB2009.nb 1 Solutions to the 2009 AP Calculus AB Free Response Questions Louis A. Talman, Ph.D. Department of Mathematical & Computer Sciences Metropolitan State College of Denver Problem 1. ü a. 1 At time t = 7.5, the acceleration of Caren's bicycle is - ÅÅÅÅÅÅ miles per minute per minute. 10 ü b. The integral Ÿ0 » vHtL » „ t gives, in miles, the total distance that Caren traveled during the period 0 § t § 12. The value of 9 this integral is ÅÅÅÅ miles. 5 12 ü c. Her turn-around time corre
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  Solutions to the2009 AP Calculus ABFree Response Questions Louis A. Talman, Ph.D. Department of Mathematical & Computer SciencesMetropolitan State College of Denver Problem 1. ü a. At time t = 7.5, the acceleration of Caren's bicycle is - 1 ÅÅÅÅÅÅÅ 10 miles per minute per minute. ü b. The integral Ÿ  012 » v H t  L » „  t  gives, in miles, the total distance that Caren traveled during the period 0 § t  § 12. The value of this integral is 9 ÅÅÅÅ 5 miles. ü c. Her turn-around time corresponds to the point on the graph where the sign of her velocity changes from positive to negative.That's t  = 2 minutes. ü d. Caren lives Ÿ  512 v H t  L „  t  = 7 ÅÅÅÅ 5 miles from school because she left home at t  = 5, arrived at school at t  = 12, traveled in onedirection only, and the distance she traveled during that time is given by the integral. Larry's distance is the integral of hisvelocity over the interval @ 0,12 D , or    AB2009.nb 1  p ÅÅÅÅÅÅÅ 15   ‡  012 Sin A p ÅÅÅÅÅÅÅ 12   t E „ t 8 ÅÅÅÅ 5 At 8 ÅÅÅÅ 5 miles, Larry lives farther than Caren, at 7 ÅÅÅÅ 5 miles from the school. Thus, Caren lives closer. Problem 2. ü a. At time t  = 2, the auditorium contains Ÿ  02 H 1380 t  2 - 675 t  3 L „  t  people. ‡  02 H 1380 t 2 - 675 t 3 L „ t 980 That's 980 people. ü b. We are given  R H t  L = 1380 t  2 - 675 t  3 , so that  R '   H t  L = 2760 t  - 2025 t  2 = 15 t  H 184 - 135 t  L . Thus,  R H t  L is increasing onthe interval @ 0, 184 ÅÅÅÅÅÅÅÅÅ 135 D and decreasing on the interval @ 184 ÅÅÅÅÅÅÅÅÅ 135 ,2 D , because  R '   H t  L is positive on the first of these intervals andnegative on the second. It follows that the maximal rate at which people enter the auditorium is at t  = 184 ÅÅÅÅÅÅÅÅÅ 135 hours. ü c. We have R @ t_  D = 1380 t 2 - 675 t 3 1380 t 2 - 675 t 3 Dw  @ t_  D = H 2 - t L   R @ t D H 2 - t L H 1380 t 2 - 675 t 3 L By the Fundamental Theorem of Calculus, the difference w H 2 L - w H 1 L is given by ‡  12 Dw  @ t D „t 775 ÅÅÅÅÅÅÅÅÅÅ 2 Total wait time for those who enter the auditorium after t  = 1 is 775 ÅÅÅÅÅÅÅÅÅ 2 hours.  AB2009.nb 2  ü d. From part (a) of this problem, above, we know that there are 980 people in the auditorium at time t  = 2. We also know thatthe total wait time for these 980 people is ‡  02 Dw  @ t D „t 760 760 hours. Consequently, average waiting time is 760 ÅÅÅÅÅÅÅÅÅ 980 = 38 ÅÅÅÅÅÅÅ 49 hours. Problem 3. ü a. Mighty's profit on a cable of length k  meters is given by P H k  L = 120 k  - Ÿ  0 k  6 è!!!  x „   x , in dollars. P @ k_  D = 120 k - 6   ‡  0k è!!!! x „ x 120 k - 4k 3 ê 2 P @ 25 D 2500 Thus, profit on a 25-meter cable is $2500 ü b. The integral Ÿ  2530 6 è!!!  x „   x gives the cost, in dollars, to Mighty for building the last five meters of a 30-meter cable. ü c. As we saw in part (a) of this problem, above, Mighty's profit, in dollars, on the sale of a cable that is k  meters long is120 k  - 6 Ÿ  0 k  è!!!  x „   x , or 120   k  - 4 k  3 ê 2 = 4 k  I 30 - è!!! k  M .  AB2009.nb 3  ü d. We first observe that profit for a cable k  meters long, which is given by 4   k  I 30 - è!!! k  M , is non-negative only when0 § k  § 900 and is zero at the endpoints of this interval. Thus, the maximum does not occur at an endpoint of the interval (oroutside the interval). However, the profit function is continuous, so there must be a maximum ocurring at some k  œ @ 0,900 D .As we have observed, we must actually have k  œ H 0,900 L , so k  must be a critical number for the profit function. Noting thatthe derivative P' @ k D 120 - 6 è!!! k of the profit function vanishes only when k  = 400, we conclude that the maximum profit must come at k  = 400. Themaximal profit is therefore P H 400 L = 16,000 dollars.We also notice that if cables longer than 900 meters ever become very popular, Mighty should restructure something. Problem 4. ü a. The desired area is ‡  02 H 2 x - x 2 L „ x 4 ÅÅÅÅ 3 ü b. The desired volume is ‡  02 Sin A p ÅÅÅÅ 2   x E „ x 4 ÅÅÅÅp ü c. Solving both equations for  y in terms of   x gives us  x = 1 ÅÅÅÅ 2    y for the first and  x = è!!!  y for the second. The length of the baseof the square corresponding to  y = y 0 is therefore è!!!!!  y 0 - 1 ÅÅÅÅ 2    y 0 , and the desired volume is therefore ‡  04 Iè!!!  y - 1 ÅÅÅÅ 2    y M 2 „   y .(For those who are unable to comply with the instruction no to evaluate the integral,, this is 8 ÅÅÅÅÅÅÅ 15 .)  AB2009.nb 4
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