MATH 115 FINAL EXAM. April 25, 2005


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1 MATH 115 FINAL EXAM April 25, 2005 NAME: Solution Key INSTRUCTOR: SECTION NO: 1. Do not open this exm until you re told to begin. 2. This exm hs 9 pges including this cover. There re 9 questions. 3. Do not seprte the pges of the exm. If ny pges do become seprted, write your nme on them nd point them out to your instructor when you turn in the exm.. Plese red the instructions for ech individul exercise crefully. One of the skills being tested on this exm is your bility to interpret questions, so instructors will not nswer questions bout exm problems during the exm. 5. Show n pproprite mount of work for ech exercise so tht the grders cn see not only the nswer but lso how you obtined it. Include units in your nswers where pproprite. 6. You my use your clcultor. You re lso llowed two sides of one 3 by 5 note crd. 7. If you use grphs or tbles to obtin n nswer, be certin to provide n explntion nd sketch of the grph to mke cler how you rrived t your solution. 8. Plese turn off ll cell phones nd other sound devices, nd remove ll hedphones. PROBLEM POINTS SCORE TOTAL 100
2 2 1. ( points) The figure below shows the tngent line pproximtion of f(x) ner x =. y = 3x + 9 f(x) 2 x () Wht re, f(), nd f ()? = 2 f() = 3 f () = 3 (b) Estimte f(2.1). Is this n overestimte or n underestimte? Why? f(2.1) 2.7 is n underestimte becuse the tngent line pproximtion of f(x) for x > 2 lies below the grph of f(x). (c) Estimte f(1.98). Is this n overestimte or n underestimte? Why? f(1.98) 3.06 is n overestimte becuse the tngent line pproximtion of f(x) lies bove the grph of f(x) for x < 2. (d) Would you expect your estimtion for f(2.1) or f(1.98) to be more ccurte? Why? The tngent line pproximtion is incresingly more ccurte the closer one gets to x = 2. Since = 0.1 nd = 0.02, we would expect f(1.98) to be more ccurte.
3 3 2. (5 points) Suppose 9 (f(x) + 7)dx = 315. Find 9 f(x)dx f(x)dx + 9 7dx = 315 f(x)dx + 35 = 315 f(x)dx = 280 f(x)dx = (5 points) Use the Fundmentl Theorem to determine the positive vlue of b if the re under the grph of f(x) = x + 1 between x = 2 nd x = b is equl to (x + 1)dx = 11 x 2 2 b 2 + x b 2 = 11 (2b 2 8) + (b 2) = 11 2b 2 + b 21 = 0 (2b + 7)(b 3) = 0 b = 7 2, 3 Since b is positive, b = 3.
4 . (2 points ech no prtil credit) Suppose f(x)dx = 2 nd g(x)dx = 6. Evlute the following expressions, if possible. If the expression cnnot be evluted with wht is given, simply indicte Insufficient informtion. Assume tht ll functions re continuous on the intervl [, b]. () (g(x)) 2 dx ( g(x)dx) 2 Insufficient informtion: we re not given the vlue of (b) (h(x))dx (g(x)) 2 dx. This integrl is equl to 0, since the integrl of ny function from point to itself is zero. (c) f(x 2)dx This integrl is equl to 2 since it is equl to (d) (f(x)g(x))dx f(x)dx. Insufficient informtion. We do not know how to evlute the integrl of product of two rbitrry functions. (e) b (g(x))dx This equls g(x)dx = 6.
5 5 5. (12 points) Using the grph of h in the figure below nd the fct tht h(0) = 20, sketch the grph of h(x). Give the coordintes of ll criticl points, inflection points, nd end points of h. Py ttention to the concvity of the grph. (36, 5) (10, 2) h (x) x (22, 2) y = h(x) (50, 0) (20, 0) (36, 5) x (0, 20) (10, 10) (30, 10) (22, 2) The points (10, 10), (22, 2) nd (36, 5) re inflection points. The grph of h(x) hs locl nd globl min t (0, 20), locl min t (30, 10), locl mx t (20, 0), nd locl nd globl mx t (50, 0).
6 6 6. (++ points) Hrry Potter, Ron, nd Hermione decide to ttend the Wizrd Fir. The newest ride t the fir, clled The Coil of Doom TM, is spinoff on bungee jumping. Riders re ttched to specil bungee cord which oscilltes up nd down. The riders position bove the ground, in feet, is given s function of time, t, in seconds, by y = y 0 cos(ωt) + C, with y 0, ω, nd C constnts. () The riders bord from pltform 15 feet bove the ground, re pulled upwrd until, 6 seconds lter, they rech mximum height of 165 feet. In nother 6 seconds, riders re bck t the initil position. The cycle repets for one minute, t which point the ride ends. Using this informtion, determine n explicit formul for y. [Show ll constnts in exct form.] The riders strt from pltform 15 feet bove the ground nd rech mximum height of 165 ft. The midline is C = 90 nd the mplitude must be = 75 feet. Since the ride strts t the bottom, 2, y 0 = 75 The period is the time it tkes the riders to return to their originl position. So, the period equls 12 seconds. Since ω = 2π period, ω = π 6. This mens tht y = 75 cos(π t) (b) Find formuls for the velocity nd ccelertion of the riders s function of t. v(t) = y = 75( π 6 )sin(π 6 t) (t) = y = 75( π 6 )2 cos( π 6 t) (c) Show tht the function y stisfies the eqution d2 y dt 2 + ω 2 y = K, where K is constnt. Wht is the vlue of K? d 2 y dt 2 + ω2 y = (t) + ω 2 y nd from prt () we know ω = π 6 = 75( π 6 )2 cos( π 6 t) (π 6 )2 ((75)cos( π 6 ) + 90) = 75( π 6 )2 cos( π 6 t) (π 6 )2 (75)cos( π 6 ) + 90(π 6 )2 So K = 5π2 2. = 90π2 36 = 5π2 2
7 7 7. (17 points) At the Wizrd Fir, there is booth where wizrds win Bertie Bott s Every Flvor Bens. To determine how mny bens one gets, contestnt is given string 50 inches long. From this string, contestnts cn cut lengths to form n equilterl tringle nd rectngle whose length is twice its width. The number of Bertie Bott s bens one wins depends on the combined res of the tringle nd rectngle. Hrry, knowing clculus, goes immeditely to work setting up function, finding criticl points, etc. () Use your knowledge of clculus to determine the res of the tringle nd rectngle tht will mximize the number of bens tht Hrry cn win. Show your work. We need to find formul for the totl re of the tringle nd rectngle. Let s begin by finding formul for the re of the tringle. Sy we cut the string x inches from the left end of the string, nd suppose the left piece is used to mke the equilterl tringle nd the right piece is used to mke the rectngle. So, ech side of the tringle must hve length x. An equilterl tringle hs 3 ll of its ngles equl to π 3. Let h be the height if the tringle. Then sin(π 3 ) = h x 3 So re of tringle = ( 1 2 )(x 3 )(x 3 6 ). nd h = x 3 6. Now we need to find formul for the re of the rectngle. Let the two sides of the rectngle be l nd w. Then the perimeter of the rectngle must equl the length of the right piece of string. So, 2l + 2w = 50 x. We know though tht the length is twice the width, which mens 2(2w) + 2w = 6w = 50 x, nd therefore, w = 50 x. Since the re of the rectngle equls lw, we know tht the re equls 2w 2 = 2 Thus we hve A totl = x (50 x)2 36 (50 x)2. 36 = 1 36 (( 3 + 2) 2 200x + 2(50) 2 ). So da dt = 1 36 (2( 3 + 2)x 200). Setting the derivtive equl to zero to find the criticl point gives 0 = ( 3 + 2)x 100 nd thus x Notice though tht d2 A dt 2 = 1 18 ( 3 + 2) > 0, which mens tht A hs minimum t x = nd the mximum must occur t one of the endpoints  ie. when x = 0 or x = 50. x = 0: 2w + 2l = 6w = 50 w = 8.3. Thus, A totl = lw = 2(8.3) 2 = squre inches. x = 50: re = 1 2 (50 3 )(50 3 ) squre inches. 6 This nlysis sys tht Hrry should not cut the string nd the mximum occurs when the re of the tringle is zero nd the re of the rectngle is squre inches. (b) If the number of bens won is 9 times the combined re, wht is the gretest number of bens contestnt cn win? This is just nine times the re we found in (). So, the gretest number of bens contestnt cn win is = 1232 bens. 6
8 8 8. (3 points ech) Hrry, Ron, nd Hermione re ll thrilled bout their bundnce of Bertie Bott s Every Flvor Bens; however, they prefer Chocolte Frogs to Bertie Bott s Bens. Luckily, t the wizrd fir there is booth where wizrds re ble to exchnge Bertie Bott s Bens for Chocolte Frogs. The number of bens, N, needed to purchse F chocolte frogs is given by the function N = C(F). Using complete sentences, give the prcticl interprettions of ech of the following sttements in the context of this problem. () C(3) C(3) is the number of bens needed to purchse 3 chocolte frogs. (b) C (3) = 18 If you purchse 3 chocolte frogs, you need pproximtely 18 more bens to purchse n dditionl frog. (c) C 1 (91) C 1 (91) is the number of chocolte frogs you ll receive for exchnging 91 bens. (d) (C 1 ) (91) = 0.05 If you exchnge 91 bens for chocolte frogs, you ll get pproximtely 1 20 exchnging one more ben. of chocolte frog by (e) 10 (C (F))dF The expression 10 (C (F))dF represents the difference between the number of bens needed to purchse 10 chocolte frogs nd the number of bens needed to purchse chocolte frogs i.e., the dditionl bens needed to go from to 10 frogs.
9 9 9. ( points) The three hppy wizrds leve the fir nd go home to wtch the Simpsons. In this episode, Homer needs to deliver Lis s homework to her t school, nd he must do so before Principl Skinner rrives. Suppose Homer strts from the Simpson home in his cr nd trvels with velocity given by the figure below. Suppose tht Principl Skinner psses the Simpson home on his bicycle 2 minutes fter Homer hs left, following him to the school. Principl Skinner is ble to sil through ll the trffic nd trvels with constnt velocity 10 miles per hour. v (miles/hr) t (minutes) () How fr does Homer trvel during the 10 minutes shown in the grph?. To clculte how fr Homer trvels during the 10 minutes shown in the grph we find the re under the grph of Homer s velocity. Note tht we must multiply by constnt so tht the units re correct! This gives tht distnce 1 1 (re under curve)= (97.5) = miles. (b) Wht is the verge of Homer s velocity during the 10 minute drive? verge = ( 1 10 ) 10 0 v(t)dt = 9.75 miles/hr. (c) At wht time, t > 0, is Homer the gretest distnce hed of Principl Skinner? As long s Homer s velocity is greter thn Principl Skinner s velocity, Homer is becoming frther wy from Principl Skinner. Since Principl Skinner is trveling t constnt velocity of 10 miles/hr, Homer is the gretest distnce hed of Skinner t t 5.5 minutes. (d) Does Principl Skinner overtke Homer, nd if so, when? Explin. Principl Skinner will overtke Homer when the distnce he hs trveled is equl to the distnce tht Homer hs trveled. Notice though tht the re under Homer s velocity curve nd the re under Principl Skinner s velocity curve overlp. So, they will hve trveled the sme distnce when the re between Homer s velocity curve nd Skinner s velocity curve from t = 0 to t = 5.5 equls the re between the two velocity curves from t = 5.5 to some time t > 5.5. Notice though tht the re between the two curves from t = 0 to = 5.5 is greter thn the re between the two curves from t = 5.5 to t = 10. So, Skinner does not overtke Homer. Or, more precisely, Principl Skinner trvels 1 60 (10)(8) = 1.33 miles. This is less thn the miles tht Homer trveled in the 10 minutes.
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