Mth 6 Finl Exm April 26, 23 Nme: EXAM SOLUTIONS Instructor: Section:. Do not open this exm until you re told to do so. 2. This exm hs 5 pges including this cover. There re problems. Note tht the problems re not of equl difficulty, so you my wnt to skip over nd return to problem on which you re stuck. 3. Do not seprte the pges of this exm. If they do become seprted, write your nme on every pge nd point this out to your instructor when you hnd in the exm. 4. Plese red the instructions for ech individul problem crefully. One of the skills being tested on this exm is your bility to interpret mthemticl questions, so instructors will not nswer questions bout exm problems during the exm. 5. Show n pproprite mount of work (including pproprite explntion) for ech problem, so tht grders cn see not only your nswer but how you obtined it. Include units in your nswer where tht is pproprite. 6. You my use ny clcultor except TI-92 (or other clcultor with full lphnumeric keypd). However, you must show work for ny clcultion which we hve lerned how to do in this course. You re lso llowed two sides of 3 5 note crd. 7. If you use grphs or tbles to find n nswer, be sure to include n explntion nd sketch of the grph, nd to write out the entries of the tble tht you use. 8. Turn off ll cell phones nd pgers, nd remove ll hedphones. Problem Points Score 2 6 3 4 5 4 6 7 9 8 8 9 4 9 Totl
Mth 6 / Finl (April 26, 23) pge 2 You my find the following expressions useful. Known Tylor series (ll round x = ): sin(x) = n= ( ) n x 2n+ (2n + )! = x x3 3! + + ( )n x 2n+ + for ll vlues of x (2n + )! ( ) n x 2n cos(x) = (2n)! n= = x2 2! + + ( )n x 2n + for ll vlues of x (2n)! e x = n= x n n! = + x + x2 2! + + xn + for ll vlues of x n! ( ) n+ x n ln( + x) = n n= = x x2 2 + x3 3 + ( )n+ x n + for < x n ( + x) p = + px + p(p ) 2! x 2 + p(p )(p 2) 3! x 3 + for < x < x = x n = + x + x 2 + x 3 + + x n + for < x < n=
Mth 6 / Finl (April 26, 23) pge 3. [ points] Indicte if ech of the following is true or flse by circling the correct nswer. No justifiction is required.. [2 points] Let < q <, then q n = q + q 2 + q 3 + + q n + = n= q q. series n= True Flse Since n= qn = q ( n= qn ), then using the formul for geometric r n = r with r = q yields the result. b. [2 points] Let F (t) be n ntiderivtive of continuous function f(t). If the units of f(t) re meters nd t is in seconds, then the units of F (t) re meters per second. True Flse The Second Fundmentl Theorem of Clculus sys tht if F (t) is n ntiderivtive of f(t), then F (t) = t f(x)dx. The units of definite integrl re the unites of f(t) times the units of t. In this cse, the units of F (t) re meters times seconds. c. [2 points] If the motion of prticle is given by the prmetric equtions x = t + t 3, y = t2 + t 3 for >, then the prticle pproches the origin s t goes to infinity. True Flse Since lim x(t) = lim y(t) =, then the prticle pproches the origin t t s t goes to infinity. d. [2 points] Let n be sequence of positive numbers stisfying lim n =. Then n the series converges. n n= If n = n, then lim n n =, but n= e. [2 points] Let f(x) be continuous function. Then n True diverges by p-series test. Flse f(2x)dx = 2 f(x)dx.
Mth 6 / Finl (April 26, 23) pge 4 Using the substitution u = 2x, you get f(2x)dx = 2 2 f(u)du. True Flse
Mth 6 / Finl (April 26, 23) pge 5 2. [6 points] Let the sequence n be given by =, 2 =. [ point] Find 7. 2 3 3, 3 = 5, 4 = 7 = 7 3. 4 5 7, 5 = 9, 6 = 6 b. [3 points] Write formul for n. n n = ( ) n 2n. c. [2 points] Does the sequence n converge? If so, find its limit. Yes, it converges to.
Mth 6 / Finl (April 26, 23) pge 6 3. [ points] A bot s initil vlue is $, ; it loses 5% of its vlue ech yer. The bot s mintennce cost is $5 the first yer nd increses by % nnully. In the following questions, your formuls should not be recursive.. [2 points] Let B n be the vlue of the bot n yers fter it ws purchsed. Find B nd B 2. B = $, (.85). B 2 = $, (.85) 2. b. [3 points] Find formul for B n. B n =, (.85) n c. [2 points] Let M n be the totl mount of money spent on the mintennce of the bot during the first n yers. Find M 2 nd M 3. M 2 = 5( +.) M 3 = 5( +. + (.) 2 ) d. [3 points] Find closed form formul for M n. M n = 5 ( (.)n ).
Mth 6 / Finl (April 26, 23) pge 7 4. [ points] The lifetime t (in yers) of tree hs probbility density function for t. (t + ) p f(t) = for t <. where > nd p >.. [4 points] Use the comprison method to find the vlues of p for which the verge lifetime M is finite (M < ). Properly justify your nswer. The verge lifetime M is given by the formul M = Since then We know tht t (t + ) t p t = for t >, p t p t (t + ) dt p t p dt t (t + ) dt. p converges precisely when p > ( p > 2) by the p-test, tp so the first integrl converges precisely when p > 2. This implies tht the verge lifetime M is finite for p > 2. Note: We use the inequlity t (t + ) dt dt since the inequlity p tp ( ) dt is not useful dt = for ll vlues of p. tp tp You do not need to discuss the convergence of the integrl this integrl is not n improper integrl. t (t + ) p t dt since (t + ) p
Mth 6 / Finl (April 26, 23) pge 8 b. [4 points] Find formul for in terms of p. Show ll your work. We know tht = (t + ) p dt. We use u-substition with u = t + to clculte the integrl: Therefore = dt = lim (t + ) p b b b+ = lim b = lim b (since p > ) = p., so = p. p (t + ) p dt b+ du = lim u p du up b u p+ ( p + ) b+ = lim b+ b ( p + )up c. [2 points] Let C(t) be the cumultive distribution function of f(t). For given tree, wht is the prcticl interprettion of the expression C(3)? C(3) is the probbility tht given tree lives t lest 3 yers.
Mth 6 / Finl (April 26, 23) pge 9 5. [4 points] A skydiver jumps from plne t height of 2, meters bove the ground. After some time in free-fll, he opens his prchute, reducing his speed, nd lnds sfely on the ground.. [5 points] The grph of the skydiver s downwrd velocity v(t) (in meters per second) t seconds fter he jumped is shown below. Sketch the grph of the ntiderivtive y(t) of v(t) stisfying y() =. Mke sure your grph reflects the regions t which the function is incresing, decresing, concve up or concve down. It is importnt to notice tht y (t) exist for ll vlues of t since y (t) = v(t). b. [3 points] Write down right-hnd sum with 4 subintervls in order to pproximte the verge downwrd velocity of the skydiver during the time the skydiver is in free-fll. Show ll the terms in your sum. The verge downwrd velocity is 2 v(t)dt. We pproximte this 2 s 2 2 v(t)dt 5(35 + 5 + 55 + 6) 2 c. [2 points] Is your estimte in (b) gurnteed to be n underestimte or overestimte of the verge velocity of the skydiver, or there is not enough informtion to decide? Justify. It s gurnteed to be n overstimte, becuse v(t) is incresing throughout [, 2]. d. [4 points] Find formul for the height H(t) (in meters) bove the ground of the skydiver t seconds fter he jumped. H(t) = 2, t v(s)ds = 2, y(t).
Mth 6 / Finl (April 26, 23) pge 6. [ points] At hospitl, ptient is given drug intrvenously t constnt rte of r mg/dy s prt of new tretment. The ptient s body depletes the drug t rte proportionl to the mount of drug present in his body t tht time. Let M(t) be the mount of drug (in mg) in the ptient s body t dys fter the tretment strted. The function M(t) stisfies the differentil eqution dm dt = r M with M() =. 4. [7 points] Find formul for M(t). Your nswer should depend on r. We use seprtion of vribles dm r 4 M = dt. Using u-substition with u = r /4M, du = /4dM for the left-hnd-side, we nti-differentite: 4 ln r 4 M = t + C. Therefore, ln r 4 M = t/4 + C 2 nd r 4 M = e t/4+c 2 = C 3 e t/4. Therefore /4M = r C 3 e t/4 nd M(t) = 4r C 4 e t/4. With M() =, we conclude tht C 4 = 4r, so we get M(t) = 4r 4re t/4. b. [ point] Find ll the equilibrium solutions of the differentil eqution. M = 4r. c. [2 points] The tretment s gol is to stbilize in the long run the mount of drug in the ptient t level of 2 mg. At wht rte r should the drug be dministered? You need 4r = 2, then r = 5 mg/dy.
Mth 6 / Finl (April 26, 23) pge 7. [9 points] A tnk hs the shpe of circulr cone. The cone hs rdius 2 m nd height 7 m (s shown below). The tnk contins liquid up to depth of 4 m. The density of the liquid is δ(y) = y 2 kg/m 3, where y mesures the distnce in meters from the bottom of the tnk. Use the vlue g = 9.8 m/s 2 for the ccelertion due to grvity.. [6 points] Find definite integrl tht computes the mss of the liquid in the tnk. Show ll your work. Let r(y) be the rdius t height y. By similr tringles, 2/7 = r/y, so r = 2y. The pproximte mss of thin slice t height y is 7 π(2/7y)2 ( y) 2 y, so the nswer is 4 π(2/7y) 2 ( y 2 )dy. b. [3 points] Find definite integrl tht computes the work required to pump the liquid 2 meters bove the top of the tnk. Show ll your work. We wnt to lift ech thin slice (9 y) feet. The work to lift slice is 9.8(9 y)π(2/7y) 2 ( y 2 ) y, so the integrl is 4 9.8(9 y)π(2/7y) 2 ( y 2 )dy.
Mth 6 / Finl (April 26, 23) pge 2 8. [8 points] Consider the power series n= 2 n n (x 5)n. In the following questions, you need to support your nswers by stting nd properly justifying the use of the test(s) or fcts you used to prove the convergence or divergence of the series. Show ll your work.. [2 points] Does the series converge or diverge t x = 3? At x = 3, the series is n= series test, since / n is decresing nd converges to. ( ) n n, which converges by the lternting b. [2 points] Wht does your nswer from prt () imply bout the rdius of convergence of the series? R 2. Becuse it converges t x = 3, we know tht the rdius of convergence c. [4 points] Find the intervl of convergence of the power series. Using the rtio test, we hve lim n 2 n+ n+ x 5 n+ 2 n x = n 5 n 2 x 5 = L, so the rdius of convergence is 2. Now we hve to check the endpoints. We know from prt () tht it converges t x = 3. For x = 7, we get, which diverges. n Thus, the intervl of convergence is 3 x < 7. n=
Mth 6 / Finl (April 26, 23) pge 3 9. [4 points] Determine the convergence or divergence of the following series. In questions () nd (b) you need to support your nswers by stting nd properly justifying the use of the test(s) or fcts you used to prove the convergence or divergence of the series. Circle your nswer. Show ll your work.. [4 points] n= 2n n5 + Converges Diverges You cn use either the limit comprison test or the comprison test. We simply use the comprison test. We know tht < 2n n5 + 2n n 2 5/2 n. 3/2 Becuse n converges by the p-series, the series 2n 3/2 n5 + n= the comprison test. converges by b. [4 points] n 2 e n3 Converges Diverges n= Since the function f(x) = x 2 e x3 is positive nd decresing for x >, we cn use the integrl test to determine the convergence or divergence of n 2 e n3. To do this, we use u-substitution. Let u = x 3, du = 3x 2 dx. Therefore n= Hence n= x 2 e x3 dx = lim x 2 e x3 dx = lim b 3 = lim b 3 eu b = 3 3. b b n 2 e n3 converges by the integrl test. b 3 e u du c. [6 points] Determine if the following series converge bsolutely, conditionlly or diverge. Circle your nswers. No justifiction is required. ). n= sin(3n) n 6 + Converges bsolutely Converges conditionlly Diverges b). ( ) n+ n 3n + n= Converges bsolutely Converges conditionlly Diverges
Mth 6 / Finl (April 26, 23) pge 4. [9 points]. [3 points] Find the first three nonzero terms in the Tylor series of f(y) = bout y =. Show ll your work. Using the binomil expnsion, this is ( + y) 3 2 ( + y) 3 ( 3/2) ( 5/2) y + y 2 = 3 3/2 2 2 2 y + 5 8 y2 b. [2 points] Use your nswer in () to find the first three nonzero terms in the Tylor series of g(x) = bout x =. Show ll your work. ( 2 + x 2 ) 3 2 Fctoring, we hve ( 2 + x 2 ) 3 2 = ( 2 ( + ( x )2 )) 3 2 = ( 2 ) 3 2 ( + ( x )2 ) 3 2 ( x ) 2, Therefore, letting y = we hve 3 ( + ( x )2 ) = ( ) ( 3 3/2 3 ( + y) 3/2 3 2 y + 5 ) 8 y2 = 3 ( + ( x )2 ) 3/2. ( 2 + x 2 ) 3 2 = ( 3 ( x ) 2 5 ( ) x 4 + = 3 2 8 ) 3 3 2 5 x2 + 5 8 7 x4. 3 3 2 5 x2 + 5 8 7 x4. c. [2 points] For which vlues of x is the Tylor series for g(x) bout x = expected to converge? The Binomil series in ) converges for y <. This implies tht the series for g(x) converges for ll vlues of x stisfying x <, so < x <. Problem continues on the next pge
Mth 6 / Finl (April 26, 23) pge 5 Continution of problem. The force of grvittionl ttrction F between rod of length 2L nd prticle t distnce is given by L F = k dx, ( 2 + x 2 ) 3 2 where k is positive constnt. d. [2 points] Use your nswer in (b) to obtin n pproximtion for the force of grvittionl ttrction F between the rod nd the prticle. Your nswer should depend on the constnts k, nd L. Show ll your work. We wnt F = k = k L L dx k 3/2 ) ( 2 + x 2 ) ( L 3 2 5 L3 + 3 8 7 L5 Hence F kl 3 k 2 5 L3 + 3k 8 7 L5.. 3 3 2 5 x2 + 5 8 7 x4 dx