Math 6 Secod Midterm November 3, 7 EXAM SOLUTIONS. Do ot ope this exam util you are told to do so.. Do ot write your ame aywhere o this exam. 3. This exam has pages icludig this cover. There are problems. Note that the problems are ot of equal difficulty, so you may wat to skip over ad retur to a problem o which you are stuck. 4. Do ot separate the pages of this exam. If they do become separated, write your UMID (ot ame) o every page ad poit this out to your istructor whe you had i the exam. 5. Note that the back of every page of the exam is blak, ad, if eeded, you may use this space for scratchwork. Clearly idetify ay of this work that you would like to have graded. 6. Please read the istructios for each idividual problem carefully. Oe of the skills beig tested o this exam is your ability to iterpret mathematical questios, so istructors will ot aswer questios about exam problems durig the exam. 7. Show a appropriate amout of work (icludig appropriate explaatio) for each problem, so that graders ca see ot oly your aswer but how you obtaied it. 8. The use of ay etworked device while workig o this exam is ot permitted. 9. You may use ay oe calculator that does ot have a iteret or data coectio except a TI-9 (or other calculator with a qwerty keypad). However, you must show work for ay calculatio which we have leared how to do i this course. You are also allowed two sides of a sigle 3 5 otecard.. For ay graph or table that you use to fid a aswer, be sure to sketch the graph or write out the etries of the table. I either case, iclude a explaatio of how you used the graph or table to fid the aswer.. Iclude uits i your aswer where that is appropriate.. Problems may ask for aswers i exact form. Recall that x = is a solutio i exact form to the equatio x =, but x =.4435637 is ot. 3. Tur off all cell phoes, smartphoes, ad other electroic devices, ad remove all headphoes, earbuds, ad smartwatches. Put all of these items away. 4. You must use the methods leared i this course to solve all problems. Problem Poits Score 9 3 4 3 5 5 Problem Poits Score 6 7 7 8 9 9 9 Total
Math 6 / Exam (November 3, 7) page. [ poits] Cosider the ifiite sequeces c, d, j ad l, defied for as follows: ( ) k c = k! d = arcta(. ) j = 3 e x dx l = si(x ) for some fixed value of x satisfyig < x <. a. [8 poits] Decide whether each of these sequeces is bouded, ubouded, always icreasig, ad/or always decreasig. Record your coclusios by clearly circlig the correct descriptios below. Cotradictory coclusios will be marked icorrect. i. The sequece c is bouded ubouded icreasig decreasig ii. The sequece d is bouded ubouded icreasig decreasig iii. The sequece j is bouded ubouded icreasig decreasig iv. The sequece l is bouded ubouded icreasig decreasig b. [4 poits] For parts i ad ii below, decide whether the sequece coverges or diverges. If the sequece coverges, circle coverges, fid the value to which it coverges, ad write this value o the aswer blak provided. If the sequece diverges, circle diverges. i. The sequece d Coverges to π Diverges Note that the geometric sequece. diverges to. So sice lim arcta(x) = π/, we have lim x arcta(. ) = π/. That is, the sequece d coverges to π/. ii. The sequece j Coverges to Diverges the improper itegral x ). The sequece j diverges because it is ubouded or, alteratively, because e x dx diverges (e.g. by direct compariso sice e x for
Math 6 / Exam (November 3, 7) page 3. [9 poits] Cosider the graph of y = e 4x for x. a. [3 poits] Let R be the regio i the first quadrat betwee the graph of y = e 4x ad the x-axis. Which of the followig improper itegrals best expresses the volume of the solid that is obtaied by rotatig R aroud the x-axis? Circle oe: πe 4x dx xe 4x dx xe 8x dx πe 8x dx π 7 l(y) dy π 4 y l(y) dy 4 y l(y ) dy A thi slice of this solid of thickess x take perpedicular to the x-axis at the poit x has approximate volume π(e 4x ) = πe 8x cubic uits. If b is a positive umber, the volume of the portio of this solid betwee x = ad x = b is So the volume of the etire solid would be lim b b πe 8x dx = b πe 8x dx. πe 8x dx. b. [6 poits] Determie whether the improper itegral you circled i part a coverges or diverges. If the itegral coverges, circle coverges, fid its exact value (i.e. o decimal approximatios), ad write the exact value o the aswer blak provided. If the itegral diverges, circle diverges ad justify your aswer. I either case, you must show all your work carefully usig correct otatio. Ay direct evaluatio of itegrals must be doe without usig a calculator. Coverges to π 8 Diverges We use the defiitio of this type of improper itegral to fid its value. b πe 8x dx = lim b [ = lim b = lim b [ π 8 = π 8 πe 8x dx π 8 e 8x] x=b x= ( e 8b )] (sice lim b e 8b = ). So πe 8x dx coverges to π 8.
Math 6 / Exam (November 3, 7) page 4 3. [ poits] Defie a sequece a for by a = { if is odd /3 if is eve. Note that this meas the first three terms of this sequece are a =, a = /3, a 3 =. a. [ poits] Does the sequece {a } coverge or diverge? Circle oe: coverges diverges Briefly explai your aswer. This sequece does ot coverge because the terms of the sequece alterate forever betwee ad /3, ad, i particular, ever approach a sigle value. I other words, lim a does ot exist. b. [4 poits] Cosider the power series = (3 a ) 3 x. i. Write out the partial sum with three terms for this power series. Note that is eve for all itegers, so a is always equal to /3. Therefore, x the power series ca be rewritte as 3. = The partial sum with three terms is thus x 3 + x 9 + x3 3 7 = x 3 + x 8 + x3 8. Aswer: x 3 + x 8 + x3 8 ii. Give the exact value of this partial sum whe x =. Substitutig x = / ito the partial sum above gives ( /) 3 + ( /) 8 + ( /)3 8 = 6 + 7 648 = 5 6. Aswer: 5 6
Math 6 / Exam (November 3, 7) page 5 x c. [6 poits] Determie the iterval of covergece of the power series 3. = Show every step of ay calculatios ad fully justify your aswer with careful reasoig. Write your fial aswer o the aswer blak provided. To compute the radius of covergece we apply the Ratio Test to the geeral term of the power series, which we will call b = x 3. b + lim = lim b 3 x = lim ( + )3+ + x 3 = x 3. By the Ratio Test, this power series will therefore coverge if x /3 <, i.e. x < 3, ad diverge if x > 3. This shows that the radius of covergece is R = 3, ad the iterval of covergece is ( 3, 3) together with possibly oe or both of the edpoits. We still eed to check covergece at the edpoits. At x = 3 the power series reduces to = ( 3) 3 = ( ) = 3 3 = = ( ). This series coverges by the alteratig series test. (It is the alteratig harmoic series.) This implies that x = 3 is i the iterval of covergece. Fially, at x = 3, the power series reduces to = (3) 3 = = 3 3 = This series diverges by p-test with p =. (It is the harmoic series.) This implies that x = 3 is NOT i the iterval of covergece. Hece, the iterval of covergece is [ 3, 3). =. Aswer: [ 3, 3)
Math 6 / Exam (November 3, 7) page 6 4. [3 poits] The orbit of a sigle electro aroud the ucleus of a atom is determied by the eergy level of that electro ad by the other electros orbitig the ucleus. We ca model oe electro s orbital i two-dimesios as follows. Suppose that the ucleus of a atom is cetered at the origi. The the (so-called p ) orbital has the shape show below. This shape is made up of two regios that we call lobes. The outer edge of the lobes are described by the polar equatio r = k si(θ) for some positive costat k. Note that oly the relevat portio of the polar curve r = k si(θ) is show. The top lobe is the portio i the first quadrat (show i bold). y x a. [ poits] For what values of θ with θ π does the polar curve r = k si(θ) pass through the origi? The curve passes through the origi whe r =, i.e. whe si(θ) =, or θ =, π, π, 3π, ad π. b. [3 poits] For what values of θ does the polar curve r = k si(θ) trace out the top lobe? Give your aswer as a iterval of θ values. From part (a), we are at the origi whe θ = ; as θ icreases from, the radius k si(θ) icreases ad the decreases (sice k is positive). So we fiish the lobe whe we get to the secod value of θ at which the curve itersects the origi, θ = π. The fial aswer is thus θ π/ or, equivaletly, the closed iterval [, π/]. c. [4 poits] Write, but do ot evaluate, a itegral that gives the area of the top lobe. Usig the formula for polar area ad the values of θ from part, area = π/ π/ (k si(θ)) (f(θ)) dθ = dθ. d. [4 poits] Imagie that a electro lies withi the top lobe of this orbital, but is as far away from the origi as possible. What are the polar coordiates of this poit of greatest distace from the origi? Your aswer may ivolve the costat k. Maximizig distace from the origi meas maximizig r, so wat si(θ) = ±. For θ i the iterval θ π/, this implies that θ = π/ so θ = π/4. Whe θ = π/4, we have r = k si(π/) = k. ( Therefore, the polar coordiates for this poit are (r, θ) = k, π ). 4
Math 6 / Exam (November 3, 7) page 7 5. [5 poits] Cosider the improper itegral Note that for x >, we have 3x + 5e x < 3x 3x + 5e x dx. ad 3x + 5e x <.4e x. Use this iformatio together with the (Direct) Compariso Test for Itegrals to determie whether dx coverges or diverges. 3x + 5ex Write the compariso fuctio you use o the blak below ad circle your coclusio for the improper itegral. The briefly explai your reasoig. Aswer: Usig (direct) compariso of 3x + 5e x with the fuctio.4e x, the improper itegral dx 3x + 5ex Coverges Diverges Briefly explai your reasoig. Note that the improper itegral e x dx is oe of the useful itegrals for compariso from the textbook. This itegral is kow to coverge (expoetial decay), so the improper itegral.4e x dx =.4 e x dx also coverges. Together with the give iequality 3x + 5e x <.4e x this implies that the improper itegral also coverge by the (Direct) Compariso Test for Improper Itegrals. 6. [7 poits] Cosider the series = + 4 4 3. dx must 3x + 5ex Use the Limit Compariso Test to determie whether this series coverges or diverges. Circle your aswer (either coverges or diverges ) clearly. + The series 4 4 3 Coverges Diverges = Give full evidece to support you aswer below. Be sure to clearly state your choice of compariso series, show each step of ay computatio, ad carefully justify your coclusios. By cosiderig the expoets i the umerator ad the deomiator of the geeral term of this series, we decide to compare this series to =. 4 ( + )/(4 4 3 ) (4 ) ( + ) lim /4 = lim 4 4 3 =. Sice this limit exists ad is o-zero, we ca apply the Limit Compariso Test. So the origial series ad the series = either both coverge or both diverge. The series 4 = 4 is /4 times a p-series (p = ) with the first term ( = ) omitted. Omittig a sigle term ad multiplyig by a o-zero costat do ot affect the covergece of a series, so sice the p-series with p = coverges, so too will the series = coverge. 4 + By the Limit Compariso Test, we ca therefore coclude that the origial series 4 4 3 = must also coverge.
Math 6 / Exam (November 3, 7) page 8 7. [ poits] A boucy ball is lauched up feet from the floor ad the begis boucig. Each time the ball bouces up from floor, it bouces up agai to a height that is 6% the height of the previous bouce. (For example, whe it bouces up from the floor after fallig ft, the ball will bouce up to a height of.6() = feet.) Cosider the followig sequeces, defied for : Let h be the height, i feet, to which the ball rises whe the ball leaves the groud for the th time. So h = ad h = Let f be the total distace, i feet, that the ball has traveled (both up ad dow) whe it bouces o the groud for the th time. For example, f = 4 ad f = 4 + 4 = 64. a. [ poits] Fid the values of h 3 ad f 3. h 3 =.6() = 7. ad f 3 = 64 + 4.4 = 78.4. Aswer: h 3 = 7. ad f 3 = 78.4 b. [6 poits] Fid a closed form expressio for h ad f. ( Closed form here meas that your aswers should ot iclude sigma otatio or ellipses ( ). Your aswers should also ot ivolve recursive formulas.) h =.6h is a recursive relatioship that holds betwee the terms of the sequece h for >, ad this recursive formula meas that h is a geometric sequece. The (costat) ratio of successive terms is equal to.6 ad first term is h =. So we see that h = (.6). Note that the term f is twice the sum of the first terms of the h sequece. (Twice because the boucy ball travels both up ad dow.) We use the formula for a partial sum of a geometric series (i.e. a fiite geometric series) to fid f = (h + h +... + h ) = ( +... + (.6) ) = ()( (.6) ).6 = 4( (.6) ).4 = ( (.6) ). Aswer: h = (.6) ad f = 4( (.6) ).4 = ( (.6) )
Math 6 / Exam (November 3, 7) page 9 c. [4 poits] Decide whether the give sequece or series coverges or diverges. If it diverges, circle diverges. If it coverges, circle coverges ad write the value to which it coverges i the blak. i. The sequece f Coverges to Diverges The limit of the sequece f is lim f 4( (.6) ) = lim = 4.4.4 =. Sice this limit exists, the sequece f coverges, ad this computatio shows that it coverges to. Alteratively, as we saw i part b, the sequece f is the sequece of partial sums of the geometric series h k = 4(.6) k. Sice r =.6 ad.6 <, we kow that 4 this geometric series coverges to =. By defiitio of series covergece, this.6 sum is the limit of the sequece of partial sums f, i.e. lim f =. ii. The series = h Coverges to 5 Diverges Next, we cosider the series h, which we kow is geometric from part = b. Sice the commo ratio betwee successive terms is.6, the series coverges, ad the formula for the sum of a coverget geometric series gives us h = = = (.6) =.6 = 5, Alteratively, sice the sequece f is the sequece of partial sums of the series we have = h = lim f = = 5. h k,
Math 6 / Exam (November 3, 7) page 8. [9 poits] For each of parts a through c below, circle all of the statemets that must be true. Circle oe of these if oe of the statemets must be true. You must circle at least oe choice to receive ay credit. No credit will be awarded for uclear markigs. No justificatio is ecessary. a. [3 poits] Suppose f(x) is a cotiuous ad decreasig fuctio o the iterval [, ] with f() = ad f() =. Let a be a costat with < a <. Cosider the itegral f(x) dx. i. This itegral is ot improper. ii. This itegral coverges by direct compariso with the costat fuctio. iii. This itegral coverges by direct compariso with the fuctio f(x). iv. This itegral coverges for some values of a betwee ad but diverges for other values of a betwee ad. v. oe of these b. [3 poits] Suppose g(x) is a positive ad decreasig fuctio that is defied ad cotiuous o the ope iterval (5, ) such that g(x) dx coverges i. The series ad g() coverges. = 8 5 g(x) dx diverges. a ii. The series = g() diverges. iii. The sequece c = iv. The itegral 7 v. oe of these 5 5 g(x) dx, 5, coverges. g(x) dx diverges. c. [3 poits] Cosider the sequece a = l(),. Note: Due to a typo o the origial exam (corrected here), all studets received full credit for part c. i. lim a =. ii. The series a coverges. = iii. The series a diverges. = iv. The series ( ) a coverges. = v. oe of these
Math 6 / Exam (November 3, 7) page 9. [ poits] For each of parts a through c below, determie the radius of covergece of the power series. Show your work carefully. e a. [3 poits] (x )! = We apply the Ratio Test. (e(x ) + )/( + )! e lim (e(x ) = lim )/! e! x = lim ( + )! x =. + This limit is always less tha oe, so, by the Ratio Test, this power series will coverge for every value of x. Hece the radius of covergece is. Aswer: radius of covergece = b. [3 poits] 5(x + π) + 5 4(x + π) + 5 9(x + π) 3 + 5 6(x + π) 4 + We apply the Ratio Test. 5( + ) (x + π) + lim 5 (x + π) = lim + x + π = x + π. The Ratio Test guaratees covergece whe this limit is less tha oe (ad divergece whe the limit is greater tha oe). Now x + π < meas x is withi uit of π (or π < x < π + ), so the radius of covergece is. c. [3 poits] = Aswer: radius of covergece = π (x + )3 8 Note that this series is geometric with first term π ad ratio of successive terms (x+)3 8, so the series coverges if ad oly if (x+)3 8 <. Alteratively, we ca apply the Ratio Test: π(x + ) 3(+) /8 + 8 lim π(x + ) 3 /8 = lim 8 + x + x + 3 3 = 8 The Ratio Test guaratees covergece whe this limit is less tha oe (ad divergece whe it is greater tha oe). x + 3 Usig either approach, we see that the power series coverges whe <, i.e. 8 whe x + <. Aswer: radius of covergece = d. [3 poits] Cosider the power series j= C j(x 5) j, where each C j is a costat. Suppose this power series coverges whe x = ad diverges whe x =. Based o this iformatio, which of the followig values could be equal to the radius of covergece of the power series? Circle all possibilities from the list below. 3 4 5 6 7 8 9 oe of these
Math 6 / Exam (November 3, 7) page. [9 poits] The sequece {γ } is defied accordig to the formula γ = l() + (You may recall this sequece from team homework 5.) This sequece coverges to a positive umber γ (which happes to be γ.57756649). a. [ poits] Does the sequece {γ } coverge or diverge? If this sequece coverges, compute the value to which this sequece coverges, either i terms of the costat γ or with five decimal places of accuracy. b. [3 poits] Does the series Yes, this sequece coverges. k. ( ) lim γ = lim γ = γ.3338. γ coverge or diverge? Briefly explai your aswer, ad if = this series coverges, compute the value to which the series coverges either i terms of the costat γ or with five decimal places of accuracy. This series diverges by the th term test for divergece. The terms of this series do ot approach ; istead the terms approach γ, i.e. c. [4 poits] Let h = lim γ = γ.57756649. e h. Fid the value of lim k. You may give your aswer either i terms of the costat γ or with five decimal places of accuracy. ( e h Hit: First cosider lim l ). Followig the hit, we first compute ( e h ) (e ) /k l = l = l(e /k ) l() = l() + k = γ. Next, we take the limit of this expressio. ( e h lim l ) = lim γ = γ.57756649. Fially, the limit we are lookig for is foud by expoetiatig this result (i order to udo the atural log from before). lim e h = eγ e.57756649.787.