Math 116 Second Exam

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1 Math 6 Secod Exam November, 6 Name: Exam Solutios Istructor: Sectio:. Do ot ope this exam util you are told to do so.. This exam has 9 pages icludig this cover. There are 8 problems. Note that the problems are ot of equal difficulty, ad it may be to your advatage to skip over ad come back to a problem o which you are stuck. 3. Do ot separate the pages of this exam. If they do become separated, write your ame o every page ad poit this out to your istructor whe you had i the exam. 4. 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.. 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. Iclude uits i your aswer where that is appropriate. 6. You may use ay calculator except a TI-9 (or other calculator with a full keyboard). 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 3 ote card. 7. If you use graphs or tables to fid a aswer, be sure to iclude a explaatio ad sketch of the graph, ad to write out the etries of the table that you use. 8. Tur off all cell phoes ad pagers, ad remove all headphoes. Problem Poits Score Total

2 Math 6 / Exam (November, 6) page. [ poits] While at home for Thaksgivig, Alex fids a forgotte ca of cor that has bee sittig o the shelf for a umber of years. The cotets have started to settle towards the bottom of the ca, ad the desity of cor iside the ca is therefore a fuctio, δ(h), of the height h (measured i cm) from the bottom of the ca. δ is measured i g/cm 3. The ca has a radius of 4 cm, ad a height of cm. (a) [3 poits of ] Write a expressio that approximates the mass of cor i the cylidrical cross-sectio from height h to height h + h. The cylidrical cross-sectio is a disk with height h ad radius 4 cm, so its volume is V = π(4) h. The mass of the cross-sectio is the M = δ(h) V = 6π δ(h) h. (b) [3 poits of ] Write a defiite itegral that gives the total mass of cor i the ca. We let h ad add the cotributios from each disk by itegratig, to get M = 6π δ(h) dh. (c) [3 poits of ] If δ(h) = 4e.3h, what is the total mass of cor iside the ca? We have M = 6π δ(h) dh = 6π 4e.3h dh. Thus M = 64π e.3h dh = 64.3 π e.3h = 64 3 π ( e.36 ) 6. g. (d) [3 poits of ] Write, but do ot evaluate, a expressio for the ca s ceter of mass i the h directio. Would you expect the ceter of mass to be i the top or bottom half of the ca? Do ot solve for the ceter of mass, but i oe setece, justify your aswer. The ceter of mass is h = 6π h δ(h) dh, M where M is the mass we foud before. We expect this to be i the bottom half of the ca, because the desity decreases with icreasig h. (Obviously, we could also write h = pluggig i δ(h) from (c) is fie too.) 6π h δ(h) dh 6π δ(h) ;

3 Math 6 / Exam (November, 6) page 3. [ poits] Alex ad Chris decide to ivest moey i a savigs accout to prepare for their expeses after they lad a posh mathematical cosultig job followig their success i calculus. They deposit $ o the first of each moth ito a accout that pays.467% iterest at the ed of each moth (a aual yield of about %). Let B be the amout i their accout immediately after their th deposit. (a) [ poits of ] B is a sequece. Give the first four terms i this sequece. After the first deposit, Alex ad Chris have B = $. Immediately before the secod deposit, they get.467% iterest o this, ad so have $.4, to which they add $. Thus B = $.4(= (.467)() + ). Similarly, B 3 = $.4(.467) + $ = $3.6, B 4 = $3.6(.467) + $ = $4.. ad (b) [ poits of ] Write a geeral, closed-form, formula for B (your expressio should ivolve oe of the symbols Σ,, or ). If we rewrite the precedig slightly, we ca see that B is just the sum of a geometric series. We have B = $, B = $(.467) + $, B 3 = ($(.467) + $)(.467) + $ = $(.467) + $(.467) + $, etc. Thus B = $(.467 ) + + $(.467) + $. This is a fiite geometric series with terms, ad so ( ) (.467) B = $ $3, (.467 )..467 (Either of these is fie as the correct aswer.)

4 Math 6 / Exam (November, 6) page 4 3. [4 poits] For the graduatig class of from a major uiversity (its ame cocealed so as to protect its idetity), the probability desity fuctio, p(x), for the umber of job offers, x, obtaied by a graduate is show i the figure to the right. The value a appearig i the values o the y-axis of this figure is a costat. (a) [3 poits of 4] What is the value of a? We kow that p(x) dx =, so we must have p(x) dx =. Calculatig the value of the itegral usig the area uder the curve, p(x) dx = 6a, so that 6a =, or a = 6. 3a a a p(x) x (b) [3 poits of 4] offers? What is the probability that a graduate will get at least 4 but o more tha 8 job This is just 8 p(x) dx. We ca fid the actual value for this probability by evalutig the area 4 uder the curve, which is 7 6. (c) [4 poits of 4] Write, but do ot evaluate, a expressio givig the mea umber of job offers obtaied by a graduate. Explai i oe setece how you would evaluate your expressio. The mea umber of offers is x = x p(x) dx. To fid this we would have to fid a piecewise expressio for p(x) (for < x <, p(x) = 6, etc.), multiply each by x, ad evaluate the resultig itegral(s). (d) [4 poits of 4] Write a expressio that gives the media umber of job offers obtaied by a graduate. Use your expressio to fid the media. The media umber of offers, T, is the umber such that half of the graduates get less tha or equal to T offers. This is T so that. = T p(x) dx, which requires us to fid the value T such that the area uder p(x) from x = to x = T is the same as the area uder p(x) from x = T to x =. By ispectio of the areas show i the figure, this is x = 4.

5 Math 6 / Exam (November, 6) page 4. [ poits] The followig three parts of this problem have to do with the covergece of series. (a) [4 poits of ] For Σ ++e : i. [ poits of 4] What is a good test to determie the covergece of this series? Explai, i seteces oly, why this is. The Compariso Test is a good test for this series. This is because we kow the covergece of Σ e, ad ote that ++e < e, so we have a ready made compariso to test the covergece of Σ ++e. Limit compariso will also work (which is reasoable because there s a obvious compariso series), as will the ratio test (which is reasoable because of the expoetial term). ii. [ poits of 4] Determie if this series coverges, diverges, or if we ca t tell. Because ++e < e, ad because we kow that Σ ++e coverges, we kow that Σ a must also coverge. Alterately, limit compariso with the same two series gives lim lim ++e b = e ++e =, a fiite o-zero umber, so both must coverge. Ad also alterately, the ratio test gives lim cosistet. ++e = + e <, so, agai, the series must coverge. Math is astoishigly ++e (b) [4 poits of ] For Σ + : i. [ poits of 4] What is a good test to determie the covergece of this series? Explai, i seteces oly, why this is. x The Itegral Test is a good test for this series. This is because it is easy to itegrate x +, so the itegral test is a dady choice. The limit compariso test is aother good optio, because + = whe, which makes us thik this series must diverge, but because + < the compariso test does t work whe comparig to. It s possible to use the compariso test, but that requires quite a bit of cuig to use correctly. ii. [ poits of 4] Determie if this series coverges, diverges, or if we ca t tell. Itegratig, test the series Σ x x + + series Σ + ad Σ have that our series diverges. ( dx = lim b l(b + ) l(6) ), which diverges as b, so by the itegral must also diverge. Alterately, usig the limit compariso test with the two, we have lim + =, so that, kowig that Σ diverges, we must this problem is cotiued o the followig page...

6 Math 6 / Exam (November, 6) page 6... problem cotiued from the previous page. (c) [4 poits of ] For Σ 3 : i. [ poits of 4] What is a good test to determie the covergece of this series? Explai, i seteces oly, why this is. The Limit Compariso Test is a good test for this series. This is because as, we ote that 3 = 3, which makes us thik this series must coverge. However, 3 >, so the compariso test does t work whe comparig to, which leads us to try the limit compariso test. It s possible to use the compariso test, but requires quite a bit of cuig to use correctly. ii. [ poits of 4] Determie if this series coverges, diverges, or if we ca t tell. We choose to compare with Σ. The ratio of the terms 3 ad is 3 3, so that as the ratio is. Thus, kowig that Σ coverges, we coclude by the limit compariso test that Σ must also coverge. 3. [8 poits] Let a ad b be the two sequeces show i the figure to the right. The sequece a = is show with solid dots ( ) ad the sequece b is show with crosses ( ). For <, < b < a. (a) [4 poits of 8] Does the sequece b coverge, diverge, or ca we ot tell? Explai i oe or two seteces. If it coverges, idicate the value to which it coverges. We kow that the sequece a = coverges to zero, ad for large the b are trapped betwee a ad zero, so b must also coverge to zero. 3 4 (b) [4 poits of 8] Does the series Σ b coverge, diverge, or ca we ot tell? seteces. If it coverges, idicate the value to which it coverges. Explai i oe or two We kow that < b < a for large, but all this tells us is that partial sums of Σ a, which is a diverget series, are larger tha those of Σ b. Thus we are uable to determie whether Σ b coverges or diverges.

7 Math 6 / Exam (November, 6) page 7 6. [6 poits] Chris has decided to take flyig lessos, ad otices that the cross-sectio of the airplae wig is give approximately by the figure to the right. The frot-to-back legth of the wig, as show i the figure, is m. The ed-to-ed legth of the wig is m (that is, its legth alog a axis comig out of this page is m), ad its eds are flat. (a) [3 poits of 6] If this cross-sectio is described by the polar equatio r = a cos(3θ), what is a? m The idicated frot-to-back legth occurs whe θ =. r() = a, so we must have a = m. (b) [4 poits of 6] What rage of values for θ geerate this figure? The curve starts ad eds at r =, which requires that cos(3θ) =, so that 3θ = ± π is a good solutio. Thus π 6 θ π 6. There are may other sets of θ values that give the regio as well (e.g., ay iterval that gives the top half of the wig, e.g., [, π 6 ], [ π 3, π 6 ], [ 4π 3, 3π 3π ], [π, 6 ], etc., plus ay iterval that gives the bottom half, e.g., [ π 6, ], [ π, π 3 ], [ 7π 6, 4π π 3 ], [ 6, π], etc.). (c) [9 poits of 6] Airplaes frequetly have fuel taks i their wigs. If 7% of the wig s volume is available space for a fuel tak, what volume of fuel could be stored i this wig? The volume of the wig is give by the area of its cross-sectio times m. Slicig with polar slices, a slice of the cross-sectioal area is give by A r θ = ( cos(3θ)) θ, so that the total cross-sectioal area is give by π/6 π/6 cos (3θ) dθ = π/6 π/6 +cos(6θ) dθ = (θ+ 6 si(6θ)) π/6 = π π/6 3. Thus the total volume is π m 3, ad the total volume available for fuel storage is π 4 m 3, or about.8 m 3.

8 Math 6 / Exam (November, 6) page 8 7. [ poits] A mysterious three-dimesioal abstract sculpture has appeared o the major uiversity s cetral campus. Alex, beig a particularly astute calculus studet, otes that the volume is give by V = (e x + ) dx, where x is i meters. (a) [4 poits of ] What does the itegrad of Alex itegral tell you about the shape of the sculpture? It appears that Alex cosidered a slice of the sculpture that has volume V = (e x + ) x. Thus (e x + ) appears to be the cross-sectioal area of the slice suggestig that the crosssectios of the sculpture perpedicular to the x-axis are square. There are several other correct iterpretatios of this, the most obvious of which is that the object has rotatioal symmetry about the x-axis ad has circular cross-sectios with radius r = π (e x + ). (b) [4 poits of ] Suppose that the sculpture was placed o a set of x-y axes. Sketch the base of the sculpture, labelig all importat dimesios ad features. If we follow the first slicig idicated i (a), with square cross-sectios, it is reasoable to guess that the base is bouded by x ad y e x +, as show i the figure to the left, below. Alterately, if we regard x as a dummy variable that measures up the object, we may cosider the base to be as show i the figure to the right. y y = e x + e+ e e+ x e Similarly, if we thik about the object havig circular slices, we ca thik of the base of the object i ay of the followig three ways: the first two look at the projectio of the object ito a xy-plae, thikig of the x i the itegral as that measured alog the idicated x-axis, ad the last cosiders x to be a dummy variable that measures up the object. y y y = e x + y = e x r = (e+) + πe x - y = e x x (c) [4 poits of ] Sketch ad/or carefully explai what the shape of the sculpture is. Sketched below are, from left to right, the figure with square cross-sectios o the base x, y e x + ; the figure with square cross-sectios ad x measurig up the figure; The figure with circular cross-sectios rotated aroud the x-axis, ad the figure with circular cross sectios ad x measurig height. r = e x + π x e x + x

9 Math 6 / Exam (November, 6) page 9 8. [6 poits] Chris is stadig at the edge of a swimmig pool, holdig a chai that is partially submerged i the water of the pool, as show i the figure to the right. The chai is six feet log ad weighs lb/ft. Whe it is i the water, however, the buoyat force of the water makes the effective weight of the chai less i the water, it weighs oly 3 lb/ft. If the chai is iitially half submerged i the pool ad Chris lifts it straight up util it is etirely out of the water, how much work does Chris do? We ca fid the total work by cosiderig the work to move from a give positio, x (measured as the distace that the chai has bee raised), to the positio x + x. The required force is the weight of the chai, F = (weight of chai above the water) + (weight of chai i the water). The legth of chai above the water is 3 + x ft, ad the legth below is 3 x ft. Thus F = (3 + x)() + (3 x)(3) = 4 + x lb. The work to lift the chai through this distace x is the W = (4 + x) x. The total work is foud by itegratig over the 3 ft that it is lifted, so W = 3 (4 + x) dx = (4x + x ) 3 = = 8 ft lb. A alterate solutio is to cosider the top half ad bottom half of the chai separately. The top half all moves 3 ft ad has a costat weight, so the work is W t = ((3 ft)( lb/ft))(3 ft) = 4 ft lb. The, a piece of the bottom part of chai that has legth x ad is x feet from the bottom of the chai s iitial positio has a weight 3 lb/ft for the distace 3 x ad a weight lb/ft for the distace x. Thus the work to lift it is W = (3 x)(3 x) + ( x)x = (9 + x) x. The total work to lift the bottom half of the chai is the W b = x dx = 9x + x 3 = = 36 ft lb. The total work is the sum of W t ad W b, which is ot surprisigly still 8 ft lb.

Math 116 Practice for Exam 3

Math 116 Practice for Exam 3 Math 6 Practice for Exam Geerated October 0, 207 Name: SOLUTIONS Istructor: Sectio Number:. This exam has 7 questios. Note that the problems are ot of equal difficulty, so you may wat to skip over ad retur