as best you can in any three (3) of a f. [15 = 3 5 each] e. y = sec 2 (arctan(x)) f. y = sin (e x )

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Lorraine Gardner
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1 Mathematics Y Calculus I: Calculus of oe variable Tret Uiversity, Summer Solutios to the Fial Examiatio Time: 9: :, o Weesay, August,. Brought to you by Stefa. Istructios: Show all your work a justify all your aswers. If i oubt, ask! Ais: Calculator; two ai sheets [all sies]; oe brai [may be caffeiate]. Part I. Do all three of.. Compute y as best you ca i ay three of a f. [5 5 each] a. x e x+y b. y. y x cosx x te t t c. y x lx e. y sec arctax f. y si e x Solutio i to a. We will use implicit ifferetiatio. The alterative is to solve for y first, which is t too har i this case; see solutio ii below. Differetiatig both sies of x e x+y gives from which it follows that ex+y e x+y e x+y + y x + y ex+y + y e x+y x+y y + e, x+y y e ex+y y e x+y ex+y e x+y e x+y. Solutio ii to a. We will solve for y as a fuctio of x first, a the ifferetiate. x e x+y e x e y e y x e x xe x y l e y l xe x l x + l e x l x xl e l x x It follows that y lx x x. If you plug y lx x ito y e x+y it works out to x too. Solutio to b. We will use the Chai Rule a the Fuametal Theorem of Calculus. Let u x; the y u te t t a so y y u u u te t t u x ue u xe x xe x.

2 Oe coul also o this by computig the efiite itegral first usig itegratio by parts, a the ifferetiatig. Solutio to c. We will o this oe usig the Prouct Rule as the mai tool: y x lx x lx + x lx xlx + x x xlx + x Solutio to. We will o this oe usig the Quotiet Rule as the mai tool: y x cosx x cosx cosx cos x cosx x six cos x cosx + x six cos x Give the multitue of trig ietities, which coul be applie before or after ifferetiatig, y there are lots of equivalet ways of writig the aswer. For oe ice example, secx + x secx tax. Solutio to e. This ca be oe usig the Chai Rule a followig up with a trig ietity or two, but it s easier if oe simplifies y usig a trig ietity or two first: y sec arctax + ta arctax + ta arctax + x It the follows that y + x + x x. Solutio to f. There is o avoiig the Chai Rule i this oe: y si ex cos e x ex cos e x e x e x cos e x. Evaluate ay three of the itegrals a f. [5 5 each] x π/ a. b. siz cosz z c. x lx 4 x. l e s s e. f. + x w + w w Solutio to a. This ca be oe usig a trig substitutio, amely x siθ, but it s faster to use the substitutio u 4 x, so that u x a x u: x u u / u 4 x u u/ / + C u + C 4 x + C

3 Solutio to b. This is probably easiest usig the substitutio u siz, so u cosz z a z π/ u. π/ siz cosz z u u u Solutio to c. We will use itegratio by parts with u lx a v x, so u x a v x : x lx x lx x x x x lx x x lx 9 + C Solutio to. This is a improper itegral, so we ee to set up a compute a it: l e s s sice e t as t. t t l t e s s e l e t t es l t t e t, Solutio to e. We will use the trig substitutio x taθ, so sec θ θ. + x + ta θ sec θ θ sec θ sec θ θ secθ sec θ θ l x + + x + C secθ θ l taθ + secθ + C Note that secθ sec θ + ta θ + x if x taθ. Solutio to f. This itegral requires the use of partial fractios. First, if w + w ww + A w + B w + Aw + + Bw ww + A + Bw + A w + w the, comparig umerators, we must have A + B a A, so B. It follows that w + w w w + w w + w w w + w lw lw + [l l] [l + l + ] [l ] [l l] l l + l l l,

4 Those who feel compelle to simplify further may o so: l l l l l4 l l 4. Do ay five 5 of a i. [5 5 5 ea.] a. Determie whether the series coverges absolutely, coverges coitioally, or iverges. b. Why must the arc-legth of y arctax, x, be less tha + π? c. Fi a power series equal to fx x whe the series coverges without + x usig Taylor s formula.. Fi the area of the regio betwee the origi a the polar curve r π + θ, where θ π. e. Fi the iterval of covergece of the power series x. f. Use the it efiitio of the erivative to compute f for fx x. g. Compute the area of the surface obtaie by rotatig the the curve y x, where x, about the y-axis. h. Use the Right-ha Rule to compute the efiite itegral i. Use the ε δ efiitio of its to verify that x x +. Solutio to a. We will apply the Ratio Test: a + a x Sice a + a <, the give series coverges absolutely by the Ratio Test. 4

5 Solutio to b. The easiest way to see a explai why the arc-legth of y arctax, x, must be less tha + π is to raw a picture; eve a crue sketch will suffice: Note that sice arcta is pretty close to π, the poit, π is pretty close to the poit, arcta. It shoul be pretty clear from the picture that goig from the origi by way of the y-axis to the poit, π a the o to the poit, π by way of the lie y π is a loger trip tha goig from the origi to, arcta by way of the curve y arctax. It follows that π + is greater tha the arc-legth of y arctax for x. Solutio to c. Usig the formula for the sum of a geometric series i reverse, fx x + x x x x x x x x + x x + x x 4 + x 5 Solutio to. We plug the polar curve r π + θ, θ π, ito the area formula i polar cooriates: π π π r θ + θ π θ π 4 θ + π θ + π θ π 4 π + π π + π π π π + θ + θ θ 4 + π + 4 π Solutio to e. We will first fi the raius of covergece of the power series 5 x

6 usig the Ratio Test. a + a + x + + x + x + + x x + x + x It follows by the Ratio Test that the series coverges absolutely whe x <, i.e. whe < x <, a iverges whe x >, i.e. whe x < or x >. Thus the raius of covergece of the give power series is R. It remais to etermie what happes at the epoits of the iterval, amely at x a x. Sice a, the Divergece Test tells us that the series iverges for both x a x. The iterval of covergece is therefore,. Solutio to f. We will plug fx x ito the it efiitio of the erivative to compute f : f h f + h f h h h h h + h h h h h h Solutio to g. We will plug the curve y x, x, ito the formula for the area of a surface of revolutio, πr s. First, ote that sice we are rotatig the curve about the y-axis, r x x. Seco, sice y x x x, we have that s + y + x. Thus SA 4 π πr s πx + x Let u + x, so u x a x u 4. 4 u u π u / u 4 u/ 6 4 / / 8 4

7 Solutio to h. We plug the efiite itegral b b a formula fx f sice 9 a x + i as. i a + b a i [ + [ ] i + i [ ] i + i [ ] + + x + ito the Right-ha Rule a chug away: ] i + [ i i [ ] i + i ] + i [ + ] [ 5 + ] + [ ] 5 Solutio to i. We ee to show that for ay ε >, there is a correspoig δ > such that wheever x < δ, we have x + < ε. Suppose, the, that ε >. As usual, we will reverse-egieer the correspoig δ > : x + < ε x < ε, so δ ε will o the job. Oe step of reverse-egieerig is as easy as it gets! Thus x x + by the ε δ efiitio of its. Part II. Do ay three of Fi the omai, all maximum, miimum, a iflectio poits, a all vertical a horizotal asymptotes of fx x x, a sketch its graph. [5] + Solutio. We ll ru through the usual check list of items a the sketch the graph: i. Domai Sice x + > for all x, the eomiator is ever. It follows that the x ratioal fuctio fx x is efie a cotiuous a ifferetiable, too for all + x, i.e. its omai is R,. ii. Itercepts f, so the y-itercept is the origi, i.e. at y. O the + other ha, fx x x + is oly possible if the umerator x, i.e. x. It follows that the origi is also the oly x-itercept. 7

8 iii. Vertical asymptotes Sice fx is efie a cotiuous for all x, it caot have ay vertical asymptotes. iv. Horizotal asymptotes We compute the relevat its to check for horizotal asymptotes: x x + x x + x x + x x + x x + x x + x x x x x x x + + x Note that x both as x a as x +. It follows that fx has the horizotal asymptote y i both irectios. v. Maxima a miima First, with a little help from the Quotiet Rule, f x x x + x x + x x + x + x x + x x x + x + x x x + x x +. Seco, ote that f x is also a ratioal fuctio that is efie a cotiuous for all x, usig reasoig very similar to that use i i above. This meas that the oly ki of critical poit that ca occur is the sort where f x. f x x + ca oly occur whe the umerator, x, is, which happes oly whe x. Moreover, sice the eomiator, x + > for all x a >, too f x < whe x < a f x > whe x >. We ca summarize this iformatio a its effect o fx with the usual sort of table: x,, + f x + fx mi Note that x gives a local a absolute miimum poit for fx a that fx has o maximum poit. This makes particular sese if you o iv above a little more carefully a otice that fx approaches the horizotal asymptote from below i both irectios. vi. Graph We cheat ever so slightly a let Maple o the rawig: > plotx^/x^+,x-..,y ; 8 Whew!

9 5. Do both of a a b. x a. Verify that x x l x + x + C. [7] b. Fi the arc-legth of y x x x l + x for x. [8] Solutio to a. We ca compute the iefiite itegral usig the trig substitutio x secθ, so secθ taθ θ, but it s ofte easier to ifferetiate the atierivative a check that the result is equal to the itegra. Tryig this here [ [ x ] x [ x + x [ x + x x x l x + x l ] + C x + x + C ] x x + x x x ] x + x + x x [ x ] x + x x x + x + x x [ x x ] + x x + x x + x [ ] x x x + x x x + x x x + x + x x x + x x + x x x x x + x x + x x x x x x x x + x + x x x x x x x... makes for a algebra-fest of epic proportios. Maybe it woul have bee easier to itegrate... :- 9

Solutio to b. It follows from a a the Fuametal Theorem of Calculus that y x. Pluggig this ito the arc-legth formula gives: arc-legth + y + x x x x 9 + x 8 4 6.

10 Solutio to b. It follows from a a the Fuametal Theorem of Calculus that y x. Pluggig this ito the arc-legth formula gives: arc-legth + y + x x x x 9 + x Sketch the soli obtaie by rotatig the square with corers at,,,,,, a, about the y-axis a fi its volume a surface area. [5] Solutio. Here is sketch of the soli: Note that the origial regio has as its borers pieces of the vertical lies x a x a the horizotal lies y a y, i.e. the regio cosists of all poits x, y with x a y. We ca compute the volume of the soli i several easy ways: i. Shells Note that sice we revolve the origial regio about the y-axis a are usig shells, we will have to itegrate with respect to x. The cylirical shell at x, for some x, has raius r x x a height h. Thus the volume of the soli is: V πrh πx π x πx π π ii. Washers Note that sice we revolve the origial regio about the y-axis a are usig washers, we will have to itegrate with respect to y. The washer at y, for some y, has outer raius R a ier raius r. Note that the washers are all ietical... Thus the volume of the soli is: V π R r y π y π y πy π π

11 iii. Look, Ma! No calculus! The soli is a cylier of raius r a height h a hece volume πr h 4π with a cylier of raius r a height h a hece volume πr h π remove from it. It follows that the soli has volume 4π π π. Ay correct metho, correctly a completely worke out, woul o, of course. :- It remais to fi the surface area of the soli. The complicatio is that the surface of the soli cosists of four istict pieces: the upper a lower faces of the soli, which are both washers with outsie raius R a isie raius r, the outsie face, which is a cylier of raius R a height h, a the isie face i.e. the hole i the mile, which is a cylier of raius r a height h. While the area of each of these ca be compute pretty quickly as the area of a surface of revolutio, it is poitless to work so har whe we shoul have formulas for the areas of these objects at our figertips from our kowlege of the washer a shell methos for computig volume: The areas of the upper a lower faces are each π R r π π, the area of the outsie face is πr h π 4π, a the area of the isie face is πr h π π. The total surface area of the soli is therefore π + 4π + π π. 7. Do all three of a c. a. Use Taylor s formula to fi the Taylor series at of fx lx +. [7] b. Determie the raius a iterval of covergece of this Taylor series. [4] c. Use your aswer to part a to fi the Taylor series at of without usig x + Taylor s formula. [4] Solutio to a. We first ifferetiate a evaluate away to figure out what f must be for each. f x fx lx +, so f l + ; f x f x lx + x+ x + x+ x+, so f + ; f x f x x+ x+ x + x+ x+, so f + ; f x f x x+ x+ x+ x+ x+, so f + ; a so o: f x lx x+ x+ x+ x+ 4 x+ 5 x+ 6 f A little reflectio o this patter shows us that at stage >, f x! x+ a so f! +!. Applyig Taylor s formula, a otig that is the exceptio to the patter ote above, the Taylor series at of fx lx + is therefore: f! x f x +! f x! +! x! x x x + x x4 4 + x5 5

12 Solutio to b. To fi the raius of covergece of the power series obtaie i a, we use the Ratio Test: a + a + + x + x x+ x + x x x + + x + / x / + x + x It follows that the series coverges absolutely whe x < a iverges whe x >, so the raius of covergece is R. To fi the iterval of covergece, we have to etermie whether the series coverges or iverges at each of x a x. That is, we have to etermie whether each of the series a coverges or ot. The first iverges: , so it is a o-zero multiple of the harmoic series, which iverges. We showe this i class usig the Itegral Test, though it is eve quicker to use the p-test. The seco coverges: it is just the alteratig harmoic series, which coverges coitioally. We showe this i class usig the Alteratig Series Test. It follows that the iterval of covergece is, {}, ]. Solutio to c. Sice lx + as ote i the solutio to a above, we x + ca get the Taylor series at of f x x+ by ifferetiatig the Taylor series at of fx lx + term-by-term: x x + x x4 x x x5 x 5 x + x 4x 4 + 5x4 5 x + x x + x 4 x x 5 5 This is the geometric series with first term a a commo ratio r x.

13 8. A spherical balloo is beig iflate at a rate of m /s. How is its surface area chagig at the istat that its volume is 6 m? [5] [Recall that a sphere of raius r has volume 4 πr a surface area 4πr.] Solutio. We ee to relate V t First, observe that A t t 4πr r 4πr r t This meas that we will ee to kow r t Seco, we have so r t 4πr. V t t to A t, where A is the surface area of the balloo. 8πr r t. at the istat i questio. 4 πr 4 r πr r t 4 πr r t 4πr r t, It follows that A t 8πr 4πr r. Thir, it still remais to etermie what the value of r is at the istat i questio. Sice V 4 / / 6 7 πr 6 at the istat i questio, r 4 π π π. / Thus, at the istat that the volume is 6 m, the surface area is chagig at a rate of A t π/ m /s. π / Part MMXI - Bous problems.. Show that l secx tax l secx + tax. [] [Total ] Solutio. Here goes: secx + tax l secx tax l [secx tax] l [secx tax] secx + tax sec x ta x + ta x ta x l l secx + tax secx + tax l l [secx + tax] secx + tax l secx + tax l secx + tax Note that the key trick is the same oe we use i class to compute secx. 4. Write a origial poem touchig o calculus or mathematics i geeral. [] Solutio. Write your ow! I hope that you ha some fu with this! Get some rest ow...

1. Do the following sequences converge or diverge? If convergent, give the limit. Explicitly show your reasoning. 2n + 1 n ( 1) n+1.

1. Do the following sequences converge or diverge? If convergent, give the limit. Explicitly show your reasoning. 2n + 1 n ( 1) n+1. Solutio: APPM 36 Review #3 Summer 4. Do the followig sequeces coverge or iverge? If coverget, give the limit. Eplicitly show your reasoig. a a = si b a = { } + + + 6 c a = e Solutio: a Note si a so, si