Math 1272 Solutions for Spring 2005 Final Exam. asked to find the limit of the sequence. This is equivalent to evaluating lim. lim.

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1 Math 7 Solutions for Spring 5 Final Exam ) We are gien an infinite sequence for which the general term is a n 3 + 5n n + n an are 3 + 5n aske to fin the limit of the sequence. This is equialent to ealuating lim n + n, which can be foun by iiing numerator an enominator by n law lim n p ineterminate (twice): an using the limit for p >, or by applying L Hôpital s Rule, since the ratio becomes 3 + 5n lim n + n n (3 + 5n ) n (n + n ) n + n (n) n ( + n) 5. n (D) ) The calculation of this integral calls for integration by parts. In orer that the new integral which will be prouce be no more ifficult than the one we starte with, we choose u x an e -x x. Because it is also a Type I improper integral, we will escribe the result in terms of a limit. x e x x u x u x ; e x x u e x lim x t u ( ) e x t e x x u lim t x e x + lim t t e x t t ( ) lim t x e x 4 e x ( ) ( e 4 ) e. t t e t 4 e t The first term uner the limit operation gies the ineterminate prouct ; we sole this by re-writing it as an ineterminate ratio an applying L Hôpital s Rule: lim t e t t t t e t t t t t et t e t " ". Therefore, our original integral is ( ) ( 4 ) x e x x 4 4. (B)

2 3) In orer to ecie on a ecomposition for the metho of partial fractions, we shoul look at the enominator of the rational function. If we were to multiply out the factors, we woul obtain a polynomial with a leaing term of x (x ) x 6 ; this is a higher power of x than in the leaing term of the polynomial in the numerator, so no polynomial iision is neee. The enominator contains two quaratic factors, one of which is factorable [ x ( x + ) ( x ) ] an the other is irreucible (not factorable), but is repeate [ ( x + ) ]. For a repeate factor, the ecomposition must contain terms haing enominators with eery power of that factor up to the power appearing in the rational function. As these are quaratic factor terms, the numerators must be linear polynomials. So there nee to be two terms of the form: C x + C ( x + ) + C 3x + C 4 ( x + ). The other quaratic factor, which coul be factore, is represente by two terms containing its linear factors in the enominators. Since these are linear factor terms, their numerators are just constants. This gies us terms in the ecomposition which resemble C 5 x + + C 6 x. The complete partial fraction ecomposition can thus be written as x 4 + 3x + x + 5 ( x ) ( x + ) A x + B x + + C x + D ( x + ) + E x + F. (B) ( x + ) 4) The key to soling this problem is in recognizing the resemblance of the gien series to the Maclaurin series for e x, We can thus conclue that + x + x! + x 3 + x 4 n (ln 5) n n! 4! + K x n n! e (ln 5) 5. (C) It is legitimate to use the Maclaurin series for e x, which is centere on x, in this way, since its raius of conergence is infinite. n

3 5) It will be easiest to ecie among the choices for this question by looking at the ( ) n conitions for conergence of the original series, n p, an its absolute series, n n p. By the p-test, a corollary of the Integral Test, we know that the absolute n series conerges for p > ; we thus say that the original series is absolutely conergent for p >. The original series is an alternating series, for which the general term is b n n p. Accoring to the Alternating Series Test, the original series conerges if b n b n+ an lim b n. Since n is positie, both of these conitions are met for p >. In summary then, we can say that the original series is absolutely conergent for p >, conitionally conergent for < p, an iergent for p. Among the aailable choices, the only one that is incorrect is (B). 6) The istance between two parallel planes is measure along a line perpenicular to each plane, that is, a share normal line. The normal ector to the planes x y + z 5 an x y + z is <,, >. To gauge the istance between the planes along this irection, we will nee to construct a normal line. We may choose any point in the plane x y + z, say, (,, ), an create the line parallel to <,, > which passes through it. The parametric equations for this line are x t, y t, z t x + t, y t, z t. We next fin where this line, which is also perpenicular to the plane x y + z 5, intersects that plane at the alue of the parameter t gien by x y + z ( + t) ( t) + (t) + 9t 5 t 5 9. The intersection point of this chosen normal line with the secon plane is then ( + 5 9, 5 9, 5 9 ) ( 5 9, 9, ). The two points are on a line 9 perpenicular to both planes, so the istance between them is the efine istance between the two parallel planes: ( 5 9 ) + ( 9 ) + ( 9 ) ( 5 9 ) + ( 9 ) + ( 9 ) (A)

4 7) The basic integral for fining the area boune by a polar cure r f(θ) between two angles is " A # [ f (")] " ", which represents the area of a wege haing the origin as its ertex. For the cure r e θ an the angles specifie, our area integral, which can be ealuate pretty irectly, will be A (eθ ) θ eθ θ ( eθ ) 4 (e e ) 4 (e ). (B) 8) For a parametric cure, the eriatie at a point gien by ( x(t), y(t) ) is For our cure, the eriaties are x t + ln t x t + t an y + t y t. t y x t Thus, we obtain the formula for the slope of the tangent line, + t y x y t x t t t +.. To fin the slope of the tangent line to a specific point ( x, y ), we will nee to fin the alue of the parameter t associate with that point. For the particular functions we hae, it will be easier to start from the equation for the y-coorinate: y + t t t ±, x() + ln +. The alue t is exclue, since it is not in the omain of t + ln t. Thus, the alue of the parameter at the point (, ) is t, where the slope of the tangent line is y x t t t + t +. (A) 9) The surface area integral for a soli of reolution can always be written as A π r s, where r is the perpenicular istance from the axis of reolution to the cure being reole an s is the infinitesimal element of arclength along the cure. For this problem, the axis of reolution is the x-axis, so the perpenicular istance will be r y. (continue)

5 The surface area integral for the soli escribe is thus A π r s π y x + y x π y x + y x π y + y x x ( ) x For our cure y 4 x ( 4 x ) /, y x / ( 4 x ) / ( / x) x 4 x A π ( 4 x ) / x + x π ( 4 x ) / 4 x (4 / ) + / 4 x x / x / x π 4 (4 x ) / x π 4 x x 4π x 4π ( ) 4π. (C) In fact, the full cure y 4 x is a semi-circle of raius centere at the origin; when rotate about the x-axis, it woul sweep out a sphere of that raius. The surface area we hae foun correspons to the portion of a globe between the equator an, say, 3º north latitue; this works out to be one-quarter of the total surface area of the globe. This gies the result ¼ 4π 4π. ) Pretty much the way to eal with this problem is to sift through the aailable choices irectly. We note, howeer, that there are two types of series liste. Choice (A) is a geometric series, which can be written as ( 3 4 )n, which means the ratio between successie terms in the series is r ¾. Since a geometric series conerges for r <, this specific series is conergent. All of the other choices are infinite series with general terms which are rational functions of polynomials. For the purposes of comparison, they can be likene to general terms using only the leaing terms of the polynomials, a n nα n α β n β n β α. We know, from the p-test, that an infinite series of the form only conerges for β α >. Through the Limit Comparison Test, we n β α n can show that the series with rational functions of polynomials will behae similarly (this woul take rather more space to show in etail). This means that only the infinite series with such general terms will conerge if the egree of the polynomial in the enominator is more than larger than the egree of the numerator s polynomial. Among the aailable choices, the only series for which this is not the case is (E). n

6 ) This trigonometric-powers integral inoling the tangent an secant functions separates nicely when we choose the tangent function as the basis for breaking up the integran. We can write sec 4 x tan x x sec x tan x sec x x. u u The Pythagorean Ientity can then be applie to eal with the outstaning factor of sec x, since sin x + cos x tan x + sec x : sec x tan x u sec x x u (tan x + ) u + tan x u sec x x u 5 u5 + 3 u3 + C 5 tan5 x + 3 tan3 x + C. (E) u 4 + u u This integral will not break up well if we choose u sec x instea: separating out the ifferential u sec x tan x x still leaes a single factor of tan x behin, with no tiy way of resoling it. ) In the conention use in our textbook (Stewart) for spherical coorinates, θ is the azimuthal angle measure along the equator of the sphere counter-clockwise from the positie x-axis an φ is the polar angle measure from the upper or north pole of the sphere. In this system, the equations for the transformation from spherical coorinates ( r, θ, φ ) to rectangular coorinates ( x, y, z ) are x r cosθ sinφ, y r sinθ sinφ, z r cosφ. For the point ( r, θ, φ ) ( 4, π/3, π/6 ), we thus fin ( x, y, z) ( 4 cos π 3 sin π 6, 4 sin π 3 sin π 6, 4 cos π 6 ) ( 4, 4 3, 4 3 ) (, 3, 3 ). (B) Be aware that other labeling conentions exist: in many other books an in other fiels, such as physics, θ enotes the polar angle an φ, the azimuthal angle, using the coorinate orer ( r, θ, φ ). 3) An infinite geometric series a r n has the sum s a r n. For the series gien in this problem, it will first be necessary to put it into this form, in orer to accurately etermine the alues of a an r. We can write ( 3) n + n n ( 3) n ( 3) ( ) n n 9 ( 3) n 9 ( 3 4 )n. n 4 n n We can now rea off that a 9 an r ¾, from which we fin that s 9 ( 3 4) (A)

7 4) For this question, we will nee to work through the aailable choices in orer to eliminate all but one of them (in this case, we are looking for the single incorrect statement). It is clear that this series is an alternating series ( eliminating choice (A) ). ln(n+3) The limit lim n+3 prouces an ineterminate ratio of the form may apply L Hôpital s Rule to it:, so we ln(n+3) lim n+3 n ln(n+3) n (n+3) n+3. So the statement in choice (B) is correct. The most irect way to etermine whether the sequence of absolute terms for this series is monotonically ecreasing is to look at the eriatie of the function they represent: x ln(x + 3) [ (x + 3) x + 3 ] [ ln(x + 3) ] x + 3 (x + 3) ln(x + 3) (x + 3) ; since ln 3 >, the eriatie of this function is negatie for x, so this function is ecreasing for x. The sequence of terms is thus monotonically ecreasing, making statement (C) correct. The truth of statements (B) an (C) mean that the gien series satisfies the conitions of the Alternating Series Test. Therefore the series conerges an statement (D) is correct. On the other han, if we integrate the function represente by the general terms oer the interal [, ), we see that ln(x + 3) x u x + 3 u x x + 3 lnu u u x : u x+3 : 3 3 lnu u u u : 3 ln u : ln 3 ln 3 t t ln 3 (lim t ) ( [ln 3] ). The absolute series is thus iergent, making the statement incorrect in choice (E).

8 5) a) This integral will require a combination of techniques, since it is the argument of the sine function that woul be inole in a u-substitution. Once we break up the integran, we hae x 3 sin(+ x ) x u + x u x x x x sin(+ x ) x x u sin(+ x ) sin u (u ) sinu u u sinu u sinu u x x u We know the anti-eriatie for the secon term. The first term nees an integration by parts: u u u ; w sinu u w cosu u ( cosu) w sinu u w ( cos u) u u cosu + cosu u w u cosu + sinu + C. The complete integral is then (u ) sinu u ( u cosu + sinu) ( cosu) + C ( / + x ) cos(+ x ) + sin(+ x ) + c o s ( + x ) + C sin(+ x ) x cos(+ x ) + C. b) This integral is a bit tricky, since it is not immeiately clear how a substitution will help an integration by parts woul be nightmarish an inefficient. Because the numerator an enominator of the rational function in the integran are of the same egree (the highest power of x being ½), it will be helpful to perform a polynomial iision first: 4 4 x x x + x x. + x The integral of the first term is simple, so we will focus our attention on the secon term. If we now try the reasonable substitution u + x, we fin that its ifferential is u x. Unfortunately, that factor of x is in the numerator of our x integran, rather than in the enominator, so it s not obious how this will help us. But if we insert the neee factor of x into the enominator ourseles, we obtain 4 x x + x x x 4 4 x + x x x. (continue)

9 This gies us more hope of progress, since we may now write 4 4 x + x x x 4 4 x + x x x u + x (u ) x x : 4 u + x : (u ) u u 4 3 u u + 3 u u 4 u + u u. We then nee to assemble our original integral, in orer to ealuate it: 4 x 4 3 x x 4 u + u u + x ( x 4 ) 4 ( u u + ln u ) [( ln 3) ( + ln ) ] 4 4 ( ln 3) ( + ) [ ] 4 4 (ln3) 4 ( ln3). The result is a small negatie number: a graph of the integran function reeals that it is in fact negatie an close to zero oer the interal [, 4 ]. 6) This ifferential equation looks pretty nasty, but it is separable: we can write ( x + ) y x y 3 y y 3 x x + We then integrate both sies of the equation the left-han integral will gie a logarithmic function, the right-han integral we recognize as yieling the arctangent function: y y 3 We can then exponentiate both sies to obtain x + x ln y 3 tan x + C.. e ln y 3 e tan x + C y 3 e tan x e C A e tan x exponentiation makes the aitie constant, C, a multiplicatie constant, A. This gies us two possible general solutions, epening on the sign of y 3 : y 3 A e tan x y 3 + A e tan x, y 3 or 3 y B e tan x y 3 B e tan x, y < 3. (continue)

10 y There is also what is calle a triial solution, which is foun by setting x in the original ifferential equation: ( x + ) y x y 3 ( x + ) y 3 y 3. b) We now are gien an initial-alue problem, in which we must fin the specific solutions of the ifferential equation with y() 5. Since tan(), it follows that tan (). We will only nee to sole this problem for y 3, so y 3 + A e tan 3 + A e 3 + A 5 A. The solution to the ifferential equation satisfying y() 5 is thus y 3 + e tan x. 7) a) The Funamental Theorem of (Integral) Calculus states that if a function g(x) is x efine by the integral g(x) f (t) t, where x is in the interal [ a, b ] an f(x) is a continuous on that interal, then g(x) f (x). For our function y f(x) efine x by f (x) e t s x t, then, we hae the eriatie y x ex. b) The arclength of a cure escribe by y f(x) on the interal [ a, b ] is gien by b b s x + y, where s x + y is the infinitesimal element of a a arclength in the (Eucliean) plane. Since we are to integrate along the x-irection, we will nee to factor out the ifferential x, making our arclength integral s b b x + y a a x x + y x ( ) x. x Since we are looking for the arclength of the cure escribe by the integral function f (x) x e t t on the interal [, ], we can use the eriatie we foun in part (a) in the arclength integral, giing us s ( ) + y x + e x x ( ) x / + e x / x e x x e x x e x e.

11 8) a) The Maclaurin series for this function will nee to be constructe from Maclaurin series for the basic functions y e x an y sin x. For the first of these, we hae e x + x + x! + x 3 + x 4 4! + K e x + ( x ) + ( x )! + ( x ) 3 + ( x ) 4 4! + K for the secon, we fin x + x 4! x 6 + x 8 4! + K ; sin x x x 3 + x 5 5! x 7 7! + K sin(x 3 ) (x 3 ) (x 3 ) 3 + (x 3 ) 5 5! (x 3 ) 7 7! + K x 3 x 9 + x5 5! x 7! + K. We now create the series for our function by multiplying these two series together: f (x) e x sin(x 3 ) x + x 4! x 6 + x 8 4! + K x 3 x 9 + x5 5! x 7! + K x 3 x 9 + x5 5! K x 5 x + x7 5! K + x 7! x3! + x9! 5! K x 9 x5 + x 5! K + x 4! x 7 4! + x 3 4! 5! K K. Upon sureying the terms present, we fin that we will be able to consoliate the first fie non-zero terms by collecting only the terms with powers of x up to x, yieling e x sin(x 3 ) x 3 x 5 + x 7! x 9 + x 9 + x + x 4! K x 3 x 5 + x 7 x x 4 K.

12 b) The general term in a Maclaurin series is hae for k 7, f (7) () 7! f (k ) () k! x k, hence, in our series, we x 7 x 7 f (7) () 7! 5. 9) The general term for this power series is a n n (x +) n n + conergence for the series is foun by application of the Ratio Test:. The raius of lim a n + a n lim n + (x+) n + (n+)+ n (x+) n n+ n + (x+)n + n (x+) n n+ n+ (x +) < x + <. So the raius of conergence of our series is R ½ (an the interal of conergence is centere on x. To etermine completely the interal of conergence, we must examine the behaior of the series at each enpoint: x : n ( +)n n + n n ( )n n + n n n + n, n + n for which the general term is b n n+ ; this may be compare with the term a n n a from which we fin that lim n b n n n n / n n n+ n+ n ; since ierges by the p-test ( p ½ < ), then our series also ierges by the Limit Comparison Test. x 3 : n ( 3 +)n n + n n ( )n n + n ( ) n, n + n which is an alternating series with b n lim n + n + ; since (n +) + < n + an, our series satisfies the Alternating Series Test, hence it conerges. The interal of conergence for our series is then [ 3, ).

13 ) a) For the points gien, the esire ectors are OA <, 3, 4 > an OB < 3, 4, >. The ot prouct of two ectors can be expresse in two ways. The first is by the efinition r a b a b cos θ, where θ is the angle between the two ectors an represents the length of a ector. The secon proies the means of computing the alue of the ot prouct from the Cartesian (rectangular) components of the ectors: a b a x b x + a y b y + a z b z. The (shorter) angle between the ectors can then be etermine from cos θ a b r a b a x b x + a y b y + a z b z a x + a ( y + a z ) b x + b y + b z ( ) For our ectors, we calculate OA OB ,. OA OB ()(3) + (3)(4) + (4)() 6, giing cos θ an so θ cos (6/9) b) The magnitue of the cross prouct of two ectors gies the area of the parallelogram efine by these ectors. A triangle can be forme by the two ectors an a thir ector passing from the tip of one of these ectors to the other, thus making a iagonal of the parallelogram. The area of this triangle is thus one-half of the area of the parallelogram. For the three points gien in the problem, we may choose any two pairs, for which we will construct ectors. We may pick, for instance, a BC (4 3), ( 4), (3 ),, an b BA ( 3), (3 4), (4 ),,. The cross prouct ector is then foun by the calculational eice a b ""* ˆ i ˆ j ˆ k i ˆ ˆ j + *this 3 x 3 eterminant is a means of computing the cross prouct, but is not a efinition ˆ k ( [ 4] [ ] ) i ( [ ] ) j + ( [ ] ) k < -3, -3, -3 >. The magnitue (or length) of this ector is so the area of triangle ABC is 7 or 3 a b ( 3) + ( 3) + ( 3) 7 3 3, 3.

14 c) A parallelopipe is a three-imensional figure with six faces, where the faces opposite each other are ientical parallelograms. For three ectors, not all in the same plane, which exten from a single point, two of the ectors can be chosen to efine a base parallelogram an the thir ector will then establish the olume. The orer in which these ectors are chosen turns out to make no ifference in fining the olume, which we obtain by taking the absolute alue of a scalar triple prouct of the three ectors. The three ectors chosen here all emanate from the origin O, so they are gien simply by a OA, 3, 4, b OB 3, 4,, an c OC 4,, 3. It oesn t matter for calculating the olume of the parallelopipe in what orer we take the ectors in the scalar triple prouct. We will choose, say, triple prouct can be foun using the eterminant ai r b ( c a ) ; the scalar r b ( c a ) "" b x b y b z c x c y c z a x a y a z (3)( 8 9 ) (4)( 6 6 ) + ()( 4 ) ( 3) The olume of the parallelopipe is the absolute alue of this scalar triple prouct, which is 7. G. Ruffa /9