MATH 6 FINAL EXAM ANSWERS December 7, 6 Part A. ( points) Find the volume of the solid obtained by rotating about the y-axis the region under the curve y x, for / x. Using the shell method, the radius of a shell is x, the height is y x, and so the volume of the shell is πx x. Integrating for / x, V / πx x. Let u x and then du x. When x /, u 3/ and when x, u. Therefore, V π 3/ 3 πu3/ 3/ π 3. u du The method of cylinders can also be used. We slice the solid perpendicular to the y-axis and integrate with respect to y. A cross-section is an annulus with outer radius x y and inner radius /. Then the volume is given by the integral V 3/ [ π ( y ) ] dy [ ] 3 π y y3 3/ 3 π 3.
. ( points) A large bathtub has the shape of a hemisphere (half of a sphere) of radius 5 feet, with the center at ground level. It is full of water from the bottom to ground level. How much work is done in pumping the water to the top? Remember that the weight of water is 6. pounds per cubic foot. Let x be the distance below ground level, in feet. A slice of water which is at level x, with thickness x, would have a circular shape with radius r 5 x. Therefore, its volume would be πr x π(5 x ) x. The weight of the water in that slice would be 6.πr x 6.π(5 x ) x. Since weight is given in pounds in the British system, and pounds is a measure of force, this is also the force on the slice of water. The work done to raise that slice to the top of the tank would be x times the force, or 6.πx(5 x ) x. So, the total work done to empty the tank would be 5 The work is measured in foot-pounds. 6.πx ( 5 x ) 6.π 3. ( points) Solve this indefinite integral: x 3/ ln x 5 ( 5x x 3 ) ( ) 5x 6.π x 5 ( 5 5 6.π 65 ) 6.π 65 We use integration by parts with u ln x dv x 3/ du x v 5 x5/ and we get x 3/ ln x 5 x5/ ln x 5 x 3/ 5 x5/ ln x ( ) x 5/ + C. 5
. ( points) Solve this integral: x 9x We use the substitution x (/3) cos(u), so that (/3) sin(u)du. Then x 9x (/9) cos (u) (/3) sin(u) du cos (u) 3 cos (u) du 3 sec (u) du 3 tan(u) + C Drawing a triangle, we see that 3 tan(u) + C reduces to 9x + C x The problem can also be solved using the substitution x (/3) sin(u), since the integral of csc (u) is cot(u) + C. 5. ( points) Solve this integral: t ( t) dt We use partial fractions: Then bringing to a common denominator, t ( t) A t + B t + C t At( t) + B( t) + Ct (C A)t + (A B)t + B, and it follows that A B C. So the integral becomes ( t ( t) dt t + t + ) t dt ln t t ln t + C 3
6. ( points) Find You must justify your answer. (n)! lim n n n We might expect the answer to be, since n! does not grow as quickly as n n. However, it might be different because we have (n)!. In fact, we can pair up the factors as follows. (n)! n n (n + ) (n) 3 n n n n! n + n n n n! So, (n)!/n n is at least as large as n!, which tends to as n. Therefore, (n)! lim n n n 7. ( points) Does the following series converge or diverge? + n + 3n + n + 6n 3 You must justify your answer. n Keeping only the leading terms, we would get n 3n 6n 3 n n We still need to justify our reasoning. It seems logical to use the limit comparison test for the divergent series with terms b n /(n). Let a n be the terms of the original series. Then, factoring out the leading terms in a n and taking the ratio with b n, we get we get / a n + n + 3n b n + n + 6n 3 n /n + /n + 3 /n 3 + /n + 6 n n 3 n 3 6
Since a n /b n tends to a limit which is not or, the limit comparison tests says that a n and b n must have the same convergence properties. Since b n diverges, it follows that an diverges. Thus, the original series diverges. 8. ( points) Is the following series absolutely convergent, conditionally convergent or divergent? You must justify your answer. n5 ( ) n n ln n First we note this is an alternating series. We use the Alternating Series Test to see if the series is convergent. It is clear that lim n n ln n, (n + ) ln(n + ) < n ln n Hence the series converges. Now we want to check if it is absolutely convergent, that is, if n5 n ln n converges. We use the Integral Test: t 5 x ln x lim t 5 x ln x lim ln(ln x) t Since the integral diverges, so does convergent. n5. n lnn Thus, the original series is conditionally convergent. t 5 and the original series is NOT absolutely 5
9. ( points) Does the following series converge or diverge? cos(n) n You must justify your answer. We can use the Root Test: lim cos(n) n n /n n lim n cos(n) /n /n lim n. Since the Root Test give a number less than one, the series is absolutely convergent. We can also use the Comparison Test to show the series cos(n), n cos(n) n n n, n cos(n) n is convergent. Since and the series on the right hand side is a geometric series with r /, hence convergent. We conclude the original series is absolutely convergent. Part B. (3 points) Find a power series representation for + x You must show your reasoning. Secondly, find the radius of convergence. We recall that the geometric series has the following sum. r + r + r + valid for r <. Substituting r x /, we get + x / x + x x6 8 + valid for x / <, or in other words, valid for x. Now we just have to divide the equation by, which doesn t change the radius, so + x x + x 8 x6 6 + 6
also valid for x.. (3 points) Let f(x) (x 3). (a) Find the Taylor series for f(x), centered at a 3. (b) Find f (5) (3) (c) Find f () (3) (a) f(x) is already in the form of its Taylor series, centered at a 3. All terms except (x 3) have coefficient zero. That is, if we write f(x) then c and c n for all n. c n (x 3) n, (b) Recall that c 5 f(5) (3) 5! and since c 5, we have f (5) (3) (c) Similarly to (b), f () (3)!c! n. (3 points) Suppose we approximate e x with its Taylor series up to x 3. T 3 (x) + x! + x! + x3 3! Find a number A > such that for x [ A, ], e x T 3 (x) The error estimate in Taylor s formula states that for x [ A, ], e x T 3 (x) M! x where M is the largest value of (e x ) e x for x [ A, ]. However, if x [ A, ] then x A and the largest value of e x occurs at x, which gives M. Therefore, we must find A such that! A 7
This gives A (!)/ / 3. ( points) Find the length of the curve y ln x x, for x. First, dy x x, and the length of the curve is then given by ( ) dy L + ( + x + x x + + x ( x + x ) ( x + x ) ) ( ln x + x ln + 3 ). ( points) Find the area of the surface generated by rotating the curve y x 3 about the x-axis, where x lies between and 3. Using the formula for the area of a surface rotated about the x-axis, we find A 3 3 3 πyds πy + dy πy + ( ) dy 8
Since y x 3 and dy/ 3x, A π π 3 3 x 3 + (3x ) x 3 + 9x Using the substitution u + 9x, du 36x 3 we get A π 36 3 6 + u / du π 36 36+ 3 u3/ π ( [3 6 + ] /3 3/) 7 5. ( points) Suppose that a parametric curve is defined by the following equation. x(t) te t y(t) sin(t) Write an integral which represents the length of the curve between t and t. DO NOT SOLVE THE INTEGRAL. Your integral must be written in terms of the functions t, e t, ln(t) and trig functions. It should not explicitly involve x(t) or y(t). We have L ds + dy ( ) ( ) dy + dt dt dt (e t + te t ) + cos t dt 6. (3 points) Find the points on the polar curve r sin θ where the tangent line is horizontal. 9
Recall that y r sin θ, and thus y which implies dy. In other words, the tangent to this curve is always horizontal. If we did not observe y, using the slope formula for polar curves, and r, we have sin θ dy sin θ + r cosθ dr dθ dr dθ cos θ cosθ r sin θ sinθ cosθ sin sin θ + θ cos θ cos θ sin θ sin θ sin θ 7. ( points) Consider the polar curve r 3θ 3, θ π Find the area of the region bounded by the curve and the ray θ, r. Using the area formula for polar curves, we get A 9 π π 9 θ7 7 ( 3θ 3 ) dθ θ 6 dθ π 9 (π)7