MATH 1372, SECTION 33, MIDTERM 3 REVIEW ANSWERS

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MATH 1372, SECTION 33, MIDTERM 3 REVIEW ANSWERS 1. We have one theorem whose conclusion says an alternating series converges. We have another theorem whose conclusion says an alternating series diverges. Substitute b n = into these theorems. Is the following 2n 2 +8 series convergent of is it divergent? Justify your answer. 2n 2 + 8 The Alternating Series Theorem says that for the alternating series 2n 2 + 8 ; if (1) lim n 2n 2 +8 = 0 and (2) 3n+10 2(n+1) 2 +8 < 2n 2 +8, then 2n 2 + 8 converges. The Divergence Theorem for Alternating Series says that for the alternating series 2n 2 + 8 ; if lim n = L and L 0, then 2n 2 +8 2n 2 + 8 diverges. The series is convergent: lim n 2n 2 +8 = lim n 3/n+7/n2 = 0. So, (1) 2+8/n 2 holds. 3n + 10 2(n + 1) 2 + 8 < 3n + 7 2n 2 (3n + 10)(2n 2 + 8) < (3n + 7)(2(n + 1) 2 + 8) + 8 6n 3 + 20n 2 + 24n + 80 < (3n + 7)(2n 2 + 4n + 10) 6n 3 + 20n 2 + 24n + 80 < 6n 3 + 29n 2 + 58n + 70 10 < 9n 2 + 34n Since 10 < 9n 2 + 34n is clearly true for n 1, we conclude that (2) is true. Since (1) and (2) are true, we conclude that 2n 2 + 8 converges. 2. Is the following series convergent or divergent? Start by substituting b n = 2n+8 into one of the theorems about alternating series. Justify your answer to the question using this statement. 2n + 8

2 MATH 1372, SECTION 33, MIDTERM 3 REVIEW ANSWERS The Divergence Theorem for Alternating Series says that if 2n+8 = L and L 0, then lim n lim n 2n+8 = lim n 3+7/n 2+8/n = 3 2 0. So, we conclude that 3. Is the following series convergent or divergent? Start by substituting b n = 3n2 +7 2n+8 into one of the theorems about alternating series. Justify your answer to the question using this statement. n 1 3n2 + 7 2n + 8 3n A theorem says that if lim 2 +7 n 2n+8 does not exist, then n 1 3n2 + 7 3n lim 2 +7 n 2n+8 = lim n 3+7/n2 which does not exist. 2/n+8/n 2 So, we conclude that n 1 3n2 + 7 4. Find the value of L in the ratio test for the following series. Write a (2n + 1)x n 3 n (5n 2) L = lim (2n + 3)x n+1 n 2 n+1 (5n + 3) 3n (5n 2) (2n + 1)x n = lim x(2n + 3)(5n 2) n 3(2n + 1)(5n + 3) = lim x(10n 2 + 4n 6) n 3(10n 2 + 8n + 3) = x 3 lim 10 + 4/n 6/n 2 n 10 + 8/n + 3/n 2 = x 3 The ratio test says that if 0 x 3 < 1, then (2n + 1)x n 3 n (5n 2) converges.

MATH 1372, SECTION 33, MIDTERM 3 REVIEW ANSWERS 3 So, we conclude that the series converges for x such that x < 3. 5. Find the value of L in the ratio test for the following series. Write a n (2n + 1)x2n 3 n (5n 2) L = x 2 /3. The series converges for x such that x < 3. 6. Find the value of L in the ratio test for the following series. Write a (2n + 1)x n 3 n n! L = 0. The series converges for all x. 7. Find the value of L in the ratio test for the following series. Write a (2n + 1)x 2n 3 n (2n)! L = 0. The series converges for all x. 8. Find the value of L in the ratio test for the following series. Write a n!x n 3 n (5n 2) L does not exist if x 0. L = 0 if x = 0. The series converges only for x = 0. 2n 6 9. Find constants c and p such that n(n 2 2) < for all n 1. Show that the following series is convergent using the comparison test. 2n 6 n(n 2 2) c = 3, p = 2. Other values of c and p would also work. c n p

4 MATH 1372, SECTION 33, MIDTERM 3 REVIEW ANSWERS 2n 6 10. Find constants c and p such that n(n 2) > c n for all n 2. Show p that the following series is divergent using the comparison test. 2n 6 n(n 2) c = 1, p = 1. Other values of c and p would also work. 11. You can assume that the following alternating series satisfies the hypotheses of the alternating series test. If x 0.4 find the smallest value of N such that k 1 x 2k k(k + 1)! < 10 9. k=1 N = 5 12. Suppose we use N+1 N k=1 k 1 to approximate the sum of the series (k 2) 3 k 1. What is the smallest value of N for which we can be sure that (k 2) 3 the error is less than 10 4? N = 23 13. Assume it is known that the following series satisfies the hypotheses of the alternating series theorem for all x. If x 0.5 and N = 4 find a bound for k xk 2k!. 0.00013 k=n+1 14. Find the Taylor polynomial T 4 (x) about a = 0 for the function f(x) = e 2x + (x + 9) 1/2. 4 11 6 x + 431 216 x2 5183 3888 x3 + 186619 279936 x4 15. The Taylor polynomial T 4 (x) for (8 + 3x) 2/3 is (8 + 3x) 2/3 4 + x 1 16 x2 + 1 96 x3 7 3072 x4. (a) Using differentiation find the Taylor polynomial for (8 + 3x) 1/3. (b) Using integraion find the Taylor polynomial for (8 + 3x) 5/3.

MATH 1372, SECTION 33, MIDTERM 3 REVIEW ANSWERS 5 (a) 1 2 + 1 16 x + 1 64 x2 7 (b) 32 + 20x + 5 2 x2 5 48 x3 + 5 384 x4 1536 x3 + 35 24576 x4 16. The Taylor polynomial T 4 (x) for (1 + 2x) 1/2 is Find T 4 (x) for (9 + 2x) 1/2. (1 + 2x) 1/2 1 x + 3 2 x2 5 2 x3 + 35 8 x4. 1 3 1 27 x + 1 162 x2 5 4374 x3 + 35 157464 x4 17. The power series for sin x converges for x < 1. For what range of values of x is x3 sin x [x 3! + x5 5! ] < 10 7? x < 0.3385 18. Write the power series (3n 2)x n n=3 (3n + 7)x n+3 in standard position. 19. The graphs of r(t) and R(s) are lines. Find the point where these lines intersect. (1, 4, 3) r(t) = (3t 2) i + ( 2t + 6) j + (7t 4) k R(s) = (2s + 2) i + (4s + 6) j + ( 2s + 2) k 20. Consider the graphs of the following two vector functions. Show that the two graphs are actually the same line. r(t) = ( 5t + 2) i + (2t 3) j + (t + 1) k R(s) = (10s 23) i + ( 4s + 7) j + ( 2s + 6) k r(0) = (2, 3, 1) = R(5/2) and r(1) = ( 3, 1, 2) = R(2). Since the two lines share 2 points, the lines must be the same line. 21. Let r(t) = (3t 2) i + (2t) j + ( 3t + 1) k. (a) Find a line parallel to r(t) that contains the point (1,1,2).

6 MATH 1372, SECTION 33, MIDTERM 3 REVIEW ANSWERS (b) Find a line perpendicular to r(t) that contains the point (1,1,2). (a) R(t) = (3t + 1) i + (2t + 1) j + ( 3t + 2) k (b) R(t) = (2t + 1) i + ( 3t + 1) j + 2 k 22. Find a vector perpendicular to the plane containing the three points (2,1,3), (0,-2,1), and (-1,2,5). 4 i + 10 j 11 k 23. Let v = 2 i k and w = i + 3 j 2 k. (a) Find the angle between the vectors v and w. (b) Find P roj w v. (c) Find P roj v w. (a) 1.072 (b) 2 7 ( i + 3 j 2 k) (c) 4 5 (2 i k) 24. Find the arc length of the curve r(t) = 4t i 2 2t 2 j +( 4 3 t3 4) k from the point where t = 1 to the point where t = 3. 128 3 25. Find the area of the parallelogram with vertices (1, 2, 3), (4, 5, 2), (2, 2, 2), and (3, 1, 3). 38

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