MTH 33 Solutions to Exam 2 April 9, 207 Name: Section: Recitation Instructor: INSTRUCTIONS Fill in your name, etc. on this first page. Without fully opening the exam, check that you have pages through 2. Show all your work on the standard response questions. Write your answers clearly! Include enough steps for the grader to be able to follow your work. Don t skip limits or equal signs, etc. Include words to clarify your reasoning. Do first all of the problems you know how to do immediately. Do not spend too much time on any particular problem. Return to difficult problems later. If you have any questions please raise your hand and a proctor will come to you. You will be given exactly 90 minutes for this exam. Remove and utilize the formula sheet provided to you at the end of this exam. ACADEMIC HONESTY Do not open the exam booklet until you are instructed to do so. Do not seek or obtain any kind of help from anyone to answer questions on this exam. If you have questions, consult only the proctor(s). Books, notes, calculators, phones, or any other electronic devices are not allowed on the exam. Students should store them in their backpacks. No scratch paper is permitted. If you need more room use the back of a page. Anyone who violates these instructions will have committed an act of academic dishonesty. Penalties for academic dishonesty can be very severe. All cases of academic dishonesty will be reported immediately to the Dean of Undergraduate Studies and added to the student s academic record. I have read and understand the above instructions and statements regarding academic honesty:. SIGNATURE Page of 2
MTH 33 Solutions to Exam 2 April 9, 207 Standard Response Questions. Show all work to receive credit. Please BOX your final answer.. (8 points) Find the length L of the curve given by x = 3 y3/2 from y = 4 to y = 8. Solution: dx dy = 3 3 2 y/2 = 2 y/2 ( ) 2 dx = dy 4 y so using the arc length formula we have L = 8 4 New lower bound y = 4 u = + 4 4 = 2 New upper bound y = 8 u = + 4 8 = 3 3 L = 4 u /2 du 2 = 4 2 3 3 u3/2 2 = 8 ) (3 3/2 2 3/2 3 + 4 y dy u = + 4 y 4du = dy Page 2 of 2
MTH 33 Solutions to Exam 2 April 9, 207 2. (8 points) Determine whether the sequence converges or diverges. If it converges, find the limit. (a) a k = k + ln k k Solution: lim k k + ln k k indeterminate case. So we apply L Hopital s Rule + /k lim k Therefore the sequence converges. ( ) (b) b k = k 2 ln cos k ( Solution: lim k 2 ln cos ) indeterminate 0 case. So we rewrite as k k ( lim k k2 ln cos ) ( ) ln cos k = lim (indeterminate k k k 2 = 0 0 case) LH = lim k sin k cos k k 2 2 k 3 = lim k tan k k 2 k3 2 = lim k k 2 tan k (indeterminate 0 case. so rewrite) = lim k k 2 tan k = lim k 2 tan k LH = lim k 2 sec2 k k 2 k 2 = 2 k (indeterminate 0 0 case) Therefore the sequence converges. Page 3 of 2
MTH 33 Solutions to Exam 2 April 9, 207 3. (8 points) Determine if the following series converge or diverge. You must justify your answer with work and explicitly state which test(s) you use and whether the conditions for the test are met! (a) (b) 2n + n Solution: Compare with b n = 2n which is a divergent p-series, p = lim n 2n + n 2n 2n = lim n 2n + n = 2 2 = Therefore, the series diverges by Limit Comparison Test. 9 n (2n)! Solution: Use the Ratio Test So therefore a n+ = 9 n+ (2(n + ))! Since 0 < the series converges by Ratio Test lim a n+ n a n = lim 9 n+ n (2n + 2)! (2n)! 9 n = lim 9 n 9 n (2n + 2)(2n + )(2n)! (2n)! 9 n 9 = lim n (2n + 2)(2n + ) = 0 Page 4 of 2
MTH 33 Solutions to Exam 2 April 9, 207 4. (8 points) Consider the function the power series (3x 2) n n2 n+. (a) Find the open interval of convergence for the power series. Solution: Use the Ratio Test to find open interval of convergence lim (3x 2) n+ n (n + )2 n2 n+ n+2 (3x 2) n = lim (3x 2) n (3x 2) n (n + )2 n+ 2 The series converges when The open interval of convergence is ( 0, 4 3 3x 2 2 < 3x 2 < 2 2 < 3x 2 < 2 0 < 3x < 4 0 < x < 4/3 a n+ = n2 n+ (3x 2) n = lim n (b) Test the left end point of the interval for convergence or divergence. Solution: The left end point is x = 0. ) (0 2) n n2 n+ = This series converges by the Alternating Series Test. ( 2) n n2 = n+ ( ) n (c) Test the right end point of the interval for convergence or divergence. Solution: The right end point is x = 4 3 This series is a divergent p-series, p =. (3 4 3 2)n n2 n+ = 2n 2 n n2 = n+ 2n (3x 2)n+ (n + )2 n+2 3x 2 2 = 3x 2 2 Page 5 of 2
MTH 33 Solutions to Exam 2 April 9, 207 5. (8 points) Find the 3rd degree Taylor polynomial of f(x) = (x + ) 3/2 centered at x = 3. Solution: Giving us the final answer of f(x) = (x + ) 3/2 f(3) = 8 f (x) = 3 2 (x + )/2 f (3) = 3 f (x) = 2 3 2 (x + ) /2 f (3) = 3 8 f (3) (x) = 2 2 3 2 (x + ) 3/2 f (3) (3) = 3 64 T 3 (x) = 8 + 3(x 3) + 3 8 2! (x 3)2 3 64 (x 3)3 3! Page 6 of 2
MTH 33 Solutions to Exam 2 April 9, 207 Multiple Choice. Circle the best answer. No work needed. No partial credit available. No credit will be given for choices not clearly marked. 6. (7 points) Which of the following statements is true? () If lim a n = 0, then a n converges. n (2) If (3) If a n converges, then a n diverges, then A. () only B. (2) only C. (3) only D. none a n converges absolutely. a n diverges. E. () and (3) only x( cos x) 7. (7 points) Evaluate the limit using Taylor series: lim x 0 + ln ( + x 2 ) A. /2 B. /4 C. 0 D. /3 E. 8. (7 points) Which of the following series converge? n 2 () + n (2) (3) ( ) n 9n 4 n+ 2 2 + 3 3 + 4 4 + 5 5 +... A. none B. (2) only C. (3) only D. (2) and (3) only E. () and (2) only Page 7 of 2
MTH 33 Solutions to Exam 2 April 9, 207 9. (7 points) Determine which of the following statements is true and which are false. () is convergent by the ratio test. n 2 + (2) (3) n 2 + is convergent by the limit comparison test with n. 2 n 2 + is convergent by the direct comparison test with n. 2 A. () is true; (2) and (3) are false. B. (2) is true; () and (3) are false. C. (2) and (3) are true, () is false. D. () and (2) are true; (3) is false. E. (), (2), and (3) are false. 0. (7 points) Find the sum of the series A. 2 B. 32/3 C. 4 D. 7/2 E. It diverges 2 n 3 n+ 4 n. (7 points) Suppose the power series c n (x 2) n converges when x = but diverges when x = 5. Which of the following statements are true? () The radius of convergence is R = 3. (2) c n4 n converges. (3) c n2 n converges. A. () and (2) only. B. () and (3) only. C. (2) and (3) only. D. All of them. E. None of them. Page 8 of 2
MTH 33 Solutions to Exam 2 April 9, 207 2. (7 points) The series A. x + B. x + x C. x + D. x x + x 2 E. x + ( ) n x n converges for x < to 3. (7 points) Let sin x = A. /2 B. /4 C. 0 D. /4 E. /2 ( c n x π ) n be the Taylor series for sin x at a = π 6 6. Then c 2= sin x 4. (7 points) Evaluate the indefinite integral x x 2n+ A. (2n + )(2n + )! + C B. C. D. E. ( ) n x 2n (2n)(2n)! + C ( ) n x 2n+ (2n + )(2n + )! + C ( ) n x2n (2n)! + C x 2n+ (2n + )! + C dx as a power series. Page 9 of 2
MTH 33 Solutions to Exam 2 April 9, 207 Congratulations you are now done with the exam! Go back and check your solutions for accuracy and clarity. Make sure your final answers are BOXED. When you are completely happy with your work please bring your exam to the front to be handed in. Please have your MSU student ID ready so that is can be checked. DO NOT WRITE BELOW THIS LINE. Page Points Score 2 8 3 8 4 8 5 8 6 8 7 2 8 2 9 2 Total: 53 No more than 50 points may be earned on the exam. Page 0 of 2
MTH 33 Solutions to Exam 2 April 9, 207 Integrals FORMULA SHEET PAGE Derivatives Volume: Suppose A(x) is the cross-sectional area of the solid S perpendicular to the x-axis, then the volume of S is given by d d (sinh x) = cosh x dx Inverse Trigonometric Functions: (cosh x) = sinh x dx V = b a A(x) dx d dx (sin x) = x 2 d dx (csc x) = x x 2 Work: Suppose f(x) is a force function. The work in moving an object from a to b is given by: W = b dx = ln x + C x tan x dx = ln sec x + C a f(x) dx sec x dx = ln sec x + tan x + C a x dx = ax ln a + C for a Integration by Parts: u dv = uv Arc Length Formula: v du d dx (cos x) = x 2 d dx (tan x) = d dx (sec x) = d + x 2 dx (cot x) = x x 2 + x 2 If f is a one-to-one differentiable function with inverse function f and f (f (a)) 0, then the inverse function is differentiable at a and (f ) (a) = f (f (a)) Hyperbolic and Trig Identities Hyperbolic Functions sinh(x) = ex e x 2 cosh(x) = ex + e x 2 csch(x) = sinh x sech(x) = cosh x L = b a + [f (x)] 2 dx tanh(x) = sinh x cosh x cosh 2 x sinh 2 x = sin 2 x = 2 ( cos 2x) cos 2 x = 2 ( + cos 2x) sin(2x) = 2 sin x cos x coth(x) = cosh x sinh x sin A cos B = 2 [sin(a B) + sin(a + B)] sin A sin B = 2 [cos(a B) cos(a + B)] cos A cos B = 2 [cos(a B) + cos(a + B)] Page of 2
MTH 33 Solutions to Exam 2 April 9, 207 Series nth term test for divergence: If lim a n does n not exist or if lim a n 0, then the series a n n is divergent. The p-series: and divergent if p. Geometric: If r < then FORMULA SHEET PAGE 2 is convergent if p > np ar n = a r The Integral Test: Suppose f is a continuous, positive, decreasing function on [, ) and let a n = f(n). Then (i) If (ii) If then then f(x) dx is convergent, a n is convergent. f(x) dx is divergent, a n is divergent. The Comparison Test: Suppose that a n and b n are series with positive terms. (i) If b n is convergent and a n b n for all n, then a n is also convergent. (ii) If b n is divergent and a n b n for all n, then a n is also divergent. The Limit Comparison Test: Suppose that an and b n are series with positive terms. If a n lim = c n b n where c is a finite number and c > 0, then either both series converge or both diverge. Alternating Series Test: If the alternating series ( ) n b n satisfies (i) 0 < b n+ b n (ii) lim b n = 0 n for all n then the series is convergent. The Ratio Test (i) If lim a n+ n a n = L <, then the series a n is absolutely convergent. (ii) If lim a n+ n a n = L > or lim a n+ n a n =, then the series a n is divergent. (iii) If lim a n+ n a n =, the Ratio Test is inconclusive. f (n) (0) Maclaurin Series: f(x) = n! x n Taylor s Inequality If f (n+) (x) M for x a d, then the remainder R n (x) of the Taylor series satisfies the inequality R n (x) M (n + )! x a n+ for x a d Some Power Series e x x n = n! sin x = cos x = ( ) n x2n+ (2n + )! ln( + x) = ( ) n x2n (2n)! ( ) n xn n R = R = R = R = x = x n R = Page 2 of 2