Exam 3 Solutions (Form A) Fall 2000.pdf

Student (Print) Section Last, First Middle Student (Sign) Student ID Instructor MATH 152 Exam 3 Fall 2000 Test Form A Solutions Part I is multiple choice. There is no partial credit. Part II is work out. Show all your work. Partial credit will be given. You may not use a calculator. 1-9 /45 10 /10 11 /15 12 /10 13 /10 14 /10 TOTAL 1 Part I: Multiple Choice (5 points each) There is no partial credit. You may not use a calculator. 1. Compute G21 nG1D0 G2E 3 n 5 2G22n a. 25 3 b. 125 3 c. 125 2 correctchoice d. 75 2 e. G2E Geometric: a G1D 3 0 5 2 G1D 25, r G1D 3 5 G1C 1 S G1D 25 1 G22 3 5 G1D 125 2 2. The series G21 nG1D1 G2E G9FG221GA0 n n is a. absolutely convergent b. convergent but not absolutely convergent correctchoice c. divergent d. none of the above The series is convergent by the Alternating Series Test since it is (1) alternating due to the G9FG221GA0 n ,(2) decreasing in absolute value since 1 n gets smaller, and lim nG76G2E 1 n G1D 0. However it is not absolutely convergent because the related absolute series is G21 nG1D1 G2E 1 n which is the harmonic series which is divergent. 3. Determine the interval of convergence of the power series G21 nG1D0 G2E G9Fx G0E 3GA0 n n G0E 1 . a. G9F2,4GA2 b. GA12,4GA0 c. G9FG224,G222GA2 d. GA1G224,G222GA0 correctchoice e. G9FG22G2E,G2EGA0 Ratio Test: L G1D lim nG76G2E a nG0E1 a n G1D lim nG76G2E |x G0E 3| nG0E1 n G0E 2 n G0E 1 |x G0E 3| n G1D |x G0E 3| G1C 1 or G224 G1C x G1C G222 At x G1D G224 the series is alternating and decreasing and so convergent. At x G1D G222 the series is harmonic and so divergent. So the interval of convergence is G224 G74 x G1C G222. 2 4. Find the Maclaurin series for fG9FxGA0 G1D sin2x. a. 2x G22 2 3 x 3 3! G0E 2 5 x 5 5! G22 2 7 x 7 7! G0E G43 correctchoice b. 2x G22 2x 3 3! G0E 2x 5 5! G22 2x 7 7! G0E G43 c. x G22 x 3 3! G0E x 5 5! G22 x 7 7! G0E G43 d. x G0E x 3 3! G0E x 5 5! G0E x 7 7! G0E G43 e. 2x G0E 2x 3 3! G0E 2x 5 5! G0E 2x 7 7! G0E G43 The Maclaurin series for sinx is sinx G1D x G22 x 3 3! G0E x 5 5! G22 x 7 7! G0E G43 Substitute x G76 2x: sin2x G1D 2x G22 2 3 x 3 3! G0E 2 5 x 5 5! G22 2 7 x 7 7! G0E G43 5. Find the 3 rd degree Taylor polynomial for fG9FxGA0 G1D e G222x about x G1D 1. a. e G222 G0E 2e G222 G9Fx G22 1GA0 G0E 2e G222 G9Fx G22 1GA0 2 G0E 4 3 e G222 G9Fx G22 1GA0 3 b. e G222 G22 2e G222 G9Fx G22 1GA0 G0E 2e G222 G9Fx G22 1GA0 2 G22 4 3 e G222 G9Fx G22 1GA0 3 correctchoice c. e G222 G0E 2e G222 G9Fx G22 1GA0 G0E 4e G222 G9Fx G22 1GA0 2 G0E 8e G222 G9Fx G22 1GA0 3 d. e G222 G22 2e G222 G9Fx G22 1GA0 G0E 4e G222 G9Fx G22 1GA0 2 G22 8e G222 G9Fx G22 1GA0 3 e. e G222 G22 2e G222 G9Fx G22 1GA0 G22 4e G222 G9Fx G22 1GA0 2 G22 8e G222 G9Fx G22 1GA0 3 The function, 3 derivatives and their values at x G1D 1 are fG9FxGA0 G1D e G222x f G55 G9FxGA0 G1D G222e G222x f G55G55 G9FxGA0 G1D 4e G222x f G55G55G55 G9FxGA0 G1D G228e G222x fG9F1GA0 G1D e G222 f G55 G9F1GA0 G1D G222e G222 f G55G55 G9F1GA0 G1D 4e G222 f G55G55G55 G9F1GA0 G1D G228e G222 So the Taylor polynomial is T 3 G1D fG9F1GA0 G0E f G55 G9F1GA0G9Fx G22 1GA0 G0E f G55G55 G9F1GA0 2 G9Fx G22 1GA0 2 G0E f G55G55G55 G9F1GA0 6 G9Fx G22 1GA0 3 G1D e G222 G22 2e G222 G9Fx G22 1GA0 G0E 4e G222 2 G9Fx G22 1GA0 2 G22 8e G222 6 G9Fx G22 1GA0 3 6. Compute G21 nG1D0 G2E n 1 2 nG221 . HINT: Differentiate G21 nG1D0 G2E x n . a. G224 b. G222 c. 4 9 d. 2 e. 4 correctchoice G21 nG1D0 G2E x n G1D 1 1 G22 x provided |x| G1C 1 Differentiate: G21 nG1D0 G2E nx nG221 G1D 1 G9F1 G22 xGA0 2 provided |x| G1C 1 Evaluate at x G1D 1 2 : G21 nG1D0 G2E n 1 2 nG221 G1D 1 1 G22 1 2 2 G1D 4 3 7. Find a unit vector which is orthogonal to the plane containing the points P G1D G9F1,0,0GA0, Q G1D G9F0,2,0GA0, and R G1D G9F0,0,3GA0. a. 6 11 , G223 11 , 2 11 b. G9F6,G223,2GA0 c. G9F6,3,2GA0 d. 6 7 , G223 7 , 2 7 e. 6 7 , 3 7 , 2 7 correctchoice PQ G1D Q G22 P G1D G9FG221,2,0GA0, PR G1D R G22 P G1D G9FG221,0,3GA0 N G1D PQ G95 PR G1D ijk G22120 G22103 G1DG9F6,3,2GA0 N G1D 36 G0E 9 G0E 4 G1D 49 G1D 7 N N G1D 6 7 , 3 7 , 2 7 8. Find the equation of the circle for which one diameter has endpoints P G1D G9F1,1GA0 and Q G1D G9F7,9GA0. a. G9Fx G0E 3GA0 2 G0E G9Fy G0E 4GA0 2 G1D 25 b. G9Fx G0E 3GA0 2 G0E G9Fy G0E 4GA0 2 G1D 5 c. G9Fx G22 4GA0 2 G0E G9Fy G22 5GA0 2 G1D 25 correctchoice d. G9Fx G22 3GA0 2 G0E G9Fy G22 4GA0 2 G1D 5 e. G9Fx G0E 4GA0 2 G0E G9Fy G0E 5GA0 2 G1D 25 The center is at C G1D P G0E Q 2 G1D G9F4,5GA0 and the radius is r G1D CP G1D |G9F3,4GA0| G1D 5. 9. Consider the triangle with vertices A G1D G9F0,0,0GA0, B G1D G9F1,1,0GA0 and C G1D G9F0,2,2GA0. Find the angle at A. a. 0° b. 30° c. 45° d. 60° correctchoice e. 90° AB G1D 1 G0E 1 G1D 2 AC G1D 4 G0E 4 G1D 8 G1D 22 AB G19 AC G1D 2 cosG32 G1D AB G19 AC AB AC G1D 2 2 22 G1D 1 2 G32G1D60° 4 Part II: Work Out Show all your work. Partial credit will be given. You may not use a calculator. 10. (10 points) a. Find the power series representation for fG9FxGA0 G1D 1 1 G0E x 2 centered at x G1D 0 and its radius of convergence. 1 1 G22 x G1D G21 nG1D0 G2E x n provided |x| G1C 1 Substitute x G76 G22x 2 1 1 G0E x 2 G1D G21 nG1D0 G2E G9FG22x 2 GA0 n G1D G21 nG1D0 G2E G9FG221GA0 n x 2n provided |x| G1C 1 So R G1D 1 b. Find the power series representation for fG9FxGA0 G1D xtan G221 x centered at x G1D 0 and its radius of convergence. HINT: tan G221 x G1D G3B 0 x 1 1 G0E t 2 dt tan G221 x G1D G3B 0 x 1 1 G0E t 2 dt G1D G3B 0 x G21 nG1D0 G2E G9FG221GA0 n t 2n dt G1D G21 nG1D0 G2E G9FG221GA0 n x 2nG0E1 2n G0E 1 and R G1D 1 fG9FxGA0 G1D xtan G221 x G1D G21 nG1D0 G2E G9FG221GA0 n x 2nG0E2 2n G0E 1 and R G1D 1 5 11. (15 points) Determine if each of the following series converges or diverges. Say why. Be sure to name or quote the test(s) you use and check out all requirements of the test. a. G21 nG1D2 G2E G9FG221GA0 n nG9FlnnGA0 2 Circle one: Converges Diverges Explain: The related absolute series is G21 nG1D2 G2E 1 nG9FlnnGA0 2 . Apply the Integral Test: 1 nG9FlnnGA0 2 is continuous, decreasing and G3B 2 G2E 1 nG9FlnnGA0 2 G1D G221 lnn 2 G2E G1D 0 G22 G221 ln2 G1D 1 ln2 So G21 nG1D2 G2E G9FG221GA0 n nG9FlnnGA0 2 also converges. b. G21 nG1D0 G2E G9FG223GA0 n n! Circle one: Converges Diverges Explain: Apply Alternating Series Test: It is alternating because of the G9FG223GA0 n . It is decreasing in absolute value because 3 n n! gets smaller as n gets larger provided n G1E 3. Finally, lim nG76G2E 3 n n! G1D lim nG76G2E 3 G19 3 G19 3 G19G43G193 1G192G193G19G43G19n G1D0 because beyond n G1D 3 we keep multiplying by fractions less than 1.SoG21 nG1D0 G2E G9FG223GA0 n n! converges. OR Apply the Ratio Test: |a n | G1D 3 n n! |a nG0E1 | G1D 3 nG0E1 G9Fn G0E 1GA0! L G1D lim nG76G2E a nG0E1 a n G1D lim nG76G2E 3 nG0E1 G9Fn G0E 1GA0! n! 3 n G1D lim nG76G2E 3 n G0E 1 G1D 0 G1C 1 So G21 nG1D0 G2E G9FG223GA0 n n! converges. c. G21 nG1D1 G2E G9FG221GA0 nG0E1 5 nG221 n 2 4 n Circle one: Converges Diverges Explain: Apply the Ratio Test: |a n | G1D 5 nG221 n 2 4 n |a nG0E1 | G1D 5 n G9Fn G0E 1GA0 2 4 nG0E1 L G1D lim nG76G2E a nG0E1 a n G1D lim nG76G2E 5 n G9Fn G0E 1GA0 2 4 nG0E1 n 2 4 n 5 nG221 G1D 5 4 lim nG76G2E n 2 G9Fn G0E 1GA0 2 G1D 5 4 G1E 1 So G21 nG1D0 G2E G9FG223GA0 n n! diverges. 6 12. (10 points) Determine if each of the following series converges or diverges. Say why. Be sure to name or quote the test(s) you use and check out all requirements of the test. If it converges, find the sum. If it diverges, does it diverge to G0EG2E, G22G2E or neither? a. G21 nG1D1 G2E 1 1 G0E 3 n Circle one: Converges Diverges Explain: G21 nG1D1 G2E 1 3 n diverges because it is a p-series with p G1D 1 3 G1C 1. Apply the Limit Comparison Test: a n G1D 1 1 G0E 3 n b n G1D 1 3 n L G1D lim nG76G2E a n b n G1D lim nG76G2E 3 n 1 G0E 3 n G1D 1 and 0 G1C L G1C G2E So G21 nG1D1 G2E 1 1 G0E 3 n also diverges. Since 1 1 G0E 3 n G1E 0, it diverges to G0EG2E. b. G21 nG1D1 G2E n G22 1 n G22 n n G0E 1 Circle one: Converges Diverges Explain: S k G1D G21 nG1D1 k n G22 1 n G22 n n G0E 1 G1D 0 G22 1 2 G0E 1 2 G22 2 3 G0E G43 G0E k G22 1 k G22 k k G0E 1 G1D G22k k G0E 1 S G1D lim kG76G2E S k G1D lim kG76G2E G22k k G0E 1 G1D G221 7 13. (10 points) Find the volume of the parallelepiped with adjacent edges PQ, PR and PS, where P G1D G9F1,1,1GA0, Q G1D G9F2,3,1GA0, R G1D G9F4,1,5GA0 and S G1D G9F1,3,4GA0. PQ G1D Q G22 P G1D G9F1,2,0GA0, PR G1D R G22 P G1D G9F3,0,4GA0 and PS G1D S G22 P G1D G9F0,2,3GA0 V G1D PQ G19 PR G95 PS G1D det 120 304 023 G1D|G228G2218| G1D 26 14. (10 points) Determine the interval of convergence of the power series G21 nG1D1 G2E G9Fx G22 2GA0 n n 1/3 3 n . Be sure to show how you checked the convergence at the endpoints. Apply the Ratio Test: |a n | G1D |x G22 2| n n 1/3 3 n |a nG0E1 | G1D |x G22 2| nG0E1 G9Fn G0E 1GA0 1/3 3 nG0E1 L G1D lim nG76G2E a nG0E1 a n G1D lim nG76G2E |x G22 2| nG0E1 G9Fn G0E 1GA0 1/3 3 nG0E1 n 1/3 3 n |x G22 2| n G1D |x G22 2| 3 lim nG76G2E n n G0E 1 1/3 G1D |x G22 2| 3 The series converges when L G1D |x G22 2| 3 G1C 1 or |x G22 2| G1C 3 or G22 1 G1C x G1C 5 We check the endpoint at x G1D G221. The series is G21 nG1D1 G2E G9FG223GA0 n n 1/3 3 n G1D G21 nG1D1 G2E G9FG221GA0 n n 1/3 which is convergent because it is an alternating, decreasing series. We check the endpoint at x G1D 5. The series is G21 nG1D1 G2E G9F3GA0 n n 1/3 3 n G1D G21 nG1D1 G2E 1 n 1/3 which is a divergent p-series since p G1D 1 3 G1C 1. Thus the interval of convergence is G221 G74 x G1C 5 or GA1G221,5GA0. 8