MATH 2433-006 Exam II ANSWERS INSTRUCTIONS Show your work on all problems. 1. Find a power series representation for the following function. [Use the geometric series.] 2 π (π₯) = 3−π₯ Answer: The geometric series is ∞ ∑ 1 = ππ for β£ π β£< 1 1 − π π=0 So ∞ ∞ 2 2 1 2∑ π₯ π ∑ 2 π = ( )= ( ) = π₯ 3−π₯ 3 1 − (π₯/3) 3 π=0 3 3π+1 π=0 2. Find the Taylor series for π (π₯) centered at the given value of π. π (π₯) = ππ₯ , π = 3 Answer: The Taylor series is ∞ ∑ π (π) (π) (π₯ − π)π π! π=0 Since for π (π₯) = ππ₯ , π (π) (π₯) = ππ₯ for all π, the Taylor series is ∞ ∑ π3 (π₯ − 3)π π! π=0 3. Evaluate the following indeο¬nite integral as an inο¬nite series ∫ π₯ cos(π₯3 )ππ₯ using cos π₯ = ∞ ∑ (−1)π π=0 1 π₯2π (2π)! Answer: From the given series for the cosine, π₯ cos(π₯3 ) = ∞ ∑ π₯(−1)π π=0 ∞ ∑ (π₯3 )2π π₯6π+1 = (−1)π (2π)! (2π)! π=0 Thus ∫ 3 π₯ cos(π₯ )ππ₯ = ∞ ∫ ∑ π=0 ππ₯ (−1) 6π+1 ππ₯ = πΆ + (2π)! ∞ ∑ (−1)π π=0 π₯6π+2 (6π + 2)(2π)! 4. Approximate π by a Taylor polynomial with degree π at the number π: π (π₯) = π₯2/3 , π = 1, π = 3 Answer: π (π₯) and its ο¬rst three derivatives are π (0) (π₯) = π₯2/3 π (1) (π₯) = (2/3)π₯−1/3 π (2) (π₯) = (−1/3)(2/3)π₯−4/3 π (3) (π₯) = (−4/3)(−1/3)(2/3)π₯−7/3 (1) so π (0) (1) = 1 π (1) (1) = 2/3 π (2) (1) = −2/9 π (3) (1) = 8/27 thus 1 1 1 (π₯ − 1) + (−2/9) (π₯ − 1)2 + (8/27) (π₯ − 1)3 1! 2! 3! 2 1 4 2 3 = 1 + (π₯ − 1) − (π₯ − 1) + (π₯ − 1) 3 9 81 π3 (π₯) = 1 + (2/3) 5. Eliminate the parameter to ο¬nd a Cartesian equation of the curve π₯ = π2π‘ , π¦ = π‘ + 1 Answer: Since π2π‘ = π₯, 2π‘ = ln π₯ so π‘ = 1 2 2 ln π₯ and so π¦ = 1 2 ln π₯ + 1 6. Find an equation of the tangent line to the curve at the point corresponding to the given value of the parameter π₯ = π‘4 + 1, π¦ = π‘3 + π‘, π‘ = −1 Answer: when π‘ = −1, π₯ = 2 and π¦ = −2. Also ππ₯ = 4π‘3 ππ‘ ππ¦ = 3π‘2 + 1 ππ‘ so for π‘ = −1 ππ₯ = −4 ππ‘ ππ¦ =4 ππ‘ So the equation of the tangent line is 4(π₯ − 2) − (−4)(π¦ − −2) = 0, or π¦ = −π₯ 7. Find the exact length of the curve π₯ = 1 + 3π‘2 , π¦ = 4 + 2π‘3 , 0 ≤ π‘ ≤ 1 Answer: The length is given by the integral ∫ 1 √ ( 0 ππ₯ 2 ππ¦ 2 ) + ) ππ‘ ππ‘ ππ‘ Since here ππ₯ = 6π‘ ππ‘ ππ¦ = 6π‘2 ππ‘ the integral is 1 ∫ ∫ √ 2 4 36π‘ + 36π‘ ππ‘ = 6 0 2 √ 1 + π‘2 π‘ππ‘ 0 1 2 ππ’ If π’ = 1 + π‘ so π‘ππ‘ = ∫ √ 1 then 1+ π‘2 π‘ππ‘ ∫ = 1 1 π’1/2 ππ’ = π’3/2 + πΆ 2 3 3 so ∫ 6 0 1 √ 1 1 + π‘2 π‘ππ‘ = 6 (1 + π‘2 )3/2 β£10 = 2(23/2 − 1) 3 8. Set up an integral that represents the exact area of the surface obtained by rotating the given curve about the π₯-axis π₯ = π‘3 , π¦ = π‘2 , 0 ≤ π‘ ≤ 1 Answer: The area is given by the integral formula √ π ∫ 2ππ¦ ( π ππ₯ 2 ππ¦ 2 ) + ) ππ‘ ππ‘ ππ‘ Here ππ₯ = 3π‘2 ππ‘ ππ¦ = 2π‘ ππ‘ so the formula becomes ∫ 1 2ππ‘2 √ 9π‘4 + 4π‘2 ππ‘ 0 9. Find the slope of the tangent line to the given polar curve at the point speciο¬ed by the value of π π π = 2 sin π, π = 6 Answer: π₯ = π cos π = 2 sin π cos π and π¦ = π sin π = 2 sin2 π. Then ππ₯ = 2(− sin π sin π + cos π cos π) ππ and ππ¦ = 4 sin π cos π ππ √ √ For π = (π/6), cos π = ( 3/2) and sin π = (1/2) so ππ₯/ππ = 1 and ππ¦/ππ = 3 and the tangent slope is given by √ ππ¦ ππ¦ 3 ππ = ππ₯ = ππ₯ 1 ππ 10. Find the area of the region that is bounded by the given curve and lies in the speciο¬ed sector π = sin π, π/3 ≤ π ≤ 2π/3 4 Answer: The area is given by the integral formula ∫ π½ πΌ which here is 2π 3 ∫ π 3 1 2 π ππ 2 1 (sin π)2 ππ 2 using sin2 π = 1 − cos(2π) 2 the integral becomes 2π 3 ∫ π 3 2π 1 1 1 (1 − cos(2π))ππ = π − sin(2π)β£ π3 3 4 4 8 which gives 1 2π 1 4π π 1 2π ( − sin( ) − + sin( )) 4 3 2 3 3 2 3 5