exam 2 sol'n

exam 2 sol'n - C HECKMAN Solutions 12:15 class MAT 266 Test...

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C. HECKMAN MAT 266 Solutions, 12:15 class Test 2 (1) [10 points] Use a convergence test to determine whether X n =0 7 n · (2 n 2 + 1) ( n + 1)! converges. Solution: If a n = 7 n · (2 n 2 + 1) ( n + 1)! , then | a n | = a n , and so L = lim n + | a n +1 | | a n | = lim n + 7 n +1 · ( 2( n + 1) 2 + 1 ) ( n + 2)! · ( n + 1)! 7 n · (2 n 2 + 1) = lim n + 7 n + 2 · 2 n 2 + 4 n + 3 2 n 2 + 1 = 0 · 1 < 1 , so the Ratio Test states that the series converges. Grading: +3 points for setting up the Ratio Test, +2 points for cancelling, +2 points for taking the limit, +3 points for the Ratio Test’s conclusion. Grading for partial credit: - 1 point for ( n + 1)! ( n + 2)! = 1 n + 1 ; +8 points (total) for bad can- cellation with the correct result (assuming the cancellation was the correct way); +5 points (total) for bad cancellation with the incorrect result (even assuming the cancellation was the correct way). 1
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(2) Let R be the region bounded by the curves y = x 5 and x = y 10 . (a) [10 points] Find the area of R . [0.742] Solution: Either Z LEFT RIGHT (TOP - BOTTOM) dx = Z 1 0 x 1 / 10 - x 5 dx or Z TOP BOTTOM (RIGHT - LEFT) dy = Z 1 0 y 1 / 5 - y 10 dy , both of which equal 49 66 . Grading: +3 points for the form, +3 points for finding the limits, +4 points for substitution. (b) [15 points] Find the centroid of R . [(0.449,0.500)] Solution: x, ¯ y ) = Z 1 0 x ( x 1 / 10 - x 5 ) dx Z 1 0 x 1 / 10 - x 5 dx , Z 1 0 1 2 h ( x 1 / 10 ) 2 - ( x 5 ) 2 i dx Z 1 0 x 1 / 10 - x 5 dx = ± 22 49 , 1 2 ² . Grading: +5 points for each integral. (c) [10 points] The region R is rotated around the line y = 1. Find the volume of the resulting solid, stating which method you used. [2.332] Solution: If the washer method was used, the integral is Z RIGHT LEFT πR 2 - πr 2 dx = Z 1 0 π ( (1 - x 5 ) 2 - (1 - x 1 / 10 ) 2 ) dx, and if the cylindrical shell method was used, the integral is Z TOP BOTTOM 2 πR (RIGHT - LEFT) dy = Z 1 0 2 π (1 - y )( y 1 / 5 - y 10 ) dy, both of which equal 49 π 66 . Grading: +3 points for the form, +4 points for substitution, +3 points for the name of the method. Grading for common mistakes: - 2 points for the incorrect method. = 2
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R is submerged in water so that the upper tip is one meter below the surface of the water, and the plate is vertical. Find the hydrostatic force acting on the plate. [10 . 913 · 10 3 N] Solution: The easiest place to put y = 0 is at the bottom of the plate, since it is described using equations. Then, the depth of the water above height y is 2 - y , and the width of the plate at height y is y 1 / 5 - y 10 . The hydrostatic force is then Z TOP BOTTOM g · ± Density of liquid ² · ± Distance from y to water level ² · ± Width of plate at height y ² dy or Z 1 0 (9 . 8)(1000)(2 - y )( y 1 / 5 - y 10 ) dy = 120050 11 . Grading:
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This note was uploaded on 10/05/2008 for the course MAT 266 taught by Professor Farisodish during the Spring '08 term at ASU.

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exam 2 sol'n - C HECKMAN Solutions 12:15 class MAT 266 Test...

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