2005a_x2b_sols

2005a_x2b_sols - Spring 2005 Math 152 Exam 2B Solutions c...

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Spring 2005 Math 152 Exam 2B: Solutions Mon, 28/Mar c ± 2005, Art Belmonte For specificity, lengths are in centimeters unless stated otherwise. 1. (c) For non-STEPS folks: We have ¯ x = 3 X k = 1 m k x k 3 X k = 1 m k = ( 3 )( 1 ) + ( 5 - 2 ) + ( 7 3 ) 3 + 5 + 7 = 14 15 . (Note that this only gives you the x -coordinate of the center of mass, not the entire center of mass.) For STEPS folks: Let p = [3 , 5 , 7] be the row vector of masses and r = 12 - 25 31 be a matrix whose rows are position vectors of the points. Then m = sum ( p ) = 3 + 5 + 7 = 15 is the total mass. The center of mass is [ ¯ x , ¯ y ] = 1 m ( pr ) = 1 15 [3 - 10 + 21 , 6 + 25 + 7] = ± 14 15 , 38 15 ² . So the x -coordinate of the center of mass is ¯ x = 14 15 . ( Remark: pr represents matrix multiplication, realized by taking the dot products of rows with columns. This immediately extends to 3-D center of mass problems in Calc 3.) 2. (c) Let f ( x ) = ln x . We’ll determine K = max 2 x 5 ³ ³ f 00 ( x ) ³ ³ , then employ the Trapezoidal error estimate. Now f 0 ( x ) = 1 x = x - 1 and f 00 ( x ) =- x - 2 . Thus K = max 2 x 5 ´ 1 x 2 µ = 1 4 . Therefore, | E T | ≤ K ( b - a ) 3 12 n 2 = 1 4 ( 5 - 2 ) 3 12 ( 4 ) 2 = 27 3 ( 4 ) 4 = 9 256 . 3. (d) The improper integral converges to π/ 4. Z 1 1 x 2 + 1 dx = lim t →∞ Z t 1 1 x 2 + 1 = lim t →∞ tan - 1 x · ³ ³ ³ t 1 = lim t →∞ tan - 1 t - tan - 1 1 · = π 2 - π 4 = π 4 4. (a) The step size is h = b - a n = 2 - 1 4 = 1 4 . Partition points are n 1 , 5 4 , 3 2 , 7 4 , 2 o . The Trapezoidal Rule gives T n = step size × (
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This note was uploaded on 03/25/2008 for the course MATH 152 taught by Professor Teitler during the Spring '08 term at Texas A&M.

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2005a_x2b_sols - Spring 2005 Math 152 Exam 2B Solutions c...

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