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Math_137_Winter_2010_Solution_5

# Math_137_Winter_2010_Solution_5 - Math 137 Winter 2010...

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Math 137 Winter 2010 Assignment 5 Due Friday, February 26 All solutions must be clearly stated and fully justified. Use the format given on UW-Ace under Content, in the folder Assignments; it is the file Math 137 Assignment Templates . Text problems: Section 3.4: 18, 28, 38, 48, 76, 80 Section 3.5: 8, 18, 28, 42, 54, 63, 64a Section 3.6: 4, 14, 24, 34, 44, 54 Section 3.4: 18. Find the derivative of h(t) = (t 4 – 1) 3 (t 3 + 1) 4 28. Find the derivative of u u u u e e e e y + = 2 2 2 2 2 2 2 ) ( 4 ) ( ) 2 ( ) 2 ( ) ( ) )( ( ) )( ( u u u u u u u u u u u u u u u u u u e e e e e e e e e e e e e e e e e e y + = + + + + = + + + = 38. Find the derivative of y = e k tan x x x k e y x k 2 1 sec 2 tan = 48. Find the first and second derivatives of y = xe cx y = cxe cx + e cx = (cx + 1) e cx y = c(cx + 1) e cx + ce cx = (c 2 x + 2c) e cx 76. Find the 1000 th derivative of x e -x . By 48, we detect a pattern: the nth derivative of y = xe cx is (c n x + nc) e cx . If c = – 1 and n = 1000, then the 1000 th derivative of x e -x is ((– 1) 1000 x + (1000(– 1)) e -x = (x – 1000) e -x 80. A model for the length of daylight (in hours) in Philadelphia on the t th day of the year is L(t) = 12 + 2.8 sin[(2 π /365)(t – 80)]. Use this model to compare how the number of hours of daylight is increasing in Philadelphia on March 21 and May 21. L (t) = 2.8 cos[(2 π /365)(t – 80)] (2 π /365) On March 21 t = 80; L (80) = 2.8 cos[(2 π /365)(80 – 80)] (2 π /365) = 2.8 cos(0) (2 π /365) = 2.8(2 π /365) 0.042 hours per day. On May 21 t = 141; L (141) = 2.8 cos[(2 π /365)(141 – 80)] (2 π /365) 0.02398 hours per day.

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Math_137_Winter_2010_Solution_5 - Math 137 Winter 2010...

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