Homework 5 solutions

Homework 5 solutions - AMS 361: Applied Calculus IV by Prof...

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Unformatted text preview: AMS 361: Applied Calculus IV by Prof Y. Deng Homework 5 Assignment Date: Collection Date: Grade: Wednesday (03/11/2009) Wednesday (03/18/2009) Each problem is worth 10 points. Problem 5.1 (Prob. 27, P. 107, Modified) Suppose that a projectile is fired straight upward from the surface of the earth with initial velocity v0 2GM / R . Then its height y(t) above the surface satisfies the initial value problem d2y GM , y(0) 0 , y(0) v0 . Compute the maximum altitude the projectile 2 dt ( y R) 2 reaches and the time (from launch) for the projectile to reach the maximum altitude. Solution: The maximum altitude ( ): 2 2 2 = (+)2 ; = - ; = ( +)2 0 - = = ( ) 2 = 0 - ( + )2 ()2 0 2 - = - 2 2 + => ()2 =0 2 - ( 2 +) 2 = 0 => = (2- 2 - 1) 0 The time (from launch) for the projectile to reach the maximum altitude: = = 0 2 - 2 , ( + ) 1 1 = 0 = 0 0 1 , where T is the time the projectile going up. 0 1 0 2 - 2 + = , Let = + , = + 2 2-0 2 -0 0 2 -2 =( 0 )2 - 0 2 0 2 - 2 2 -2 3 2 Problem 5.2 (Prof Deng's fun problem, modified) On a straight stretch of road of length there were two stop signs at the two ends. Assume you follow traffic laws: stop at these signs. Your poor car can accelerate at a constant 1 and decelerate at another constant 2 . Compute the shortest time to travel from one stop sign to another if there is no speed limit in between. Do it again if the speed limit between the signs is . In both cases, calculate the speed of your car at every point during the stretch of road of length L. Solution: Let 1 be the time you accelerate, 2 be the time you decelerate. 1 + 2 = , which is the time you travel from one stop sign to another. Note that 1 1 = 2 2 since your initial velocity and final velocity are both zero. So, 1 1 2 + 2 2 2 = L, plug in 2 = 1 = If 1 1 2 2 1 2 + 1 2 1 2 1 2 1 2 1 , we have 2 1 1 2 + 2 1 1 1 2 2 1 = L 2 , 2 = 2 1 1 2 + 2 2 , , 1 + 2 = = 2 2 , the result stays the same. If < 1 1 , then you should accelerate to (need time 1 ), then travel a constant velocity for time 3 , finally decelerate from to velocity 0 (need time 2 ). And total time = 1 + 2 + 3 . 1 = 3 = 1 3 1 = = 2 1 2 2 1 , 2 = 2 2 = 2 2 2 , 3 = - 1 - 2 1 2 2 - - 2 2 , and = 1 + 2 + 3 = + 2 + 2 2 - - 2 1 2 2 3 In both cases, calculate the speed of your car at every point during the stretch of road of length L. If 1 1 = 1 If < 1 1 , = 1 = 2 2 2 1 2 + 1 2 2 2 1 2 2 2 , , at every point at 1 , at 2 , 2 = 3 ( - 2 2 1 - 2 2 2 ) at 3 Problem 5.3 (Prob 34, P. 156, 3E) Find the general solution of the following equation: + 2 - 15 = 0 Solution: The characteristic equation is 2 + 2 - 15 = 0 D= 3 or -5 So 1 = 3 , 2 = -5 Therefore, = 1 3 + 2 -5 , 1 , 2 : Problem 5.4 (Prob 41, P. 156, 3E) Find the general solution of the following equation: 6 - 7 - 20 = 0 Solution: The characteristic equation is 6 2 - 7 - 20 = 0 4 5 D= - or 3 So 1 = 2 4 - 3 , 2 = 2 - 4 3 5 Therefore, = 1 + 2 2 , 1 , 2 : 5 Problem 5.5 (Prob 53, P. 157, 3E) Find the general solution of the following equation: 2 + 2 - 12 = 0 2 2 Solution: Let = ln => = 2 => = , 2 = - 2 + 2 2 1 1 2 So, - + 2 +2 - 12 = 0 => 2 + - 12 = 0 The characteristic equation is D= 3 or -4 2 + - 12 = 0 So 1 = 3 , 2 = -4 Therefore, = 1 3 + 2 -4 , 1 , 2 : Problem 5.6 (Prob 1, P. 167, 3E) Check if the following three functions are linearly independent: 2, 3 2 , 5 - 8 2 Solution: Let 1 = 15, 2 = -16, 3 = -6, then we have 15*2 - 16 3 2 -6 (5 - 8 2 ) = 0 So these functions are linearly dependent. 3 ...
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This note was uploaded on 05/11/2009 for the course AMS 361 taught by Professor Staff during the Spring '08 term at SUNY Stony Brook.

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