Quiz-2-f10_Sol

# Quiz-2-f10_Sol - AMS 361: Applied Calculus IV by Prof. Y....

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Unformatted text preview: AMS 361: Applied Calculus IV by Prof. Y. Deng Quiz 2 Thursday (10/28/2010) at 5:20-6:40PM Open Book and Open Own Lecture Notes but no Electronics (1) Do Any Two of the Three Problems; Mark the Two Problems You Attempt; Will Only Grade the First Two if You Don’t Mark the Problems You Attempted; (2) Each Problem Is Worth 5 Points; (3) No Points for Guessing Work and for Solutions without Appropriate Intermediate Steps; Partial Credits are given only for Steps that are Relevant to the Solutions. SB ID: Name: Problems Q2-1 Q2-2 Q2-3 Total Score To Grade? Class ID: Score Remarks Q2-1 (5 Points): Find the general solution of the following non-homogeneous DE: First, solve the characteristic equation ( )( ) Second, find the particular solution. Trial solution () () () () ( () () ) () () ( ) ( ( ( () () ) ) ( )) () 1 () () () () ( ) Q2-2 (5 Points): Find the general solution of DE: ( is a real constant, and ) () where integer , is the usual derivative operator. First, solve the characteristic equation: ( ( ) ) () We have got 2010 identical solutions. In order to get the general solution, we must get 2010 linearly independent solutions. To do so, we prove that () ( ) satisfy the original DE and are linearly independent. To prove the satisfaction, we may use induction method. For For we have we have () () ( ( Assume the equation holds for ( ) ) {( , i.e., () ( The final general solution is () (∑ ) ( ) ( ) )* () ( )+ ( ( ) ) and it’s obvious. ), () )* ( ( )* )+} ( ( ) )+ *+ We can prove the equation also holds for as follows, i.e., 2 ) jumps out of a horizontally flying plane of speed at height . The horizontal component of the jumper’s initial velocity is equal to that of the plane and the jumper’s vertical component of initial velocity is The air resistance is ⃗ Q2-3 (5 Points): A guy (of body mass with the parachute open where is a given constant and ⃗ √ is the magnitude of jumper’s velocity. Please compute the total horizontal distances the guy has moved from where he left the plane to the points he lands if he the parachute at heights and . You may express your solution implicitly. For ) Before opening √ () √ After opening: For vertical direction: () (√ ) () ∫ () (√ ) (√ ) () For horizontal direction: () ( ) Compute the horizontal distance 3 √ ∫ For ) Before opening: √ () √ After opening: For vertical direction: () (√ ) () ∫ () (√ ) (√ ) () For horizontal direction: () Compute the horizontal distance ( ) 4 √ ∫ 5 ...
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## This note was uploaded on 01/31/2011 for the course AMS 361 taught by Professor Staff during the Fall '08 term at SUNY Stony Brook.

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