solution to 2-3-13

# solution to 2-3-13 - ENGR 200 HW 28 Ryan Franz Part A: L...

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ENGR 200 Ryan Franz HW 28

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Part A: The formula to use here is ( ) ( ) ( ) 0 L Nx dx ExAx δ = (worth 1 point) ( ) Nx P = (constant) (worth 1 point) ( ) Ex E = (constant) (worth 1 point) A(x) is a linear function with form y mx C = + where ( ) ( ) 21 yy m xx = ( ) ( ) 1 2 0 A bt A L = = ( ) ( ) 0 bt bt m L = ( ) 1 0 C A = ( ) ( ) ( ) ( ) 2 1 11 0 b b Ax xb tt LL −−  = +=   (worth 4 points) ( ) 0 1 1 L P dx Et bb L = + This integral is of the form ( ) 1 ln du u u = . ( ) ( ) 1 0 ln L PL Et b b L  = +   Note that the ( ) L expression is added by inspection to account for the chain rule. This is a great place to differentiate in order to check your work. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ln ln 0 ln ln L Lb b E tbb L L PL Et b b  = +− +   = Since ( ) ( ) ln ln ln Q QR R −= ,
( ) 2 21 1 ln b PL Et b b b δ  =   (worth 2 points – since this was given in the problem, a logical progression to the answer
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## This note was uploaded on 09/18/2011 for the course ENGR 200 taught by Professor Mullen during the Spring '08 term at Case Western.

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solution to 2-3-13 - ENGR 200 HW 28 Ryan Franz Part A: L...

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