MATH111-200630-PS08-Solutions

MATH111-200630-PS08-Solutions - MATH 111 Problem Set 8...

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MATH 111 Problem Set 8 Solutions DRAFT Edward Doolittle November 23, 2006 1. (a) Taking the derivative, we have y x 1 4 x Squaring, we have y 2 x 2 1 2 1 16 x 2 It follows that 1 y 2 x 2 1 2 1 16 x 2 x 1 4 x 2 so L 4 2 x 1 4 x 2 dx 4 2 x 1 4 x dx x 2 2 ln x 4 4 2 8 1 4 ln4 2 1 4 ln2 6 1 4 ln2 (b) The hard way. We have y 2 1 2 4 x 1 2 4 2 x 1 so L 2 0 1 1 x dx Now we make the rationalizing substitution u 1 1 x , du 1 2 1 1 x 1 2 1 x 2 dx , 2 udu u 1 2 dx to obtain 1 1 x dx u 2 u u 1 2 du 2 u 2 u 2 2 u 1 du 2 4 u 2 u 1 2 du We continue with the partial fractions decomposition, etc. The easy way. We integrate with respect to y instead of x . We have dx dy y 2 so 1 dx dy 2 1 y 2 4 1
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so 1 dx dy 2 dy 1 y 2 4 dy Making the substitution y 1 2 tan θ , dy 1 2 sec 2 θ d θ we have 1 dx dy 2 dy 1 tan 2 θ 1 2 sec 2 θ d θ 1 2 sec 3 θ d θ We then look up the integral for sec 3 θ in a table and proceed from there. (c) We have y d dx ln e x 1 d dx ln e x 1 e x e x 1 e x e x 1 e 2 x e x e 2 x e x e x 1 e x 1 2 e x e 2 x 1 Squaring, y 2 4 e 2 x e 2 x 1 2 4 e 2 x e 4 x 2 e 2 x 1 so 1 y 2 e 4 x 2 e 2 x 1 4 e 2 x e 4 x 2 e 2 x 1 e 4 x 2 e 2 x 1 e 4 x 2 e 2 x 1 e 2 x 1 2 e 2 x 1 2 e 2 x 1 e 2 x 1 (assuming x 0; if x 0, a slight change to the final equality is required; what do you think that might be?). It follows that the arc length is L b a e 2 x 1
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