MATH111-200630-PS08-Solutions

Math111 200630 ps08 solutions

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Unformatted text preview: MATH 111 Problem Set 8 Solutions DRAFT Edward Doolittle November 23, 2006 1. (a) Taking the derivative, we have Squaring, we have It follows that 1 so y 2 x2 x 2 2 2 (b) The hard way. We have y 2 1 2 4 x so 0 We continue with the partial fractions decomposition, etc. The easy way. We integrate with respect to y instead of x. We have so 1 1 ¨ ¡ ¡ dx dy y 2 dx dy 2 y2 4 1 £ £ 2 u2 ¤ ¦ £ ¨ ¦ 1 u du 2 ¥ ¨ £ © ¡ £ © ¡ £ £ ¥  © ¡ 1 dx x 2u u 1 2u2 du 2u 1 4u 2 du u 12 £  £  ¨   Now we make the rationalizing substitution u 1 2 dx to obtain 1 1 x, du 1 21 1 x 1 2 1 x2 dx, 2udu u ¥ £   ¦ ¨ ¥ ¡ ¨ ¡ ¤ L 1 ¨ 2 1 dx x  ¡   ¦  1 4x 2 2 1 ¤ £ £ L x dx x 8 2 6 ¨ ¡ ¨ ¡ ¨ ¡ ¨ © ¡ ¨ 4 1 4x 2 4 ¨ ¡ ¨ 1 2 1 16x2 ¤ ¨ ¥ £ y 2 x2 ¨ ¤ 1 2 1 16x2 1 4x 2 ¡ ¦ ¥ ¨ ¨ £ y x 1 4x ¡ § ¥ ¦ ¡ ¦ ¥ © ¡ © ¡ ¡ ¢ ¨ 1 dx 4x x2 2 ln x 4 4 1 ln 4 4 1 ln 2 4 1 ln 2 4 © ¡ ¦ so We then look up the integral for sec3 θ in a table and proceed from there. (c) We have ex 1 ex 1 e2x d ln ex dx d ln ex dx ex ex e2x ex e2x ex ex 1 ex 1 2ex 1 Squaring, e2x 2 so 1 (assuming x 0; if x 0, a slight change to the final equality...
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This note was uploaded on 01/12/2010 for the course MATH 111 taught by Professor Doolittle during the Fall '06 term at Berkeley.

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