MATH111-200630-PS08-Solutions

10 we have y l 4 3 x1 3 y 1 0 2 16 9 x2 3 so this is a

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Unformatted text preview: tiating implicitly, we have Integrating over the branch of the curve in quadrant I we have 0 0 0 0 Since there are four branches each with the same length by symmetry, we have the total length of the curve is 4 3 2 6. 10. We have y L 4 3 x1 3 , y 1 0 2 16 9 x2 3, so This is a tough integral, but not impossible. Let u 1 3 x1 3 . Then du 1 2 3 3x dx, dx Probably the best way to tackle the above integral is to change all the tans to secs using the Pythagorean identity and then use a reduction formula (or integration by parts) for the integral of powers of sec. The integral can be evaluated in closed form so the arc length can be calculated exactly. (If anyone has completed the calculation, I’d appreciate it if I could borrow your notes to complete this solution set.) 4 ¤ 1 3 dx 1 tan2 θ © ¡ ¨ © ¡ 16 2 x 9 81 2 tan θ sec2 θ dθ 64  Now make the trig substitution 4 3 u tan θ, du 3 4 sec2 θ dθ, 81 64 tan2 θ sec3 θ dθ ¦ ¥ ¡ ¤ dx 1 ¡ ¦ ¨  ¥ © ¡ 16 2 x 9 16 2 2 u 3u du 9 ¡   ¡  ¡ ¤    ¨ 1 16 2 x 9 3 dx ¤ L 1 y 2 dx x 2 3 dx x dx  1 3 3 ¡ ¡   © ¡  ¦    ¥ ¡ ¦ ¥ © ¡ ¦ ¥ ¨ 1 1 1 3 2 x 2 1   3 3 x2 3 3 2 3u2 du, ¤  £ 1 3 1 3 y 0 y y 2 x £   ¡ £ ¡ ¡ ¦ ¥ & '¡ ¡ ( & '¡ ¡ ¢  ¦  ¥ ¡   ¨   2 x 3 2 y 3 1   y1 x1 3 y2 x2 3 x2 3 2 3 1  ¡  ¡ ¦ ¥ ¡  ¦ % ¡   £  ¦ ¦  ¥ "¥ ¡ £ ¥ ¡ ¡ ¦ ¥    ¡  ¦ ¡   ¥ ¡ £¡ ¦ ¥ ¡ ¦ ¥   ¦ £ ¥ ¡ ¨ ¨ © ¡ © ¡ ¡ ¦ © ©  ¥...
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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 University of California, Berkeley.

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