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find the exact length of the polar curve, r=(theta)^2,  0<= theta<=2pi please be...

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dawks | Student, Undergraduate | (Level 1) Honors

Posted October 6, 2012 at 1:11 AM via web

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find the exact length of the polar curve, r=(theta)^2,  0<= theta<=2pi

please be as thorough as possible as you solve, especially towards end

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lfryerda | High School Teacher | (Level 2) Educator

Posted October 6, 2012 at 2:32 AM (Answer #1)

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The length of a polar function `r=f(theta)` is:

`s=int_{theta_1}^{theta_2}sqrt{r^2+({dr}/{d theta})^2}d theta`  sub in `r=theta^2`  so `{dr}/{d theta}=2 theta`

`=int_0^{2pi} sqrt{theta^4+4theta^2}d theta`   factor `theta`

`=int_0^{2pi}theta sqrt{theta^2+4}d theta`   let `u=theta^2+4`  so `du=2 theta d theta` .

`=1/2 int_4^{4(pi^2+1)}u^{1/2}du`

`=1/2(2/3)(u^{3/2})_4^{4(pi^2+1)}`

`=1/3(4^{3/2}(pi^2+1)^{3/2}-4^{3/2})`

`=4^{3/2}/3((pi^2+1)^{3/2}-1)`

The length of the curve is `4^{3/2}/3((pi^2+1)^{3/2}-1)` .

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