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
1 Answer
lfryerda | High School Teacher | (Level 2) Educator
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)` .