Question

Calcule o comprimento de arco da curva r = 3θ^2 de θ = 0 até θ = 2π/3.

Answer 1

O comprimento de arco de uma curva descrita por uma equação polar é:

                                          $L=\int\limits^a_b {\sqrt{r^2+(\frac{dr}{d\theta})^2}} \, d\theta$

$L=\int\limits^\frac{2\pi}{3}_0 {\sqrt{(3\theta^2)^2+(6\theta)^2}} \, d\theta\\\\\\L=\int\limits^\frac{2\pi}{3}_0 {\sqrt{9\theta^4+36\theta^2}} \, d\theta\\\\\\L=\int\limits^\frac{2\pi}{3}_0 {3\theta\sqrt{\theta^2+4}} \, d\theta\\\\\\\\x=\theta^2+4\\\\dx=2\theta.d\theta\\\\d\theta=\frac{dx}{2\theta}\\\\\\\\L=\int\limits^{\frac{4\pi^2}{9}+4}_4 {3\theta\sqrt{x}} \, \frac{dx}{2\theta}\\\\\\L=\frac{3}{2}.\int\limits^{\frac{4\pi^2}{9}+4}_4 {\sqrt{x}} \, dx}$

$L=\frac{3}{2}.(\frac{2}{3}x^\frac{3}{2})\Big|_4^{\frac{4\pi^2}{9}+4}\\\\\\L=(\frac{4\pi^2}{9}+4)^{\frac{3}{2}}-(4)^{\frac{3}{2}}=16.28680668$