Comprimentos de uma curva polar

One special form of parametrization, called polar coordinates, shifts the original $(x,y)$-system into the $(r,\theta)$-system. This transformation is by the equations $x=r\cos{\theta}$ and $y=r\sin{\theta}$, where the function is now represented by the equation $r=f(\theta)$. In this section, we will learn how to find the length of such a polar curve.

Let's start with an example:

Remember that $r$ in polar coordinates represents the distance from the origin. and so $r=3$ represents the set of all points that are $3$ units away from the origin.

So, the question just asked us to find the perimeter of this circle, which is equal to $$P=2\pi r=2\pi\cdot 3=6\pi.$$

This was a simple example, but what about more complicated curves like $r=f(\theta)=\theta^2$, as shown below?

Thankfully, there's a way to calculate this arc length, by using the formula derived in the previous section $$L=\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\, dt,$$ to the new expression $$L=\int_{\alpha}^{\beta} \sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\, d\theta,$$ where $\theta$ ranges from $\alpha$ to $\beta$.

Let's look at the first problem from above, and calculate the full arc length of the circle defined by the equation $r=3$. Since we are working with the entire circle, $\theta$ ranges from $0$ to $2\pi$. Therefore, by the formula above, the desired arc length is equal to $$L=\int_{\alpha}^{\beta} \sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\, d\theta=\int_0^{2\pi} \sqrt{(3)^2+\left(\frac{d}{d\theta}(3)\right)^2}\, d\theta$$ $$=\int_0^{2\pi} \sqrt{9+0^2}\, d\theta=\int_0^{2\pi} 3\, d\theta=3\theta\bigg|_0^{2\pi}=6\pi,$$ as expected.

Now you try! This time, let's look at the second curve shown above.

Since $r=\theta^2$, we have $\frac{dr}{d\theta}=2\theta$. So, by our formula above, $$L=\int_0^1 \sqrt{(\theta^2)^2+(2\theta)^2}\, d\theta=\int_0^1 \sqrt{\theta^4+4\theta^2}\, d\theta=\int_0^1 \theta\sqrt{\theta^2+4}\, d\theta.$$ Now, we make a $u$-substitution: let $u=\theta^2+4$. Then $du=2\theta\, d\theta$, and so we get $$\theta\, d\theta=\frac{du}{2}.$$ We can plug this into the integral above to find $$L=\int_0^1 \sqrt{u}\cdot \frac{du}{2}=\frac{1}{2}\int_0^1 \sqrt{u}\, du.$$ Since $\theta$ ranged from $0$ to $1$, $u$ ranges from $0^2+4=4$ to $1^2+4=5$. So, this evaluates to $$L=\frac{1}{2}\left(\frac{2}{3}u^{\frac{3}{2}}\bigg|_4^5\right)=\frac{5\sqrt{5}-8}{3}.$$

Recall that $x=r\cos{\theta}=f(\theta)\cos{\theta}$ and $y=r\sin{\theta}=f(\theta)\sin{\theta}$. Then by the Product Rule, $$\frac{dx}{d\theta}=\frac{d}{d\theta}(r\cos{\theta})=\frac{dr}{d\theta}\cos{\theta}-r\sin{\theta}$$ and $$\frac{dy}{d\theta}=\frac{d}{d\theta}(r\sin{\theta})=\frac{dr}{d\theta}\sin{\theta}+r\cos{\theta}.$$ Therefore, we have $$\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2=(\frac{dr}{d\theta}\cos{\theta}-r\sin{\theta})^2+(\frac{dr}{d\theta}\sin{\theta}+r\cos{\theta})^2$$ $$=\left[\left(\frac{dr}{d\theta}\right)^2\cos^2{\theta}-2r\frac{dr}{d\theta}\sin{\theta}\cos{\theta}+r^2\sin^2{\theta}\right]$$ $$+\left[\left(\frac{dr}{d\theta}\right)^2\sin^2{\theta}+2r\frac{dr}{d\theta}\sin{\theta}\cos{\theta}+r^2\cos^2{\theta}\right]$$ $$=\left(\frac{dr}{d\theta}\right)^2(\cos^2{\theta}+\sin^2{\theta})+r^2(\cos^2{\theta}+\sin^2{\theta})=r^2+\left(\frac{dr}{d\theta}\right)^2.$$ Therefore, we have $$L=\int_{\alpha}^{\beta} \sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}\, d\theta=\int_{\alpha}^{\beta} \sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\, d\theta,$$ as desired.