# Arc Length polar curve

$$r=a\sin^3\left(\frac{\theta}{3}\right)$$ I tried solving it using the equation for arc length with $dr/d\theta$ and $r^2$. Comes out messy and complicated.

• Perhaps you could post your work, because it doesn't seem to me that it is very ugly at all. – rogerl May 18 '14 at 0:37
• Be careful to apply the Chain Rule fully when calculating $\ \frac{dr}{d\theta} \$ . The Pythagorean Identity will help with simplifying the arclength element $\ \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \$ . You will get a "perfect square" under the radical. – colormegone May 18 '14 at 1:23

The portion of the arclength calculation in which we determine the infinitesimal arclength element $\ ds \$ just requires applying the Chain Rule carefully and using the Pythagorean Identity:

$$\frac{dr}{d\theta} \ = \ \frac{d}{d\theta} \left[ \ a \ \sin^3 \left(\frac{\theta}{3}\right) \right] \ = \ a \ \cdot \ 3 \ \sin^2\left(\frac{\theta}{3}\right) \cdot \cos\left(\frac{\theta}{3}\right) \cdot \frac{1}{3} \ = \ a \ \sin^2\left(\frac{\theta}{3}\right) \cos\left(\frac{\theta}{3}\right)$$

$$\Rightarrow \ \ ds \ = \ \sqrt{ \ r^2 + \left(\frac{dr}{d\theta}\right)^2} \ \ d\theta \ = \ \sqrt{ \ \left[ \ a \ \sin^3 \left(\frac{\theta}{3}\right) \right]^2 + \left[ \ a \ \sin^2\left(\frac{\theta}{3}\right) \cos\left(\frac{\theta}{3}\right) \right]^2} \ \ d\theta$$

$$= \ \sqrt{ \ a^2 \ \sin^6 \left(\frac{\theta}{3}\right) + \ \ a^2 \ \sin^4\left(\frac{\theta}{3}\right) \cos^2\left(\frac{\theta}{3}\right)} \ \ d\theta$$

$$= \ \sqrt{ \ a^2 \ \sin^4 \left(\frac{\theta}{3}\right) \left[ \ \sin^2 \left(\frac{\theta}{3}\right) + \ \cos^2\left(\frac{\theta}{3}\right) \right]} \ \ d\theta$$

$$= \ \sqrt{ \ a^2 \ \sin^4 \left(\frac{\theta}{3}\right) } \ \ d\theta \ = \ a \ \sin^2 \left(\frac{\theta}{3}\right) \ d\theta \ \ .$$

What turns out to be the tricky part is in finding the limits of integration in order to cover this curve fully. Here is a graph of the curve described by this polar equation (for $\ a \ = \ 1 ) \$ : A graph of $\ r \$ as a function of $\ \theta \$ shows that, while the period of the function is $\ 6 \pi \$ , the radius is negative in the interval $\ ( \ 3 \pi \ , \ 6 \ \pi \ ) \$ . The curve resembles a limaçon , but the interval of negative radii retraces the curve already "swept out" by positive radii in $\ ( \ 0 \ , \ 3 \pi \ ) \$ . Thus, we will find the arclength correctly by integrating arclength over this interval, rather than only up to $\ 2 \pi \$ or all the way to $\ 6 \pi \$ .

Our arclength integral is therefore

$$\int_0^{3 \pi} \ ds \ \ = \ \ \int_0^{3 \pi} a \ \sin^2 \left(\frac{\theta}{3}\right) \ \ d\theta \ \ = \ \ a \ \int_0^{3 \pi} \ \frac{1}{2} \left[ \ 1 \ - \ \cos \left(\frac{2\theta}{3}\right) \right] \ \ d\theta$$

$$= \ \ \frac{a}{2} \left[ \ \theta \ - \ \frac{3}{2} \sin \left(\frac{2\theta}{3}\right) \right] \vert_0^{3 \pi} \ = \ \frac{a}{2} \left[ \ \left(3 \pi \ - \ \frac{3}{2} \sin \left[\frac{6 \pi}{3}\right] \ \right) \ - \ \left(0 \ - \ \frac{3}{2} \sin \left[\frac{0}{3}\right] \right) \right]$$

$$= \ \frac{a}{2} \ \left(3 \pi \ - \ \frac{3}{2} \sin \ 2 \pi \ \ - \ 0 \ + \ 0 \right) \ = \ \frac{3 \pi}{2} a \ \ .$$

As a check, we can see from the graph above that the arclength of the curve will be somewhat larger than the circumference of a circle of diameter $\ \frac{5}{4} a \$ , which is $\ \frac{5 \pi}{4} a \$ (the small "self-crossing" loop of this curve does not add much to the total arclength).

This solution borders on trivial when solved in the complex plane. First, the range of $\theta$ must be determined. It was pointed out in an earlier that $\theta\in[0,3\pi]$. I found the same thing by plotting the equation. Now, the equation for the arc length in the complex plane is given by

$$s=\int|\dot z|du$$

Thus, for this problem we have (here we take $a=1$ without any loss in generality)

z=\sin^3(\theta/3)e^{i\theta}\\ \dot z=\left[3\sin^2\left(\frac{\theta}{3}\right)\cos\left(\frac{\theta}{3}\right)\frac{1}{3}+i\sin^3\left(\frac{\theta}{3}\right)\right]e^{i\theta}\\ \begin{align} |\dot z| &=\sqrt{\sin^4\left(\frac{\theta}{3}\right)\cos^2\left(\frac{\theta}{3}\right)+\sin^6\left(\frac{\theta}{3}\right)}\\ &=\sin^2\left(\frac{\theta}{3}\right)\sqrt{\cos^2\left(\frac{\theta}{3}\right)+\sin^2\left(\frac{\theta}{3}\right)}\\ &=\sin^2\left(\frac{\theta}{3}\right)\\ \end{align}\\

And finally, $$s=\int_0^{3\pi}\sin^2\left(\frac{\theta}{3}\right)d\theta=\frac{3\pi}{2}$$

I have verified this result by numerical integration.