Thread: arc length problem

1. arc length problem

This is actually an arc length problem for the polar equation r=1+sin(x) (x being theta) After some work you get:
L=integral from 0 to 2 times pi (2+2*sin(x))^(1/2) dx. It gets reduced to

L=2^(1/2) times integral from 0 to 2 times pi (cos(x)/(1-sin(x))^1/2) dx

But if u=1-sin(x) don't the limits of integration both equal 1?

The book says the answer is L=8 The figure is a cardioid.

2. Re: arc length problem

Instead of simplifying $\sqrt{\cos^2{x}}=\cos{x}$, you should have used $\sqrt{\cos^2{x}}=|\cos{x}|,$ and then the integral splits into three parts: integrating from 0 to $\frac{\pi}{2}$, from $\frac{\pi}{2}$ to $\frac{3\pi}{2}$, and from $\frac{3\pi}{2}$ to $2\pi$. The answer 8 is correct.

- Hollywood

3. Re: arc length problem

To make it a little more clear what you are saying:

$L = \int_0^{2\pi} \sqrt{2+2\sin \theta}d\theta = \sqrt{2} \int_0^{2\pi} \left| \cos \theta \right| \dfrac{1}{\sqrt{1-\sin \theta}} d\theta$

Note the absolute value around $\cos \theta$ since $\sqrt{\cos^2 \theta} \ge 0$ for all $\theta$.

Since $\cos \theta \ge 0$ on $\left[0,\dfrac{\pi}{2} \right]\cup \left[\dfrac{3\pi}{2}, 2\pi \right]$ and $\cos \theta \le 0$ on $\left[ \dfrac{\pi}{2},\dfrac{3\pi}{2} \right]$, you need to break up the integral:

$L & = \sqrt{2}\left[ \int_0^{\tfrac{\pi}{2}} \dfrac{\cos \theta}{\sqrt{1-\sin \theta}}d\theta - \int_{\tfrac{\pi}{2}}^{\tfrac{3\pi}{2}} \dfrac{\cos \theta}{\sqrt{1-\sin \theta}}d\theta + \int_{\tfrac{3\pi}{2}}^{2\pi} \dfrac{\cos \theta}{\sqrt{1-\sin \theta}}d\theta \right]$

Using the substitution $u = 1-\sin \theta, du = -\cos \theta d\theta$, you get:

\begin{align*}L & = \sqrt{2}\left[ -\int_1^0 u^{-1/2}du + \int_0^2 u^{-1/2}du - \int_2^1 u^{-1/2}du \right] \\ & = \sqrt{2}\left[ \int_0^1 u^{-1/2}du + \int_0^2 u^{-1/2}du + \int_1^2 u^{-1/2}du \right] \\ & = 2\sqrt{2}\int_0^2 u^{-1/2}du \\ & = 4\sqrt{2}(\sqrt{2}-0) = 8\end{align*}

Edit: Someone else beat me to it.

4. Re: arc length problem

Wow ! thanks to both of you. I never considered that.