Differential Geometry of Curves and Surfaces - Chapter 1 Section 3 Exercise 2

Problem:

Solution:

We will skip part (a) because it is basically the same as this previous exercise. The solution is $(t - \sin t, 1 - \cos t)$

For part (b), we compute the arc length as

$\begin{eqnarray*} & & \int\limits_{0}^{2\pi}{\left|\alpha'(t)\right|dt} \\ &=& \int\limits_{0}^{2\pi}{\sqrt{\alpha'(t) \cdot \alpha'(t)}dt} \\ &=& \int\limits_{0}^{2\pi}{\sqrt{(1-\cos t)^2 + \sin^2 t}dt} \\ &=& \int\limits_{0}^{2\pi}{\sqrt{1 - 2\cos t + \cos^2 t + \sin^2 t}dt} \\ &=& \int\limits_{0}^{2\pi}{\sqrt{2 - 2\cos t}dt} \\ \end{eqnarray*}$

The rest is really just an integration problem, let $x = \cos t$, so

$\begin{eqnarray*} dx &=& -\sin t dt \\ &=& -\sqrt{1 - \cos^2 t} dt \\ &=& -\sqrt{1 - x^2}dt \\ dt &=& \frac{dx}{-\sqrt{1-x^2}} \end{eqnarray*}$

When $t = 0$, $x = \cos 0 = 1$
When $t = \pi$, $x = \cos \pi = 0$

$\begin{eqnarray*} & & \int\limits_{0}^{2\pi}{\sqrt{2 - 2\cos t}dt} \\ &=& 2\int\limits_{0}^{\pi}{\sqrt{2 - 2\cos t}dt} \\ &=& 2\int\limits_{1}^{-1}{\sqrt{2 - 2x}\frac{dx}{-\sqrt{1-x^2}}} \\ &=& 2\int\limits_{-1}^{1}{\sqrt{\frac{2 - 2x}{1-x^2}}dx} \\ &=& 2\sqrt{2}\int\limits_{-1}^{1}{\sqrt{\frac{1}{1 + x}}dx} \\ &=& 4\sqrt{2}\sqrt{1+x}|_{-1}^{1} \\ &=& 8 \\ \end{eqnarray*}$

Just as an aside, I used numerical integration and arc length approximation to make sure the answer is correct.