Mathematical Immaturity

14.19 Exercises

• In each of Exercises 8 through 11, make a sketch of the plane curve having the given polar equation and compute its arc length. $$r = 1 - \cos \theta,\quad 0 \leq \theta \leq 2\pi.$$
• We know from Exercise 4 that the arc length $s$ of the curve given by the polar equation $r = f(\theta)$ from $\theta = a$ to $\theta = b,$ where $a \leq \theta \leq b \leq a + 2\pi$ is: \begin{align*} \\ s &= \int_a^b \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta. \end{align*}
• The arc length $s$ of the curve given by the polar equation $r = f(\theta)$ from $\theta = a$ to $\theta = b,$ where $a \leq \theta \leq b \leq a + 2\pi$ is: \begin{align*} \\ s &= \int_a^b \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta. \end{align*} This means that the arc length of the polar function with radius $r = 1 - \cos \theta$ from $\theta = 0$ to $\theta = \pi$ is \begin{align*} \\ s &= \int_0^{2\pi}\sqrt{(1 - \cos\theta)^2 + \sin^2\theta}\,d\theta \\ \\ &= \int_0^{2\pi}\sqrt{2 - 2\cos\theta}\,d\theta \\ \\ &= \sqrt{2}\int_0^{2\pi}\sqrt{1 - \cos\theta}\,d\theta \end{align*} Using the double-angle identity of the cosine $(\cos \theta = 1 - 2\sin^2\frac{\theta}{2}),$ and noting that the sine function is non-negative on the interval $[0, \pi],$ the above integral then becomes \begin{align*} \\ s &= \sqrt{2}\int_0^{2\pi}\sqrt{1 - \cos\theta}\,d\theta \\ \\ &= \sqrt{2}\int_0^{2\pi}\sqrt{1 - \left[1 - 2\sin^2\left(\frac{\theta}{2}\right)\right]}\,d\theta \\ \\ &= \sqrt{2}\int_0^{2\pi}\sqrt{2\sin^2\left(\frac{\theta}{2}\right)}\,d\theta \\ \\ &= 2\int_0^{2\pi}\sin\left(\frac{\theta}{2}\right)\,d\theta \\ \\ &= 4\cos\left(\frac{\theta}{2}\right)\Biggr|_{2\pi}^0 \\ \\ &= 8. \quad \blacksquare \end{align*}