- Calculus, Volume 1: One Variable Calculus, with an Introduction to Linear Algebra
- Tom M. Apostol
- Second Edition
- 1967
- 978-1-119-49673-1
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. $$ \begin{align*} r = \theta, \quad 0 \leq \theta \leq \pi. \end{align*} $$
- 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: $$ \begin{align*} \\ s &= \int_a^b \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta. \end{align*} $$ For integrals of the form $$ \begin{align*} \\ \int R\left(\theta, \sqrt{a^2 + (c\theta + d)^2}\right)\,d\theta, \end{align*} $$ we can use the trigonometric substitution $c\theta + d = a\tan x$ and $d\theta = a\sec^2 x\,dx.$
- 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: $$ \begin{align*} \\ s &= \int_a^b \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta. \end{align*} $$ As such, for $r = \theta,$ from $\theta = 0$ to $\theta = \pi,$ we get $$ s = \int_0^{\pi}\sqrt{1 + \theta^2}\,d\theta. $$ Now, we saw in Exercise 7 that integrals of the form $$ \begin{align*} \\ \int R\left(\theta, \sqrt{a^2 + (c\theta + d)^2}\right)\,d\theta \end{align*} $$ can be evaluated using the trigonometric substitution $c\theta + d = a\tan x$ and $d\theta = a\sec^2 x\,dx$ to give us $$ \begin{align*} \\ s &= \int_{\arctan 0}^{\arctan{\pi}}\sec^2 t\sqrt{1 + \tan^2 t}\,dt \\ \\ &= \int_{\arctan 0}^{\arctan{\pi}}\sec^3 t\,dt \\ \\ &= \frac{1}{2}\Big[\sec t \tan t + \log(\sec t + \tan t )\Big]_{\arctan 0}^{\arctan \pi} \end{align*} $$ Using the identity $\sec^2 t = 1 + \tan^2 t,$ we can deduce that for $0 \leq t < \pi/2,$ $\sec t = \sqrt{1 + \tan^2 t}$ giving us $\sec(\arctan \pi) = \sqrt{1 + \pi^2}$ and: $$ \begin{align*} \\ s &= \frac{1}{2}\Big[\sec t \tan t + \log(\sec t + \tan t )\Big]_{\arctan 0}^{\arctan \pi} \\ \\ &= \frac{1}{2}\pi (\pi^2 + 1)^{1/2} + \frac{1}{2}\,\log\left(\pi + \sqrt{\pi^2 + 1}\right) \quad \blacksquare \end{align*} $$