Mathematical Immaturity

14.19 Exercises

• The curve described by the polar equation $r = a(1 + \cos \theta),$ where $a > 0$ and $0 \leq \theta \leq 2\pi,$ is called a cardioid. Draw a graph of the cardioid $r = 4(1 + \cos \theta)$ and compute its arc length.
• We proved in Exercise 4 that the arc length of a curve given by the polar equation $r = f(\theta)$ from $a$ to $b,$ where $0 \leq a \leq b \leq a + 2\pi$ is $$ \begin{align*} \\ \int_a^b \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta. \end{align*} $$
• Recall from Exercise 4 that the arc length of a curve given by the polar equation $r = f(\theta)$ from $a$ to $b,$ where $0 \leq a \leq b \leq a + 2\pi$ is $$ \begin{align*} \\ \int_a^b \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta. \end{align*} $$ Thus, the arc length $s$ of the curve with polar equation $r = 4(1 + \cos\theta)$ from $\theta = 0$ to $\theta = 2\pi$ is $$ \begin{align*} \\ s &= \int_0^{2\pi} \sqrt{16(1 + \cos\theta)^2 + 16\sin^2\theta}\,d\theta \\ \\ &= 4 \int_0^{2\pi} \sqrt{2 + 2 \cos\theta}\,d\theta. \\ \\ \end{align*} $$ Applying the the double angle identity for cosine $(\cos 2t = 1 - 2\sin^2 t)$ we get $$ \begin{align*} \\ 4 \int_0^{2\pi} \sqrt{2 + 2 \cos\theta}\,d\theta &= 4\sqrt{2} \int_0^{2\pi}\sqrt{2 - 2\sin^2\left(\frac{\theta}{2}\right)} \\ \\ &= 8\int_0^{2\pi}\sqrt{\cos^2\left(\frac{\theta}{2}\right)}\,d\theta \\ \\ &= 8\int_0^{2\pi}\left|\cos\left(\frac{\theta}{2}\right)\right|\,d\theta \\ \\ &= 8\int_0^{\pi}\cos\left(\frac{\theta}{2}\right)\,d\theta - 8\int_{\pi}^{2\pi}\cos\left(\frac{\theta}{2}\right)\,d\theta \\ \\ &= 16\sin\left(\frac{\theta}{2}\right) \, \Biggr|_0^{\pi} - 16\sin\left(\frac{\theta}{2}\right)\, \Biggr|_{\pi}^{2\pi} \\ \\ &= 32 \quad \blacksquare \end{align*} $$