Mathematical Immaturity

14.19 Exercises

• For each of the curves in Exercises 8 through 11, compute the radius of curvature for the value of $\theta$ indicated. (a) Arbitrary $\theta$ in Exercise 8. (b) Arbitrary $\theta$ in Exercise 9. (c) $\theta = \frac{1}{4}\pi$ in Exercise 10. (d) $\theta = \frac{1}{4}\pi$ in Exercise 11.
• In Exercise 12, we proved that for a curve with polar equation $r = f(\theta),$ its radius of curvature $\rho$ is \begin{align*} \\ \rho &= \frac{\left(r^2 + r'^{2}\right)^{3/2}}{\left|r^2 + 2r'^2 - rr''\right|}. \end{align*}
• (a) Recall from Exercise 8 that $r = \theta$ where $0\leq \theta \leq \pi.$ This gives us $r' = 1$ and $r'' = 0$ for all $\theta$ in the given interval. The radius of curvature is then: \begin{align*} \\ \rho &= \frac{\left(r^2 + r'^{2}\right)^{3/2}}{\left|r^2 + 2r'^2 - rr''\right|} \\ \\ &= \frac{\left(\theta^2 + 1\right)^{3/2}}{\theta^2 + 2}. \quad \blacksquare \end{align*} (b) Recall from Exercise 9 that $r = e^{\theta},$ where $0 \leq \theta \leq \pi.$ The radius of curvature is then: \begin{align*} \\ \rho &= \frac{\left(r^2 + r'^{2}\right)^{3/2}}{\left|r^2 + 2r'^2 - rr''\right|} \\ \\ &= \frac{\left(2e^{2\theta}\right)^{3/2}}{2e^{2\theta}} \\ \\ &= \sqrt{2}\,e^\theta. \quad \blacksquare \end{align*} (c) Recall from Exercise 10 that $r = 1 + \cos \theta,$ where $0 \leq \theta \leq \pi.$ The radius of curvature at $\theta = \pi/4$ is then: \begin{align*} \\ \rho &= \frac{\left(r^2 + r'^{2}\right)^{3/2}}{\left|r^2 + 2r'^2 - rr''\right|} \\ \\ &= \frac{\left[\left(1 + \cos\frac{\pi}{4}\right)^2 + \sin^2\frac{\pi}{4}\right]^{3/2}}{\left|\left(1 + \cos\frac{\pi}{4}\right)^2 + 2\sin^2\frac{\pi}{4} + \left(1 + \cos\frac{\pi}{4}\right)\cos\frac{\pi}{4}\right|} \\ \\ &= \frac{\left[\left(1 + \frac{\sqrt{2}}{2}\right)^2 + \frac{1}{2}\right]^{3/2}}{\left|\left(1 + \frac{\sqrt{2}}{2}\right)^2 + 1 + \left(1 + \frac{\sqrt{2}}{2}\right)\frac{\sqrt{2}}{2}\right|} \\ \\ &= \frac{\left(2 + \sqrt{2}\right)^{3/2}}{(1 + \sqrt{2} + \frac{1}{2}) + (1 + \frac{\sqrt{2}}{2} + \frac{1}{2})} \\ \\ &= \frac{\left(2 + \sqrt{2}\right)^{3/2}}{\frac{3}{2}(2 + \sqrt{2})} \\ \\ &= \frac{2}{3}\sqrt{2 + \sqrt{2}}\,. \quad \blacksquare \end{align*} (d) Recall from Exercise 11 that $r = 1 - \cos \theta,$ where $0\leq \theta \leq 2\pi.$ The radius of curvature at $\theta = \pi/2$ is then: \begin{align*} \\ \rho &= \frac{\left(r^2 + r'^{2}\right)^{3/2}}{\left|r^2 + 2r'^2 - rr''\right|} \\ \\ &= \frac{\left[\left(1 - \cos\frac{\pi}{2}\right)^2 + \sin^2\frac{\pi}{2}\right]^{3/2}}{\left|\left(1 - \cos\frac{\pi}{2}\right)^2 + 2\sin^2\frac{\pi}{2} - \left(1 - \cos\frac{\pi}{2}\right)\cos\frac{\pi}{2}\right|} \\ \\ &= \frac{\left(1^2 + 1^2\right)^{3/2}}{\left|1^2 + 2(1^2)\right|} \\ \\ &= \frac{2^{3/2}}{3} \\ \\ &= \frac{2}{3}\sqrt{2}\,. \quad \blacksquare \end{align*}