Mathematical Immaturity

14.13 Exercises

• If $0 < b < 4a,$ let $\mathbf{r}(t) = a(t - \sin t)\mathbf{i} + a(1 - \cos t)\mathbf{j} + b\sin \frac{t}{2}\mathbf{k}.$ Show that the length of the path traced out from $t = 0$ to $t = 2\pi$ is $8aE(k),$ where $E(k)$ has the meaning given in Exercise 17 and $k^2 = 1 - (b/4a)^2.$
• Recall the definition of $E(k)$ from exercise 17: $$E(k) = \int_0^{\pi/2}\sqrt{1 - k^2\sin^2 t}\,dt$$ This is known as an elliptic integral of the second kind, where $0 \leq k < 1.$ To turn the equation for arc length into a multiple of $E(k),$ use Theorem 1.7 (Invariance under translation) $$ \int_{a}^{b}f(x)\,dx = \int_{a + c}^{b + c}f(x - c)\,dx $$ Theorem 1.8 (Expansion or contraction of the interval of integration): $$ \int_{ka}^{kb}f\left(\frac{x}{k}\right)\,dx = k\int_{a}^{b}f(x)\,dx $$ and the trigonometric identity $\sin(x) = \cos(x - \frac{\pi}{2}).$
• We have velocity: $$ \begin{align*} \mathbf{v}(t) &= a\left(1 - \cos t\right)\mathbf{i} + a\left(\sin t\right)\mathbf{j} + \frac{b}{2}\cos\frac{t}{2}\mathbf{k} \end{align*} $$ and speed $$ \begin{align*} v(t) &= \sqrt{a^2(1 - \cos t)^2 + a^2\sin^2 t + \frac{b^2}{4}\cos^2\frac{t}{2}} \\ &= \sqrt{a^2(2 - 2\cos t) + \frac{b^2}{4}\cos^2\frac{t}{2}} \\ &= \sqrt{a^2\left[2 - 2\left(1 - \sin^2 \frac{t}{2}\right)\right] + \frac{b^2}{4}\cos^2\frac{t}{2}} \\ &= \sqrt{4a^2\sin^2\frac{t}{2} + \frac{b^2}{4}\cos^2 \frac{t}{2}} \\ &= 2a\sqrt{\sin^2\frac{t}{2} + \frac{b^2}{16a^2}\cos^2 \frac{t}{2}} \\ &= 2a\sqrt{1 -\left(1 - \frac{b^2}{16a^2}\right)\cos^2 \frac{t}{2}} \\ &= 2a\sqrt{1 - k^2\cos^2 \frac{t}{2}} \end{align*} $$ where $k^2 = 1 - (b/4a)^2$ Then, the arc length traced out by $f$ from $t = 0$ to $t = 2\pi$ is: $$ \begin{align*} s &= 2a\int_0^{2\pi}\sqrt{1 - k^2\cos^2\frac{t}{2}}\,dt \end{align*} $$ We can use Theorem 1.8 (Expansion or contraction of the interval of integration) to turn the above integral into: $$ s = 4a\int_0^\pi \sqrt{1 - k^2\cos^2 t}\,dt $$ Then, we use Theorem 1.7 (Integral invariance under translation) with $c = \pi/2$ to give us: $$ s = 4a\int_{-\pi/2}^{\pi/2} \sqrt{1 - k^2\cos^2\left(t - \frac{\pi}{2}\right)}\,dt $$ By trigonometric identity, $\cos(t - \frac{\pi}{2}) = \sin(t),$ giving us $$ s = 4a\int_{-\pi/2}^{\pi/2} \sqrt{1 - k^2\sin^2t}\,dt $$ Using Theorem 1.8 once more with $k = -1$ and noting that $\sin^2(-t) = \sin^2(t)$ for all $t,$ we see that the integral from $t = -\pi/2$ to $t = 0$ is equal the integral from $t = 0$ to $t = \pi/2,$ giving us: $$ \begin{align*} s &= 8a\int_{0}^{\pi/2} \sqrt{1 - k^2\sin^2t}\,dt \\ &= 8aE(k) \quad \blacksquare \end{align*} $$