Mathematical Immaturity

14.13 Exercises

• Find the length of the path traced out by a particle moving on a curve according to the given equation during the time interval specified: $$\mathbf{r}(t) = \frac{c^2}{a}\cos^3 t\mathbf{i} + \frac{c^2}{b}\sin^3 t\mathbf{j}, \quad 0 \leq t \leq 2\pi, \quad c^2 = a^2 - b^2, \quad 0 < b < a$$
• Velocity $\mathbf{v}(t)$ is $$ \begin{align*} \mathbf{v}(t) &= \frac{c^2}{a}\left(-3 \cos^2 t \sin t\right)\mathbf{i} + \frac{c^2}{b}\left(3 \sin^2 t \cos t\right)\mathbf{j} \end{align*} $$ Speed $v(t) = \|\mathbf{v}(t)\|$ is $$ \begin{align*} v(t) &= \left[\frac{c^4}{a^2}\left(9 \cos^4 t \sin^2 t \right) + \frac{c^4}{b^2}\left(9 \sin^4 t \cos ^2 t \right)\right]^{1/2} \\ &= \frac{3c^2}{ab}\left[\sin^2 t \cos^2 t\left(b^2 \cos^2 t + a^2 \sin^2 t\right)\right]^{1/2} \\ &= \frac{3c^2}{ab}\left[\sin^2 t \cos^2 t\left(b^2 + c^2 \sin^2 t\right)\right]^{1/2} \\ &= \frac{3c^2}{ab}\left[\sin^2 t \cos^2 t\left(b^2 + \left(a^2 - b^2\right) \sin^2 t\right)\right]^{1/2} \end{align*} $$ Making the substitution $u = \sin t,\ \frac{du}{dt} = \cos t,\ (0 \leq t \leq \pi/2),$ and noting that the arc traced out over $0 \leq t \leq 2\pi$ is the same length as four times the arc traced out over $0 \leq t \leq \pi/2,$ we get: $$ \begin{align*} v(t) &= \frac{3c^2}{ab}u\,\frac{du}{dt}\left[b^2 + \left(a^2 - b^2\right)u^2\right]^{1/2} \end{align*} $$ Arc length $s$ is $$ \begin{align*} s &= \int_0^{2\pi}v(t)\,dt \\ &= 4\int_0^1 u\left[b^2 + \left(a^2 - b^2\right)u^2\right]^{1/2}\,du \\ &= 4\frac{c^2}{(a^2 - b^2)ab}\left[b^2 + \left(a^2 - b^2\right)u^2\right]^{3/2}\Biggr|_0^{1} \\ &= \frac{4}{ab}\left[\left(a^2\right)^{3/2} -\left(b^2\right)^{3/2}\right] \\ &= \frac{4}{ab}\left(a^3 - b^3\right) \quad \blacksquare \end{align*} $$