Mathematical Immaturity

14.13 Exercises

• Use the vector equation $\mathbf{r}(t) = a\sin t\mathbf{i} + b\cos t\mathbf{j},$ where $0 < b < a,$ to show that the circumference $L$ of an ellipse is given by the integral: $$L = 4a\int_0^{\pi/2}\sqrt{1 - e^2\sin^2 t}\,dt$$ where $e = \sqrt{a^2 - b^2}/a$ (The number $e$ is the eccentricity of the ellipse.) This is a special case of an integral of the form: $$E(k) = \int_0^{\pi/2}\sqrt{1 - k^2\sin^2 t}\,dt$$ called an elliptic integral of the second kind, where $0 \leq k < 1.$ The numbers $E(k)$ have been tabulated for various values of $k.$
• Noting that $0 < b < a,$ we have: $$ \begin{align*} \mathbf{v}(t) &= a \cos t\,\mathbf{i} - b \sin t\,\mathbf{j} \\ \\ v(t) &= \sqrt{a^2 \cos^2 t + b^2\sin^2 t} \\ &= \sqrt{a^2 - (a^2-b^2)\sin^2t} \\ &= a\sqrt{1 - e^2 \sin^2 t} \end{align*} $$ where $e = \sqrt{a^2 - b^2}/a$ Now, the circumference of the ellipse $C_e$ is the arc length traced by $f(t)$ from $t = 0$ to $t = 2\pi.$ But this value is equal to four times the arc length traced by $f$ from $t = 0$ to $t = \pi/2.$ In other words: $$ \begin{align*} C_e &= \int_0^{2\pi} v(t)\,dt \\ &= 4\int_0^{\pi/2} v(t)\,dt \\ &= 4a\int_0^{\pi/2} \sqrt{1 - e^2 \sin^2 t}\,dt \end{align*} $$ where $e = \sqrt{a^2 - b^2}/a.$ This completes the proof.