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Featured Proof

Length of Arc of Astroid Theorem The total length of the arcs of an astroid constructed within a stator of radius $a$ is given by: $\mathcal L = 6 a$ Proof Let $H$ be embedded in a cartesian coordinate plane with its center at the origin and its cusps positioned on the axes. We have that $\mathcal L$ is $4$ times the length of one arc of the astroid. From Arc Length for Parametric Equations: $\displaystyle \mathcal L = 4 \int_{\theta \mathop = 0}^{\theta \mathop = \pi/2} \sqrt {\left({\frac{\mathrm d x}{\mathrm d \theta}}\right)^2 + \left({\frac{\mathrm d y}{\mathrm d \theta}}\right)^2} \mathrm d \theta$ where, from Equation of Astroid: $\begin{cases} x & = a \cos^3 \theta \\ y & = a \sin^3 \theta \end{cases}$ We have: \(\displaystyle \frac {\mathrm d x} {\mathrm d \theta}\) \(=\) \(\displaystyle -3 a \cos^2 \theta \sin \theta\) $\quad$ $\quad$ \(\displaystyle \frac {\mathrm d y} {\mathrm d \theta}\) \(=\) \(\displaystyle 3 a \sin^2 \theta \cos \theta\) $\quad$ $\quad$ Thus: \(\displaystyle \sqrt {\left({\frac {\mathrm d x} {\mathrm d \theta} }\right)^2 + \left({\frac {\mathrm d y} {\mathrm d \theta} }\right)^2}\) \(=\) \(\displaystyle \sqrt {9 a^2 \left({\sin^4 \theta \cos^2 \theta + \cos^4 \theta \sin^2 \theta}\right)}\) $\quad$ $\quad$ \(\displaystyle \) \(=\) \(\displaystyle 3 a \sqrt {\sin^2 \theta \cos^2 \theta \left({\sin^2 \theta + \cos^2 \theta}\right)}\) $\quad$ $\quad$ \(\displaystyle \) \(=\) \(\displaystyle 3 a \sqrt {\sin^2 \theta \cos^2 \theta}\) $\quad$ Sum of Squares of Sine and Cosine $\quad$ \(\displaystyle \) \(=\) \(\displaystyle 3 a \sin \theta \cos \theta\) $\quad$ $\quad$ \(\displaystyle \) \(=\) \(\displaystyle \frac {3 a \sin 2 \theta} 2\) $\quad$ Double Angle Formula for Sine $\quad$ Thus: \(\displaystyle \mathcal L\) \(=\) \(\displaystyle 4 \frac {3 a} 2 \int_0^{\pi / 2} \sin 2 \theta \, \mathrm d \theta\) $\quad$ $\quad$ \(\displaystyle \) \(=\) \(\displaystyle 6 a \left[{\frac {-\cos 2 \theta} 2}\right]_0^{\pi / 2}\) $\quad$ Primitive of $\sin a x$ $\quad$ \(\displaystyle \) \(=\) \(\displaystyle 6 a \left({-\frac {\cos \pi} 2 + \frac {\cos 0} 2}\right)\) $\quad$ evaluating limits of integration $\quad$ \(\displaystyle \) \(=\) \(\displaystyle 6 a \frac {- \left({-1}\right) + 1} 2\) $\quad$ $\quad$ \(\displaystyle \) \(=\) \(\displaystyle 6 a\) $\quad$ $\quad$ $\blacksquare$ Sources 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.21$: The Cycloid: Problem $7$