<s>[INST] Let C be the graph of hypocycloid $x^{2/3} + y^{2/3} = 1$ oriented clockwise. Parametrize the curve and find its arc length. Let C be the graph of hypocycloid $x^{2/3} + y^{2/3} = 1$ oriented clockwise. Parametrize the curve and find its arc length.
Attempt:
let $x = \sin^{3}t, y = \cos^3({t})$
$r(t) = (\sin^{3}(t), \cos^3(t)), 0 \le t \le 2\pi$
$r'(t) = (3 \sin^2(t) \cos(t), -3 \cos^2(t) \sin(t))$
\begin{align*}
||r'(t)||
& = \sqrt{3 \sin^2(t) + \cos(t))^2 + (3 \cos^2(t) + \sin(t)^2} \\
& = \sqrt{9 \sin^4 + \cos^2(t)+ (\sin^2(t) + \cos^2(t)) } \\
& = \sqrt{9 \sin^2(t) + \cos^2(t)}
= \sqrt{(3 \sin(t) \cos(t))^2} \\
& = 3 \sin(t) \cos(t)
= \frac{3}{2} (2 \sin(t) \cos(t))
= \frac{3}{2} \sin(2t).
\end{align*}
Now arc length:
$$
L = \int_{0}^{2\pi} \frac{3}{2} \sin(2t)dt
= 4 \cdot \frac{3}{2} \cdot \frac{1}{2}\bigg[-\cos(2t) \bigg]_{0}^{2\pi}
= 3(1+1)
= 6.$$
Is this correct?
 [/INST] Now $||r'(t)||=\sqrt{(3 \sin^2(t) \cos (t))^2+(-3 \cos^2(t) \sin (t))^2}=\sqrt{9\sin^4(t)\cos^2(t)+9\cos^4(t)\sin^2(t)}=\sqrt{9\sin^2(t)\cos^2(t)(\sin^2(t)+\cos^2(t)}=\sqrt{9\sin^2(t)\cos^2(t)}=|3\sin(t)\cos(t)|=|\frac{3}{2}\sin(2t)|$ 
$$L=\int_0^{2\pi}\frac{3}{2} |\sin(2t)| \ dt=\frac{3}{2}\int_0^{2\pi}|\sin(2t)| \ dt=\frac{3}{2}\big(\int_0^{\frac{\pi}{2}}\sin(2t) \ dt-\int_{\frac{\pi}{2}}^{\pi}\sin(2t) \ dt+\int_\pi^{\frac{3\pi}{2}}\sin(2t) \ dt-\int_{\frac{3\pi}{2}}^{2\pi}\sin(2t) \ dt\big)=\frac{3}{2}\cdot \frac{1}{2}[(-\cos(2t))_0^{ \frac{\pi}{2}}+(\cos(2t))_{\frac{\pi}{2}}^{ \pi}+(-\cos(2t))_{\pi}^{\frac{ 3\pi}{2}}+(\cos(2t))_{\frac{3\pi}{2}}^{2 \pi}]=\frac{3}{4}(1+1+1+1+1+1+1+1)=6$$
</s>