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\bf{Question}

\quad  In each case sketch the curve $\gamma$ and find its total
path length:

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\begin{tabular}{rl}
\hspace{1.5cm}(i)&amp;\ $\gamma(t)=(2\cos t+\cos2t,2\sin t-\sin2t)$,
\quad$0\le t\le2\pi$;\\ (ii)&amp;\
$\gamma(t)=(\sin^3t,\cos^3t)$,\quad$0\le t\le2\pi$.
\end{tabular}



\bf{Answer}

\begin{description}
\item{(i)}
Hypocycloid: point of radius 1 rolling inside a circle of radius
3.

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\begin{eqnarray*}
\gamma'(t) &amp; = &amp; (-2\sin t - 2 \sin 2t, 2\cos t - 2\cos 2t)\\ \|
\gamma'(t) \|^2 &amp; = &amp; 4(1 + \sin t \sin 2t - 2\cos t \cos 2t +
1)\\ &amp; = &amp; 8(1-\cos 3t)\\ \Rightarrow  \textrm{length} &amp; = &amp;
\int_0^{2\pi} \sqrt{8} (1-\cos 3t)^{\frac{1}{2}} \,dt\\ &amp; = &amp;
\sqrt{8} \sqrt{2} \int_0^{2\pi} \underbrace{\left | \sin
\frac{3t}{2} \right | }_{note!} \,dt\\ &amp; = &amp; 4 \times 3
\int_0^{\frac{2\pi}{3}} \sin \frac{3t}{2} \,dt =16
\end{eqnarray*}

\item{(ii)}
\begin{eqnarray*}
x &amp; = &amp; s^3\\ y &amp; = &amp; c^3\\ \| \gamma'(t) \| &amp; = &amp; \| ( 3s^2c,
-3c^2s) \|\\ &amp; = &amp; 3 |sc|\\ \Rightarrow  \textrm{length} &amp; = &amp; 4
\int_0^{2\pi} \frac{3}{2} \sin 2t \,dt\\ &amp; = &amp; 4 \left [ -
\frac{3}{4} \cos 2t \right ]_0^{2\pi} = 6
\end{eqnarray*}
\end{description}


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