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\section{Astroid:}
\noindent Last week on the homework, there was a problem that asked:\\
Calculate the length of the astroid of \(x^{\frac{2}{3}} + y^{\frac{2}{3}}= 2\).

\solution We will find the arclength in the first quadrant, and then by symmetry we  can take our solution and multiply by four. Using implicit differentiation, 
\[\frac{2}{3}x^{-\frac{1}{3}} dx+\frac{2}{3}y^{-\frac{1}{3}}dy =0\]
Solving for $y'$ and squaring it (and substituting so the solution is in terms of $x$ we have,
\[(y')^2+1=\frac{2}{x^{\frac{2}{3}}}\]
Therefore, the arclength (of one quarter) is given by 
\begin{align*}
s=&\int_0^{2^{\frac{3}{2}}} \sqrt{((y')^2+1)}dx\\
=&\int_0^{2^{\frac{3}{2}}}\frac{\sqrt{2}}{x^{\frac13}}dx\\
=&\sqrt{2} \frac{3x^{\frac{2}{3}}}2\bigg|_0^{2^{\frac{3}{2}}}\\
=&\frac{3}{2}\sqrt{2}\times 2=\frac{3}{2} 2^{\frac{3}{2}}
\end{align*}
Therefore, the total arclength is 

\[6\cdot2^{\frac{3}{2}}\]


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