Calculate the length of the portion of the hypocycloid $x^{\frac{2}{3}}+y^{\frac{2}{3}}=1$ Calculate the length of the portion of the hypocycloid $x^{\frac{2}{3}}+y^{\frac{2}{3}}=1$ in the first quadrant from the point $(\frac{1}{8},\frac{3\sqrt{3}}{8})$ to the point (1.0)
$x^{\frac{2}{3}}+y^{\frac{2}{3}}=1$
$y^{\frac{2}{3}}=1-x^{\frac{2}{3}}$
$y=(1-x^{\frac{2}{3}})^{\frac{3}{2}}$
$\frac{dy}{dx}=(1-x^{\frac{2}{3}})^{\frac{3}{2}}$
$\frac{dy}{dx}= \frac {3}{2}(1-x^{\frac{2}{3}})^{\frac{1}{2}}(0-\frac{2}{3}x)$
i am not sure where to go from there

Formula for arc length is
$$L=\int_{\frac18}^1\sqrt{1+\left(y'\right)^2}dx$$
It is easy to find the derivative:
$$y'=-\frac{\sqrt{1-x^{\frac23}}}{x^\frac13}$$
The length is now:
$$\begin{align}L&=\int_{\frac18}^1\sqrt{1+\left(-\frac{\sqrt{1-x^{\frac23}}}{x^\frac13}\right)^2}dx\\&=\int_{\frac18}^1x^{-\frac13}dx\\&=\left.\frac32x^\frac23\right|_\frac18^1\\&=\frac98\end{align}$$

