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: $$\begin{align}L&=\int_{\frac18}^1\sqrt{1+\left(-\frac{\sqrt{1-x^{\frac23}}}{x^\frac13}\right)^2}dx\\\&=\int_{\frac18}^1x^{-\frac13}dx\\\&=\left.\frac32x^\frac23\right|_\frac18^1\\\&=\frac98\end{align}$$

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