# How do you find the length of the curve x^(2/3)+y^(2/3)=1 for the first quadrant?

Oct 10, 2017

$L = \frac{3}{2}$

#### Explanation:

We can write the curve in parametric form by posing:

${x}^{\frac{2}{3}} = t$

so that:

${y}^{\frac{2}{3}} = 1 - t$

As in the first quadrant $x$ and $y$ are positive:

$\left\{\begin{matrix}x = {t}^{\frac{3}{2}} \\ y = {\left(1 - t\right)}^{\frac{3}{2}}\end{matrix}\right.$

where $0 \le t \le 1$.

The length of the curve is therefore:

$L = {\int}_{0}^{1} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

$L = {\int}_{0}^{1} \sqrt{{\left(\frac{3}{2} {t}^{\frac{1}{2}}\right)}^{2} + {\left(- \frac{3}{2} {\left(1 - t\right)}^{\frac{1}{2}}\right)}^{2}} \mathrm{dt}$

$L = {\int}_{0}^{1} \frac{3}{2} \sqrt{t + \left(1 - t\right)} \mathrm{dt}$

$L = \frac{3}{2} {\int}_{0}^{1} \mathrm{dt} = \frac{3}{2}$

graph{x^(2/3)+y^(2/3) = 1 [-2.5, 2.5, -1.25, 1.25]}