# How do you find the lengths of the curve y=(x-1)^(2/3) for 1<=x<=9?

Mar 23, 2018

Work in terms of $y$.

#### Explanation:

y=(x−1)^(2/3) with 1≤x≤9

Invert:

$x = {y}^{\frac{3}{2}} + 1$ with 0≤y≤4

$x ' = \frac{3}{2} \sqrt{y}$

Arc length is given by:

$L = {\int}_{0}^{4} \sqrt{1 + \frac{9}{4} y} \mathrm{dy}$

Integrate directly:

$L = \frac{8}{27} {\left[{\left(1 + \frac{9}{4} y\right)}^{\frac{3}{2}}\right]}_{0}^{4}$

Hence

$L = \frac{8}{27} \left(10 \sqrt{10} - 1\right)$