How do you find the arc length of the curve y=(5sqrt7)/3x^(3/2)-9 over the interval [0,5]?

Jun 16, 2018

$s = {\int}_{0}^{5} \sqrt{1 + \frac{175}{4} x} \textcolor{w h i t e}{.} \mathrm{dx} = \frac{1}{525} \left({879}^{\frac{3}{2}} - 8\right)$

Explanation:

The arc length of a function $y$ on the interval $\left[a , b\right]$ is given by:

$s = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \textcolor{w h i t e}{.} \mathrm{dx}$

Here:

$y = \frac{5 \sqrt{7}}{3} {x}^{\frac{3}{2}} - 9$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5 \sqrt{7}}{3} \left(\frac{3}{2} {x}^{\frac{1}{2}}\right) = \frac{5 \sqrt{7}}{2} {x}^{\frac{1}{2}}$

Then the arc length desired is:

$s = {\int}_{0}^{5} \sqrt{1 + {\left(\frac{5 \sqrt{7}}{2} {x}^{\frac{1}{2}}\right)}^{2}} \textcolor{w h i t e}{.} \mathrm{dx}$

$s = {\int}_{0}^{5} \sqrt{1 + \frac{175}{4} x} \textcolor{w h i t e}{.} \mathrm{dx}$

Let $u = 1 + \frac{175}{4} x$, which implies that $\mathrm{du} = \frac{175}{4} \mathrm{dx}$. Moreover, note the change of bounds of the integral under this substitution: $x = 0 \implies u = 1$ and $x = 5 \implies u = \frac{879}{4}$. The integral is then:

$s = \frac{4}{175} {\int}_{1}^{879 / 4} {u}^{\frac{1}{2}} \textcolor{w h i t e}{.} \mathrm{du}$

Integrating using the power rule for integration:

$s = \frac{4}{175} \left(\frac{2}{3} {u}^{\frac{3}{2}}\right) {|}_{1}^{879 / 4}$

$s = \frac{8}{525} \left({\left(\frac{879}{4}\right)}^{\frac{3}{2}} - {1}^{\frac{3}{2}}\right)$

$s = \frac{8}{525} \left({879}^{\frac{3}{2}} / 8 - 1\right)$

$s = \frac{1}{525} \left({879}^{\frac{3}{2}} - 8\right)$