# How do you find the arc length of the curve  y = (3/2)x^(2/3) from [1,8]?

Jul 16, 2015

The arc length can be quickly derived by realizing that it's just the distance formula in combination with the derivative over some interval $\Delta x$. Thus:

$s \left(x\right) = {\sum}_{a}^{b} \sqrt{{\left(\Delta x\right)}^{2} + {\left(\Delta y\right)}^{2} / {\left(\Delta x\right)}^{2} {\left(\Delta x\right)}^{2}}$

$= {\sum}_{a}^{b} \sqrt{1 + {\left(\Delta y\right)}^{2} / {\left(\Delta x\right)}^{2}} \Delta x$

$s \left(x\right) = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

Therefore:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{2} \cdot \frac{2}{3} {x}^{\text{-1/3" = x^"-1/3}}$
${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = {x}^{\text{-2/3}}$

$s \left(x\right) = {\int}_{1}^{8} \sqrt{1 + \frac{1}{x} ^ \text{2/3}} \mathrm{dx}$

What normally what we would do is try to find a way to complete the square. Let's just make this look nicer for now.

$= {\int}_{1}^{8} \frac{1}{\sqrt{{x}^{\text{2/3")sqrt(x^"2/3}} + 1}} \mathrm{dx}$

= int_1^8 1/x^"1/3"sqrt((x^"1/3")^2 + 1)dx

Now let's make the substitution:
$u = {x}^{\text{1/3}} \implies {u}^{3} = x$
$\mathrm{dx} = 3 {u}^{2} \mathrm{du}$

$= 3 {\int}_{1}^{8} {u}^{2} \cdot \frac{1}{u} \sqrt{{u}^{2} + 1} \mathrm{du}$

$= 3 {\int}_{1}^{8} \sqrt{{u}^{2} + 1} \cdot u \mathrm{du}$

The way I would proceed is with another substitution. Let:
$w = {u}^{2} + 1$
$\mathrm{dw} = 2 u \mathrm{du}$
$u \mathrm{du} = \frac{\mathrm{dw}}{2}$

$\implies \frac{3}{2} {\int}_{1}^{8} \sqrt{w} \mathrm{dw}$

$= {w}^{\text{3/2" = (u^2 + 1)^"3/2}}$

= [(x^"2/3" + 1)^"3/2"]|_(1)^(8)

= [(8^"2/3" + 1)^"3/2"] - [(1^"2/3" + 1)^"3/2"]

$= {\left(4 + 1\right)}^{\text{3/2" - (1 + 1)^"3/2}}$

$= {5}^{\text{3/2" - 2^"3/2}}$

$= \sqrt{125} - \sqrt{8}$

$= \textcolor{b l u e}{5 \sqrt{5} - 2 \sqrt{2}} \approx 8.3519 \text{u}$