# How do you find the length of the curve for y=x^(3/2)  for (0,6)?

Mar 10, 2018

See below.

#### Explanation:

We need ${\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$.

${\int}_{0}^{6} \sqrt{1 + \frac{9 x}{4}} \mathrm{dx} = \frac{4}{9} {\int}_{0}^{6} \sqrt{1 + \frac{9 x}{4}} \frac{9}{4} \mathrm{dx}$

$= \frac{4}{9} {\left[\frac{2}{3} {\left(1 + \frac{9 x}{4}\right)}^{\frac{3}{2}}\right]}_{0}^{6}$

$= \frac{8}{27} {\left[{\left(4 + 9 x\right)}^{\frac{3}{2}} / 8\right]}_{0}^{6}$

$= \frac{1}{27} \left(58 \sqrt{58} - 8\right)$

Mar 10, 2018

16.064 (to 3 places of decimals)

#### Explanation:

Denoting some short segment of the curve by $\mathrm{ds}$, and abusing terminology a little, note that it is possible to approximate $\mathrm{ds}$ using the Pythagorean relationship

${\mathrm{ds}}^{2} = {\mathrm{dx}}^{2} + {\mathrm{dy}}^{2}$

Extracting ${\mathrm{dx}}^{2}$ as a factor from the right hand side

${\mathrm{ds}}^{2} = \left(1 + {\mathrm{dy}}^{2} / {\mathrm{dx}}^{2}\right) {\mathrm{dx}}^{2}$

Taking square roots of both sides

$\mathrm{ds} = \sqrt{\left(1 + {\mathrm{dy}}^{2} / {\mathrm{dx}}^{2}\right)} \mathrm{dx}$

from which

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

Now noting

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

implies

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{2} {x}^{\frac{1}{2}}$

so

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

that is

$s = \int \left(\sqrt{1 + \frac{9}{4} x}\right) \mathrm{dx}$

so the required solution is

${\int}_{0}^{6} \left(\sqrt{1 + \frac{9}{4} x}\right) \mathrm{dx}$

$= \frac{1}{2} {\int}_{0}^{6} \left(\sqrt{4 + 9 x}\right) \mathrm{dx}$ (after rearrangement)

Gritting teeth and going for integration by substitution with

$u = 4 + 9 x$

so that

$\frac{\mathrm{du}}{\mathrm{dx}} = 9$

from which

$\frac{1}{9} \int \mathrm{du} = \int \mathrm{dx}$

and noting required limits under the substitution

$u \left(0\right) = 4 \mathmr{and} u \left(6\right) = 58$

so

$s = \left(\frac{1}{2}\right) \left(\frac{1}{9}\right) {\int}_{4}^{58} \left(\sqrt{u}\right) \mathrm{du}$

that is

$s = \left(\frac{1}{2}\right) \left(\frac{1}{9}\right) \left(\frac{2}{3}\right) {\left[{u}^{\frac{3}{2}}\right]}_{4}^{58}$

$= \left(\frac{1}{27}\right) \left({58}^{\frac{3}{2}} - {4}^{\frac{3}{2}}\right)$

$= 16.064$ (3 places of decimals)