# How do you find the lengths of the curve y=(4/5)x^(5/4) for 0<=x<=1?

Jul 17, 2018

$\frac{8}{15} \left(1 + \sqrt{2}\right)$

#### Explanation:

We Need the formula $s = {\int}_{a}^{b} \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}} \mathrm{dx}$.

given $f \left(x\right) = \frac{4}{5} \cdot {x}^{\frac{5}{4}}$
using that $\left({x}^{n}\right) ' = n {x}^{n - 1}$ we get

$f ' \left(x\right) = \frac{4}{5} \cdot \frac{5}{4} \cdot {x}^{\frac{5}{4} - 1}$

$f ' \left(x\right) = {x}^{\frac{1}{4}}$ so we have to integrate

${\int}_{0}^{1} \sqrt{1 + \sqrt{x}} \mathrm{dx}$
to calculate the indefinite integtral we Substitute

$t = \sqrt{1 + \sqrt{x}}$

so $x = {\left({t}^{2} - 1\right)}^{2}$

$\mathrm{dx} = 4 t \left({t}^{2} - 1\right) \mathrm{dt}$

and we have 4*int t^2(t^2-1)dt=4int(t^4-t^2dt

this is $4 \left({t}^{5} / 5 - {t}^{3} / 3\right) + C = \frac{4}{15} {\left(1 + \sqrt{x}\right)}^{\frac{3}{2}} \left(3 \sqrt{x} - 2\right) + C$
and we have

${\left[\frac{4}{5} {\sqrt{1 + \sqrt{x}}}^{5} - \frac{4}{3} {\sqrt{1 + \sqrt{x}}}^{3}\right]}_{0}^{1} = \frac{8}{15} \left(1 + \sqrt{2}\right)$

Jul 19, 2018

This is where the formula that is in Sonnhard's solution comes from.

#### Explanation:

Set the total length of the curve $C$ over the interval $a \mathmr{and} b$ as ${L}_{c}$

Let us consider a minuscule portion of the curve that is so short and greatly magnified that it looks as though it is straight.

Set its length as $\delta s$ Using Pythagoras we have ${\left(\delta s\right)}^{2} = {\left(\delta x\right)}^{2} + {\left(\delta y\right)}^{2}$

${\lim}_{\delta s \to 0} {\left(\delta s\right)}^{2} \to {\left(\mathrm{ds}\right)}^{2} = {\left(\mathrm{dx}\right)}^{2} + {\left(\mathrm{dy}\right)}^{2}$

Take the square root of both sides

$\mathrm{ds} = \sqrt{{\left(\mathrm{dx}\right)}^{2} + {\left(s y\right)}^{2}}$

Multiply by 1 and you do not change the value. However 1 comes in many forms.

$\mathrm{ds} = 1 \times \sqrt{{\left(\mathrm{dx}\right)}^{2} + {\left(s y\right)}^{2}} \textcolor{w h i t e}{\text{dd")->color(white)("dd}} \mathrm{ds} = \frac{\mathrm{dx}}{\mathrm{dx}} \times \sqrt{{\left(\mathrm{dx}\right)}^{2} + {\left(s y\right)}^{2}}$

$\textcolor{w h i t e}{\text{ddddddddddddddddddddd")->color(white)("d}} \mathrm{ds} = \mathrm{dx} \times \sqrt{{\left(\mathrm{dx}\right)}^{2} / {\left(\mathrm{dx}\right)}^{2} + {\left(\mathrm{dy}\right)}^{2} / {\left(\mathrm{dx}\right)}^{2}}$

$\textcolor{w h i t e}{\text{ddddddddddddddddddddd")->color(white)("d}} \mathrm{ds} = \mathrm{dx} \times \sqrt{1 + {\left(\mathrm{dy}\right)}^{2} / {\left(\mathrm{dx}\right)}^{2}}$

But ${\left(\mathrm{dy}\right)}^{2} / {\left(\mathrm{dx}\right)}^{2}$ may be written as ${\left(f ' \left(x\right)\right)}^{2}$

$\textcolor{w h i t e}{\text{ddddddddddddddddddddd")->color(white)("d}} \mathrm{ds} = \mathrm{dx} \times \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}}$

$\textcolor{w h i t e}{\text{ddddddddddddddddddddd")->color(white)("d")ds=color(white)("dd}} \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}} \times \mathrm{dx}$

Multiply $\mathrm{ds}$ by 1 and use the dot instead of $\times$

$\textcolor{w h i t e}{\text{ddddddddddddddddddddd")->color(white)("d")1.ds=color(white)("dd}} \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}} . \mathrm{dx}$

Sum up all the $\mathrm{ds} ' s$ using integration

$\textcolor{w h i t e}{\text{ddddddddddd")->color(white)("d")int_a^b color(white)(.)1.ds=L_c=color(white)("dd}} {\int}_{a}^{b} \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}} . \mathrm{dx}$

As in Sonnhard's solution