Question

Find the length of the curve: ${\left(\frac{x}{a}\right)}^{2/3}+{\left(\frac{y}{b}\right)}^{2/3}=1$.

Answer 1

$Lengthofacurvey-f\left(x\right)fromatobis$

$L=\underset{x=a}{\overset{b}{\int }}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx$

$Given$

${\left(\frac{x}{a}\right)}^{2/3}+{\left(\frac{y}{b}\right)}^{2/3}=1$

$L=4\times lengthofcurvefromx=0toa$

$L=4\times l$

$l=\underset{0}{\overset{a}{\int }}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx$

${\left(\frac{y}{b}\right)}^{2/3}=1-{\left(\frac{x}{a}\right)}^{2/3}$

$\Rightarrow \frac{y}{b}={[1-{\left(\frac{x}{a}\right)}^{2/3}]}^{3/2}$

$\Rightarrow y=b{[1-{\left(\frac{x}{a}\right)}^{2/3}]}^{3/2}$

$\Rightarrow \frac{dy}{dx}=b\times \frac{3}{2}{[1-{\left(\frac{x}{a}\right)}^{2/3}]}^{1/2}\left[\frac{2}{3}{\left(\frac{x}{a}\right)}^{-1/3}\right]\left[\frac{-1}{a}\right]$

$\Rightarrow \frac{dy}{dx}=\frac{-b}{a}{\left(\frac{x}{a}\right)}^{-1/3}{[1-{\left(\frac{x}{a}\right)}^{2/3}]}^{1/2}$

$\Rightarrow {\left(\frac{dy}{dx}\right)}^2=\frac{b^2}{a^2}{\left(\frac{x}{a}\right)}^{-2/3}[1-{\left(\frac{x}{a}\right)}^{2/3}]$

$\Rightarrow {\left(\frac{dy}{dx}\right)}^2=\frac{b^2}{a^2}[{\left(\frac{x}{a}\right)}^{-2/3}-1]$

$l=\underset{0}{\overset{a}{\int }}\sqrt{1+\frac{b^2}{a^2}[{\left(\frac{x}{a}\right)}^{-2/3}-1]}dx$

$\Rightarrow letsassumea=b$

$l=\underset{0}{\overset{a}{\int }}\sqrt{1+{\left(\frac{x}{a}\right)}^{-2/3}-1}dx$

$\Rightarrow l=\underset{0}{\overset{a}{\int }}{\left(\frac{x}{a}\right)}^{-1/3}dx$

$\Rightarrow l={\left[\frac{3a}{2}{\left(\frac{x}{a}\right)}^{2/3}\right]}_0^a$

$\Rightarrow l=\frac{3a}{2}$

$L=4\times l$

$\Rightarrow L=4\times \frac{3a}{2}$

$\Rightarrow L=6a$