Question

differentiation if X = A cos cube theta y = a sin cube theta prove that DY by DX is equal to minus 3 root of y by x ​

Answer 1

Given:

$x=a\;cos^3\theta\;\;and\;\;y=a\;sin^3\theta$

First we eliminate $\theta$ from these equations

$\frac{x}{a}=cos^3\theta$

$(\frac{x}{a})^\frac{1}{3}=cos\theta$

$(\frac{x}{a})^\frac{2}{3}=cos^2\theta$

similarly

$(\frac{y}{a})^\frac{2}{3}=sin^2\theta$

Adding these equations, we get

$(\frac{x}{a})^\frac{2}{3}+(\frac{y}{a})^\frac{2}{3}=cos^2\theta+sin^2\theta=1$

$\implies\bf\;x^\frac{2}{3}+y^\frac{2}{3}=a^\frac{2}{3}$

Differentiate with respect to x

$\frac{2}{3}x^\frac{-1}{3}+\frac{2}{3}y^\frac{-1}{3}\;\frac{dy}{dx}=0$

$\implies\;x^\frac{-1}{3}+y^\frac{-1}{3}\;\frac{dy}{dx}=0$

$\implies\;y^\frac{-1}{3}\;\frac{dy}{dx}=-x^\frac{-1}{3}$

$\implies\frac{dy}{dx}=-\frac{x^\frac{-1}{3}}{y^\frac{-1}{3}}$

$\implies\frac{dy}{dx}=-\frac{y^\frac{1}{3}}{x^\frac{1}{3}}$

$\implies\boxed{\bf\frac{dy}{dx}=-\sqrt[^3]{\frac{y}{x}}}$