Question

Find dy/dx if x=acos^3theta and y=bsin^3theta​

Answer 1

Answer:

Step-by-step explanation:

I Hope You Understand :-)

Answer 2

Answer

$\rm \dashrightarrow -\dfrac{b}{a}\tan\theta$

Solution

Given ,

$\rm \implies x = a\cos^{3}\theta\\\\ \rm \implies \dfrac{dx}{d\theta} = \dfrac{d}{d\theta}(a\cos^{3}\theta )\\\\ \implies\rm \dfrac{dx}{d\theta} = 3a\cos^{2}\theta\sin\theta$

And we have

$\rm \implies y = b\sin^{3}\theta \\\\ \rm \implies \dfrac{dy}{d\theta}= \dfrac{d}{d\theta}(b\sin^{3}\theta) \\\\ \rm \implies \dfrac{dy}{d\theta} =- 3b\sin^{2}\theta \cos\theta$

Therefore ,

$\rm \implies \dfrac{dy}{dx} = \dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta} }\\\\ \rm \implies \dfrac{dx}{dy} = \dfrac{-3b\sin^{2}\theta\cos\theta}{3a\cos^{2}\theta\sin\theta } \\\\ \implies \rm \dfrac{dy}{dx} = - \dfrac{b}{a} \tan^{2}\theta\cot\theta \\\\ \implies\bf \dfrac{dy}{dx} = - \dfrac{b}{a}\tan\theta$