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Find $\large \frac{dy}{dx}$ of the function expressed in parametric form $x=3\cos \theta-2\cos^3\theta,y=3\sin \theta-2\sin^3\theta$

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Toolbox:
  • To find $\large\frac{dy}{dx}$ in the case of parametric functions,if $x=\phi(t)$ and $y=\psi(t)$,then $\large\frac{dy}{dx}$$=\large\frac{\Large\frac{dy}{dt}}{\Large\frac{dx}{dt}}$
Step 1:
Given : $x=3\cos\theta-2\cos^3\theta$ and $y=3\sin\theta-2\sin^3\theta$
Consider $x=3\cos \theta-2\cos^3\theta$
Differentiating w.r.t $\theta$ we get,
$\large\frac{dx}{d\theta}$$=-3\sin\theta-6\cos^2\theta(-\sin\theta)$
$\quad\;\;=-3\sin\theta+6\sin^2\theta\sin\theta$
$\quad\;\;=-3\sin\theta(2\cos^2\theta-1)$
Step 2:
Consider $y=3\sin\theta-2\sin^3\theta$
Now differentiating w.r.t $\theta$ we get,
$\large\frac{dy}{d\theta}$$=3\cos\theta-6\sin^2\theta.\cos\theta$
$\quad\quad=3\cos\theta(1-2\sin^2\theta)$
Step 3:
Therefore $\large\frac{dy}{dx}=\frac{dy}{d\theta}\times \frac{d\theta}{dx}$
$\qquad\qquad\;\;\;\;=\large\frac{3\cos\theta(1-2\sin^2\theta)}{3\sin\theta(2\cos^2\theta-1)}$
$\sin^2\theta=1-\cos^2\theta$
$\qquad\qquad\;\;\;\;=\cot\theta\bigg[\large\frac{1-2(1-\cos^2\theta)}{2\cos^2\theta-1}\bigg]$
$\qquad\qquad\;\;\;\;=\cot\theta\bigg[\large\frac{2\cos^2\theta-1}{2\cos^2\theta-1}\bigg]$
$\qquad\qquad\;\;\;\;=\cot\theta$
answered Jul 1, 2013 by sreemathi.v
 

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