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If $x$ and $y$ are connected parametrically by the equations given in $ x = a \cos \theta, y = b \cos \theta $ without eliminating the parameter, Find $\frac{\large dy}{\large dx}$

$\begin{array}{1 1} \large\frac{3b}{2a} \\\large\frac{b}{2a} \\ \large\frac{-b}{a} \\ \large\frac{b}{a} \end{array} $

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  • By chain rule we have $\large\frac{dy}{dx}=\frac{dy}{d\theta}$$\times$$\large\frac{d\theta}{dx}$
Step 1:
Given :
$x=a\cos\theta$
Differentiate with respect to $\theta$
$\large\frac{dx}{d\theta}$$=a(-\sin\theta)$
$f'(t)=-a\sin\theta$
Step 2:
$y=b\cos\theta$
Differentiate with respect to $\theta$
$\large\frac{dy}{d\theta}$$=b(-\sin\theta)$
$g'(t)=-b\sin\theta$
Step 3:
$\large\frac{dy}{dx}=\frac{dy}{d\theta}$$\times$$\large\frac{d\theta}{dx}$
$\quad\;=\large\frac{dy}{d\theta}$$\times\large\frac{d\theta}{dx}$
$\quad\;=-b\sin\theta\times \large\frac{1}{-a\sin\theta}$
$\quad\;=\large\frac{b}{a}$
answered May 10, 2013 by sreemathi.v
 

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