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If $x$ and $y$ are connected parametrically by the equations given in $ x = \sin t, y = \cos 2t $ without eliminating the parameter, Find $\large\frac{dy}{dx}$

$\begin{array}{1 1} -4\cos t \\ -5\sin t \\ 4\sin t \\ -4\sin t \end{array} $

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  • By chain rule we have $\large\frac{dy}{dx}=\frac{dy}{dt}$$\times\large\frac{dt}{dx}$
Step 1:
Given :
$x=\sin t$
Differentiating with respect to $t$
$\large\frac{dx}{dt}$$=\cos t$
$f'(t)=\cos t$
Step 2:
$y=\cos 2t$
Differentiating with respect to $t$
$\large\frac{dy}{dt}$$=-\sin 2t.2$
$\large\frac{dy}{dt}$$=-2\sin 2t$
$g'(t)=-2\sin 2t$
Step 3:
$\large\frac{dy}{dx}=\Large\frac{\Large\frac{dy}{dt}}{\Large\frac{dx}{dt}}$
$\large\frac{dy}{dx}=\frac{dy}{dt}$$\times\large\frac{dt}{dx}$
$\quad\;=-2\sin 2t\times \large\frac{1}{\cos t}$
We know that $\sin 2A=2\sin A\cos A$
Hence $\sin 2t=2\sin t\cos t$
Replacing $\sin 2t$ we get
$\large\frac{dy}{dt}$$=-4\sin t\cos t$
Thus $\large\frac{dy}{dx}=\frac{dy}{dt}$$\times\large\frac{dt}{dx}$
$\qquad\;\;\;\;\;\;=-4\sin t\cos t\times\large\frac{1}{\cos t}$
$\qquad\;\;\;\;\;\;=-4\sin t$
answered May 10, 2013 by sreemathi.v
 

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