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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the equation of tangent to the curve \( x-\sin\: 3t, y=\cos\: 2t \: at \: t=\large\frac{\pi}{4}.\)

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  • Equation of the tangent at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=m(x-x_1)$
Step 1:
$x=\sin 3t$
Differentiating w.r.t $t$ we get,
$\large\frac{dx}{dt}$$=3\cos 3t$
$y=\cos 2t$
Differentiating w.r.t $t$ we get,
$\large\frac{dy}{dt}$$=-2\sin 2t$
$\large\frac{dy}{dx}=\frac{-2\sin 2t}{3\cos 3t}$
$\big(\large\frac{dy}{dx}\big)_{t=\Large\frac{\pi}{4}}=\frac{-2\sin 2(\pi/4)}{3\cos 3(\pi/4)}$
$\qquad\qquad=\large\frac{-2\sin 2(\pi/2)}{3\cos 3(\pi/4)}$
But $\sin\large\frac{\pi}{2}$$=1$
$\cos\large\frac{3\pi}{4}$$=\large\frac{-1}{\sqrt 2}$
$\therefore \large\frac{dy}{dx}$$=m$
$\Rightarrow \large\frac{-2}{3\times -1/\sqrt 2}$
$\Rightarrow \large\frac{2\sqrt 2}{3}$
Step 2:
Equation of the tangent is
$y-y_1=m(x-x_1)$
Here $x_1=\sin 3t$
$\sin 3(\large\frac{\pi}{4})=\large\frac{-1}{\sqrt 2}$
$y_1=\cos 2\big(\large\frac{\pi}{4}\big)$$=0$
Equation of the tangent is $y-0=\large\frac{2\sqrt 2}{3}$$(x+\large\frac{1}{\sqrt 2}\big)$
$\Rightarrow 3y=2\sqrt 2(\sqrt x+1)$
$\therefore 4x-3y+2\sqrt 2=0$
answered Sep 26, 2013 by sreemathi.v
 

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