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 Q3   If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx .    x = \sin t , y = \cos 2 t

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Given equations are
x = \sin t , y = \cos 2 t
Now, differentiate both w.r.t  t
We get,
\frac{dx}{dt}=\frac{d(\sin t)}{dt}= \cos t
Similarly,
\frac{dy}{dt}=\frac{d(\cos 2t)}{dt}= -2\sin 2t = -4\sin t \cos t \ \ \ \ \ (\because \sin 2x = \sin x\cos x)
Now, \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{-4\sin t \cos t }{\cos t} = -4\sin t
Therefore, the answer is \frac{dy}{dx} = -4\sin t

 

Posted by

Gautam harsolia

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