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  • If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx x= a cos t+log tan t/ 2 y = a sin t

Q8    If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx   x = a ( \cos t + \log \tan t/2 ),y = a \sin t

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Given equations are
x = a ( \cos t + \log \tan \frac{t}{2} ),y = a \sin t
Now, differentiate both w.r.t  t
We get,
\frac{dx}{dt}=\frac{d(a ( \cos t + \log \tan \frac{t}{2} ))}{dt}= a(-\sin t + \frac{1}{\tan\frac{t}{2}}.\sec^2\frac{t}{2}.\frac{1}{2})
                                                            = a(-\sin t+\frac{1}{2}.\frac{\cos \frac{t}{2}}{\sin\frac{t}{2}}.\frac{1}{\cos^2\frac{t}{2}}) = a(-\sin t+\frac{1}{2\sin \frac{t}{2}\cos \frac{t}{2}})
                                                            =a(-\sin t+\frac{1}{\sin 2.\frac{t}{2}} ) = a(\frac{-\sin^2t+1}{\sin t})= a(\frac{\cos^2t}{\sin t})
Similarly,
\frac{dy}{dt}=\frac{d(a\sin t)}{dt}= a\cos t
Now, \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{a \cos t }{ a(\frac{\cos^2t}{\sin t})} = \frac{\sin t}{\cos t} = \tan t
Therefore, the answer is \frac{dy}{dx} = \tan t
 

Posted by

Gautam harsolia

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