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 Q12  Find  dy/dx    if y = 12 (1 - \cos t), x = 10 (t - \sin t), -\frac{\pi }{2} <t< \frac{\pi }{2}

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Given  equations are
y = 12 (1 - \cos t), x = 10 (t - \sin t),
Now, differentiate both y and x w.r.t t independently
\frac{dy}{dt}=\frac{d(12(1-\cos t))}{dt}= -12(-\sin t)=12\sin t
And
\frac{dx}{dt}=\frac{d(10(t-\sin t))}{dt}= 10-10\cos t
Now
\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{12\sin t}{10(1-\cos t)} = \frac{6}{5}.\frac{2\sin \frac{t}{2}\cos \frac{t}{2}}{2\sin^2\frac{t}{2}} = \frac{6}{5}.\frac{\cos \frac{t}{2}}{\sin \frac{t}{2}}\\ \\
                                                                                                                  (\because \sin 2x = 2\sin x\cos x \ and \ 1-\cos 2x = 2\sin^2x)
\frac{dy}{dx}=\frac{6}{5}.\cot \frac{t}{2}
Therefore, differentiation w.r.t x is \frac{6}{5}.\cot \frac{t}{2}

Posted by

Gautam harsolia

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