Q7  If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx x = \frac{\sin ^3 t }{\sqrt {\cos 2t }} , y = \frac{\cos ^3 t }{\sqrt {\cos 2t }}

Answers (1)

Given equations are
x = \frac{\sin ^3 t }{\sqrt {\cos 2t }} , y = \frac{\cos ^3 t }{\sqrt {\cos 2t }}
Now, differentiate both w.r.t  t
We get,
\frac{dx}{dt}=\frac{d(\frac{\sin ^3 t }{\sqrt {\cos 2t }})}{dt}=\frac{\sqrt{\cos 2t}.\frac{d(\sin^3t)}{dt}-\sin^3t.\frac{d(\sqrt{\cos2t})}{dt}}{(\sqrt{\cos2t})^2} =\frac{3\sin^2 t\cos t.\sqrt{\cos 2t}-\sin^3t.\frac{1}{2\sqrt{\cos 2t}}.(-2\sin 2t)}{{\cos 2t}}
                                                                                                             =\frac{3\sin^2t\cos t . \cos 2t+sin^3t\sin 2t}{\cos2t\sqrt{\cos2t}}
                                                                                                             =\frac{\sin^3t\sin2t(3\cot t \cot 2t+1)}{\cos2t\sqrt{\cos2t}} \ \ \ \ \ (\because \frac{\cos x}{\sin x}=\cot x)
Similarly,
\frac{dy}{dt}=\frac{d( \frac{\cos ^3 t }{\sqrt {\cos 2t }})}{dt}=\frac{\sqrt{\cos 2t}.\frac{d(\cos^3t)}{dt}-\cos^3t.\frac{d(\sqrt{\cos2t})}{dt}}{(\sqrt{\cos2t})^2} =\frac{3\cos^2 t(-\sin t).\sqrt{\cos 2t}-\cos^3t.\frac{1}{2\sqrt{\cos 2t}}.(-2\sin 2t)}{(\sqrt{\cos 2t})^2}
                                                                                                             =\frac{-3\cos^2t\sin t\cos2t+\cos^3t\sin 2t}{\cos2t\sqrt{\cos2t}}
                                                                                                             =\frac{ \sin2t\cos^3t(1-3\tan t \cot 2t)}{\cos2t\sqrt{\cos2t}}
Now, \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{\frac{ \sin2t\cos^3t(1-3\tan t \cot 2t)}{\cos2t\sqrt{\cos2t}} }{\frac{\sin^3t\sin2t(3\cot t \cot 2t+1)}{\cos2t\sqrt{\cos2t}}} = \frac{\cot^3t(1-3\tan t \cot 2t)}{(3\cot t \cot 2t+1)}
                                                                                 = \frac{\cos^3t(1-3.\frac{\sin t}{\cos t}.\frac{\cos2t}{\sin 2t})}{\sin^3t(3.\frac{\cos t}{\sin t}.\frac{\cos 2t}{\sin 2t}+1)} = \frac{\cos^2t(\cos t\sin2t -3\sin t \cos 2t)}{\sin^2t(3\cos t \cos2t+\sin t \sin 2t)}
                                                                                 =\frac{\cos^2t(\cos t .2\sin t \cos t - 3\sin t (2\cos^2t-1))}{\sin^2t(3\cos t(1-2\sin^2 2t)+\sin t.2\sin t \cos t)}
                                                                                                                                                                                                                                                                                                                                                                              (\because \sin 2x = 2\sin x\cos x \ and \ \cos 2x = 2\cos^2x-1 \ and \ \cos 2x = 1-2\sin^2x)
                                                                                =\frac{\cos^2t(2\sin t\cos^2 t-6\sin t\cos^2t+3\sin t)}{\sin^2t(3\cos t-6\cos t \sin^2t+2\sin^2\cos t)}\\=\frac{sint cost(-4cos^3t+3cost)}{sintcost(3sint-4sin^3t)}

\frac{dy}{dx} = \frac{-4\cos^3t+3\cos t}{3\sin t -4\sin^3 t}= \frac{-\cos 3t}{\sin 3t} = -\cot 3t                   \left ( \because \sin3t = 3\sin t-4\sin^3t \\ \ and \ \cos3t = 4\cos^3t - 3\cos t \right )
 
Therefore, the answer is \frac{dy}{dx} = -\cot 3t

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