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If x=2\sin \theta -\sin 2\theta and y=2\cos\theta -\cos 2\theta ,\: \theta \equiv \left [ 0,2\pi \right ], then \frac{\mathrm{d^{2}y} }{\mathrm{d} x^{2}}\: at\: \theta =\pi is :  
Option: 1 \frac{3}{8}
Option: 2 \frac{3}{4}
Option: 3 \frac{3}{2}
Option: 4 -\frac{3}{4}
 

Answers (1)

best_answer

 

 

Differentiation of Function in Parametric Form -

Differentiation of Function in Parametric Form

Sometimes, x and y are given as functions of a single variable, i.e., x = f(t) and y = g(t) are two functions and t is a variable. In such cases x and y are called parametric functions or parametric equations and t is called the parameter. 

To find \frac{dy}{dx} in such cases, first find the relationship between x and y by eliminating the parameter t and then differentiate with respect to t. 

But sometimes it is not possible to eliminate t, then in that case use

\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{f'(t)}{g'(t)}

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{\frac{d x}{d \theta}=2 \cos \theta-2 \cos 2 \theta} \\ {\frac{d y}{d \theta}=-2 \sin \theta+2 \sin 2 \theta} \\ {\therefore \frac{d y}{d x}=\frac{\sin 2 \theta-\sin \theta}{\cos \theta-\cos 2 \theta}} \\ {=\frac{2 \sin \frac{\theta}{2} \cdot \cos \frac{3 \theta}{2}}{2 \sin \frac{\theta}{2} \sin \frac{3 \theta}{2}}=\cot \frac{3 \theta}{2}}

{\frac{d^{2} y}{d x^{2}}=\frac{-3}{2} \csc ^{2} \frac{3 \theta}{2} \cdot \frac{d \theta}{d x}} \\ {\frac{d^{2} y}{d x^{2}}=\frac{-\frac{3}{2} \csc ^{2} \frac{3 \theta}{2}}{2(\cos \theta-\cos 2 \theta)}} \\ {\frac{d^{2} y}{d x^{2}}|_{\theta=\pi}=-\frac{3}{4(-1-1)}=\frac{3}{8}}

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