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If x = a (\theta - \sin \theta), y = a (1 - \cos \theta), a > 0, then find \frac{d^{2}y}{dx^{2}} at \theta = \frac{\pi}{3}

 

 
 
 
 
 

Answers (1)

Given equations are

x = a (\theta - \sin \theta), y = a (1 - \cos \theta), a > 0

Now, differentiate both w.r.t \theta
We get,
\frac{dx}{d\theta}=\frac{d(a(\theta- \sin \theta))}{d\theta}= a(1-\cos \theta)
Similarly,
\frac{dy}{d\theta}=\frac{d(a(1-\cos \theta))}{d\theta}=a\sin \theta
Now, 

\\ \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \\ \\= \frac{a\sin \theta}{a(1-\cos \theta)} \\\\ = \frac{\sin }{1-\cos \theta} \\\\ = \cot \frac{\theta}{2} \ \ \ \ \ \ \ (\cot \frac{x}{2}=\frac{\sin x}{1-\cos x})

\\ \frac{d^2y}{dx^2}= -\csc^2 \frac{\theta}{2} \times \frac{1}{2} \times \frac{d\theta }{dx} \\\\ = -\csc^2 \frac{\theta}{2} \times \frac{1}{2} \times \frac{1}{a(1-\cos \theta )} \\\\ \frac{d^2y}{dx^2}_{at \frac{\pi}{3}}= -4 \times \frac{1}{2} \times \frac{1}{a(1-\frac{1}{2} )} = - \frac{4}{a}\\\\

 

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Safeer PP

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