Question

If $ x=a(1+\cos \theta ),y=a(\theta +\sin \theta ), $ then $ \frac{{{d}^{2}}y}{d{{x}^{2}}} $ at $ \theta =\frac{\pi }{2} $ is

• A. $ -\frac{1}{a} $
• B. $ \frac{1}{a} $
• C. $ -1 $
• D. $ -2 $
• E. $ -\frac{2}{a} $

Answer 1

Answer: (A)

Given that, $ x=a(1+\cos \theta ) $ ...(i) and $ y=a(\theta +\sin \theta ) $ ...(ii) Differentiating Eqs. (i) and (ii) w.r.t.
$ \theta , $ we get $ \frac{dx}{d\theta }=-a\sin \theta $ ...(iii) and
$ \frac{dy}{d\theta }=a+a\cos \theta $ ...(iv)
Dividing Eq. (iii) by Eq. (iv), we get $ {\frac{dy}{d\theta }}/{\frac{dx}{d\theta }}\;=\frac{a(1+\cos \theta )}{-a\sin \theta } $
$ \Rightarrow $ $ \frac{dy}{dx}=\frac{1+\cos \theta }{-\sin \theta } $
$=-\frac{1+2{{\cos }^{2}}\frac{\theta }{2}-1}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}=-\frac{\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}} $
$=-\cot \frac{\theta }{2} $
$ \therefore $ $ \frac{{{d}^{2}}y}{d{{x}^{2}}}=\cos e{{c}^{2}}\frac{\theta }{2}.\frac{1}{2}\frac{d\theta }{dx} $
$=-\frac{1}{2}\frac{1}{a}\frac{\cos e{{c}^{2}}\frac{\theta }{2}}{\sin \theta } $
$ \therefore $ $ {{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{\theta =\frac{\pi }{2}}}=-\frac{1}{a} $