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KEAM 2010 Math Paper

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Question : 33 of 120

Marks: +1, -0

If

x=a(1+\cos\theta), y=a(\theta+\sin\theta),

\;\frac{d^{2} y}{d x^{2}}

\theta=\;\frac{\pi}{2}

is

$-\;\frac{1}{a}$
$\;\frac{1}{a}$
-1
-2
$-\;\frac{2}{a}$

Solution:

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{d x}{d \theta} = -a \sin\theta

....(iii)

and

\;\frac{d y}{d \theta} = a + a \cos\theta

.....(iv)

Dividing Eq. (iii) by Eq. (iv), we get

\;\frac{\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}} = \csc^{2} \;\frac{\theta}{2} \cdot \;\frac{1}{2} \;\frac{d\theta}{dx}

=-\;\frac{1}{2} \;\frac{1}{a} \;\frac{\csc^{2} \frac{\theta}{2}}{\sin \theta}

\therefore \left(\;\frac{d^{2} y}{d x^{2}}\right)_{\theta=\;\frac{\pi}{2}} = -\;\frac{1}{a}

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