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UPSC NDA II 2014 Mathematics Paper

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For the next two (02) items that follow:

Consider the curve

x=a(\cos\theta+\theta\sin\theta)

and

y=a(\sin\theta-\theta\cos\theta)

[NDA II 2014]

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Question : 70 of 120

Marks: +1, -0

What is

\frac{d^2y}{dx^2}

$\sec^2\theta$
$-\csc^2\theta$
$\frac{\sec^3\theta}{a\theta}$
None of the above

Solution:

x=a(\cos\theta+\theta\sin\theta)

and

y=a(\sin\theta-\theta\cos\theta)

\Rightarrow\frac{dx}{d\theta}=-a\sin\theta+a\sin\theta+a\theta\cos\theta

and

\frac{dy}{d\theta}=a\cos\theta-a\cos\theta+a\theta\sin\theta

\Rightarrow\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{a\theta\sin\theta}{a\theta\cos\theta}=\tan\theta

\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{\frac{d}{d\theta}\left(\frac{dy}{dx}\right)}{\frac{dx}{d\theta}}=\frac{\sec^2\theta}{a\theta\cos\theta}=\frac{\sec^3\theta}{a\theta}

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