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NDA Math Differentiation

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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 : 58 of 64

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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