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UPSC NDA Math Model Paper 7

Question : 92 of 120

Marks: +1, -0

x=a \sec^{2} \theta, y=a \tan^{2} \theta

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

Solution:

This is the case of both Composite Functions and Parametric Functions. Both x and y are given as composite functions in terms of the parameter θ.

Let us first find out

\frac{dx}{d \theta}

and

\frac{dy}{d \theta}

by using the Chain Rule for Composite Functions.

x=a \sec^{2} \theta

\Rightarrow \frac{dx}{d\theta}=a \frac{d}{d\theta}(\sec^{2}\theta)

=a \times \frac{d}{d(\sec\theta)}(\sec^{2}\theta) \times \frac{d}{d\theta} \sec\theta

\Rightarrow \frac{dx}{d\theta}=a \times 2 \sec\theta \times \tan\theta \sec\theta

=2 a \tan\theta \sec^{2}\theta

y=a \tan^{2}\theta

\Rightarrow \frac{dy}{d\theta}=a \frac{d}{d\theta}(\tan^{2}\theta)=a \times \frac{d}{d(\tan\theta)}(\tan^{2}\theta)

\times \frac{d}{d\theta} \tan\theta

\Rightarrow \frac{dy}{d\theta}=a \times 2 \tan\theta \times \sec^{2}\theta

=2 a \tan\theta \sec^{2}\theta

\therefore \frac{dy}{dx}=\frac{dy}{d\theta} \times \frac{d\theta}{dx}

\Rightarrow \frac{dy}{dx}=2 a \tan\theta \sec^{2}\theta

\times \frac{1}{2 a \tan\theta \sec^{2}\theta}=1

And finally,

\frac{d^{2} y}{dx^{2}}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(1)=0

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