If x = a θ and y = a θ, then d² y/d x² = {1cm}{0.15mm} \, (a ≠ 0; θ ≠ kπ, k Z).

Correct Answer is : −1acsc⁡3θ-1/a ³ θ−a1csc3θ

Correct Answer is : −1acsc⁡3θ-1/a ³ θ−a1csc3θ

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GUJCET 2025 Math Paper

Question : 21 of 40

Marks: +1, -0

x = a \cos \theta

y = a \sin \theta

\frac{d^2 y}{d x^2} = \rule{1cm}{0.15mm} \, (a \neq 0; \theta \neq k\pi, k \in Z)

Solution:

Concept:

For parametric equations, first find

\frac{dy}{dx}

using

\theta

, then differentiate again with respect to

x

Explanation:

Given

x=a\cos\theta

and

y=a\sin\theta

Differentiate with respect to

\theta

\frac{dx}{d\theta}=-a\sin\theta

and

\frac{dy}{d\theta}=a\cos\theta

\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=-\cot\theta

Now use the chain rule:

\frac{d^2y}{dx^2}=\frac{d}{d\theta}(-\cot\theta)\cdot\frac{d\theta}{dx}

Since

\frac{d}{d\theta}(-\cot\theta)=\csc^2\theta

and

\frac{d\theta}{dx}=\frac{1}{-a\sin\theta}

Thus

\frac{d^2y}{dx^2}=\csc^2\theta\cdot\frac{1}{-a\sin\theta}

\frac{d^2y}{dx^2}=-\frac{1}{a}\csc^3\theta

Answer:

-\frac{1}{a}\csc^3\theta

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