If x=a ³ t and y=a ³ t, then the value of {d² y}{d x²} at t=π/3 is equal to

Correct Answer is : 3227a32/27 a27a32

Correct Answer is : 3227a32/27 a27a32

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MHTCET 20 Apr 2026 Shift 2 Paper

Question : 124 of 150

Marks: +1, -0

x=a \sin ^{3} t

y=a \cos ^{3} t

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

t=\frac{\pi}{3}

is equal to

Solution:

Concept:

For parametric equations, use

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

and then differentiate again with respect to

x

Explanation:

Given

x = a\sin^3 t

and

y = a\cos^3 t

Differentiate both with respect to

t

\frac{dx}{dt} = 3a\sin^2 t \cos t, \quad \frac{dy}{dt} = -3a\cos^2 t \sin t.

Therefore,

\frac{dy}{dx} = \frac{-3a\cos^2 t \sin t}{3a\sin^2 t \cos t} = -\cot t.

Now,

\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}.

Since

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

\frac{d^2y}{dx^2} = \frac{\csc^2 t}{3a\sin^2 t \cos t} = \frac{1}{3a\sin^4 t \cos t}.

t = \frac{\pi}{3}

\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}, \quad \cos\frac{\pi}{3} = \frac{1}{2}.

Thus,

\frac{d^2y}{dx^2} = \frac{1}{3a\left(\frac{\sqrt{3}}{2}\right)^4 \cdot \frac{1}{2}} = \frac{1}{3a \cdot \frac{9}{16} \cdot \frac{1}{2}} = \frac{32}{27a}.

Answer:

\frac{32}{27a}

Option C is correct.

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