NIMCET 2013 — Mathematics PYQ

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Question

NIMCET 2013 — Mathematics PYQ

NIMCET | Mathematics | 2013

If $x = a \cos t$ , $y = b \sin t$ , then $\frac{d^2y}{dx^2}$ is:

Choose the correct answer:

Explanation

Solution

Step 1: Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$

$x = a \cos t \implies \frac{dx}{dt} = -a \sin t$

$y = b \sin t \implies \frac{dy}{dt} = b \cos t$

Step 2: Find the first derivative $\frac{dy}{dx}$

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{b \cos t}{-a \sin t} = -\frac{b}{a} \cot t

Step 3: Find the second derivative $\frac{d^2y}{dx^2}$

\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d}{dt} \left( -\frac{b}{a} \cot t \right) \cdot \frac{dt}{dx}

\frac{d^2y}{dx^2} = \left( -\frac{b}{a} (-\csc^2 t) \right) \cdot \frac{1}{-a \sin t}

\frac{d^2y}{dx^2} = \frac{b \csc^2 t}{a} \cdot \frac{1}{-a \sin t} = -\frac{b}{a^2 \sin^3 t}

Step 4: Convert the result into terms of $y$

Since $y = b \sin t$ , we have $\sin t = \frac{y}{b}$ .

Substituting $\sin t$ into the equation:

\frac{d^2y}{dx^2} = -\frac{b}{a^2 (\frac{y}{b})^3}

\frac{d^2y}{dx^2} = -\frac{b}{a^2 (\frac{y^3}{b^3})}

\frac{d^2y}{dx^2} = -\frac{b \cdot b^3}{a^2 y^3}

\frac{d^2y}{dx^2} = -\frac{b^4}{a^2 y^3}

Final Answer:

The second derivative is $-\frac{b^4}{a^2y^3}$ , which corresponds to Option 1.