Question

If \(x = a\sin^3 t\) and \(y = b\cos^3 t\), then the value of \(\frac{d^2y}{dx^2}\) is given by

• A. \(\frac{b}{3a^2}\sin^4 t\cos t\)
• B. \(\frac{b}{3a^2}\csc^4 t\sec t\)
• C. \(\frac{a}{3b^2}\sin^4 t\cos t\)
• D. \(\frac{a}{3b^2}\csc^4 t\sec t\)

Hint:

A common mistake in parametric second derivatives is forgetting to multiply by \( \frac{dt}{dx} \) at the end.
Always write the formula as:
\[ \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} \] to ensure you include the \( dx/dt \) term in the denominator.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem requires finding the second derivative of a function defined parametrically in terms of a parameter \( t \).
Key Formula or Approach:
For parametric equations \( x = f(t) \) and \( y = g(t) \):
- First derivative:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \] - Second derivative:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} \]

Step 2: Detailed Explanation:
Let us perform the calculations step-by-step:
-

Step 1: Differentiate \( x \) and \( y \) with respect to \( t \):
Using the chain rule:
\[ \frac{dx}{dt} = \frac{d}{dt}(a\sin^3 t) = 3a\sin^2 t \cos t \] \[ \frac{dy}{dt} = \frac{d}{dt}(b\cos^3 t) = 3b\cos^2 t (-\sin t) = -3b\cos^2 t \sin t \] - Find the first derivative \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{-3b\cos^2 t \sin t}{3a\sin^2 t \cos t} = -\frac{b}{a} \frac{\cos t}{\sin t} = -\frac{b}{a} \cot t \] -

Step 2: Find the second derivative \( \frac{d^2y}{dx^2} \):
Differentiate \( \frac{dy}{dx} \) with respect to \( t \):
\[ \frac{d}{dt}\left(-\frac{b}{a} \cot t\right) = -\frac{b}{a} (-\csc^2 t) = \frac{b}{a} \csc^2 t \] Now, divide this expression by \( \frac{dx}{dt} \):
\[ \frac{d^2y}{dx^2} = \frac{\frac{b}{a} \csc^2 t}{3a\sin^2 t \cos t} \] Simplify the expression by grouping the constants and trigonometric terms:
\[ \frac{d^2y}{dx^2} = \frac{b}{3a^2} \cdot \frac{\csc^2 t}{\sin^2 t \cos t} \] We know that \( \frac{1}{\sin^2 t} = \csc^2 t \) and \( \frac{1}{\cos t} = \sec t \). Substitute these identities:
\[ \frac{d^2y}{dx^2} = \frac{b}{3a^2} \cdot \csc^2 t \cdot \csc^2 t \cdot \sec t \] \[ \frac{d^2y}{dx^2} = \frac{b}{3a^2} \csc^4 t \sec t \] Note: In standard Indian exam prints, cosecant is represented as "cosec". This matches Option (B).

Step 3: Final Answer:
The second derivative is \( \frac{b}{3a^2}\csc^4 t\sec t \).
Therefore, the correct choice is Option (B).