Question

The second derivative of \(a\sin ^{3}t\) with respect to \(a\cos ^{3}t\) at \(t=\frac{\pi }{4}\) is...

• A. \(\frac{4\sqrt{2}}{3a}\)
• B. \(\frac{3a}{4\sqrt{2}}\)
• C. \(\frac{4a}{3\sqrt{2}}\)
• D. \(\frac{3\sqrt{2}}{4a}\)

Hint:

Treat both as functions of t, find dy/dx and then differentiate it again with respect to x using the chain rule.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Let \(y=a\sin^3t\) and \(x=a\cos^3t\). We need \(\dfrac{d^2y}{dx^2}\) at \(t=\dfrac{\pi}{4}\).

Step 2: First derivative
\(\dfrac{dy}{dt}=3a\sin^2t\cos t\) and \(\dfrac{dx}{dt}=-3a\cos^2t\sin t\).
So \(\dfrac{dy}{dx}=-\tan t\).

Step 3: Second derivative
\(\dfrac{d^2y}{dx^2}=\dfrac{d}{dt}(-\tan t)\cdot\dfrac{dt}{dx}=-\sec^2t\cdot\dfrac{1}{-3a\cos^2t\sin t}=\dfrac{1}{3a\cos^4t\sin t}\).

Step 4: Put t = pi/4
\(\cos^4t=\dfrac14\) and \(\sin t=\dfrac{1}{\sqrt2}\), so the value is \(\dfrac{1}{3a\cdot\frac14\cdot\frac{1}{\sqrt2}}=\dfrac{4\sqrt2}{3a}\).

Final Answer:
The second derivative is \(\dfrac{4\sqrt2}{3a}\). \[ \boxed{\text{(A) }\dfrac{4\sqrt{2}}{3a}} \]