Question

If \(x = asin^3t\) and \(y = acos^3t\), then the value of \(\frac{d^2y}{dx^2}\) at \(t = \frac{π}{3}\) is equal to

• A. \(\frac{-32}{27a}\)
• B. \(\frac{-32}{3\sqrt{3}a}\)
• C. \(\frac{32}{27a}\)
• D. \(\frac{-32}{9a}\)

Hint:

Find dy/dx = -cot t, then differentiate with respect to t and divide by dx/dt.

Answer 1

The Correct Option is C

Step 1: Approach
Convert to the Cartesian curve $x^{2/3}+y^{2/3}=a^{2/3}$ idea is long; instead use the chain rule with $t$.

Step 2: Chain rule
$y'=-\cot t$. Then $y''=\dfrac{dy'/dt}{dx/dt}$ with $dy'/dt=\csc^2t$ and $dx/dt=3a\sin^2t\cos t$.

Step 3: Numbers
At $t=\pi/3$: $\csc^2t=\dfrac43$, $\sin^2t=\dfrac34$, $\cos t=\dfrac12$, so $dx/dt=3a\cdot\dfrac34\cdot\dfrac12=\dfrac{9a}{8}$.
\[ y''=\frac{4/3}{9a/8}=\frac{32}{27a} \]
Option (C).

Final Answer:
At t = pi/3 the second derivative equals 32/(27a), option (C). \[ \boxed{\frac{32}{27a}} \]