Question

If \(x = a\sin^{3}t,\ y = b\cos^{3}t\), then find \(\frac{dy}{dx}\) at \(t = \frac{\pi}{4}\).

Hint:

Use dy/dx = (dy/dt)/(dx/dt) to get -(b/a) cot t, then put t = pi/4.

Answer 1

Step 1: Find the derivatives at the given point first:
We can put $t = \frac{\pi}{4}$ before simplifying.
We need $\frac{dx}{dt}$ and $\frac{dy}{dt}$.
$\frac{dx}{dt} = 3a\sin^{2}t\cos t$ and $\frac{dy}{dt} = -3b\cos^{2}t\sin t$.

Step 2: Substitute t = pi/4:
Here $\sin\frac{\pi}{4} = \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$.
So $\sin^{2}t\cos t = \frac{1}{2}\cdot\frac{1}{\sqrt{2}} = \frac{1}{2\sqrt{2}}$.
The same value holds for $\cos^{2}t\sin t$.
\[ \frac{dx}{dt} = \frac{3a}{2\sqrt{2}},\qquad \frac{dy}{dt} = -\frac{3b}{2\sqrt{2}} \]

Step 3: Divide:
\[ \frac{dy}{dx} = \frac{-3b/(2\sqrt{2})}{3a/(2\sqrt{2})} = -\frac{b}{a} \]
The common factor $\frac{3}{2\sqrt{2}}$ cancels.

Final Answer:
The value of $\frac{dy}{dx}$ at $t = \frac{\pi}{4}$ is $-\frac{b}{a}$. \[ \boxed{-\frac{b}{a}} \]