Question

If \(x=a\sin^3t,\ y=b\cos^3t\), then find \(\dfrac{dy}{dx}\) at the point \(t=\dfrac{\pi}{2}\).

Hint:

Differentiate both parametrically, form dy/dx = −(b/a)cot t, then plug in t = π/2.

Answer 1

Step 1: Differentiating x and y w.r.t. t:
\(\dfrac{dx}{dt}=3a\sin^2t\cos t\), \(\dfrac{dy}{dt}=-3b\cos^2t\sin t\).

Step 2: Forming dy/dx:
\(\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{-3b\cos^2t\sin t}{3a\sin^2t\cos t}=-\dfrac{b\cos t}{a\sin t}=-\dfrac{b}{a}\cot t\).

Step 3: Evaluating at t=π/2:
\(\cot\dfrac{\pi}{2}=0\), so \(\dfrac{dy}{dx}\Big|_{t=\pi/2}=-\dfrac{b}{a}(0)\).

Final Answer:
\[ \boxed{\dfrac{dy}{dx}\Big|_{t=\pi/2}=0} \]