Question

Find the equation of tangent at $t = \dfrac{\pi}{2}$ on the curve $x = a \sin^{3}t$, $y = b \cos^{3}t$.

Hint:

For parametric curves, slope of tangent = $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$.

Answer 1

We are given the parametric equations: \[ x = a \sin^{3}t, y = b \cos^{3}t \]

Step 1: Differentiate $x$ and $y$ with respect to $t$. \[ \frac{dx}{dt} = 3a \sin^{2}t \cos t, \frac{dy}{dt} = -3b \cos^{2}t \sin t \]

Step 2: Find slope of tangent. \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-3b \cos^{2}t \sin t}{3a \sin^{2}t \cos t} = -\frac{b \cos t}{a \sin t} \]

Step 3: At $t = \dfrac{\pi{2}$.} \[ x = a \sin^{3}\left(\frac{\pi}{2}\right) = a(1)^{3} = a \] \[ y = b \cos^{3}\left(\frac{\pi}{2}\right) = b(0)^{3} = 0 \] \[ \frac{dy}{dx} = -\frac{b \cos(\pi/2)}{a \sin(\pi/2)} = -\frac{b \cdot 0}{a \cdot 1} = 0 \]

Step 4: Equation of tangent. Point: $(a, 0)$, slope $m = 0$. Equation: \[ y - 0 = 0 \cdot (x - a) $\Rightarrow$ y = 0 \]

Final Answer: \[ \boxed{y = 0} \]