Find dy/dx given x = b cos³t, y = a sin³t

Find $\dfrac{dy}{dx}$ given $x = b\cos^{3}t$, $y = a\sin^{3}t$.

Step 1: Differentiate with respect to $t$

\[ x'_{t} = 3b\cos^{2}t(-\sin t) = -3b\cos^{2}t\sin t, \qquad y'_{t} = 3a\sin^{2}t\cos t. \]

Step 2: Form the ratio

The factor $3$ cancels, as do one power of $\sin t$ and one of $\cos t$: \[\frac{dy}{dx} = \frac{3a\sin^{2}t\cos t}{-3b\cos^{2}t\sin t} = -\frac{a\sin t}{b\cos t} = -\frac{a}{b}\tan t.\]

Step 3: Verification

\[t \to 0:\quad (x,y)\to(b,0),\quad \frac{dy}{dx}\to 0 \text{ — horizontal tangent at the rightmost point} \;\;\checkmark\]

\[t \to \frac{\pi}{2}:\quad (x,y)\to(0,a),\quad \tan t \to \infty \text{ — vertical tangent at the topmost point} \;\;\checkmark\]

\[t = \frac{\pi}{4},\ a=b:\quad \frac{dy}{dx} = -1, \text{ matching the symmetry about } y = x \;\;\checkmark\]

Step 4: Eliminating the parameter

Taking cube roots and using the Pythagorean identity gives the implicit equation of the astroid: $\left(\frac{x}{b}\right)^{2/3} + \left(\frac{y}{a}\right)^{2/3} = \cos^{2}t + \sin^{2}t = 1$.