If \(x = a\sin^3 t\) and \(y = a\cos^3 t\), then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → −cot t
For a curve in parametric form, \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\).
\( \displaystyle \frac{dx}{dt} = 3a\sin^2 t\cos t \)
\( \displaystyle \frac{dy}{dt} = -3a\cos^2 t\sin t \)
So
\( \displaystyle \frac{dy}{dx} = \frac{-3a\cos^2 t\sin t}{3a\sin^2 t\cos t} \)
\( \displaystyle = -\frac{\cos t}{\sin t} = -\cot t \)
Why the other options are incorrect
Option (1) tan t — Incorrect. It divides the wrong way round (giving dx/dy = −tan t) and also drops the minus sign.
Option (2) −tan t — Incorrect. This is dx/dy: it divides dx/dt by dy/dt, the wrong way round.
Option (3) cot t — Incorrect. It drops the minus sign that comes from differentiating cos t.