If \(x = 3\sin t - \sin 3t\) and \(y = 3\cos t - \cos 3t\), then the value of \(\displaystyle\frac{dy}{dx}\) at \(t = \displaystyle\frac{\pi}{6}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{1}{\sqrt{3}}\)
Differentiating each with respect to t:
\( \displaystyle \frac{dx}{dt} = 3\cos t - 3\cos 3t \)
\( \displaystyle \frac{dy}{dt} = -3\sin t + 3\sin 3t \)
At \(t = \frac{\pi}{6}\), \(3t = \frac{\pi}{2}\), so \(\sin 3t = 1\) and \(\cos 3t = 0\):
\( \displaystyle \frac{dx}{dt} = 3 \times \frac{\sqrt{3}}{2} - 0 = \frac{3\sqrt{3}}{2} \)
\( \displaystyle \frac{dy}{dt} = -3 \times \frac{1}{2} + 3 = \frac{3}{2} \)
In parametric form:
\( \displaystyle \frac{dy}{dx} = \frac{3/2}{3\sqrt{3}/2} = \frac{1}{\sqrt{3}} \)