The tangent to the curve \(x = \cos t(3 - 2\cos^2 t)\), \(y = \sin t(3 - 2\sin^2 t)\) at \(t = \displaystyle\frac{\pi}{4}\), makes with the \(x\)-axis an angle:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{\pi}{4}\)
The angle \(\theta\) that the tangent makes with the x-axis satisfies \(\tan\theta = \frac{dy}{dx}\), the slope of the tangent.
Expanding the curve:
\(x = 3\cos t - 2\cos^3 t\)
\(y = 3\sin t - 2\sin^3 t\)
Differentiating with respect to t:
\( \displaystyle \frac{dx}{dt} = -3\sin t + 6\cos^2 t\sin t \)
\( = 3\sin t\,(2\cos^2 t - 1) = 3\sin t\cos 2t\)
\( \displaystyle \frac{dy}{dt} = 3\cos t - 6\sin^2 t\cos t \)
\( = 3\cos t\,(1 - 2\sin^2 t) = 3\cos t\cos 2t\)
So, cancelling the common factor \(3\cos 2t\),
\( \displaystyle \frac{dy}{dx} = \frac{3\cos t\cos 2t}{3\sin t\cos 2t} = \cot t \)
At \(t = \frac{\pi}{4}\) the factor \(\cos 2t\) is itself 0, so the slope there is the limiting value of \(\cot t\):
\( \displaystyle \tan\theta = \cot\frac{\pi}{4} = 1 \)
\( \displaystyle \theta = \frac{\pi}{4} \)