Question

x = 3cosθ – 2cos³θ, y = 3sinθ – 2sin³θ

Answer 1

Given that: x = 3 cosθ – 2 cos³θ and y = 3sinθ – 2 sin³θ.

Differentiating both the parametric functions w.r.t. θ

${\displaystyle \frac{\text{dx}}{\text{d}\theta }=-3\sin \theta -6{\cos }^2\theta \cdot \frac{\text{d}}{\text{d}\theta }\left(\cos \theta \right)}$

= – 3 sin θ – 6 cos²θ . (– sin θ)

= – 3 sin θ + 6 cos²θ . sin θ

${\displaystyle \frac{\text{dy}}{\text{d}\theta }=3os\theta -6{\sin }^2\theta \cdot \frac{\text{d}}{\text{d}\theta }\left(\sin \theta \right)}$

= = 3 cos θ – 6 sin²θ . cos θ

∴ ${\displaystyle \frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{d}\theta }}{\frac{\text{dx}}{\text{d}\theta }}}$

= ${\displaystyle \frac{3\cos \theta -6{\sin }^2\theta \cos \theta }{-3\sin \theta +6{\cos }^2\theta \cdot \sin \theta }}$

⇒ ${\displaystyle \frac{\text{dy}}{\text{dx}}=\frac{\cos \theta (3-6{\sin }^2\theta )}{\sin \theta (-3+6{\cos }^2\theta )}}$

= ${\displaystyle \frac{\cos \theta [3-6(1-{\cos }^2\theta )]}{\sin \theta [-3+6{\cos }^2\theta ]}}$

= ${\displaystyle \cot \theta \left(\frac{3-6+6{\cos }^2\theta }{-3+6{\cos }^2\theta }\right)}$

= ${\displaystyle \cot \theta \left(\frac{-3+6{\cos }^2\theta }{-3+6{\cos }^2\theta }\right)}$

= cot θ

∴ ${\displaystyle \frac{\text{dy}}{\text{dx}}}$ = cot θ.