Question

Find the equation of the tangent and the normal to the following curves at the indicated points: 

x = 3 cos θ – cos³ θ, y = 3 sin θ – sin³θ

Answer 1

finding slope of the tangent by differentiating x and y with respect to theta

\(\frac{dx}{d\theta}=-3\sin\theta+3\cos^2\theta\sin\theta\)

\(\frac{dy}{d\theta}=3\cos\theta-3\sin^2\theta\cos\theta\)

Now dividing \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\) to obtain the slope of tangent

\(\frac{dy}{dx}\)\(=\frac{3\cos\theta-3sin^2\theta\cos\theta}{-3\sin\theta+3cos^2\theta\sin\theta}\)\(=-\tan^3\theta\)

m(tangent) at theta is\(=-\tan^3\theta\)

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at theta is cot³θ

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y - 3 sin θ + sin³ θ = -tan³ θ(x - 3cos θ + 3cos³ θ)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

y - 3 sin θ + sin³ θ = cot³ θ(x - 3cos θ + 3cos³ θ)