Find the equation of the tangent and the normal to the following curve

1.	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» 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 + sin³θ = -tan³ θ(x - 3cos θ + 3cos³ θ) equation of normal is given by y – y₁ = m(normal)(x – x₁) y - 3 sin θ + sin³θ = cot³ θ(x - 3cos θ + 3cos³ θ)

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»

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 + sin³θ = -tan³ θ(x - 3cos θ + 3cos³ θ)

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

y - 3 sin θ + sin³θ = cot³ θ(x - 3cos θ + 3cos³ θ)