Question

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

x = a cos³ θ, y = a sin³ θ at θ = π/4

Answer 1

Given:

x = acos³θ & y = asin³θ at θ = \(\frac{\pi}{4}\)

Here, To find \(\frac{dy}{dx}\), we have to find \(\frac{dy}{d\theta}\) & \(\frac{dx}{d\theta}\) and and divide \(\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\) and we get our desired \(\frac{dy}{dx}\).

 \(\therefore\frac{dy}{dx}(x^n)=n.x^{n-1}\)

⇒ x = acos³θ

⇒\(\frac{dx}{d\theta}\) = a( \(\frac{dx}{d\theta}\)(cos³θ ))

 \(\therefore\frac{d}{d\theta}\)(cos x) = -sin x

⇒\(\frac{dx}{d\theta}\) = a(3cos^{3 – 1}θ x – sin θ)

⇒\(\frac{dx}{d\theta}\) = a(3cos²θ x – sin θ)

⇒\(\frac{dx}{d\theta}\) = -3acos²θ sinθ ......(1)

⇒ y = asin³θ

⇒\(\frac{dy}{d\theta}\) = a( \(\frac{dy}{d\theta}\)(cos³θ ))

 \(\therefore\frac{d}{d\theta}\)(sin x) = cos x

⇒\(\frac{dy}{d\theta}\) = a(3sin^{3 – 1}θ x cos θ)

⇒\(\frac{dy}{d\theta}\) = a(3sin²θ x cos θ)

⇒\(\frac{dy}{d\theta}\) = -3asin²θ cosθ ......(2)

The Slope of the tangent is – tan θ

Since, \(\theta=\frac{\pi}{4}\)

\(\therefore\) tan( \(\frac{\pi}{4}\) ) = 1

\(\therefore\) The Slope of the tangent at x = \(\frac{\pi}{4}\) is – 1

⇒ The Slope of the normal =\(\frac{-1}{\text{The Slope of the tangent}}\)

⇒ The Slope of the normal = \(\frac{-1}{(\frac{dy}{dx})_{\theta=\frac{-\pi}{4}}}\)

⇒ The Slope of the normal =\(\frac{-1}{1}\)

⇒ The Slope of the normal = – 1