Question

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

x = a(θ – sin θ), y = a(1 – cos θ) at θ = π/2

Answer 1

Given:

x = a(θ – sin θ), y = a(1 – cos θ) at θ = π/2

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 = a(θ - sin θ )

⇒ y = a(1 – cos θ)

⇒\(\frac{dy}{d\theta}\) = a( \(\frac{dx}{d\theta}\)( 1 ) – \(\frac{dx}{d\theta}\)(cos θ))

\(\therefore\frac{d}{dx}(cosx)=-sinx\)

\(\therefore\frac{d}{dx}(constant)=0\)

 ⇒\(\frac{dy}{d\theta}\) = a( 0( – sin θ))

⇒ \(\frac{dy}{d\theta}\) = asin θ...(2)

The Slope of the tangent is \(\frac{-sin\theta}{(1-cos\theta)}\)

Since,\(\theta =\frac{\pi}{2}\)

\(\therefore(\frac{dy}{dx})_{\theta=\frac{\pi}{2}}\)\(=\frac{sin\frac{\pi}{2}}{(1-cos\frac{\pi}{2})}\)

\(\therefore\) sin( \(\frac{\pi}{2}\) ) = 1

\(\therefore\) cos( \(\frac{\pi}{2}\) ) = 0

\(\therefore\) The Slope of the tangent at x = \(\frac{\pi}{2}\) is 1

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

⇒ The Slope of the normal = \(\frac{-1}{(\frac{dy}{dx})_{\theta=\frac{-\pi}{2}}}\)

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

⇒ The Slope of the normal = – 1