Question

h. the curve traced by a point on a circle as it rolls on a straight line has a parametric equations ( x=\theta-sin\theta,y = 1-cos\theta ). find ( \frac{d^{2}y}{dx^{2}} ) in terms of ( \theta ).

Explanation:

Step1: Find \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\)

Using the derivative rules \((\sin\theta)^\prime=\cos\theta\), \((\cos\theta)^\prime =-\sin\theta\), and \((\theta)^\prime = 1\).
For \(y = 1-\cos\theta\), \(\frac{dy}{d\theta}=\sin\theta\).
For \(x=\theta - \sin\theta\), \(\frac{dx}{d\theta}=1-\cos\theta\).
By the formula \(\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\), we have \(\frac{dy}{dx}=\frac{\sin\theta}{1 - \cos\theta}\).

Step2: Find \(\frac{d^2y}{dx^2}\)

Use the formula \(\frac{d^2y}{dx^2}=\frac{\frac{d}{d\theta}(\frac{dy}{dx})}{\frac{dx}{d\theta}}\).
First, find \(\frac{d}{d\theta}(\frac{\sin\theta}{1 - \cos\theta})\) using the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^2}\), where \(u = \sin\theta\), \(u^\prime=\cos\theta\), \(v = 1-\cos\theta\), \(v^\prime=\sin\theta\).
\(\frac{d}{d\theta}(\frac{\sin\theta}{1 - \cos\theta})=\frac{\cos\theta(1 - \cos\theta)-\sin\theta(\sin\theta)}{(1 - \cos\theta)^2}=\frac{\cos\theta-1}{(1 - \cos\theta)^2}=-\frac{1}{1 - \cos\theta}\).
Since \(\frac{dx}{d\theta}=1-\cos\theta\), then \(\frac{d^2y}{dx^2}=\frac{-\frac{1}{1 - \cos\theta}}{1 - \cos\theta}\).

Answer:

\(\frac{d^2y}{dx^2}=-\frac{1}{(1 - \cos\theta)^2}\)