h. the curve traced by a point on a circle as it rolls on a straight line has a parametric equations (…

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})