Question

If 𝑥 = 𝑎(𝜃 − sin 𝜃 ), 𝑦 = 𝑎 (1 − cos 𝜃 ) find 𝑑2𝑦 𝑑𝑥2 .

Answer 1

$-\frac{a}{y^2}$

Answer 2

We are given the parametric equations: $x = a(\theta - \sin \theta )$ $y = a (1 − \cos \theta )$

First, find the derivatives of $x$ and $y$ with respect to $\theta$: $\frac{dx}{d\theta} = a(1 - \cos \theta)$ $\frac{dy}{d\theta} = a \sin \theta$

Now, find the first derivative $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \sin \theta}{a(1 - \cos \theta)} = \frac{\sin \theta}{1 - \cos \theta}$ Using half-angle identities, $\sin \theta = 2 \sin(\theta/2) \cos(\theta/2)$ and $1 - \cos \theta = 2 \sin^2(\theta/2)$: $\frac{dy}{dx} = \frac{2 \sin(\theta/2) \cos(\theta/2)}{2 \sin^2(\theta/2)} = \cot(\theta/2)$

To find the second derivative $\frac{d^2y}{dx^2}$, we use the formula: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{d\theta}\left(\frac{dy}{dx}\right) \cdot \frac{d\theta}{dx}$ We have $\frac{d}{d\theta}(\cot(\theta/2)) = -\frac{1}{2} \csc^2(\theta/2)$. And $\frac{d\theta}{dx} = \frac{1}{dx/d\theta} = \frac{1}{a(1 - \cos \theta)}$.

Substituting these into the formula: $\frac{d^2y}{dx^2} = \left(-\frac{1}{2} \csc^2(\theta/2)\right) \cdot \left(\frac{1}{a(1 - \cos \theta)}\right)$ Using $\csc^2(\theta/2) = \frac{1}{\sin^2(\theta/2)}$ and $1 - \cos \theta = 2 \sin^2(\theta/2)$, so $\sin^2(\theta/2) = \frac{1 - \cos \theta}{2}$: $\frac{d^2y}{dx^2} = \left(-\frac{1}{2} \cdot \frac{2}{1 - \cos \theta}\right) \cdot \left(\frac{1}{a(1 - \cos \theta)}\right) = -\frac{1}{a(1 - \cos \theta)^2}$ Since $y = a(1 - \cos \theta)$, we have $1 - \cos \theta = \frac{y}{a}$. $\frac{d^2y}{dx^2} = -\frac{1}{a \left(\frac{y}{a}\right)^2} = -\frac{1}{a \frac{y^2}{a^2}} = -\frac{a}{y^2}$