If $x=a(\theta+\sin \theta)$ and $y=a(1-\cos \theta)$ then $\left(\frac{d^2 y}{dx^2}\right)_{a t \theta=\pi…
If $x=a(\theta+\sin \theta)$ and $y=a(1-\cos \theta)$ then $\left(\frac{d^2 y}{dx^2}\right)_{a t \theta=\pi / 2}=$
- $\frac{a}{2}$
- $\frac{1}{\mathrm{a}}$
- a
- 2a
Solution
$\begin{aligned} & \mathrm{x}=\mathrm{a}(\theta+\sin \theta) \text { and } \mathrm{y}=\mathrm{a}(1-\cos \theta) \\ & \therefore \quad \frac{d x}{d \theta}=a(1+\cos \theta) \text { and } \frac{d y}{d \theta}=a(\sin \theta) \\ & \therefore \quad \frac{d y}{d x}=\frac{a \sin \theta}{a(1+\cos \theta)}=\frac{\sin \theta}{1+\cos \theta} \\ & \therefore \quad \frac{d^2 y}{d x^2}=\frac{d}{d \theta}\left(\frac{d y}{d x}\right)\left(\frac{d \theta}{d x}\right)=\frac{d}{d \theta}\left(\frac{\sin \theta}{1+\cos \theta}\right) \times \frac{1}{\left(\frac{d x}{d \theta}\right)} \\ & =\frac{(1+\cos \theta)(\cos \theta)-\sin \theta(-\sin \theta)}{(1+\cos \theta)^2 \mathrm{a}(1+\cos \theta)} \\ & =\frac{\cos \theta+\cos ^2 \theta+\sin ^2 \theta}{\mathrm{a}(1+\cos \theta)^3}=\frac{1+\cos \theta}{\mathrm{a}(1+\cos \theta)^3}=\frac{1}{\mathrm{a}(1+\cos \theta)^2} \\ & \therefore \quad\left(\frac{\mathrm{d}^2 y}{\mathrm{dx}^2}\right)_{\theta=\frac{\pi}{2}}=\frac{1}{\mathrm{a}\left(1+\cos \frac{\pi}{2}\right)^2}=\frac{1}{\mathrm{a}} \\ & \end{aligned}$
Asked in: MHT CET 2021 (24 Sep Shift 1)
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