Question

find DY by DX if x is equal to theta minus sin theta Y is equal to 1 minus cos theta at theta is equal to pi by 2​

Answer 1

❏Question :-

$\sf \: Find \: \frac{dy}{dx} \: if \: x = \theta - \sin \theta \: \: \: and \: \sf \\ \sf y \: = 1 - \cos \theta \: \: at \: \theta \: = \frac{ \pi}{2} .$

❏Answer :-

$\to \: \boxed{ \: \: \: \sf \frac{dy}{dx} \bigg|_{ \frac{ \pi}{2} } \: = 1} \:$

❏To Find :-

$\implies \sf \frac{dy}{dx} \: \: \: \: at \: \theta = \frac{ \pi}{2}$

❏Solution :-

We have ,

$\to \sf \: x = \theta - \sin \theta \\ \\ \sf \: differentiate \: with \: respect \: to \: \theta\\ \\ \to \sf \frac{dx}{d \theta} = 1 - \cos \theta \: \: \: .......eq.(1) \\ \\ \bf \: and \: we \: have \\ \\ \to \sf \: y = 1 - \cos \theta \\ \\ \: \sf \: differentiate \: with \: respect \: to \: \theta \\ \\ \to \sf \: \frac{dy}{d \theta} = 0 - ( - \sin \theta) \\ \\ \to \sf \frac{dy}{dx} = \sin \theta \: \: \: \: .......eq.(2) \\ \\ \sf now \: \: \frac{eq.(1)}{eq.(2)} \\ \\ \to \sf \frac{ \frac{dy}{d \theta} }{ \frac{dx}{ d \theta} } = \frac{ \sin \theta}{1 - \cos \theta} \\ \\ \to \sf \frac{dy}{dx} = \frac{ \sin \theta}{1 - \cos \theta} \\ \\ \sf \leadsto \frac{dy}{dx} \: \: \: at \: \theta \: = \frac{ \pi}{2} \\ \sf \: put \: \theta = \frac{ \pi}{2} \\ \sf \: \: \to \frac{dy}{dx} \bigg|_{ \frac{ \pi}{2} } \: = \frac{ \sin (\frac{ \pi}{2}) }{1 - \cos (\frac{ \pi}{2}) } \\ \\ \: \sf \to \: \frac{dy}{dx} \bigg|_{ \frac{ \pi}{2} } \: = \frac{1}{1 - 0} \\ \\ \to \: \boxed{ \: \: \: \sf \frac{dy}{dx} \bigg|_{ \frac{ \pi}{2} } \: = 1}$