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Find D Y D X ,When X = a ( 1 − Cos θ ) and Y = a ( θ + Sin θ ) at θ = π 2 ? - Mathematics

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प्रश्न

Find \[\frac{dy}{dx}\] ,When \[x = a \left( 1 - \cos \theta \right) \text{ and } y = a \left( \theta + \sin \theta \right) \text{ at } \theta  = \frac{\pi}{2}\] ?

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उत्तर

\[\text{ We have, }x = a\left( 1 - \cos\theta \right) \text{ and }y = a\left( \theta + \sin\theta \right)\]

\[ \therefore \frac{dx}{d\theta} = \frac{d}{d\theta}\left[ a\left( 1 - \cos\theta \right) \right] = a\left( \sin\theta \right)\]

\[\text{ and } \]

\[\frac{dy}{d\theta} = \frac{d}{d\theta}\left[ a\left( \theta + \sin\theta \right) \right] = a\left( 1 + \cos\theta \right)\]

\[ \therefore \left[ \frac{dy}{dx} \right]_{\theta = \frac{\pi}{2}} = \left[ \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \right]_{\theta = \frac{\pi}{2}} = \left[ \frac{a\left( 1 + \cos\theta \right)}{a\left( \sin \theta \right)} \right]_{\theta = \frac{\pi}{2}} = \frac{a\left( 1 + 0 \right)}{a} = 1\]

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अध्याय 11: Differentiation - Exercise 11.07 [पृष्ठ १०३]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 11 Differentiation
Exercise 11.07 | Q 6 | पृष्ठ १०३

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