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Find D Y D X ,When X = a E θ ( Sin θ − Cos θ ) , Y = a E θ ( Sin θ + Cos θ ) ? - Mathematics

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प्रश्न

Find \[\frac{dy}{dx}\],when \[x = a e^\theta \left( \sin \theta - \cos \theta \right), y = a e^\theta \left( \sin \theta + \cos \theta \right)\] ?

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उत्तर

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

\[\Rightarrow \frac{dx}{d\theta} = a\left[ e^\theta \frac{d}{d\theta}\left( \sin\theta - \cos\theta \right) + \left( \sin\theta - \cos\theta \right)\frac{d}{d\theta}\left( e^\theta \right) \right] \text{ and } \frac{dy}{d\theta} = a\left[ e^\theta \frac{d}{d\theta}\left( \sin\theta + \cos\theta \right) + \left( \sin\theta + \cos\theta \right)\frac{d}{d\theta}\left( e^\theta \right) \right]\]

\[ \Rightarrow \frac{dx}{d\theta} = a\left[ e^\theta \left( \cos\theta + \sin\theta \right) + \left( \sin\theta - \cos\theta \right) e^\theta \right] \text{ and } \frac{dy}{d\theta} = a\left[ e^\theta \left( \cos\theta - \sin\theta \right) + \left( \sin\theta + \cos\theta \right) e^\theta \right]\]

\[ \Rightarrow \frac{dx}{d\theta} = a\left[ 2 e^\theta \sin\theta \right] \text{ and } \frac{dy}{d\theta} = a\left[ 2 e^\theta \cos\theta \right] \]

\[\therefore \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{a\left( 2 e^\theta \cos\theta \right)}{a\left( 2 e^\theta \sin\theta \right)} = \cot\theta\]

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पाठ 11: Differentiation - Exercise 11.07 [पृष्ठ १०३]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 11 Differentiation
Exercise 11.07 | Q 4 | पृष्ठ १०३

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