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

\[\text{ We have, x } = a\left( \cos\theta + \theta \sin\theta \right) \text{ and }y = a\left( \sin\theta - \theta \cos\theta \right)\] \[ \Rightarrow \frac{dx}{d\theta} = a\left[ \frac{d}{d\theta}\cos\theta + \frac{d}{d\theta}\left( \theta \sin\theta \right) \right] \text{ and } \frac{dy}{d\theta} = a\left[ \frac{d}{d\theta}\left( \sin\theta \right) - \frac{d}{d\theta}\left( \theta \cos\theta \right) \right]\] \[ \Rightarrow \frac{dx}{d\theta} = a\left[ - \sin\theta + \theta\frac{d}{d\theta}\left( \sin\theta \right) + \sin\theta\frac{d}{d\theta}\left( \theta \right) \right] \text{ and} \frac{dy}{d\theta} = a\left[ \cos\theta - \left\{ \theta\frac{d}{d\theta}\left( \cos\theta \right) + \cos\theta\frac{d}{d\theta}\left( \theta \right) \right\} \right]\] \[ \Rightarrow \frac{dx}{d\theta} = a\left[ - \sin\theta + \theta \cos\theta \right] \text{ and } \frac{dy}{d\theta} = a\left[ \cos\theta + \theta \sin\theta - \cos\theta \right]\] \[ \Rightarrow \frac{dx}{d\theta} = a\theta \cos\theta \text{ and} \frac{dy}{d\theta} = a\theta \sin\theta\] \[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{a\theta \sin\theta}{a\theta \cos\theta} = \tan\theta\]

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

Question

Find $\frac{d y}{d x}$ , when $x = a (\cos θ + θ \sin θ) and y = a (\sin θ - θ \cos θ)$ ?

Solution

$We have, x = a (\cos θ + θ \sin θ) and y = a (\sin θ - θ \cos θ)$

$\Rightarrow \frac{d x}{d θ} = a [\frac{d}{d θ} \cos θ + \frac{d}{d θ} (θ \sin θ)] and \frac{d y}{d θ} = a [\frac{d}{d θ} (\sin θ) - \frac{d}{d θ} (θ \cos θ)]$

$\Rightarrow \frac{d x}{d θ} = a [- \sin θ + θ \frac{d}{d θ} (\sin θ) + \sin θ \frac{d}{d θ} (θ)] and \frac{d y}{d θ} = a [\cos θ - {θ \frac{d}{d θ} (\cos θ) + \cos θ \frac{d}{d θ} (θ)}]$

$\Rightarrow \frac{d x}{d θ} = a [- \sin θ + θ \cos θ] and \frac{d y}{d θ} = a [\cos θ + θ \sin θ - \cos θ]$

$\Rightarrow \frac{d x}{d θ} = a θ \cos θ and \frac{d y}{d θ} = a θ \sin θ$

$∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{a θ \sin θ}{a θ \cos θ} = \tan θ$

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Simple Problems on Applications of Derivatives

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Q 9 Q 8 Q 10

Chapter 11: Differentiation - Exercise 11.07 [Page 103]

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Exercise 11.07 | Q 9 | Page 103

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