If x = a ( cos + sin ) and y = a ( sin - cos ) , then d y d x is equal to:

Question

If $x = a (\cos θ + θ \sin θ)$ and $y = a (\sin θ - θ \cos θ)$ , then $\frac{d y}{d x}$ is equal to:

Smart Achievers · Accepted Answer

(1)

Ans. $t a n θ$

Explanation:

Given, $x = a (\cos θ + θ \sin θ), y = a (\sin θ - θ \cos θ)$

On differentiating with respect to $θ$ , we get

$\frac{d x}{d θ} = a \frac{d}{d θ} (\cos θ + θ \sin θ)$

$= a {\frac{d}{d θ} (\cos θ) + \frac{d}{d θ} (θ \sin θ)}$

$= a {- \sin θ + (θ \cos θ + \sin θ)}$

$= a θ \cos θ$

[using product rule in $\frac{d}{d θ} (θ \sin θ)$ ]

and $\frac{d y}{d θ} = a \frac{d}{d θ} (\sin θ - θ \cos θ)$

$= a {\frac{d}{d θ} (\sin θ) - \frac{d}{d θ} (θ \cos θ)}$

$= a {\cos θ - (θ (- \sin θ) + \cos θ)}$

$= a θ \sin θ$

[using product rule in $\frac{d}{d θ} (θ \cos θ)$ ]

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

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