Question

Question: Find dx/dy if x = a(cosθ+θsinθ), and y = a(sinθ−θsinθ).​

Answer 1

Answer:

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Answer 2

$\large\underline{\sf{Solution-}}$

Given that

$\rm \: x = a(cos\theta + \theta sin\theta )$

On differentiating both sides w. r. t. $\theta$, we get

$\rm \: \dfrac{d}{d\theta }x = a\dfrac{d}{d\theta }(cos\theta + \theta sin\theta )$

$\rm \: \dfrac{dx}{d\theta } = a\bigg(\dfrac{d}{d\theta }cos\theta + \dfrac{d}{d\theta }\theta sin\theta\bigg )$

$\rm \: \dfrac{dx}{d\theta } = a\bigg( - sin\theta + \theta \dfrac{d}{d\theta } sin\theta + sin\theta \dfrac{d}{d\theta }\theta \bigg )$

$\rm \: \dfrac{dx}{d\theta } = a\bigg( - sin\theta + \theta cos\theta + sin\theta \times 1 \bigg )$

$\rm \: \dfrac{dx}{d\theta } = a\bigg( - sin\theta + \theta cos\theta + sin\theta \bigg )$

$\bf\implies \:\dfrac{dx}{d\theta } = a \: \theta \: cos\theta - - - (1)$

Further given that

$\rm \: y = a(sin\theta - \theta sin\theta )$

can be rewritten as

$\rm \: y = asin\theta(1 - \theta)$

On differentiating both sides w. r. t. $\theta$, we get

$\rm \:\dfrac{d}{d\theta } y = a\dfrac{d}{d\theta }sin\theta(1 - \theta)$

$\rm \:\dfrac{dy}{d\theta } = a\bigg(sin\theta \dfrac{d}{d\theta }(1 - \theta) + (1 - \theta )\dfrac{d}{d\theta }sin\theta \bigg)$

$\rm \:\dfrac{dy}{d\theta } = a\bigg(sin\theta (0 - 1) + (1 - \theta )cos\theta \bigg)$

$\rm \:\dfrac{dy}{d\theta } = a\bigg( - sin\theta + (1 - \theta )cos\theta \bigg)$

$\bf\implies \:\dfrac{dy}{d\theta } = a\bigg( - sin\theta + cos\theta - \theta cos\theta \bigg)$

Now,

$\rm \: \dfrac{dx}{dy}$

$\rm \: = \: \dfrac{dx}{d\theta } \div \dfrac{dy}{d\theta }$

$\rm \: = \: \dfrac{a \: \theta \: cos\theta }{a( - sin\theta + cos\theta - \theta cos\theta) }$

$\rm \: = \: \dfrac{\theta \: cos\theta }{ - sin\theta + cos\theta - \theta cos\theta}$

Hence,

$\\ \bf\implies \:\dfrac{dx}{dy} = \: \dfrac{\theta \: cos\theta }{ - sin\theta + cos\theta - \theta cos\theta}$

$\rule{190pt}{2pt}$

Formulae Used :-

$\boxed{ \rm{ \:\dfrac{d}{dx}x = 1 \: }}$

$\boxed{ \rm{ \:\dfrac{d}{dx}sinx = cosx\: }}$

$\boxed{ \rm{ \:\dfrac{d}{dx}cosx = - \: sinx\: }}$

$\boxed{ \rm{ \:\dfrac{d}{dx}uv \: = \: u\dfrac{d}{dx}v \: + \: v\dfrac{d}{dx}u \: }}$