Question

x=a(cos thita+ thitha sin thitha) and y=a(sin thiyha - thitha cos thitha) find dy/dx​

Answer 1

Given :

↬ Function

$\bf{x=a(cos\theta+\theta\:cos\theta)}$

$\bf{y=a(sin\theta-\theta\:cos\theta)}$

To Find :

$\dfrac{dy}{dx}$

Formula's used :

• Differentiation Formula's :

$1) \dfrac{d(x {}^{n}) }{dx} = nx {}^{n - 1}$

$2) \dfrac{d(sinx)}{dx} = cosx$

$3) \dfrac{d(cosx)}{dx} = -sinx$

Solution :

$\bf y = \bf a(sin \theta - \theta cos \theta)$

Now Differentiate with respect to θ

$\frac{dy}{d \theta} = a(cos \theta + \theta sin \theta - cos \theta)..(1)$

$\bf x = \bf a(cos \theta + \theta sin \theta)$

Now Differentiate with respect to θ

$\dfrac{dx}{d \theta} = a( \frac{d(cos\theta)}{d \theta} + \frac{d( \theta sin \theta)}{d \theta} )$

$\dfrac{dx}{d \theta} =a( - sin \theta + \theta cos\theta +sin \theta) ..(2)$

__________________________

Now divide equation (1) and (2)

$\implies \bf \dfrac{dy}{dx} = \bf ( \frac{cos \theta + \theta sin \theta - cos \theta}{- sin \theta + \theta cos \theta + sin \theta} )$

which is the required solution!