Question

Find dy/dx when y=a sin theta and x= b cos theta.​

Answer 1

SOLUTION_☺️

Given,x= a Costheta

Differentiating w.r.t.t, we get,

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$\frac{dx}{dtheta} = a \frac{d}{dtheta} costheta \\ = = > \frac{dx}{dtheta} = a( -sintheta) \\ = = > \frac{dx}{dtheta} = - asintheta$

And,

y= b sintheta

Differentiating w.r.t.theta, we get,

$= = > \frac{dy}{dtheta} = \frac{d}{dtheta} (b \: sintheta \\ = = > \frac{dy}{dtheta} = b \frac{d}{dtheta} (sintheta) \\ = = > \frac{dy}{dtheta} = b(costheta) \\ = = > \frac{dy}{dtheta} = b \: costheta \\ thus \\ \frac{dy}{dx} = \frac{dy}{dtheta} \frac{dx}{dtheta} = \frac{ - b costheta}{a sintheta} = - \frac{b \: cottheta}{a} \\ = = > \frac{dy}{dx} = - \frac{b \: costheta}{a}$

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Answer 2

The differential equation is $\frac{dy}{dx} = \frac{- a}{b}cost\theta$ when the expression are y = asinθ and x = b cosθ.

Given that,

We have to find the $\frac{dy}{dx}$ of y = asinθ and x = bcosθ.

We know that,

What is differentiation?

Differentiation is the rate at which one quantity changes in relation to another. Speed is determined by the rate at which distance varies with time.

Take y = asinθ

Differentiate on both side with respect to θ.

$\frac{dy}{d\theta}=a\frac{d(sin\theta)}{d\theta}$                      [differentiation of sinθ is cosθ]

$\frac{dy}{d\theta}=acos\theta$------->equation(1)

Now,

Take x = bcosθ

Differentiate on both side with respect to θ.

$\frac{dx}{d\theta}=b\frac{d(cos\theta)}{d\theta}$                      [differentiation of cosθ is -sinθ]

$\frac{dx}{d\theta}=-bsin\theta$--------->equation(2)

Now

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta} }{\frac{dx}{\theta} }$

From equation(1) and equation(2)

$\frac{dy}{dx} = \frac{acos\theta}{-bsin\theta }$

$\frac{dy}{dx} = \frac{-a}{b}cost\theta$

Therefore, The differential equation is $\frac{dy}{dx} = \frac{- a}{b}cost\theta$

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