Question

find dy/dx of x=acos 2tita and y=b sin 2tita​

Answer 1

y = a cos2θ

Diff. w.r.t.x

dy/dx = d/dx [ a cos2θ]

= a (-sin2θ) . d/dx 2θ

= - a -sin2θ

•°• dy/dx = - a sin2θ

And

y = b sin2θ

Diff. w.r.t.x

dy/dx = d/dx [ b sin2θ]

= b cos2θ

•°• dy/dx = b cos2θ

Answer 2

Given , the two functions are

x = acos(2θ)

y = bsin(2θ)

Differentiating x wrt θ , we get

$\tt \frac{dx}{d \theta} = \frac{d \{acos (2\theta) \}}{d \theta}$

$\tt \frac{dx}{d \theta} =a \frac{d \{cos(2 \theta) \}}{ d \theta}$

$\tt \frac{dx}{d \theta} = - asin( \theta) \frac{d \{2 \theta \}}{ d\theta}$

$\tt \frac{dx}{d \theta} = -2 asin(2 \theta)$

Similarly , differentiating y wrt θ , we get get

$\tt \frac{dy}{d \theta} = \frac{d \{asin (2\theta) \}}{d \theta}$

$\tt \frac{dy}{d \theta} =a \frac{d \{sin(2 \theta) \}}{ d \theta}$

$\tt \frac{dy}{d \theta} = acos( \theta) \frac{d \{2 \theta \}}{ d\theta}$

$\tt \frac{dy}{d \theta} = 2 acos(2 \theta)$

Now ,

dy/dx = {dy/dθ} ÷ {dx/dθ}

Thus ,

$\tt \implies \frac{dy}{dx} = - \frac{2 acos(2 \theta)}{2 asin(2 \theta)}$

$\tt \implies \frac{dy}{dx} = -cot(2 \theta)$

Therefore , the required answer is -cot(2θ)

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