Question

x=a(θ-sinθ), y=a(1-cosθ),Find dy/dx :(Wherever y is defined as a function of x and dx/dt or dx/dθ≠0)

Answer 1

Given, $\bf{x=a(\theta-sin\theta)}$

differentiate x with respect to $\theta$,

$\frac{dx}{d\theta}=\frac{d\{a(\theta-sin\theta)\}}{dx}$

= $a(1-cos\theta)$.....(1)

again, $\bf{y=a(1-cos\theta)}$

differentiate with respect to $\theta$

$\frac{dy}{d\theta}=a\{0-(-sin\theta)\}$

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

now, dy/dx = $\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$

= $\frac{a(sin\theta)}{a(1-cos\theta}$

= $\frac{(1+cos\theta)}{sin\theta}$

=$cosec\theta+cot\theta$

Answer 2

HELLO DEAR,

GIVEN:-
x = a(θ - sinθ)

differentiating x w.r.t θ

dx/dθ = a(1 - cosθ)------( 1 )

AND,
y = a(1 - cosθ)

differentiating y w.r.t θ

dy/dθ = a(sinθ)-----( 2 )

divide-----( 1 ) by -----( 2 )

dy/dx = {dy/dθ}/{dx/dθ}

=> dy/dx = {a(sinθ)}/{a(1 - cosθ)}

=> dy/dx = (sinθ)/(1 - cosθ)

=> dy/dx = {2sin(θ/2)cos(θ/2)}/{2sin²(θ/2)}

=> dy/dx = cot(θ/2)

I HOPE ITS HELP YOU DEAR,
THANKS