Respuesta :
Given :
[tex] \: \: \: [/tex]
- [tex] \rm \large \: x = a ( \Theta + Sin \Theta )[/tex]
[tex] \: \: \: [/tex]
- [tex] \rm \large y = a ( 1 - cos \: \Theta )[/tex]
[tex] \: \: [/tex]
Now , x = a ( θ + sin θ )
[tex] \: \: [/tex]
Diff w.r.t " θ "
[tex] \: \: \: [/tex]
- [tex] \rm \large\frac{dx}{dθ } = a \: \frac{d}{dθ} (θ + \sin\theta )[/tex]
[tex] \: \: [/tex]
- [tex] \boxed{ \rm \large\underline{ \frac{dx}{d \theta} = a(1 + \cos \theta ) }}[/tex]
[tex] \: \: [/tex]
Now y = a ( 1-cosθ)
[tex] \: \: [/tex]
Diff w.r.t " θ " we get .
[tex] \: \: [/tex]
- [tex] \rm \large \frac{dy}{d \theta} = a \frac{d}{d \theta} (1 - cos \theta) [/tex]
[tex] \: \: \: [/tex]
- [tex] \boxed{ \rm \large \underline{ \frac{dy}{d \theta} = a \sin \theta}}[/tex]
[tex] \: \: \: [/tex]
From eqn ( 1 ) & ( 2 )
[tex] \: \: [/tex]
- [tex] \rm \large \frac{dy}{dx} = \frac{ \frac{dy}{d \theta} }{\frac{dx}{d \theta}} [/tex]
[tex] \: \: \: [/tex]
- [tex] \: \: \: \rm \large = \frac{ \cancel{a} \: sin \theta}{ \cancel a \: (1 + cos \theta)} [/tex]
[tex] \: \: [/tex]
- [tex] \rm \large \: = \frac{sin \theta}{1 + \cos \theta } [/tex]
[tex] \: \: \: [/tex]
- [tex] \rm \large \: \frac{2 \sin( \frac{\theta }{2} ) cos\frac{\theta }{2} }{2 \: cos ^{2} \frac{\theta }{2}} \: \: ......(sin \: a \: = 2 \: sin \frac{a}{2} \: cos \frac{a}{2} 1 + \: cos \: a \: = 2cos ^{2} \frac{a}{2} )[/tex]
[tex] \: \: [/tex]
- [tex] \rm \large \: \frac{dy}{dx} = \frac{sin \frac{ \theta}{2} }{cos \frac{ \theta}{2} } [/tex]
[tex] \: \: [/tex]
- [tex] \rm \large \: \frac{dy}{dx} = tan \frac{ \theta}{2} [/tex]
[tex] \: \: [/tex]
- [tex] \rm \large \: ( \frac{dy}{dx} ) = tan \frac{ \frac{ \theta}{2} }{2} [/tex]
[tex] \: \: [/tex]
- [tex] \rm \large \: = tan( \frac{ \theta}{4} )[/tex]
[tex] \: \: [/tex]
- [tex] \boxed{ \rm \large \underline{ ( \frac{dy}{dx} ) = \frac{\pi}{2} = 1}}[/tex]
[tex] \: \: [/tex]
Hope Helps!:)
