Question

x=a(theta+sin theta) y=a(1-cos theta) find d square y/dx square at theta =pi/2

Answer 1

$\sf Answer :$

$\sf Step-by-step explanation :$

$\sf x = a \{\theta + \sin( \theta) \} \\ \\ \sf y = a \{1 - \cos( \theta) \}$

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

$\sf \frac {dy} {d \theta} = a \sin \theta$

$\sf \frac {dx} {d \theta} = a ( 1 + \cos \theta )$

$\sf \frac {\frac {dy} {d \theta}} {\frac {dx} {d \theta}} = \frac { \sin \theta} { 1 + \cos \theta}$

$\sf At \: \theta = \frac { \pi} {2}$

$\sf \frac {dy} {dx} = \frac {1} {1 + 0}$

$\sf \frac {dy} {dx} = 1$

$\sf Hence, \: the \: answer \: is \: 1$

Answer 2

Hiiii ......friend

The answer is here,

I choose alpha instead of theta for my convenience .

Given that,

$= > \: x = a( \alpha + sin \alpha )$

Differentiate both sides w.r.t alpha .

$= > \: \frac{dx}{d \alpha } = \frac{d(a( \alpha + sin \alpha )}{d \alpha }$

$= > \: a(1 + cos \alpha )$

And also given that,

$= > \: y = a(1 - cos \alpha )$

Differentiate both sides w.r.t alpha

$= > \: \frac{dy}{d \alpha } = a(0 - ( - sin \alpha ))$

$= > \: asin \alpha$

$= > \: \frac{dy}{dx} \: = \frac{ \frac{dy}{d \alpha } }{ \frac{dx}{d \alpha } } = \frac{asin \alpha }{a(1 + cos \alpha )}$

$= > \: \frac{sin \alpha }{1 + cos \alpha } = \frac{2sin \frac{ \alpha }{2}cos \frac{ \alpha }{2} }{2 {cos}^{2} \frac{ \alpha }{2} }$

$= > \: \frac{dy}{dx} = tan \frac{ \alpha }{2}$

Differentiate both sides w.r.t x

$= > \: \frac{d }{dx} ( \frac{dy}{dx} ) = \frac{d}{dx} (tan \frac{ \alpha }{2} )$

$= > \: {sec}^{2} \frac{ \alpha }2{} \times \frac{1}{2} \times \frac{d \alpha }{dx}$

$= > \: {sec}^{2} \frac{ \alpha }{2} \times \frac{1}{2} \times \frac{1}{a(1 + cos \alpha )}$

$at \: \alpha = \frac{\pi}{2}$

$= > \: {sec}^{2} \frac{\pi}{4} \times \frac{1}{2} \times \frac{1}{a(1 + cos( \frac{\pi}{2} ))}$

$= > \: 2 \times \frac{1}{2} \times \frac{1}{a}$

=> 1/a .

:-)Hope it helps u.