# Q. If `x``` = `a`( ```theta` - sin`theta` ) and `y = a` ( 1 - cos`theta` ) find `(d^2y)/dx^2` at `theta` = `Pi /2` .

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### 1 Answer

You need to differentiate both functions `x = a(theta - sin theta)` and `y = a(1 - cos theta)` withe respect to t, such that:

`(dx)/(d theta) = a(1 - cos theta)`

`(dy)/(d theta) = a(0 - (-sin theta)) = a*sin theta`

You need to divide `(dy)/(d theta)` by `(dx)/(d theta)` such that:

`((dy)/(d theta))/((dx)/(d theta)) = (dy)/(dx) = (a*sin theta)/(a(1 - cos theta))`

Reducing duplicate factors yields:

`(dy)/(dx) = sin theta/(1 - cos theta)`

You need to find the second order derivative, `(d^2y)/(dx^2)` , such that:

`(d^2y)/(dx^2) = (d((dy)/(dx)))/(dx) = (((dy)/(dx))/(d theta))/((dx)/(d theta))`

You need to differentiate `(dy)/(dx)` with respect to theta, such that:

` ((dy)/(dx))/(d theta) = ((sin theta)'(1 - cos theta) - sin theta(1 - cos theta)')/((1 - cos theta)^2)`

`((dy)/(dx))/(d theta) = (cos theta(1 - cos theta) - sin theta*sin theta)/((1 - cos theta)^2)`

`((dy)/(dx))/(d theta) = (cos theta - cos^2 theta - sin^2 theta)/((1 - cos theta)^2)`

Using `sin^2 theta + cos^2 theta = 1` yields:

`((dy)/(dx))/(d theta) =(cos theta - 1)/((1 - cos theta)^2)`

Reducing duplicate factors yields:

`((dy)/(dx))/(d theta) =-1/(1 - cos theta)`

`(d^2y)/(dx^2) = (-1/(1 - cos theta))/(a(1 - cos theta))`

`(d^2y)/(dx^2) = -1/(a(1 - cos theta)^2)`

You need to evaluate `(d^2y)/(dx^2)` at `theta = pi/2` , such that:

`(d^2y)/(dx^2)|_(theta = pi/2) = -1/(a(1 - cos (pi/2))^2)`

Since `cos(pi/2) = 0` yields:

`(d^2y)/(dx^2)|_(theta = pi/2) = -1/(a(1 -0)^2) = -1/a`

**Hence, evaluating the second order derivative `(d^2y)/(dx^2)` at `x = pi/2` , yields `(d^2y)/(dx^2)|_(theta = pi/2) = -1/a` .**