If `x = a (cos t + t sin t) and y = a (sin t - t cos t)`, find `(d^2y)/(dx^2)`.

Explanation

We need to find `(d^2y)/(dx^2)`

First we find `dy/dx`

Here,

`dy/dx=(dy/dt)/(dx/dt)`


Calculating `dy/dt`

` y = a (sin t - t cos t)`

Differentiating both sides w.r.t. t

`dy/dt=(d(a (sin t - t cos t)))/dt`

`dy/dt=a(d( (sin t - t cos t)))/dt`

`dy/dt=a((d (sin t ))/dt-(d( t cos t))/dt)`

`dy/dt=a(cost-(d( t cos t))/dt)`

Using product rule

`dy/dt=a(cost-(dt/dt.cost+ (dcos t)/dt.t))`

`dy/dt=a(cost-(cost+(-sint).t))`

`dy/dt=a(cost-(cost-(sint).t))`

`dy/dt=a(cost-cost+t.sint)`

`dy/dt=a(0+t.sint)`

`dy/dt=a.t.sint`


Calculating `dx/dt`

`x = a (cos t + t sin t)`

Differentiating both sides w.r.t. t

`dx/dt=(d(a (cos t + t sin t)))/dt`

`dx/dt=a((d (cos t + t sin t))/dt)`

`dx/dt=a((d (cos t ))/dt+(d (t sin t))/dt)`

`dx/dt=a(-sint+(d (t sin t))/dt)`

Using product rule

`dx/dt=a(-sint+(dt/dt.sint+ (d( sin t))/dt.t))`

`dx/dt=a(-sint+(sint+ cost.t))`

`dx/dt=a(-sint+sint+ t.cost)`

`dx/dt=a.t.cost`


Finding `dy/dx`

`dy/dx=(dy/dt)/(dx/dt)`

`dy/dx=(a.t.sint)/(a.t.cost)`

`dy/dx=tan t`

Again Differentiating both sides w.r.t. x

`d/dx(dy/dx)=(d(tan t))/dx`

`(d^2y)/(dx^2)=(d(tan t))/dx`

`(d^2y)/(dx^2)=(d(tan t))/dx.dt/dt`

`(d^2y)/(dx^2)=(d(tan t))/dt.dt/dx`

`(d^2y)/(dx^2)=sec^2t.dt/dx`

`(d^2y)/(dx^2)=sec^2t/(dx/dt)`

`(d^2y)/(dx^2)=sec^2t/(a.t.cost)`

`(d^2y)/(dx^2)=sec^2t/(a.txx1/sect)`

`(d^2y)/(dx^2)=sec^3t/(a.t)`