Question

Find dy/dt and d2y/dx2 in terms of t , given x = 5 cos t and y = 4sint ?

Answer 1

`dy/dx = -4/5 \ cott`

`(d^2y)/(dx^2) = 4/25 \ csc^3t`

Explanation:

We have:

`x=5cos t`
`y=4sin t`

We can differentiate wrt `t` to get:

`dx/dt = -5sint`
`dy/dt = 4cos t`

Then we use the chain rule:

`dy/dx = dy/dt * dt/dx`

`\ \ \ \ \ = dy/dt // dx/dt`

`\ \ \ \ \ = (4cos t )/(-5sint)`

`\ \ \ \ \ = -4/5 \ cott`

For the second derivative we differentiate wrt `x` to get

`(d^2y)/(dx^2) = d/dx (-4/5 \ cott)`

And again by the chain rule, we have:

`(d^2y)/(dx^2) = d/dt (-4/5 \ cott) * dt/dx`
`\ \ \ \ \ \ \ = (-4/5 \ csc^2t) / (-5sint)`
`\ \ \ \ \ \ \ = 4/25 \ csc^3t`