Question

How do you find d²y/dx² when x=1+2cost y=1+3sint?

Answer 1

`- 3/4 csc^3 t`

Explanation:

`(dy)/(dx) = (dy/dt)/(dx/dt) = (3 cos t)/(- 2 sin t)= - 3/2 cot t`

`(d^2y)/(dx^2) = (d)/(dx) (- 3/2 cot t) = - 3/2 (d)/(dx) ( cot t)`

`= - 3/2 (d)/(dt) ( cot t) dt/dx` by the chain rule

`= - 3/2 (d)/(dt) ( cot t) 1/(dx/dt)`

`= - 3/2 (-csc^2 t ) (1/(- 2 sin t))`

`= - 3/4 csc^3 t`