Question

If x = a cosθ and y = b sinθ , find second derivative

Answer 1

I'm guessing the second derivative is for y with respect to x, i.e.

`(\mathrm d^2y)/(\mathrm dx^2)`

Compute the first derivative. By the chain rule,

`(\mathrm dy)/(\mathrm dx)=(\mathrm dy)/(\mathrm d\theta)(\mathrm d\theta)/(\mathrm dx)=((\mathrm dy)/(\mathrm d\theta))/((\mathrm dx)/(\mathrm d\theta))`

We have

`y=b\sin\theta\implies(\mathrm dy)/(\mathrm d\theta)=b\cos\theta`

`x=a\cos\theta\implies(\mathrm dx)/(\mathrm d\theta)=-a\sin\theta`

and so

`(\mathrm dy)/(\mathrm dx)=(b\cos\theta)/(-a\sin\theta)=-\frac ba\cot\theta`

Now compute the second derivative. Notice that
`(\mathrm dy)/(\mathrm dx)` is a function of
`\theta`; so denote it by
`f(\theta)`. Then

`(\mathrm d^2y)/(\mathrm dx^2)=(\mathrm df)/(\mathrm dx)`

By the chain rule,

`(\mathrm d^2y)/(\mathrm dx^2)=(\mathrm df)/(\mathrm d\theta)(\mathrm d\theta)/(\mathrm dx)=((\mathrm df)/(\mathrm d\theta))/((\mathrm dx)/(\mathrm d\theta))`

We have

`f=-\frac ba\cot\theta\implies(\mathrm df)/(\mathrm d\theta)=\frac ba\csc^2\theta`

and so the second derivative is

`(\mathrm d^2y)/(\mathrm dx^2)=(\frac ba\csc^2\theta)/(-a\sin\theta)=-\frac b{a^2}\csc^3\theta`