Question

If z = x^2 + 2y^2 , x = r cos θ , y = r sin θ , find the partial derivative (∂z/∂θ)_y ?

Answer 1

The answer is `=-r^2sin2theta`

Explanation:

Reminder

`sin2theta=2sinthetacostheta`

We are given

`z=x^2+2y^2`

`x=rcostheta`

`y=rsintheta`

`(delz)/(deltheta)=(delz)/(delx).(delx)/(deltheta)+(delz)/(dely).(dely)/(deltheta)`

`(delz)/(delx)=2x`

`(delz)/(dely)=4y`

`(delx)/(deltheta)=-rsintheta`

`(dely)/(deltheta)=rcostheta`

`(delz)/(deltheta)=-2xrsintheta+4yrcostheta`

`=-2rsinthetarcostheta+4rsinthetarcostheta`

`=2r^2sinthetacostheta`

If `y` is constant then

`(delz)/(dely)=0`

Therefore,

`((delz)/(deltheta))_y=-2r^2sinthetacostheta=-r^2sin2theta`