Question

What is the gradient of the function `p(x,y)=sqrt(24-4x^2-y^2)`?

Answer 1

It's `vec(grad)p(x,y) = langle-(4x)/(sqrt(24-4x^2-y^2)) , -(y)/(sqrt(24-4x^2-y^2)) rangle`

The gradient `vec(grad)p(x,y)` of the function `p(x,y)` is defined as the vector whose `n`-th component is the partial derivative of `p` with respect to the `n`-th variable, or:

`vec(grad)p(x,y) = langle (del p)/(del x), (del p)/(del y) rangle`

Computing the partial derivatives using the chain rule:

`(del p)/(del x) = (del)/(del x)[sqrt(24-4x^2-y^2)]`
`= 1/(2sqrt(24-4x^2-y^2)) (del)/(del x)[24-4x^2-y^2]`
`= -(4x)/(sqrt(24-4x^2-y^2))`

`(del p)/(del y) = (del)/(del y)[sqrt(24-4x^2-y^2)]`
`= 1/(2sqrt(24-4x^2-y^2)) (del)/(del y)[24-4x^2-y^2]`
`= -(y)/(sqrt(24-4x^2-y^2))`

And so,

`vec(grad)p(x,y) = langle (del p)/(del x), (del p)/(del y) rangle`
`= langle -(4x)/(sqrt(24-4x^2-y^2)) , -(y)/(sqrt(24-4x^2-y^2)) rangle`