Question

Find the gradient of the point `f(1,-2,-1)` for `f(x,y,z)=3x^2y-y^3z^2`?

Answer 1

The gradient is `=<-12, -9 ,-16>`

Explanation:

The gradient is a vector :

`gradf=((delf)/(delx), (delf)/(dely), (delf)/(delz))`

`f(x,y,z)=3x^2y-y^3z^2`

`(delf)/(delx)=6xy`

`(delf)/(dely)=3x^2-3y^2z^2`

`(delf)/(delz)=-2y^3z`

`gradf(x,y,z)=(6xy,3x^2-3y^2z^2 ,-2y^3z )`

`gradf(1,-2,-1)=(-12, -9 ,-16)`

Answer 2

Given: `f(x,y,z)=3x^2y-y^3z^2`

The gradient vector is:

`grad f(x,y,z) = (del(f(x,y,z)))/(delx)hati+(del(f(x,y,z)))/(dely)hatj+(del(f(x,y,z)))/(delz)hatk`

Compute the partial derivatives:

`grad f(x,y,z) = (6xy)hati+(3x^2-3y^2z^2)hatj+(-2y^3z)hatk`

Evaluate at the point `(1,-2,-1)`:

`grad f(1,-2,-1) = (6(1)(-2))hati+(3(1)^2-3(-2)^2(-1)^2)hatj+(-2(-2)^3(-1))hatk`

Simplify:

`grad f(1,-2,-1) = -12hati-9hatj-16hatk`