Consider f and c below. f(x, y, z) = yzexzi + exzj + xyexzk,c.r(t) = (t2 + 2)i + (t2 − 1)j + (t2 − 4t)k, 0 ≤ t ≤ 4 (a) find a function f such that f = ∇f.

[tex]\mathbf f(x,y,z)=yze^{xz}\,\mathbf i+e^{xz}\,\mathbf j+xye^{xz}\,\mathbf k[/tex]

We're looking for a scalar function [tex]f(x,y,z)[/tex] such that its gradient is equal to the given vector-valued function [tex]\mathbf f(x,y,z)[/tex]:

[tex]\nabla f(x,y,z)=\mathbf f(x,y,z)[/tex]

That is, we require

[tex]\dfrac{\partial f}{\partial x}=yze^{xz}[/tex]
[tex]\dfrac{\partial f}{\partial y}=e^{xz}[/tex]
[tex]\dfrac{\partial f}{\partial z}=xye^{xz}[/tex]

Integrating the first equation with respect to [tex]x[/tex] gives

[tex]f(x,y,z)=\dfrac{yz}xe^{xz}+g(y,z)[/tex]

Differentiating with respect to [tex]y[/tex] gives

[tex]\dfrac{\partial f}{\partial y}=e^{xz}=\dfrac zxe^{xz}+\dfrac{\partial g}{\partial y}[/tex]

but from this step we can see that it's not possible for [tex]g[/tex] to be a function of [tex]y[/tex] and [tex]z[/tex], independent of [tex]x[/tex]. So there is no such function [tex]f(x,y,z)[/tex].