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Week 5 Discussion Notes

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Potential Functions

A potential function for a vector-valued function F⃗\vec{F} is a scalar-valued function ff such that

∇f=⟨fx,fy,fz⟩=F⃗.\nabla f = \ang{f_x, f_y, f_z} = \vec{F}.
Example 1.

Find a potential function for F⃗=⟨yz,xz,y⟩\vec{F} = \ang{yz, xz, y} or show that one doesn't exist.
Solution.

When approaching these types of problems, the first thing you should do is check the mixed partials. If there were a potential function ff, then we would need to have

fz=yandfx=yz.f_z = y \quad\text{and}\quad f_x = yz.

But this would mean

fzx=0andfxz=y,f_{zx} = 0 \quad\text{and}\quad f_{xz} = y,

which contradicts Clairaut's theorem. This means that no potential function exists.
Example 2.

Find a potential function for F⃗=⟨2xzex2,0,ex2⟩\vec{F} = \ang{2xze^{x^2}, 0, e^{x^2}} or show that one doesn't exist.
Solution.

Like before, you can check all the mixed partials, and you'll see that they actually agree. For example,

∂∂xex2=2xex2and∂∂z2xzex2=2xex2.\pderiv{}{x} e^{x^2} = 2xe^{x^2} \quad\text{and}\quad \pderiv{}{z} 2xze^{x^2} = 2xe^{x^2}.

When t his happens, you'll want to start looking for a potential function, and you can usually just eyeball it. For example, fz=ex2f_z = e^{x^2}, so ff would need to look something like zex2ze^{x^2}, and this works.

f(x,y,z)=zex2\boxed{f\p{x, y, z} = ze^{x^2}}