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I have this question that says $f(x,y,z)$ such that $\nabla f(x,y,z) = <2xy,2yz+x^2,y^2>$

The way I see this is that they are asking for what can you take the partial derivative of with respect $x y$ and $z$ so that you get that $\nabla f(x,y,z)$

My answer would be $f(x,y,z) = (x^2y+y^2z+x^2y,y^2z) $

Btzzzz
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Zrot25
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2 Answers2

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$f$ is called the potential function. We have: $\nabla f = (f_x,f_y,f_z)$. Thus $f_x = 2xy\implies f = \int 2xydx + g(y,z)= x^2y+ g(y,z)\implies f_y = x^2 + \dfrac{\partial g(y,z)}{\partial y}=2yz + x^2\implies g(y,z) = \int 2yzdy + h(z)= y^2z + h(z)\implies f_z = y^2 + h'(z) = y^2\implies h'(z) = 0 \implies h(z) = C\implies f(x,y,z) = x^2y + y^2z + C$ ( constant ).

DeepSea
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  • Could you explain a little bit in words how the process goes I am confused about why the g(y,z) came from. Just a little insight on your steps. – Zrot25 Dec 04 '16 at 02:57
  • when you deal with partial derivative with respect to a variable, other variables regarded as constants, thus the $g(y,z)$ is thought of as a constant when you integrate with respect to the variable $x$ since the function is not a single variable any more, its a 3-variable function. – DeepSea Dec 04 '16 at 03:04
  • For this line 2yz+x2⟹g(y,z)=∫2yzdy+h(z)=y2z+h(z) where does the x^2 go? – Zrot25 Dec 04 '16 at 19:34
  • You can apply this type of algorithm for any desired $\mathbf{F}=(f_x,f_y,f_z)$ but it will not give you the desired result unless $\nabla \times \mathbf{F} = 0$. – Walton Jul 06 '17 at 08:42
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    @DeepSea or others, interesting! Which branch of math, physics, etc, introduces this concept of potential functions? Any good references to introduce about potential functions? The term is so overloaded that I was not able to directly find it in the same context. – user25004 Aug 28 '20 at 16:30
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I've developed a sort of algorithm for myself in taking the "anti-gradient" and it personally expedites the process for me.

For a vector valued function,

F = grad(f) = (H,G,I)

You start by sorting all of the linearly independent terms (LITs) in F into three categories (or subsets) p, q & r for each H,G and I.

That is,

{H}={p,q,r} ,{G}={p,q,r}, {I}={p,q,r}

p-terms:

  • All LITs that themselves depend on the 2 variables that were not apart of the partial differentiation (independent variables).

For the component H, p-terms would have to be a function of (y,z) as the partial derivative was with respect to x.

q-terms:

  • All LITs that depend on at least 2 variables in which one must be the dependent variable.

or

  • All LITs that depend on only one of the independent variables.

For H, q-terms would include functions of: (x,y), (x,z), (x,y,z), (z) or (y) Basically q is anything that's not p or r.

r-terms:

  • All LITs that depend only on the dependent variable.

For H, r-terms are functions of x.

Once you have these sorted out you simply plug them into the following formula.

∫ (Hp+Hq+Hr)dx + (Gq+Gr)dy + (Ir)dz

in essence,

f = C1(x,y,z) + C2(y,z) + C3(z) + C4

As an aside, I found this particularly useful when calculating f for an irrotational vector field in a line integral. It saved me a ton of time on tests.

Hayden Birch
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