My notes of the excellent lecture 30 by “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

The fundamental theorem of multivariable calculus tell us (using the symbolic $$\nabla$$-operator)

If you take the line integral of the gradient of a function, what you get back is the function.
$$\shaded{ \int_C\nabla f\cdot d\vec r = f(P_1) – f(P_0) } \label{eq:fundthm}$$

Similar to the plane, in space

When vector field $$\vec F$$ is a gradient of function $$f(x,y)$$, it is called a gradient field $$\newcommand{pdv}[1]{\tfrac{\partial}{\partial #1}} \shaded{ \vec F = \nabla f = \left\langle \pdv{x}f, \pdv{y}f, \pdv{z}f \right\rangle } \nonumber$$ where $$f(x,y,z)$$ is called the potential.

## When is a vector field a gradient field?

In the plane, to check if a vector field is a gradient field we only had to check the condition

$$\newcommand{dv}[2]{\frac{d #1}{d #2}} \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \newcommand{ppdv}[3]{\frac{\partial^2 #1}{\partial #2\partial #3}} \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \newcommand{ppdv}[2]{\frac{\partial^2 #1}{\partial #2^2}} \pdv{M}{y} = \pdv{N}{x} \nonumber$$

In space, we want to know if $$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \vec F = \left\langle P,Q,R\right\rangle \stackrel{?}{=} \nabla f = \left\langle \pdv{}{x}f, \pdv{}{y}f, \pdv{}{z}f \right\rangle \nonumber$$ where $$\vec F=\left\langle P,Q,R\right\rangle$$ be defined in a simply connected region.

If $$\vec F$$ is a gradient field, $$\vec F=\nabla f$$, then \newcommand{dv}[2]{\frac{d #1}{d #2}} \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \left\{ \begin{align*} P &= \pdv{}{x}f = f_x \\ Q &= \pdv{}{y}f = f_y \\ R &= \pdv{}{z}f = f_z \end{align*} \right.

Take the partial derivatives of $$P$$, $$Q$$ and $$R$$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \newcommand{ppdv}[3]{\frac{\partial^2 #1}{\partial #2\partial #3}} \left\{ \begin{align*} P_y=\pdv{P}{y}=\ppdv{}{x}{y}f = f_{xy} \\ P_z=\pdv{P}{z}=\ppdv{}{x}{z}f = f_{xz} \\ Q_x=\pdv{Q}{x}=\ppdv{}{y}{x}f = f_{yx} \\ Q_z=\pdv{Q}{z}=\ppdv{}{y}{z}f = f_{yz} \\ R_x=\pdv{R}{x}=\ppdv{}{z}{x}f = f_{zx} \\ R_z=\pdv{R}{y}=\ppdv{}{z}{y}f = f_{zy} \end{align*} \right.

Recall:

The mixed second derivatives are the same, no matter what order you take them $$\newcommand{dv}[2]{\frac{d #1}{d #2}} \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \newcommand{ppdv}[3]{\frac{\partial^2 #1}{\partial #2\partial #3}} \pdv{}{\color{red}x}\left(\pdv{f}{\color{blue}y}\right) =\ppdv{f}{\color{red}x}{\color{blue}y} =\ppdv{f}{\color{blue}y}{\color{red}x} =\pdv{}{\color{blue}y}\left(\pdv{f}{\color{red}x}\right) \nonumber$$

Based on this second partial derivative rule \begin{align*} \underline{P_y} = f_{xy} &= f_{yx} = \underline{Q_x} \\ \underline{P_z} = f_{xz} &= f_{zx} = \underline{R_x} \\ \underline{Q_z} = f_{yz} &= f_{zy} = \underline{R_y} \end{align*} \nonumber

Therefore, if $$\vec F=\left\langle P,Q,R\right\rangle$$, defined in a simply connected region is a gradient field, when \shaded{ \left\{ \begin{align*} P_y &= Q_x \\ P_z &= R_x \\ Q_z &= R_y \end{align*} \right. } \nonumber

### Exact differential

We can also think of it in terms of differentials.

If we have a differential of the form $$P\,dx + Q\,dy + R\,dz \nonumber$$ is an exact differential. That means it is going to equal to $$df$$ for some function $$F$$ exactly and of the same conditions. Just another way of saying it.

#### Example

For which $$a$$ and $$b$$ is the differential below exact? $$\underbrace{axy}_P\,dx + (\underbrace{x^2+z^3}_Q)dy + (\underbrace{byz^2+4z^3}_R)dz \nonumber$$

Or, for which $$a$$ and $$b$$ is $$\vec F$$ a gradient field? $$\vec F=\left\langle axy,x^2+z^3, byz^2-4z^3\right\rangle \nonumber$$

Compare $$\left. \begin{array}{lllll} P_y &= ax &= 2x &= Q_x &\Rightarrow a = 2 \\ P_z &= 0 &= 0 &= R_x \\ Q_z &= 3z^2 &= bz^2 &= R_y &\Rightarrow b=3 \end{array} \right\} \nonumber$$

For those values of $$a$$ and $$b$$ we can look for a potential. For any other values we would have to set up the line integral.

## Potential of a gradient field

Recall: for the plane

When the field is a gradient, and you know the function $$f$$, you can simplify the evaluation of the line integral for work. $$\shaded{ \int_C\nabla f\cdot d\vec r=f(P_1)-f(P_0) } \nonumber$$ where $$f(x,y)$$ is called the potential

Once again, you have two methods: computing line integrals, or using antiderivatives.

### Method 1: Computing line integrals

Similar to in the plane, apply the fundamental theorem, equation $$\eqref{eq:fundthm}$$, to find an expression for the potential at $$(x_1,y_1,z_1)$$ \begin{align} &\int_C\vec F\cdot d\vec r=f(x_1,y_1,z_1)-f(0,0,0) \nonumber \\ \Rightarrow & f(x_1,y_1,z_1) = \underbrace{\int_C\vec F\cdot d\vec r}_\text{work} + \underbrace{f(0,0,0)}_{\mathrm{constant}} \label{eq:method1} \end{align}

The work in a gradient field is path independent $$\Longrightarrow$$ find the easiest path

Apply the work differential, to find the work along $$C$$ in gradient field $$\vec F$$

Then add the line integrals together, to get the total work. $$\underline{\int_C\vec F\cdot d\vec r} = \int_{C_1}\ldots + \int_{C_2}\ldots + \int_{C_3}\ldots \nonumber$$

Substitute this back in equation $$\eqref{eq:method1}$$ and drop the subscripts $$f(x,y,z) = \underline{\int_C\vec F\cdot d\vec r} + \rm{c} \nonumber$$ If you would take the gradient of $$f(x,y,z)$$, you should get $$\vec F$$ back.

### Method 2: Using Antiderivatives

Similar to in the plane. No integrals, but you have to follow the procedure very carefully.

Continue with the earlier example, with $$a=2$$ and $$b=3$$. Solve \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align} \pdv{f}{x} &= f_x = 2xy \label{eq:potential1} \\ \pdv{f}{y} &= f_y = x^2 + z^3 \label{eq:potential2} \\ \pdv{f}{z} &= f_z = 3yz^2 – 4z^3 \label{eq:potential3} \end{align}

Integrate $$\eqref{eq:potential1}$$ in respect to $$x$$. The integration constant might depend on $$y$$ and $$z$$, so call it $$g(y,z)$$ $$\newcommand{pdv}[2]{\tfrac{\partial #1}{\partial #2}} f_x = 2xy \xrightarrow{\int dx} \underline{f = x^2y + g(y,z)} \nonumber$$

To get information on $$g(y,z)$$ we look at the other partials. Take the derivative of $$f$$ in respect to $$y$$ and compare to $$\eqref{eq:potential2}$$. The integration constant might depend on $$y$$, so call it $$h(z)$$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} x^2 + \pdv{g}{y} &= x^2 + z^3 \\ \Rightarrow \pdv{g}{y} &= z^3 \xrightarrow{\int dy} g = \underline{yz^3 + h(z)} \\ \end{align*}

Substituting $$g$$ in $$f$$ $$f = x^2y + \underline{yz^3 + h(z)} \nonumber$$

To get information on $$h(z)$$ we look at the other partials. Take the derivative of $$f$$ in respect to $$z$$ and compare to $$\eqref{eq:potential3}$$. The integration constant $$\rm{c}$$ is a true constant \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} 3yz^2 + \pdv{h}{z} &= 3yz^2 – 4z^3 \\ \Rightarrow \pdv{h}{z} &= – 4z^3 \xrightarrow{\int dz} h = \underline{-z^4 + \rm{c}} \\ \end{align*}

Substituting $$h$$ in $$f$$ $$f = x^2y + yz^3 \underline{-z^4\,(+ \rm{c})} \nonumber$$

If you want to find one potential, you can just forget about the constant. If you want to find all the potentials: they differ by this constant.

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