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17.3 Conservative Vector Fields and the Fundamental Theorem of Line Integrals

If \(\vec{F}\) is conservative, then \(\vec{F} = \vec{\nabla} \phi\), where \(\phi\) is the potential function.

Given \(\vec{F}\), how do we know if it is conservative? If so, how to find \(\phi\)?

Determining Conservativeness

Let \(\vec{F} = \langle f, g \rangle\) be conservative.

Then we know \(\vec{F} = \langle f, g \rangle = \vec{\nabla} \phi = \langle \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y} \rangle\).

So, \(f = \frac{\partial \phi}{\partial x}\) and \(g = \frac{\partial \phi}{\partial y}\).

Furthermore, we know that mixed partials are equal: \(f_{yx} = f_{xy}\).

\[ \frac{\partial}{\partial y} \left( \frac{\partial \phi}{\partial x} \right) = \frac{\partial}{\partial x} \left( \frac{\partial \phi}{\partial y} \right) \]\[ \phi_{xy} = \phi_{yx} \]

\(f_y = g_x\) if \(\vec{F} = \langle f, g \rangle\) is conservative (\(\vec{F} = \vec{\nabla} \phi\))

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Example 1

\(\vec{F} = \langle x, y \rangle\)

Note: Here \(f = x\) and \(g = y\).

Conservative if \(f_y = g_x\).

\[ \frac{\partial}{\partial y}(x) = \frac{\partial}{\partial x}(y) \quad ? \]

Yes, \(0 = 0\). So, \(\vec{F} = \langle x, y \rangle\) is conservative.

Example 2

\(\vec{F} = \langle -y, x \rangle\)

Note: Here \(f = -y\) and \(g = x\).

Is \(\frac{\partial}{\partial y}(-y) = \frac{\partial}{\partial x}(x)\)?

\[ -1 \neq 1 \]

No, \(-1 \neq 1\). So \(\vec{F} = \langle -y, x \rangle\) is NOT conservative.

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Finding the Potential Function \(\phi\)

If conservative (\(\vec{F} = \vec{\nabla} \phi\)), how to find \(\phi\)?

Example

\(\vec{F} = \langle x+y, x \rangle\)

Is \(f_y = g_x\)? Yes, so \(\vec{F} = \langle x+y, x \rangle = \langle \phi_x, \phi_y \rangle = \vec{\nabla} \phi\)

We see:

(1) \(\frac{\partial \phi}{\partial x} = x + y\)
(2) \(\frac{\partial \phi}{\partial y} = x\)

Integrate (1) with respect to \(x\) (treat \(y\) as constant):

\[ (1) \rightarrow \phi = \int (x+y) dx = \frac{1}{2}x^2 + xy + a(y) \]

Note: \(y\) is constant.

\(a(y)\) is a function that depends on \(y\) that vanishes when we take partial with \(x\) (to get (1)). It could be a constant.

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Now let's take partial with respect to \(y\) and equate with (2):

\(\phi = \frac{1}{2}x^2 + xy + a(y)\)

\(\phi_y = x + \frac{da}{dy} = x\)

\(\frac{da}{dy} = 0 \rightarrow a = C \text{ (constant)}\)

Now we know \(\phi\):

\[ \phi = \frac{1}{2}x^2 + xy + C \]

Check:

Is \(\vec{\nabla} \phi = \vec{F} = \langle x+y, x \rangle\)?

\[ \vec{\nabla} (\frac{1}{2}x^2 + xy + C) = \langle x+y, x \rangle \text{ Yes.} \]

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3D Conservative Vector Fields

3D: \( \vec{F} = \langle f, g, h \rangle \) is it conservative?

If conservative, \( \vec{F} = \langle f, g, h \rangle = \vec{\nabla} \phi = \langle \phi_x, \phi_y, \phi_z \rangle \)

\[ f = \phi_x \]\[ g = \phi_y \]\[ h = \phi_z \]

Mixed partials are equal:

\[ \phi_{xy} = \phi_{yx} \]\[ \phi_{yz} = \phi_{zy} \]\[ \phi_{zx} = \phi_{xz} \]

\[ f_y = g_x \]\[ g_z = h_y \]\[ h_x = f_z \]

To recover \( \phi \), do similar thing as in 2D case.

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Why find the Potential Function \( \phi \)?

Why bother with finding \( \phi \)?

Because if \( \vec{F} \) is conservative, then

\[ \int_C \vec{F} \cdot \vec{T} ds = \int_C \vec{F} \cdot \vec{r}' dt = \int_C \vec{F} \cdot d\vec{r} \]

is path-independent, the only things that matter are the start and end locations.

Proof / Derivation

Why? If \( \vec{F} = \vec{\nabla} \phi = \langle \phi_x, \phi_y \rangle \), \( \vec{r}(t) = \langle x(t), y(t) \rangle \)

\[ \vec{F} \cdot \vec{r}' dt = \left\langle \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y} \right\rangle \cdot \left\langle \frac{dx}{dt}, \frac{dy}{dt} \right\rangle dt \]\[ = \left( \frac{\partial \phi}{\partial x} \frac{dx}{dt} + \frac{\partial \phi}{\partial y} \frac{dy}{dt} \right) dt \]

Figure: Chain rule tree diagram showing phi branching to x and y, which both branch to t.

result of chain rule

\[ = \left( \frac{d\phi}{dt} \right) dt = d\phi \]\[ \int_C \vec{F} \cdot \vec{r}' dt = \int_C d\phi = \phi(B) - \phi(A) \]

\( \phi(B) \): end location

\( \phi(A) \): start location

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Fundamental Theorem of Line Integrals

If ⅒ is conservative (⅒ = ∇ϕ), then

\[ \int_C \vec{F} \cdot \vec{T} ds = \int_C \vec{F} \cdot \vec{r}' dt = \int_C \vec{F} \cdot d\vec{r} = \phi(B) - \phi(A) \]

Where B is the end point and A is the start point.

The theorem implies that for a conservative field, the line integral depends only on the endpoints, not the path taken.

All paths lead to the same \[ \int_C \vec{F} \cdot \vec{T} ds \] if \( \vec{F} = \nabla \phi \).

Figure: Graph showing three different paths C1, C2, and C3 connecting point A to point B in the xy-plane.

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Example

\[ \int_C \vec{F} \cdot \vec{r}' dt \quad \text{where} \quad \vec{F} = \langle x+y, x \rangle \]

\( C \): line segment from \( (1, 0) \) to \( (0, -1) \), then along the left half of \( x^2 + y^2 = 1 \) to \( (0, 1) \).

Figure: Coordinate graph showing path C1 as a line segment and C2 as a semi-circle on the left side of the y-axis.

Let's do the "old way" first and compare to the "new way".

\( C_1 \): \( \vec{r}(t) = \langle 1-t, -t \rangle \) for \( 0 \le t \le 1 \)
\( C_2 \): \( \vec{r}(t) = \langle -\sin t, -\cos t \rangle \) for \( 0 \le t \le \pi \)

\[ \underbrace{\int_0^1 \langle 1-2t, 1-t \rangle \cdot \langle -1, -1 \rangle dt}_{C_1} + \underbrace{\int_0^\pi \langle -\sin t - \cos t, -\sin t \rangle \cdot \langle -\cos t, \sin t \rangle dt}_{C_2} \]\[ = \int_0^1 (3t - 2) dt + \int_0^\pi (\sin t \cos t + \cos^2 t - \sin^2 t) dt = \dots = -\frac{1}{2} \]

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Fundamental Theorem of Line Integrals

Earlier we found \(\vec{F} = \langle x+y, x \rangle\) to be conservative (\(\vec{\nabla} \phi\)) and \(\phi = \frac{1}{2}x^2 + xy + C\).

Let's try using the Fundamental Theorem of Line Integrals:

\[ \int_C \vec{F} \cdot \vec{r}' dt = \phi(B) - \phi(A) \]

Where:

\(B\) is the end location: \((0, 1)\)
\(A\) is the start location: \((1, 0)\)

\[ = \phi(0, 1) - \phi(1, 0) \]\[ = \left[ \frac{1}{2}(0)^2 + (0)(1) + C \right] - \left[ \frac{1}{2}(1)^2 + (1)(0) + C \right] \]\[ = C - (\frac{1}{2} + C) = -\frac{1}{2} \]

Alternatively, since path doesn't matter, we could replace the original path with a simpler one but preserve start/end locations.

Figure: Coordinate graph showing a path from point A at (1,0) to point B at (0,1) along the axes.