Conservative Vector Fields and FTLI

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Objectives:

1.: Understand the mixed partials test to determine whether a vector field is conservative.
2.: Be able to determine whether a vector field is conservative and be able to find a potential function (if possible).
3.: Know the Fundamental Theorem of Line Integrals (FTLI), and the consequences of the theorem.
4.: Know the different techniques to compute line integrals.

Recap Video

First, take a look at this recap video going over the basics of conservative vector fields and the Fundamental Theorem of Line Integrals (FTLI).

To recap:

If $\textbf {F} = \left \langle F_1,F_2 \right \rangle$ is a vector field whose components are defined and have continuous partials everywhere, and $\displaystyle \frac {\partial F_2}{\partial x} = \frac {\partial F_1}{\partial y}$ , then $\textbf {F}$ is conservative.
If $\textbf {F} = \left \langle F_1,F_2,F_3 \right \rangle$ is a vector field whose components are defined and have continuous partials everywhere, and $\frac {\partial F_2}{\partial x} = \frac {\partial F_1}{\partial y} \text { and } \frac {\partial F_3}{\partial x} = \frac {\partial F_1}{\partial z} \text { and }\frac {\partial F_2}{\partial z} = \frac {\partial F_3}{\partial z},$ then $\textbf {F}$ is conservative.

FTLI If $\textbf {F}$ is a conservative vector field with $\textbf {F} = \nabla f$ , and $C$ is a curve from point $P$ to point $Q$ , then $\int _C \textbf {F} \cdot d\textbf {r} = f(Q) - f(P).$

Examples

The following video will illustrate testing whether a vector field is conservative and finding a potential function.

Determine whether the vector field $\textbf {F} = \left \langle 2xy-\sin (z),x^2+2y+ze^{yz},ye^{yz}-x\cos (z) \right \rangle$ is conservative, and if so, find a potential function.

This next video shows how to change the path for a line integral when the vector field is conservative.

If $F = \left \langle e^{x^2},e^{y^2} \right \rangle$ and $C$ is the counterclockwise portion of the curve $x^2+y^2=2$ from $(-1,1)$ to $(1,-1)$ , evaluate $\int _C \textbf {F} \cdot d\textbf {r}.$

Problems

Throughout, when referring to a vector field $\textbf {F}$ , the symbols $F_1$ , $F_2$ , $F_3$ will be the first, second, and third component functions, respectively.

Consider $\textbf {F} = \left \langle x-y, x+y \right \rangle$ . The component functions are $F_1 = \answer {x-y}\quad \text { and } \quad F_2 = \answer {x+y}.$ The cross partials are $\frac {\partial F_2}{\partial x} = \answer {1},\quad \frac {\partial F_1}{\partial y} = \answer {-1}.$ Therefore, the vector field $\textbf {F}$ isis not conservative.

If $\textbf {F} = \left \langle z,z,x \right \rangle$ , then $\frac {\partial F_2}{\partial z} = \answer {1}, \quad \frac {\partial F_3}{\partial y} = \answer {0},$ which means the vector field $\textbf {F}$ isis not conservative.

Is the vector field $\textbf {F} = \left \langle z,1,x \right \rangle$ conservative?

The cross partials are all equal. Namely, $\frac {\partial F_1}{\partial z} = \frac {\partial F_3}{\partial x} = \answer {1},$ $\frac {\partial F_1}{\partial y} = \frac {\partial F_2}{\partial x} = \answer {0},$ $\frac {\partial F_2}{\partial z} = \frac {\partial F_3}{\partial y} = \answer {0}.$ Since the components of $\textbf {F}$ are defined and have continuous partials everywhere, and all the cross-partials are equal, we find that $\textbf {F}$ is conservative. The potential function $f(x,y,z)$ with $f(0,0,0) = 0$ is $f(x,y,z) = \answer {xz+y}$ .

Is the vector field $\textbf {F} = \left \langle 2xy+5, x^2-4z, -4y \right \rangle$ conservative?

Yes No

The potential function $f(x,y,z)$ with $f(0,0,0) = 0$ is $f(x,y,z) = \answer {x^2y+5x-4yz}$ .

If $\textbf {F} = \left \langle 3,6y \right \rangle$ and $C$ is the portion of the curve $y = 2/x$ from $x=1$ to $x=4$ , evaluate $\displaystyle \int _C \textbf {F} \cdot d\textbf {r}$ .

Steps:

Is the vector field $\textbf {F}$ conservative?
Yes No
One potential function for $\textbf {F}$ is $f(x,y) = \answer {3x+3y^2}$ (for the sake of uniqueness of the answer, give the one with $f(0,0) = 0$ ).
The two endpoints of $C$ are $(1,\answer {2})$ and $(4,\answer {1/2})$ .
Plugging these endpoints into $f$ give: $f(1,\answer {2}) = \answer {15} \quad \text { and }\quad f(4,\answer {1/2}) = 51/4.$
The line integral if $\displaystyle \int _C \textbf {F} \cdot d\textbf {r} = \answer {-9/4}$ .

If $\textbf {F} = \left \langle \cos (y),-x\sin (y) \right \rangle$ and $C$ is the upper half of the unit circle oriented counterclockwise, evaluate $\displaystyle \int _C \textbf {F} \cdot d\textbf {r}$ .

Steps:

$\textbf {F}$ is conservative with potential function $f(x,y) = \answer {x\cos (y)}$ (give the one with $f(0,0) = 0$ for the sake of uniqueness).
The endpoints of $C$ are $(1,\answer {0})$ and $(-1,\answer {0})$ .
The line integral is $\int _C \textbf {F} \cdot d\textbf {r} = f(-1,\answer {0}) - f(1,\answer {0}) = \answer {-2}.$

If $\textbf {F} = \left \langle ye^z,xe^z,xye^z \right \rangle$ and $C$ is the curve with parametrization $\textbf {r}(t) = \left \langle t,t^2,\sin (\pi t) \right \rangle$ for $0 \leq t \leq 1$ , evaluate $\displaystyle \int _C \textbf {F} \cdot d\textbf {r}$ .

Steps:

The vector field $\textbf {F}$ is conservative with potential function $f(x,y,z) = \answer {xye^z}$ (give the one with $f(0,0,0) = 0$ for the sake of uniqueness).
The terminal point of $C$ is $(\answer {1},\answer {1},\answer {0})$ and the starting point is $(\answer {0},\answer {0},\answer {0})$ .
The line integral is $f(\answer {1},\answer {1},\answer {0}) - f(\answer {0},\answer {0},\answer {0}) = \answer {1}$ .

Let $\textbf {F} = \left \langle x^2,e^{y^2} \right \rangle$ .

The vector field $\textbf {F}$ isis not conservative.
If $C$ is the unit circle oriented counterclockwise, then $\displaystyle \int _C \textbf {F} \cdot d\textbf {r} = \answer {0}$ .
If is the top half of the unit circle oriented counterclockwise, evaluate .
This will be an example of changing paths. Steps:
- Since $\textbf {F}$ is conservative, we can change paths. The end point for $C$ is $(\answer {-1},\answer {0})$ and the starting point is $(\answer {1},\answer {0})$ .
- We can try using the straight line from $(1,0)$ to $(-1,0)$ instead of the unit circle, but we still need to do the line integral by hand. Let $C'$ be the line segment. A parametrization for $C'$ is $\textbf {r}(t) = \left \langle 1-2t,\answer {0} \right \rangle$ for $0 \leq t \leq 1$ .
- $\textbf {F}(\textbf {r}(t)) = \left \langle \answer {(1-2t)^2},\answer {1} \right \rangle$
- $\textbf {r}'(t) = \left \langle \answer {-2},\answer {0} \right \rangle$
- We get $\int _C \textbf {F} \cdot d\textbf {r} = \int _{C'} \textbf {F} \cdot d\textbf {r} = \answer {-2/3}.$

Determining Which Method To Use

One of the big challenges in solving line integrals is is figuring out which method to use. Evaluate the following line integrals using any method we have learned so far.

If $\textbf {F} = \left \langle x^2,y^2 \right \rangle$ and $C$ is the line starting at the origin and ending at $(2,3)$ , then $\displaystyle \int _C \textbf {F} \cdot d\textbf {r} = \answer {35/3}$ .

First, we check whether $\textbf {F}$ is conservative. Notice the components of $\textbf {F}$ have continuous partials everywhere. By checking the mixed partials, we see that $\textbf {F}$ conservative. Can we find a potential function? In this case, we can, and it is $f(x,y) = \answer {\frac {1}{3}x^3+\frac {1}{3}y^3}.$ Now we can use the FTLI to get an answer of $f(2,3) - f(0,0) = \answer {\frac {35}{3}}.$

If $\textbf {F} = \left \langle x^2+y,y^2 \right \rangle$ and $C$ is the line starting at the origin and ending at $(2,3)$ , then $\displaystyle \int _C \textbf {F} \cdot d\textbf {r} = \answer {44/3}$ .

First, we check whether $\textbf {F}$ is conservative. Notice the components of $\textbf {F}$ have continuous partials everywhere. By checking the mixed partials, we see that $\textbf {F}$ conservative. Therefore, we should do the integral directly.

Parametrize the line segment as $\textbf {r}(t) = \left \langle 2t,\answer {3t} \right \rangle$ for $\answer {0} \leq t \leq \answer {1}$ . Then $\textbf {F}(\textbf {r}(t)) = \left \langle \answer {4t^2+3t},\answer {9t^2} \right \rangle ,$ and $\textbf {r}'(t) = \left \langle \answer {2},\answer {3} \right \rangle .$ Then

$\begin{eqnarray*} \int_C \textbf{F} \cdot d\textbf{r} &=& \int_0^1 \textbf{F}(\textbf{r}(t))\cdot \textbf{r}'(t) \,dt \\ &=& \int_0^1 \answer{8t^2+6t+27t^2}\,dt\\ &=& \answer{44/3}. \end{eqnarray*}$

If $\textbf {F} = \left \langle \frac {-y}{x^2+y^2},\frac {x}{x^2+y^2} \right \rangle$ and $C$ is the portion of $x^2+y^2=1$ with $y \geq 0$ , oriented counterclockwise, then $\displaystyle \int _C \textbf {F} \cdot d\textbf {r} = \answer {\pi }$ .

First, we check whether $\textbf {F}$ is conservative. Notice the components of $\textbf {F}$ continuous partials everywhere. Therefore $\textbf {F}$ conservative. Therefore, we should do the integral directly.

Parametrize the semi-circle as $\textbf {r}(t) = \left \langle \cos (t),\sin (t) \right \rangle$ for $0 \leq t \leq \answer {\pi }$ . Then $\textbf {F}(\textbf {r}(t)) = \left \langle \answer {-\sin (t)},\answer {\cos (t)} \right \rangle ,$ and $\textbf {r}'(t) = \left \langle \answer {-\sin (t)},\answer {\cos (t)} \right \rangle .$ Then

$\begin{eqnarray*} \int_C \textbf{F} \cdot d\textbf{r} &=& \int_0^{\pi} \textbf{F}(\textbf{r}(t))\cdot \textbf{r}'(t) \,dt \\ &=& \int_0^{\pi} \answer{1}\,dt\\ &=& \answer{\pi}. \end{eqnarray*}$

If $\textbf {F} = \left \langle y+e^{\sin (x)},x+y+1 \right \rangle$ , and $C$ is the right half of the unit circle oriented clockwise, then $\displaystyle \int _C \textbf {F} \cdot d\textbf {r} = \answer {-2}$ .

First, we check whether $\textbf {F}$ is conservative. Notice the components of $\textbf {F}$ have continuous partials everywhere. By checking the mixed partials, we see that $\textbf {F}$ conservative. However, we cannot find a potential function, because that $e^{\sin (x)}$ term will be hard to integrate with respect to $x$ . So we can change the path. Let’s use the line segment, call it $C'$ , so that $\int _C \textbf {F} \cdot d\textbf {r} = \int _{C'} \textbf {F} \cdot d\textbf {r}$ by path-independence. Parametrize the line segment as $\textbf {r}(t) = \left \langle \answer {0},1-2t \right \rangle$ for $\answer {0} \leq t \leq \answer {1}$ . Then $\textbf {F}(\textbf {r}(t)) = \left \langle \answer {2-2t},\answer {2-2t} \right \rangle ,$ and $\textbf {r}'(t) = \left \langle \answer {0},\answer {-2} \right \rangle .$ Then

$\begin{eqnarray*} \int_{C'} \textbf{F} \cdot d\textbf{r} &=& \int_0^1 \textbf{F}(\textbf{r}(t))\cdot \textbf{r}'(t) \,dt \\ &=& \int_0^1 \answer{4t-4}\,dt\\ &=& \answer{-2}. \end{eqnarray*}$