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Find potential functions for conservative fields

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Recall

A vector field is an assignment of a vector $\vec{F}$ to each point (x, y) in a space, i.e., $\vec{F} = M\vec{i}+N\vec{j}$ where M and N are functions of x and y.

A vector field on a plane can be visualized as a collection of arrows, each attached to a point on the plane. These arrows represent vectors with specific magnitudes and directions.

Work is defined as the energy transferred when a force acts on an object and displaces it along a path. In the context of vector fields, we calculate the work done by a force field along a curve or trajectory C using a line integral. The work done by a force field $\vec{F}$ along a curve C is: W = $\int_{C} \vec{F}·d\vec{r} = \int_{C} Mdx + Ndy = \int_{C} \vec{F}·\hat{\mathbf{T}}ds$, where $\hat{\mathbf{T}}$ is the unit tangent vector.

A vector field is conservative if there exist a scalar function such that $\vec{F}$ = ∇f (the vector field is its gradient). This scalar function is known or referred to as the potential function associated with the vector field.

Theorem. Fundamental theorem of calculus for line integral. If $\vec{F}$ is a conservative vector field in a simply connected region of space (i.e., a region with no holes), and if f is a scalar potential function for $\vec{F}$ in that region, then $\int_{C} \vec{F}·d\vec{r} = \int_{C} ∇f·d\vec{r} = f(P_1)-f(P_0)$ where P0 and P1 are the initial and final points of the curve C, respectively.

Find Potential Functions for Conservative fields

Equivalent Properties of Conservative Vector Fields

  1. Conservative force. A force $\vec{F}$ is considered conservative if the work done by the force around any closed curve C is zero. Mathematically, this is expressed as $\int_{C} \vec{F}·d\vec{r} = 0$.

  2. Path independence. A force field is path-independent, meaning the work done by the force in moving an object from one point to another is the same, regardless of the path taken between the two points.

  3. Gradient Field. A vector field $\vec{F}$ is a gradient field if it can be expressed as the gradient of a scalar potential function f. In mathematical terms, this means $\vec{F} = ∇f$, where f is a scalar function and the vector field $\vec{F}$ has components $\vec{F} = ⟨M, N⟩ = ⟨\frac{∂f}{∂x}, \frac{∂f}{∂y}⟩ = ⟨f_x, f_y⟩$. Here, f is the potential function associated with the vector field $\vec{F}$. The lineal integral of $\vec{F}$ along a path C measures the work done by the vector field in moving an object along the path C. If $\vec{F}$ is a gradient field, then: $\int_{C} \vec{F}·d\vec{r} = f(P_1)-f(P_0).$

  4. Exact differential. In the context of differential forms, a differential expression Mdx+Ndy is called an exact differential if there exist a scalar function f(x, y) such that df = $\frac{∂f}{∂x}dx+\frac{∂f}{∂y}dy$ = Mdx + Ndy. This implies that the vector field $\vec{F} = ⟨M, N⟩$ is conservative, and there exists a potential function f such that $\vec{F} = ∇f$.

    If a differential is exact, then the line integral of $\vec{F}$ over any path C can be evaluated by simply finding the difference in the potential function values at the endpoints of the path.

These four properties —conservative force, path independence, gradient field, and exact differential— are different perspectives of the same fundamental concept: the conservativeness of a vector field.

Criterion for a Conservative Vector Field

The criterion for checking whether a vector field $\vec{F}$ is conservative can be summarized as follows: If $\frac{∂M}{∂y} = \frac{∂N}{∂x}$, and if the functions M and N have continuous first partial derivatives across the entire domain D, and the domain D is open and simply connected, then $\vec{F}$ is a gradient field, meaning it is conservative.

Finding a potential function for conservative vector fields

A vector field $\vec{F}$ is conservative if it can be expressed as the gradient of some scalar potential function f(x, y). In mathematical terms, this means:$\vec{F} = ∇f = (\frac{∂f}{∂x}, \frac{∂f}{∂y})$.

To find a potential function for a conservative vector field, we first need to ensure that the vector field is conservative. A vector field $\vec{F}$ is conservative if $\frac{∂M}{∂y} = \frac{∂N}{∂x}$, and if the functions M and N have continuous first partial derivatives across the entire domain D, and the domain D is open and simply connected, then $\vec{F}$ is a gradient field, meaning it is conservative.

For the vector field $\vec{F}= (4x^2+8xy)\vec{i}+(3y^2+4x^2)\vec{j}$, let’s check if F is conservative.

  1. Equality of Mixed Partial Derivatives: $\frac{∂M}{∂y} = \frac{∂}{∂y}(4x^2+8xy) = 8x = \frac{∂N}{∂x} = \frac{∂}{∂x}(3y^2+4x^2)$.
  2. Continuous First Partial Derivatives: The components M = 4x2 + 8xy and N = 3y2 +4x2 are polynomials, which have continuous derivatives everywhere.
  3. Open and Simply Connected Domain: The domain of $\vec{F}$ is ℝ2, which is open and simply connected (no holes or gaps).

Since these conditions are met, $\vec{F}$ is indeed conservative, meaning there exists a potential function f(x,y) such that $\vec{F}$ = ∇f.

Method 1: Finding the Potential Function via Line Integrals

We will find the potential function f(x, y) by computing a line integral along a path or trajectory C from the origin (0, 0) to a point (x1, y1) (see Figure F.a in the image below). The potential function is defined as:

$f(x_1, y_1) = \int_{C} \vec{F}·d\vec{r} + f(0, 0)$

where f(0, 0) is an arbitrary constant.

image info

To simplify the calculation, we break the path C into two segments:

Compute the line integral along C, $\vec{F}=⟨4x^2+8xy, 3y^2+4x^2⟩$, we have: $\int_{C} \vec{F}·d\vec{r} = \int_{C} (4x^2+8xy)dx + (3y^2+4x^2)dy$

This can be broken down as:

$\int_{C_1} (4x^2+8xy)dx + (3y^2+4x^2)dy + \int_{C_2} (4x^2+8xy)dx + (3y^2+4x^2)dy$

$\int_{C_1} (4x^2+8xy)dx + (3y^2+4x^2)dy$ = [x varies from 0 to x1, y = 0 ⇒ dy = 0] $\int_{0}^{x_1} 4x^2dx = \frac{4}{3}x^3\bigg|_{0}^{x_1} \frac{4}{3} = x_1^3$

$\int_{C_2} (4x^2+8xy)dx + (3y^2+4x^2)dy$ = [y varies from 0 to y1, x = x1 ⇒ dx = 0] $\int_{0}^{y_1} (3y^2+4x_1^2)dy = (y^3+4x_1^2y)\bigg|_{0}^{y_1} = y_1^3+4x_1^2y_1$

Combine the results: $f(x_1, y_1) = \int_{C} \vec{F}·d\vec{r} + f(0, 0) ⇒ f(x_1, y_1) = \int_{C} \vec{F}·d\vec{r} + c ↭ f(x_1, y_1) = \frac{4}{3}x_1^3 + y_1^3+4x_1^2y_1 + c ↭[\text{Therefore,}] f(x, y) = \int_{C} \vec{F}·d\vec{r} + c = \frac{4}{3}x^3 + y^3+4x^2 + c$ where c is a constant.

Finding the Potential Function via Antiderivatives

For the same vector field, we know that $\vec{F} = ∇f$ where F = ⟨M, N⟩ = $⟨\frac{∂f}{∂x}, \frac{∂f}{∂y}⟩ = ⟨f_x, f_y⟩$ where fx = 4x2 +8xy (i), fy = 3y2 +4x2 (ii).

Starting with the partial derivative fx (i), we integrate with respect to x:

f(x, y) = $\int (4x^2+8xy)dx = \frac{4}{3}x^3+4x^2y+g(y)$. Here, g(y) is a function of y that acts as the “constant” of integration when integrating with respect to x.

To determine g(y), we differentiate f with respect to y and match it with the given expression for fy:

Differentiate f with respect to y: fy = 4x2 + g’(y).

Matching this with (ii) fy = 3y2 +4x2, we get: 4x2 + g’(y) = 3y2 +4x2 ⇒ g’(y) = 3y2

Integrate g′(y) with respect to y: g(y) = $\int g’(y)dy = \int 3y^2dy = y^3 + c$ and c is a real constant.

Substituting g(y) =y3 +c into the expression for f(x, y), we get: f(x, y) = $\frac{4}{3}x^3+4x^2y+g(y) = \frac{4}{3}x^3+4x^2y+ y^3 +c$

Step 1. Check if $\vec{F}$ is a conservative vector field.

To find a potential function for a conservative vector field, we first need to ensure that the vector field is conservative. A vector field $\vec{F}$ is conservative if $\frac{∂M}{∂y} = \frac{∂N}{∂x}$, and if the functions M and N have continuous first partial derivatives across the entire domain D, and the domain D is open and simply connected, then $\vec{F}$ is a gradient field, meaning it is conservative.

Given: M = $ye^x+sin(y)$, N = $e^x+xcos(y)$

  1. Equality of Mixed Partial Derivatives: $\frac{∂M}{∂y} = \frac{∂}{∂y}(ye^x+sin(y)) = e^x+cos(y) = \frac{∂N}{∂x} = \frac{∂}{∂x}(e^x+xcos(y))$.
  2. Continuous First Partial Derivatives: The components M = $ye^x+sin(y)$ and N = $e^x+xcos(y)$ involve exponential, trigonometric, and polynomial functions, all of which are continuous derivatives and defined everywhere.
  3. Open and Simply Connected Domain: The domain of $\vec{F}$ is ℝ2, which is open and simply connected (no holes or gaps).

Since these conditions are met, $\vec{F}$ is indeed conservative, meaning there exists a potential function f(x,y) such that $\vec{F}$ = ∇f.

Step 2. Finding the potential function f using Antiderivatives

For the same vector field, we know that $\vec{F} = ∇f$ where F = ⟨M, N⟩ = $⟨\frac{∂f}{∂x}, \frac{∂f}{∂y}⟩ = ⟨f_x, f_y⟩$ where fx(x, y) = $ye^x+sin(y)$ (i), fy(x, y) = $e^x+xcos(y)$ (ii).

From (i), we integrate with respect to x:

f(x, y) = $\int (ye^x+sin(y))dx = ye^x+xsin(y)+g(y)$

where g(y) is a function of y (acting as the constant of integration with respect to x, but not necessarily for y), i.e., the integration constant can depend on y.

To determine g(y), we differentiate f with respect to y:

fy(x, y) = $\frac{∂f}{∂y} = \frac{∂}{∂y}(ye^x+xsin(y)+g(y))$ = ex + xcos(y) + g’(y). Matching this with (ii) fy(x, y) = $e^x+xcos(y)$, we get: $e^x+xcos(y) = e^x+cos(y)+g’(y)$ ⇒ g’(y) =0

Integrating g′(y) with respect to y: g(y) = $\int 0dy = c$ and c is a real constant.

Therefore, the potential function f is: f(x, y) = yex +xsin(y) + c. This function represents the scalar potential associated with the conservative vector field $\vec{F}$.

Evaluating a Line Integral Using the Potential Function

Given: p = yz, q = xz, r = (xy+2z).

Step 0: Check if $\vec{F}$ is conservative. The vector field is in three dimensions, so we will follow a similar process as before but adapted for three variables.

  1. Equality of Mixed Partial Derivatives: $\frac{∂q}{∂x} = \frac{∂}{∂y}(xz) = z = \frac{∂p}{∂y} = \frac{∂}{∂y}(yz), \frac{∂r}{∂y} = \frac{∂}{∂y}(xy+2z) = x = \frac{∂q}{∂z} = \frac{∂}{∂z}(xz), \frac{∂p}{∂z} = \frac{∂}{∂z}(yz) = y = \frac{∂r}{∂x} = \frac{∂}{∂x}(xy+2z)$.
  2. Continuous First Partial Derivatives: The components p = yz, q = xz, r = (xy+2z) are all polynomials, hence continuous first partial derivatives and defined everywhere.
  3. Open and Simply Connected Domain: The domain of $\vec{F}$ is ℝ3, which is open and simply connected (no holes or gaps).

Since these conditions are met, $\vec{F}$ is indeed conservative, meaning there exists a potential function f(x, y, z) such that $\vec{F}$ = ∇f.

Step 2: Find the Potential Function f(x, y, z)

We have fx(x, y, z) = yz (i), fy(x, y, z) = xz (ii), fz(x, y, z) = xy + 2z (iii).

From (i) integrate fx = yz with respect to x:

f(x, y, z) = $\int (yz)dx = yzx + g(y, z)$ where g(y, z) is a function of y and z because it acts as the “constant” of integration with respect to x, but not necessarily for y and z.

To determine g(y), we differentiate f with respect to y and z:

Differentiate with Respect to y: fy(x, y, z) = $\frac{∂}{∂y}(yzx + g(y, z))$ = xz + gy(y, z). Matching this with (ii), fy(x, y, z) = xz, we get: xz = xz + gy(y, z) ⇒ gy(y, z) = 0. Integrating gy(y, z) with respect to y, we get, g(y, z) = h(z), so g(y, z) is a function of z alone.

Differentiate with Respect to z: fz(x, y, z) = $\frac{∂}{∂z}(yzx + h(z))$ = yx + h’(z). Matching this with (iii), fz(x, y, z) = xy + 2z we get: yx + h’(z) = xy + 2z⇒ h’(z) = 2z. Integrating h(z) with respect to z, we get, h(z) = z2 + k

Therefore, the potential function f(x, y, z) = xyz + g(y, z) = xyz + h(z) = xyz + z2 + k.

Step 3. Evaluating the line integral $\int_{C} \vec{F}·d\vec{r}$

The line integral of a conservative vector field $\vec{F}$ over a curve C from point A to point B can be evaluated using the potential function f: $\int_{C} \vec{F}·d\vec{r} = f(B) - f(A)$

Here, C is the line segment from A = (1, 0, −2) to B = (4, 6, 3).

$\int_{C} \vec{F}·d\vec{r} = \int_{C} ∇f·d\vec{r} = f(4, 6, 3) -f(1, 0, -2) = 4·6·3+3^2+c-(1·0·(-2)+(-2)^2-c)=72+9-4 = 81 -4 = 77.$

Step 0: Check if $\vec{F}$ is conservative.

  1. Equality of Mixed Partial Derivatives: $\frac{∂P}{∂y} = \frac{∂}{∂y}(ycos(x)+y^2) = cos(x)+2y = \frac{∂Q}{∂x} = \frac{∂}{∂x}(sin(x)+2xy-2y)$.
  2. Continuous First Partial Derivatives. The components P(x, y) = $ycos(x)+y^2$ and Q(x, y) = $1 +sin(x)+2xy-2y$ involve polynomials, cosine, and sine functions, all of which are smooth and have continuous partial derivatives everywhere in ℝ2.
  3. Open and Simply Connected Domain: The domain of $\vec{F}$ is ℝ2, which is open and simply connected (no holes or gaps).

Since all the conditions are met, $\vec{F}$ is indeed a conservative vector field.

Step 1. Calculate the potential function

fx = ycos(x)+y2. Let’s integrate with respect to x, $f = \int f_xdx = \int ycos(x)+y^2dx = ysin(x) + y^2x + g(y)$. Here, g(y) is a function of y that acts as the “constant” of integration when integrating with respect to x.

fy = sin(x) + 2xy -2y. Let’s derivative f with respect to y, $f_y = \frac{∂f}{∂y} = \frac{∂}{∂y}(ysin(x) + y^2x + g(y)) = sin(x) +2yx +g’(y) = sin(x) + 2xy -2y ⇒g’(y) = -2y ⇒[\text{Let’s integrate g with respect to y}] g = \int -2ydy = -y^2 + k$.

Combining these results together: $f = ysin(x) + y^2x + g(y)= ysin(x) + y^2x + -y^2 + k$

Step 2. Evaluating work or the line integral $\int_{C} \vec{F}·d\vec{r}$

The line integral of a conservative vector field $\vec{F}$ over a curve C from point A to point B can be evaluated using the potential function f: Work = $\int_{C} \vec{F}·d\vec{r} = f(\frac{π}{2}, -1) - f(-π, 2)$.

$f(\frac{π}{2}, -1) = (-1)sin(\frac{π}{2})+\frac{π}{2}(-1)^2-(-1)^2+k = -1 + \frac{π}{2}-1+ k = \frac{π}{2}-2+k.$

$f(-π, 2) = 2sin(-π) +2^2(-π)-4+k = 0-4π-4+C$

Work = $\int_{C} \vec{F}·d\vec{r} = f(\frac{π}{2}, -1) - f(-π, 2) = \frac{π}{2}-2+k +4π+4-k = \frac{9π}{2}+2$

Step 0: Check if $\vec{F}$ is conservative.

  1. Equality of Mixed Partial Derivatives: $\frac{∂P}{∂y} = \frac{∂}{∂y}(e^{2y}) = 2e^{2y} = \frac{∂Q}{∂x} = \frac{∂}{∂x}(1 +2xe^{2y})$.
  2. Continuous First Partial Derivatives. The components P(x, y) = $e^{2y}$ and Q(x, y) = $1 +2xe^{2y}$ involve polynomials and exponential functions, both of which are smooth and have continuous partial derivatives everywhere in ℝ2.
  3. Open and Simply Connected Domain: The domain of $\vec{F}$ is ℝ2, which is open and simply connected (no holes or gaps).

Since all the conditions are met, $\vec{F}$ is indeed a conservative vector field.

Step 1. Calculate the potential function

fx = e2y. Let’s integrate with respect to x, $f = \int f_xdx = \int e^{2y}dx = e^{2y}x + g(y)$. Here, g(y) is a function of y that acts as the “constant” of integration when integrating with respect to x.

Similarly fy = 1 +2xe2y. Let’s integrate with respect to y, $f = \int f_ydy = \int 1 +2xe^{2y}dy = y + xe^{2y} + h(x)$.

Combining these results together: g(y) = y, h(x) = 0, f(x) = $y + xe^{2y}$

Step 2. Evaluating work or the line integral $\int_{C} \vec{F}·d\vec{r}$

The line integral of a conservative vector field $\vec{F}$ over a curve C from point A to point B can be evaluated using the potential function f: Work = $\int_{C} \vec{F}·d\vec{r} =[\text{t=1→B =(e,2),t=0→A = (0,1)}] f(e, 2) - f(0, 1) = (2+e·e^4)-(0+1) = e^5+1$

Step 1: Express the Vector Field

The given integral can be interpreted as a line integral of a vector field. We define the vector field $\vec{F}(x, y)$ as $\vec{F}(x, y) = ⟨M, N⟩ = M\vec{i}+ N\vec{j} = 2xe^{-y}\vec{i}+ (2y-x^2e^{-y})\vec{j}$

Step 2: Ensure the vector field $\vec{F}$ is Defined Everywhere

We can confirm that $\vec{F}$ is defined everywhere in R2. The components M(x, y) and N(x,y) involve polynomials and exponential functions, which are continuous (we are being very “economical” with the language, what we really mean is that the condition: continuous first partial derivatives is satisfied) and defined for all real values of x and y. Therefore, the vector field $\vec{F}$ is defined everywhere in ℝ2 and since ℝ2 which is a open and simply-connected domain, we proceed to check if $\vec{F}$ is conservative.

Step 3. Check if $\vec{F}$ is a conservative vector field.

To determine if the vector field $\vec{F} = ⟨M, N⟩$ is conservative, we check if it satisfies the condition Nx = My. This means that the mixed partial derivatives of the potential function must be equal.

Given: M = $2xe^{-y}$, N = $2y-x^2e^{-y}$

Calculate the partial derivatives:

$\frac{∂M}{∂y} = \frac{∂}{∂y}(2xe^{-y}) = -2xe^{-y}, \frac{∂N}{∂x} = \frac{∂}{∂x}(2y-x^2e^{-y}) = -2xe^{-y}$. My = Nx ⇒both conditions are satisfied, $\vec{F}$ is a conservative vector field, and therefore, the line integral over any path from (1, 0) to (2, 1) is path-independence.

Step 4. Calculate the potential function

For the same vector field, we know that $\vec{F} = ∇f$ where F = ⟨M, N⟩ = $⟨\frac{∂f}{∂x}, \frac{∂f}{∂y}⟩ = ⟨f_x, f_y⟩$

Given: M = fx(x, y) = $2xe^{-y}$ (i), N = fy(x, y) = $2y-x^2e^{-y}$ (ii)

M = fx(x, y) = $2xe^{-y}$ ⇒[To find f, we integrate fx with respect to x] f(x, y) = $\int f_x(x, y)dx = \int 2xe^{-y}dx = 2e^{-y}\int xdx = 2e^{-y}\frac{x^2}{2}+g(y) = x^2e^{-y} + g(y)$

Here, g(y) is a function of y acting as the constant of integration with respect to x.

To determine g(y), we differentiate $f(x, y) = x^2e^{-y} + g(y)$ with respect to y:

fy(x, y) = $\frac{∂}{∂y}(x^2e^{-y} + g(y)) = -x^2e^{-y}+g’(y)$

Matching this with (ii) fy(x, y) = $2y-x^2e^{-y}$, we get $2y-x^2e^{-y} = -x^2e^{-y}+g’(y)⇒ g’(y)=2y ⇒$[Let’s integrate with respect to y] $g(y)= y^2 +c$

Therefore, the potential function f is: $f(x, y) = x^2e^{-y} + g(y) = x^2e^{-y} + y^2 +c$

Step 4. Evaluating the line integral $\int_{C} \vec{F}·d\vec{r}$

The line integral of a conservative vector field $\vec{F}$ over a curve C from point A to point B can be evaluated using the potential function f: $\int_{C} \vec{F}·d\vec{r} = f(B) - f(A)$ Here, C is any path from (1, 0) to (2, 1).

$\int_{C} \vec{F}·d\vec{r} = \int_{C} ∇f·d\vec{r} = \int_{(1, 0)}^{(2, 1)} (x^2e^{-y} + y^2 +c)·d\vec{r} = f(2, 1)-f(1, 0)= (4e^{-1} + 1^2 +c)-(1^2e^{0} + 0^2 +c) = \frac{4}{e}+1+c-1-c = \frac{4}{e}.$

Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].
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