Conservative vector fields II

Enjoy the little things, for one day you may look back and realize they were the big things, Robert Brault.

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.

Path Independence and Conservative Vector Fields

Theorem. Path Independence and Conservative Vector Fields. Let $\vec{F}$ be a continuous vector field defined on an open, connected region D. If the line integral $\int_{C} \vec{F}\vec{dr}$ is independent of the path in D, then $\vec{F}$ is conservative. This means there exists a function f such that ∇f = $\vec{F}$, where ∇f represents the gradient of f.

Proof.

Suppose the line integral $\int_{C} \vec{F}\vec{dr}$ is independent of the path in D. This means that for any two arbitrary points A and B in D, the work done by $\vec{F}$ from A to B is the same no matter which path you take. $\vec{F}$ must be the gradient of some function, ∇f = $\vec{F}$.

Let (a, b) be a fixed point in D. We define the function f(x, y) for any point (x, y) in D as the value of the line integral from (a, b) to (x, y): f(x, y) = $\int_{(a, b)}^{(x, y)} \vec{F}\vec{dr}$. Claim: This function f(x, y) will be our potential function, and our goal is to show that the gradient of f is $\vec{F}$, i.e., ∇f = $\vec{F}$.

Consider a path from (a, b) to (x, y). We can break down this path into two parts:

• C₁: From (a, b) to (c, y).
• C₂: From (c, y) to (x, y) (Refer to Figure F for a visual representation and aid in understanding it) and recall that we can do so because we are working within an open, connected region.

The function f(x, y) can then be written as: f(x, y) = $\int_{(a, b)}^{(c, y)} \vec{F}\vec{dr} + \int_{(c, y)}^{(x, y)} \vec{F}\vec{dr}$. This means that f(x, y) is the sum of the work done by $\vec{F}$ along each sub-path.

Compute the partial derivative $\frac{∂f}{∂x}$

To find $f_x = \frac{∂f}{∂x}$, we focus on the path C₂ because C₁ does not involve x, so its derivative with respect to x is zero.

On the path C₂: $\frac{∂f}{∂x} = \frac{∂}{∂x}\int_{(c, y)}^{(x, y)} \vec{F}\vec{dr} =[\vec{F} = ⟨P, Q⟩] \frac{∂}{∂x}\int_{(c, y)}^{(x, y)} Pdx + Qdy$ = [On the path C₂, y is constant, hence dy = 0, and the line integral reduces to] $\frac{∂}{∂x}\int_{(c, y)}^{(x, y)} Pdx = \frac{∂}{∂x}[\hat{\mathbf{P}}(x, y)\bigg|_{(c, y)}^{(x, y)}] = $ where $\hat{\mathbf{P}}$ is the anti-derivative of P with respect to x

= $\frac{∂}{∂x}[\hat{\mathbf{P}}(x, y)-\hat{\mathbf{P}}(c, y)] =$[The derivative of an anti-derivate will give us back P, and the second term is x-independent (constant), so the derivative is zero] $P(x, y) +0$

Therefore, $f_x = \frac{∂f}{∂x} = P$. Our claim ∇f = ⟨f_x, f_y⟩ = $\vec{F}$ = ⟨P, Q⟩.

On the second hand of the proof everything is quite similar. The difference is that we are going to consider another path C composed of two parts, namely C₁ from (a, b) to (x, d) and C₂ from (x, d) to (x, y) (Refer to Figure G for a visual representation and aid in understanding it) and recall that we can do so because we are working within an open, connected region. Using a similar argument f_y = Q ∎

Since f_x = P (x, y) and f_y = Q(x,y), we have ∇f= ⟨f_x, f_y⟩ = ⟨P, Q⟩ = $\vec{F}$. Therefore, $\vec{F}$ is the gradient of f, meaning that $\vec{F}$ is conservative, which completes the proof.

Testing whether $\vec{F}$ is a gradient field

To determine if a vector field $\vec{F} = ⟨M, N⟩$ is a gradient field, we need to check whether there exists a scalar potential function f such that $\vec{F} = ∇f$. This means that the components of $\vec{F}$ can be expressed as the partial derivatives of f: $M = \frac{∂f}{∂x}$ and N = $\frac{∂f}{∂y}$.

For $\vec{F}$ to be a gradient field, f must satisfy the condition that the mixed partial derivatives are equal, i.e., $\frac{∂^2f}{∂x∂y}=\frac{∂^2f}{∂y∂x}$.

The condition that the mixed partial derivatives must be equal arises from a fundamental result in calculus known as Clairaut’s theorem (or Schwarz’s theorem). It states that if a function f(x,y) has continuous second-order partial derivatives, then the order in which you take the derivatives does not matter.

This leads to the following criterion to check that $\vec{F}$ is conservative: $\frac{∂M}{∂y}=\frac{∂N}{∂x}$

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.

Breaking Down the Criteria:

1. Equality of Mixed Partial Derivatives. The condition M_y = N_x ↭ $\frac{∂M}{∂y} = \frac{∂N}{∂x}$, ensures that the curl of the vector field $\vec{F}$ is zero. This is a necessary condition for $\vec{F}$ to be a gradient field, which means that can be expressed as the gradient of some potential function f(x, y).
2. Continuous First Partial Derivatives. The functions M and N must have continuous first partial derivatives across the domain D. This ensures that the vector field $\vec{F}$ behaves smoothly, without any abrupt changes in direction or magnitude.
3. Open and Simply Connected Domain. The domain D must be open (meaning it does not include its boundary points) and simply connected (meaning it has no holes). This condition prevents the presence of obstacles or gaps in the domain that could disrupt the path independence of the line integral. In a simply connected domain, any closed loop can be continuously shrunk to a point without leaving the domain, ensuring that the vector field $\vec{F}$ is conservative.

Conversely, if $\vec{F}$ = ⟨M, N⟩ is conservative, then it must satisfy M_y = N_x ↭ $\frac{∂M}{∂y} = \frac{∂N}{∂x}$, assuming that $\vec{F}$ is defined and differentiable everywhere.

• Example 0: Conservative Vector Field. Consider the vector field $\vec{F} = (x^2+y)\vec{i}+(y^2+x)\vec{j}$. We want to determine if this vector field is conservative. If it is, we will find its potential function.

Check if the Vector Field is Conservative.

It is conservative because:

1. Equality of Mixed Partial Derivatives: $\frac{∂M}{∂y} = \frac{∂}{∂y}(x^2+y) = 1 = \frac{∂N}{∂x} = \frac{∂}{∂x}(y^2+x)$.
2. Continuous First Partial Derivatives. The functions M(x, y) and N(x, y) are both polynomials, which means their partial derivatives are continuous everywhere in ℝ².
3. Open and Simply Connected Domain. The domain D = ℝ² is the entire plane, which is open (it does not include its boundary) and simply connected (it has no holes). This criterion is also satisfied.

Since all three conditions are met, the vector field $\vec{F}$ is indeed conservative. In the context of determining whether a vector field is conservative, conditions (2) (continuous first partial derivatives) and (3) (the domain being open and simply connected) are indeed fundamental. While it’s true that these conditions are often not explicitly mentioned in literature or problem-solving scenarios, they are nonetheless essential.

Find its potential function

To find the potential function f(x, y) such that $\vec{F} = ∇f = ⟨\frac{∂f}{∂x}, \frac{∂f}{∂y}⟩ = ⟨M(x, y), N(x, y)⟩$, we can use the following methods:

• 1st Method:

We start by integrating M(x,y) with respect to x: f_x = x²+y ⇒$f = \int (x^2+y)dx = \frac{1}{3}x^3+xy+g(y)$. Here, g(y) is a function of y alone.

Next, we integrate M(x,y) with respect to y: f_y = y²+x ⇒$f = \int (y^2+x)dy = \frac{1}{3}y^3+xy+h(x)$. Here, h(x) is a function of x alone.

Combining both results, $h(x) = \frac{1}{3}x^3, g(y) = \frac{1}{3}y^3, f(x, y) = \frac{1}{3}x^3+xy+\frac{1}{3}y^3.$

• 2nd method.

We start by integrating M(x,y) with respect to x: f_x = x²+y ⇒$f = \int (x^2+y)dx = \frac{1}{3}x^3+xy+g(y)$. Here, g(y) is a function of y alone.

Next, we differentiate this result with respect to y to find g(y): f_y = $\frac{∂}{∂y}(\frac{1}{3}x^3+xy+g(y))$ = x + g’(y).

Since f_y = $\frac{∂f}{∂y}$ must equal N(x, y) = y²+x, we have: x + g’(y) = y² + x⇒g’(y) = y². Integrating g’(y) with respect to y: g(y) = $\int y^2dy = \frac{y^3}{3}+C$. Thus, the potential function f(x, y) is: $f(x, y) = \frac{1}{3}x^3 + xy + \frac{1}{3}y^3.$

• Example 1: Non-Conservative Vector Field. Consider the vector field $\vec{F} = -y\vec{i}+x\vec{j}$ and let C be the unit circle. We want to check if this vector field is a gradient field.

Here, the components M = -y, N = x. We compute the partial derivatives: $\frac{∂M}{∂y} = -1 ≠ 1 = \frac{∂N}{∂x}$, so $\vec{F}$ does not satisfy the condition for being a gradient field. Therefore, the vector field is not conservative.

• Example 2: Another Non-Conservative Vector Field. Consider the vector field $\vec{F} = 3xy\vec{i}-x^2\vec{j}$. We will check if this vector field is a gradient field.

Here, the components are M = 3xy, N = -x². We compute the partial derivatives: $\frac{∂M}{∂y} = 3x ≠ -2x = \frac{∂N}{∂x}$, so indeed $\vec{F}$ is not a gradient field, and thus is not conservative.

• Exercise: Finding the Value of a that Makes $\vec{F}$ a gradient field.

Consider the vector field $\vec{F}=(4x^2+axy)\vec{i}+(3y^2+4x^2)\vec{j}$. We want to find the value of a that makes what value of a makes $\vec{F}$ a gradient field.

Here, the components are M = 4x²+axy and N = 3y²+4x².

We compute the partial derivatives: $M_y = \frac{∂M}{∂y} = \frac{4x^2+axy}{∂y} = ax, N_x = \frac{∂N}{∂x} = \frac{3y^2+4x^2}{∂x} = 8x$.

For $\vec{F}$ to be a gradient field, we must have $\frac{∂M}{∂y} = \frac{∂N}{∂x}$. Therefore, we require ax =8x ↭[Solving for a and assuming a ≠ 0, we get] a = 8. Thus, when a = 8, the vector field $\vec{F}=(4x^2+8xy)\vec{i}+(3y^2+4x^2)\vec{j}$ is a gradient field.

• Given the vector field $\vec{F} = ⟨2xy^3z^4 -P-, 3x^2y^2z^4 -Q-, 4x^2y^3z^3 -R-⟩$, determine if $\vec{F}$ is conservative. If it is, find the potential function and evaluate the line integral $\int_{C} \vec{F}\vec{dr}$ where C is parameterized by $d\vec{r}(t) = ⟨t, t^2, t^3⟩$ with 0 ≤ t ≤ 2.

Check if the Vector Field is Conservative.

It is conservative because:

1. Equality of Mixed Partial Derivatives: $\frac{∂P}{∂y} = 6xy^2z^4 = \frac{∂Q}{∂x}, \frac{∂P}{∂z} = 8xy^3z^3 = \frac{∂R}{∂x}, \frac{∂Q}{∂z} = 12x^2y^2z^3 = \frac{∂R}{∂y}$.
2. Continuous First Partial Derivatives. The functions P, Q and R are all polynomials in x, y, and z, which means their partial derivatives are continuous everywhere in ℝ³.
3. Open and Simply Connected Domain. The domain D = ℝ³ which is both open (it does not include its boundary) and simply connected (it has no holes). This criterion is also satisfied.

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

Find its potential function

Since $\vec{F}$ is conservative, there exists a potential function f(x, y, z) such that ∇f = $\vec{F}$. This means, $\frac{∂f}{∂x} = P = 2xy^3z^4, \frac{∂f}{∂y} = Q = 3x^2y^2z^4, \frac{∂f}{∂z} = R = 4x^2y^3z^3.$

We will find f(x, y, z) by integrating these expressions

f_x = 2xy³z⁴. Let’s integrate with respect to x, f = $\int 2xy^3z^4dx = x^2y^3z^4+g(y, z).$ Here, g(y, z) is an arbitrary function of y and z.

f_y = 3x²y²z⁴. Let’s integrate with respect to y, f = $\int 3x^2y^2z^4dy = x^2y^3z^4 + h(x, z).$

f_z = 4x²y³z³. Let’s integrate with respect to z, f = $\int 4x^2y^3z^3dz = x^2y^3z^4 + m(x, y).$

Combining all these results: $g(y, z) = h(x, z) = m(x, y) = 0, f(x, y, z) = x^2y^3z^4.$

Evaluate the Line Integral. For a conservative vector field, the line integral $\int_{C} \vec{F}\vec{dr}$ depends only on the endpoints of the path C. Specifically, $\int_{C} \vec{F}\vec{dr} = f(\text{endpoint})-f(\text{starting point})$

The curve C is parameterized by $d\vec{r}(t) = ⟨t, t^2, t^3⟩$ with 0 ≤ t ≤ 2. Therefore, the endpoints at t = 0, $d\vec{r}(0) = ⟨0, 0, 0⟩$, at t = 2, $d\vec{r}(2) = ⟨2, 4, 8⟩$.

$\int_{C} \vec{F}\vec{dr} = f(\text{endpoint})-f(\text{starting point}) = f(2, 4, 8) -f(0, 0, 0) = 2^2·4^3·8^4-0 = 4⋅64⋅4096 = 1,048,576.$

Curl, Torque, and Conservative Vector Fields

Recall.

• A vector field $\vec{F} = ⟨M, N⟩ = ∇f = ⟨\frac{∂f}{∂x}, \frac{∂f}{∂y}⟩ = ⟨f_x, f_y⟩$ is called a gradient field if it can be expressed as the gradient of a scalar potential function f(x, y).
• A vector field $\vec{F}$ is conservative if the line integral of $\vec{F}$ around any closed curvve C is zero: $\oint_C \vec{F} \cdot d\vec{r} = 0$. This is true if $\vec{F}$ is defined over the entire plane or in a simply connected region (a region with no holes).

Definition of Curl. The curl of a vector field $\vec{F}$ in two dimensions is defined as: $curl(\vec{F}) = N_x - M_y$. This scalar quantity measures the tendency of the vector field to induce rotation or swirling around a point.

Test for Conservativeness. A vector field $\vec{F}$ is conservative if its curl is zero everywhere in the region: $curl(\vec{F}) = N_x - M_y$ = 0.. When the curl is zero, the vector field has no rotational component, meaning the field is path-independent, and the work done around any closed curve is zero.

A physical interpretation of the curl is as follows. Suppose that $\vec{F}$ represents a velocity field in fluid dynamics or physics. The curl of the vector field $curl(\vec{F})$ at a point measures the rotational or swirling motion of the fluid at that point. In essence, curl tells us how much and in what direction the fluid rotates around a point.

1. Vector Field: $\vec{F} = ⟨a, b⟩$ where a and b are constants, $curl(\vec{F}) = N_x - M_y = \frac{∂b}{∂x}-\frac{∂a}{∂y} = 0 -0 = 0.$. Interpretation: The curl is zero, indicating no rotation in the field.
2. Vector Field: $\vec{F} = ⟨x, y⟩, curl(\vec{F}) = N_x - M_y = \frac{∂}{∂x}(y) - \frac{∂}{∂y}(x) = 0 -0 = 0.$ Interpretation: Again, the curl is zero, meaning there is no rotational component to this field.
3. $\vec{F} = ⟨-y, x⟩, curl(\vec{F}) = N_x - M_y = \frac{∂}{∂x}(x) - \frac{∂}{∂y}(-y) = 1 + 1 = 2$. Interpretation: The curl is 2, which indicates that the field has a rotational component. The positive curl suggests a counterclockwise rotation. The magnitude of the curl corresponds to the angular velocity of this rotational motion (Figures 1, 2, and 3 respectively).

The curl of a force field is related to the torque exerted on a object within that field. Torque is a measure of the rotational force applied to an object.The relationship between torque and angular velocity (rate of rotation) is given by: $\frac{torque}{\text{moment of inertia}} = \frac{d}{dt}(\text{angular velocity})$.

In a vector field, if the curl is non-zero, it implies that there is a rotational force acting on objects within the field, leading to a non-zero torque.

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, Andrew Misseldine, and MathMajor, YouTube’s channels.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. MIT OpenCourseWare, 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007, YouTube.
8. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.