Learn everything you can, anytime you can, from anyone you can – there will always come a time when you will be grateful you did, Sarah Caldwell.
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.
The interior of a circle (without the boundary) is an open region because you can draw a small disk around any point inside the circle that stays completely within the circle.
A filled circle is connected because you can move from any point inside the circle to any other point without leaving the circle.
A circle (a closed loop) is a simple curve because it does not intersect itself.
The interior of a circle is simply connected because any loop drawn inside the circle only encloses points within the circle, and there are no holes.
Region {(x, y) | x^{2} + y^{2} ≤ 1 or 4 ≤ x^{2} + y^{2} ≤ 9}. It is not open because it includes the boundaries x^{2} + y^{2} = 1, x^{2} + y^{2} = 4, and x^{2} + y^{2} = 9 which are not open sets (they include their boundary points). It is not connected because it consists of two disjoint parts: a small disk of radius 1 centered at the origin, and an annulus (ring) between radii 2 and 3. You cannot move from the smaller disk to the annulus without leaving the region. It is not simply connected because it is not connected and also has a hole between the two parts of the region (Refer to Figure C for a visual representation and aid in understanding it)
{(x, y) | 4 < x^{2} + y^{2} < 9} is open because it excludes the boundaries x^{2} + y^{2} = 4 and x^{2} + y^{2} = 9, so for any point in the region, you can find a small disk around it that stays entirely within the region. It is connected because you can move from any point within the annulus to any other point within it without leaving the region. It is not simply connected because it has a “hole” in the middle (the area inside the circle of radius 2), so there exist simple closed curves in the region that enclose points not in the region (Refer to Figure D for a visual representation and aid in understanding it)
The gradient of a scalar function f is the vector field (or vector-valued function) that points in the direction of the greatest rate of change (the direction in which the function increases more quickly) of the function, and its magnitude is the rate of change or increase in that direction.
The components of the gradient are the partial derivatives of the function with respect to each variable. Mathematically, the gradient is represented as: ∇f = ($\frac{∂f}{∂x_1},\frac{∂f}{∂x_2},···\frac{∂f}{∂x_n}$) where $\frac{∂f}{∂x_i}$ denotes the partial derivative of f with respect to the i-th variable.
A vector field $\vec{F}$ is called conservative if it can be expressed as the gradient of some scalar function f. In other words, there exist a function f such that: $\vec{F}$ = ∇f, then $\vec{F}$ is a conservative vector fields, and f is known or referred to as the potential function of the vector field.
When a vector field is conservative, the work done by the field along a path from one point to another $\int_{C} \vec{F}d\vec{r}$ depends only on the endpoints of the path, not on the path taken. This property simplifies the computation of line integrals significantly.
Theorem. The Fundamental theorem of Calculus for Line Integral provides a powerful tool for evaluating line integrals of conservative vector fields. The theorem states: If $\vec{F}$ is a conservative vector field defined in a simply connected region of space (a region with no holes), and if f is the potential function for $\vec{F}$ in that region, then the line integral of \vec{F} along a curve C from a point P_{0} to a point P_{1} is given by: $\int_{C} \vec{F}·d\vec{r} = \int_{C} ∇f·d\vec{r} = f(P_1)-f(P_0)$ where P_{0} and P_{1} are the initial and final points of the curve C, respectively.
This result is immensely useful because it allows us to calculate the work done by the field simply by evaluating the potential function at the endpoints of the curve, bypassing the need to evaluate the integral along the entire path.
Proof:
The line integral of the vector field $\vec{F}$ along the curve C can be written as:
$\int_{C} \vec{F}·d\vec{r} = \int_{C} ∇f·d\vec{r} = \int_{C} ⟨\frac{∂f}{∂x}, \frac{∂f}{∂y}⟩·⟨dx, dy⟩ = \int_{C}(\frac{∂f}{∂x}dx + \frac{∂f}{∂y}dy)$.
Now, let’s parametrize the curve C using a parameter t. Suppose the curve C is described by the parametric equations x = x(t), y = y(t), where t ranges from t_{0} to t_{1}: t_{0}≤ t ≤ t_{1}. The differentials are $dx = \frac{dx}{dt}dt = x’(t)dt, dy = \frac{dy}{dt}dt = y’(t)dt$.
Substituting these into the integral gives:
$\int_{C} (\frac{∂f}{∂x}dx + \frac{∂f}{∂y}dy) = \int_{C} (\frac{∂f}{∂x}x’(t)dt + \frac{∂f}{∂y}y’(t)dt) = \int_{C} (\frac{∂f}{∂x}x’(t) + \frac{∂f}{∂y}y’(t))dt$ =[By applying the chain rule, the expression inside the integral simplifies to the total derivative of f with respect to t:] $\int_{t_0}^{t_1} \frac{d}{dt}f(x(t),y(t))dt =[\text{Using the Fundamental Theorem of Calculus, we can evaluate this integral as:}] = f(x(t), y(t))\bigg|_{t_0}^{t_1} = f(P_1)-f(P_0)$ where f(P_{0}) = f(x(t_{0}), y(t_{0})), f(P_{1}) = f(x(t_{1}), y(t_{1})).
This proof shows that for a conservative vector field, the line integral between two points depends only on the values of the potential function at those points, not on the specific path taken.
Let’s dive into the problem where we analyze the work done by the vector field $\vec{F}= y\vec{i}+x\vec{j}$ = ⟨x, y⟩ as we move along a path C composed of three segments C_{1}, C_{2}, and C_{3} (C = C_{1}+ C_{2} + C_{3}) in the unit circle (Figure D)
Understanding the Vector Field. The vector field given is $\vec{F}$ =⟨y, x⟩, which assigns to each point (x, y) in the plane a vector with components y and x. This means that at any given point (x, y), the vector points in the direction where the x and y coordinates have swapped roles.
Finding the Potential Function. Given the vector field $\vec{F}= y\vec{i}+x\vec{j}$ = ⟨y, x⟩, we need to check if it can be expressed as the gradient of some scalar function f, ∇f = $\vec{F}$. Let’s verify that f(x, y) = xy is such a function: ∇f = $(\frac{∂f}{∂x}, \frac{∂f}{∂y}) = ⟨y, x⟩ = \vec{F}$. Thus, $\vec{F}$ is indeed a conservative vector field with the potential function f(x,y) = xy.
Work Done Along the Path C: $\int_{C} ∇f·dr = f(P_1)-f(P_0)$ =[Figure E, C is a closed curve starting and ending at the same point P_{0} = P_{1} = (0, 0)] 0 -0 = 0.
Let’s consider the work done along a specific segment of the path, C_{2}, which goes from (1, 0) to $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}), \int_{C_2} \vec{F}·d\vec{r} = f(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})-f(1,0) = \frac{1}{2}-0 = \frac{1}{2}.$
This results show that the work done by a conservative vector field along any closed curve C is zero. This is due to the fact that the work depends only on the potential function's values at the endpoints of the curve (and not on the path taken between them) which are identical in the case of a closed curve, $\int_{C} ∇f·d\vec{r} = f(EndPoint)-f(OriginPoint) = 0$ because both points are the same point.
The significance and main idea of conservative vector fields lies in their path independence, which means that the work done by the vector field along any path between two points depends only on the endpoints and not on the path taken. For a conservative vector field $\vec{F}= ∇f$, and any two paths or curves C_{1} and C_{2} that share the same start (P_{0}) and end points (P_{1}), the work done along both paths will be the same, $\int_{C_1} ∇f·d\vec{r} = \int_{C_2} ∇f·d\vec{r}$
In this section, we’ll explore a counterexample of a non-conservative vector field, specifically $\vec{F} = -y\vec{i}+x\vec{j}$ and demonstrate why it is not conservative by evaluating the work done by this force field along the unit circle.
Understanding the Vector Field. The given vector field assigns to each point (x, y) in the plane a vector with components −y in the x-direction (opposite to the y-coordinate) and x in the y-direction (aligned with the x-coordinate). It represents a rotational flow around the origin, where vectors point counterclockwise around the origin.
The work done by $\vec{F}$ along the curve C is given by the line integral: W = $\int_{C} \vec{F}·d\vec{r} = \int_{C} \vec{F}·\hat{\mathbf{T}}ds =$ [On the unit circle, $\vec{F}$ is tangent to the curve ↭ $\vec{F}//\hat{\mathbf{T}}⇒\vec{F}·\hat{\mathbf{T}} = |\vec{F}|$ =[The magnitude of the force in the unit circle is] $\sqrt{(-y)^2+x^2}=\sqrt{y^2+x^2}=\sqrt{1}=1$ where $\hat{\mathbf{T}}$ is the unit tangent vector to the curve] = $\int_{C} |\vec{F}|ds = \int_{C} 1·ds$ =[C is the unit circle] 2π ≠ 0.
Another way: W = $\int_{C} \vec{F}·d\vec{r}$. For a curve C parameterized by $\vec{r}(t) = ⟨x(t), y(t)⟩ =[\text{C is the unit circle}] ⟨cos(t), sin(t)⟩$ over an interval t_{0} ≤ t ≤ t_{1} ↭ t ranging from 0 to 2π (unit circle). $\frac{dx}{dt} = -sin(t), \frac{dy}{dt} = cos(t).$
W = $\int_{C} \vec{F}·d\vec{r} = \int_{t_0}^{t_1} \vec{F}·(⟨\frac{dx}{dt}, \frac{dy}{dt}⟩dt) = \int_{t_0}^{t_1} (-y(t)\frac{dx}{dt} + x(t)\frac{dy}{dt})dt = \int_{0}^{2π} (-sin(t)(-sin(t))+cos(t)cos(t))dt = \int_{0}^{2π} (sin^2(t) + cos^2(t))dt = \int_{0}^{2π}dt = 2π$
Key Observations:
If the force field $\vec{F}$ acting on an object is the gradient of a potential function (i.e., it is conservative), $\vec{F} = ∇V$, then the work done by this force on the object is indeed equal to the change in the potential energy of the system.
This relationship could be expressed using the fundamental theorem of calculus for line integrals. If an object moves along a smooth curve C in space under the influence of the force $\vec{F}$, P_{1} and P_{2} are two points on the curve C with position vectors $\vec{r_1}\text{and}\vec{r_2}$ respectively, then the work done by $\vec{F}$ as the object moves from point P_{1} to point P_{2} is given by:
$\int_{C} \vec{F}·d\vec{r} = \int_{C} ∇V·d\vec{r}$ [By the fundamental theorem of calculus for line integral, this integral can be evaluated as:] = $V\vec(r_2)-V\vec(r_1)$
This means that the work done by the conservative force $\vec{F}$ is simply the difference in the values of the potential function V evaluated at the final and initial positions of the object.
In systems governed by conservative forces, such as gravitational or electrical fields, the total energy is conserved (Conservation of energy). This principal law states that the total energy of a closed system remains constant over time. For example, in a gravitational system, the gravitational field $\vec{g}$ is related to the gravitational potential V by $\vec{g} = -∇V$ where $\vec{g}$ is the gravitational field (the force experienced by objects due to gravity, approximately 9.8 m/s^{2} directed downwards) and V is the gravitational potential. The negative sign indicates that the force points in the direction of decreasing potential.
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 $\int_{C} \vec{F}·d\vec{r} = 0$. It means that if you move an object under the influence of a conservative force along a path that eventually returns to its starting point, the total work done by the force over the entire journey is zero.
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.
This property is directly related to the first one. If $\vec{F}$ is path-independent, consider a closed curve. Path independence implies that instead of traversing the entire closed path, one could stay at the starting point, resulting in zero work done. Conversely, if $\vec{F}$ is conservative, and you have two points P_{1}, P_{2} with two different paths C_{1} and C_{2} between them, then C_{1} -C_{2} is a closed loop (returning to the original point P_{1}). Since $\vec{F}$ is conservative ⇒ $\int_{C_1-C_2} \vec{F}·d\vec{r} = 0 ⇒ \int_{C_1-C_2} \vec{F}·d\vec{r} = \int_{C_1} \vec{F}·d\vec{r} -\int_{C_2} \vec{F}·d\vec{r} =0 ⇒ \int_{C_1} \vec{F}·d\vec{r} = \int_{C_2} \vec{F}·d\vec{r}.$
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: $\int_{C} \vec{F}·d\vec{r} = f(P_1)-f(P_0)$. This result indicates that the work done by a gradient field depends only on the values of the potential function at the endpoints, further emphasizing path independence.
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.