Vector fields

Quantity has a quality of its own, Joseph Stalin.

Vector fields

A vector field is a mathematical construct where a vector $\vec{F}$ is assigned to each point (x, y) in a space. Specifically, in two-dimensional space, a vector field is often represented by the function $\vec{F}$, which can be expressed as $\vec{F} = M\vec{i}+N\vec{j}$ where M(x, y) and N(x, y) are functions of the coordinates x and y, and $\vec{i}$ and $\vec{j}$ are unit vectors in the directions of the x-axis and y-axis, respectively.

Vector fields are widely used in physics and engineering to model various phenomena, such as the speed and direction of a moving fluid or air, the direction and magnitude of electric, magnetic, or gravitational forces, etc.

Vector fields can be visualized by plotting arrows on a plane, where each arrow corresponds to a vector at a particular point. These arrows represent vectors with specific magnitudes and directions.

Illustrations of Vector Fields

We are going to illustrate the following vector fields:

1. Constant Vector Field: $\vec{F} = 2\vec{i}+\vec{j}$ (Figure 1). : This vector field does not depend on x and y. It assigns the same vector to every point in the plane. It is constant everywhere, meaning that all vectors have the same magnitude and direction (the field is uniform across the entire space).
2. Linear Vector Field: $\vec{F} = x \vec{i}$ (Figure 2). This vector field depends linearly on the x coordinate. The arrows point in the direction of the x-axis. The vector’s magnitude increases as x increases and decreases as x decreases, but it is independent of y.
3. Radial Vector Field: $\vec{F} = x\vec{i}+y\vec{j}$ (Figure 3). This vector field depends linearly on both x and y. The arrows point radially outward from the origin. The magnitude of the vectors increases as the distance from the origin increases.

   Arrows point away from the origin, growing longer as they move further away, indicating that the strength of the field increases with distance.

4. Rotational Vector Field: $\vec{F} = -y\vec{i}+x\vec{j}$ (Figure 4). This vector field is a rotational field, where each vector represents a counter-clockwise rotation around the origin. The transformation from (x, y) to (−y,x) results in vectors that are perpendicular to the radial line at each point, i.e., a 90° counter-clockwise rotation of the point (x, y) to (−y, x).

These illustrations help visualize how vector fields vary across space and provide insights into the behavior of different types of fields.

Work and line integrals

In physics, work is a fundamental concept that describes the transfer of energy when a force moves an object over a distance.

In its simplest form, when a constant force acts on an object and is aligned with the direction of motion, the work done by a force $\vec{F}$ over a displacement $Δ\vec{r}$ is calculated by the dot product of the force vector and the displacement vector: W = force · distance = $\vec{F}·Δ\vec{r}$. Here, the dot product means that work is the product of the magnitude of the force, the magnitude of the displacement, and the cosine of the angle between the force and displacement vectors.

More generally, where the force is not constant or the path is not straight, work is defined as the energy transferred when a force acts on an object and moves 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 = $\int_{C} \vec{F}·d\vec{r}$.

The line integral represents the sum of the work done by the force over each infinitesimal segment of the path.

This integral can be understood as the sum of the dot products of the force vector and the infinitesimal displacement vectors along the path C. Consider breaking the curve C into small segments $Δ\vec{r_i}$, the work done can be approximated as: $\sum_{i} \vec{F}·Δ\vec{r_i}$. The work done over each segment can be approximated by the dot product of the force vector at that segment $\vec{F}$ and the segment’s displacement vector $Δ\vec{r_i}$.

In the limit as the segment length $Δ\vec{r_i}$ approaches zero (i.e., as the segmentation becomes finer and finer), this sum becomes the line integral, W = $\int_{C} \vec{F}·d\vec{r} = \lim_{\Delta r_i \to 0} \sum_{i} \vec{F}·Δ\vec{r_i}$ =[To connect this with velocity, note that the displacement vector can also be expressed in terms of velocity: $Δ\vec{r_i} = \frac{d\vec{r}}{dt}Δt$]

Hence, the work integral can be rewritten as: W = $\lim_{\Delta t \to 0} \sum_{i} \vec{F}·(\frac{d\vec{r}}{dt}Δt)$.

Recognizing that $\frac{d\vec{r}}{dt}$ is the velocity vector $\vec{v}$, we obtain: W = $\int_{t_1}^{t_2} \vec{F}·\vec{v}dt = \int_{t_1}^{t_2} (\vec{F}·\frac{d\vec{r}}{dt}·dt)$ This shows that the work done by a force over a period from time t₁ to t₂ is the integral of the dot product of the force vector and the velocity vector over that time interval.

Solved exercises

• Consider a force field $\vec{F} = 2x\vec{i}+3y\vec{j}$ and a path C parameterized by $\vec{r}(t)= (t, t^2)$ where t ranges from 0 to 1. Calculate the work done by this force field along the path C.

Parameterization of the Path. The path C is described by the parameterization $\vec{r}(t)= (t, t^2)$. It is a parabolic curve starting at the origin (0, 0) and ending at the point (1, 1).

Derivative of the Position Vector. To compute the work done by the force field along this path, we first need the derivative of the position vector $\vec{r}(t)$ with respect to time t. This derivative represents the velocity vector along the path: $\frac{d\vec{r}}{dt} = ⟨\frac{dx}{dt}, \frac{dy}{dt}⟩ = (1, 2t)$.

Evaluate the force field along the path: $\vec{F}(\vec{r}(t)) = (2x, 3y) =[x = t, y = t^2] (2t, 3t^2)$

Work Done by the Force Field. The work W done by the force field $\vec{F}$ along the path C is given by the line integral:

W = $\int_{C} \vec{F}·d\vec{r} = \int_{t_1}^{t_2} (\vec{F}·\frac{d\vec{r}}{dt}·dt) = \int_{0}^{1} (\vec{F}·\frac{d\vec{r}}{dt}dt) = \int_{0}^{1} (2t, 3t^2)·(1, 2t)dt = \int_{0}^{1} (2t + 6t^3) = t^2 + \frac{6t^4}{4} = t^2 + \frac{3t^4}{2}\bigg|_{0}^{1} = 1 + \frac{3}{2} = \frac{5}{2}.$ So, the work done by the force field along the path C is $\frac{5}{2}.$

Visualization:

1. The path C is a parabolic path y = x² from (0, 0) to (1, 1).
2. Force Field $\vec{F}$: Visualize vectors that increase in magnitude as you move away from the origin, with x and y components scaling linearly.
3. Work Calculation: The integral calculates the total effect of the force along the curved path, accounting for both the magnitude and direction of the force and the displacement.

• Consider a force field $\vec{F} = -y\vec{i}+x\vec{j}$, and a path C parameterized by $\vec{r}(t)= (t, t^2)$, 0 ≤ t ≤ 1 (Figure 4.a). Evaluate the work done by the force field $\vec{F}$ along the given path C.

The path C is parametrized by the function $\vec{r}(t)= (t, t^2)$ where t ranges from 0 to 1. This paths is describes a parabola y = x² in the xy-plane.

Work done by the force field. The work W done by a force field $\vec{F}$ along a path C is given by the line integral:

W = $\int_{C} \vec{F}·d\vec{r} = \int_{t_1}^{t_2} (\vec{F}·\frac{d\vec{r}}{dt}·dt) = \int_{0}^{1} (\vec{F}·\frac{d\vec{r}}{dt}dt)$

Force Field Along the Path: $\vec{F} = ⟨-y, x⟩ =$[x = t, y = t²] = ⟨-t², t⟩. Besides, $\frac{dx}{dt}=1,\frac{dy}{dt}= 2t, \frac{d\vec{r}}{dt}= ⟨\frac{dx}{dt}, \frac{dy}{dt}⟩ = ⟨1,2t⟩$

W = $\int_{0}^{1} (\vec{F}·\frac{d\vec{r}}{dt}dt) =\int_{0}^{1} ⟨-t^2, t⟩ · ⟨1, 2t⟩ dt = \int_{0}^{1} (-t^2+2t^2)dt = \int_{0}^{1} t^2dt = \frac{t^3}{3}\bigg|_{0}^{1} = \frac{1}{3}$

Alternative Formulation Using Differential Forms

Another approach to understanding the work integral is to use the differential form of the line integral. If $\vec{F} = ⟨M, N⟩, \vec{dr} = ⟨dx, dy⟩ ⇒ \vec{F}·\vec{dr} = ⟨M, N⟩ · ⟨dx, dy⟩ = Mdx + Ndy$

This gives us another way to write the work integral:

W = $\int_{C} \vec{F}·d\vec{r} = \int_{C} Mdx + Ndy$

Evaluating Double Integrals

When evaluating double integrals, one effective method is to parameterize the variables x and y in terms of a single parameter (like t) and then substitute these expressions into the integral. This method simplifies the problem, especially when dealing with curves in a vector field.

Consider a vector field $\vec{F} = -y\vec{i}+x\vec{j}$, where $\vec{i}$ and $\vec{j}$ are the unit vectors in the x and y directions, respectively. Let’s parameterize the curve C as follows: x = t, y = t², with 0 ≤ t ≤ 1.

From this, we have dx = dt, dy = $\frac{dy}{dt}dt$ = 2t·dt, $\vec{dr} = ⟨dx, dy⟩ = ⟨dt, 2tdt⟩$

Calculating the Work Integral. The work W done by the force field $\vec{F}$ along the curve C can be computed using the line integral: W = $\int_{C} \vec{F}·d\vec{r} = \int_{C} Mdx + Ndy = \int_{C} -ydx + xdy =[\text{Substituting the parameterization:}] \int_{C} -t^2dt+t·2tdt = \int_{0}^{1} t^2dt = \frac{t^3}{3}\bigg|_{0}^{1} = \frac{1}{3}.$

Thus the work done by the force field $\vec{F}$ along the curce C, $\int_{C} \vec{F}·d\vec{r}= \frac{1}{3}$ units. It is specific to the path or trajectory C, but not the parametrization. This means that if we parameterize the same curve differently, such as: x = sin(θ), y = sin²(θ), 0 ≤ θ ≤ $\frac{π}{2}$ the work integral will still yield the same result.

The integral depends on the geometric shape of the curve C and the vector field $\vec{F}$, not on how the curve is parameterized.

Geometrical interpretation

1. The line integral of a vector field $\vec{F}$ along a curve C is a way to calculate the total work done by the force field as an object moves along the curve. Mathematically, it is expressed as: W = $\int_{C} \vec{F}·d\vec{r}$.
2. The unit tangent vector $\hat{\mathbf{T}}$ is a normalized (it has a length of one) vector that is tangent to the curve at any given point. It is a vector that points in the direction of the curve’s movement as it passes through that point.

For a small segment of the trajectory C, the displacement vector $Δ\vec{r}$, it will go in the same direction as the unit vector and have a length Δs (where s is the distance along the trajectory).

Thus, $d\vec{r} = ⟨dx, dy⟩ = \hat{\mathbf{T}}·ds$ -Figure A.- where ds is the differential arc length, representing the infinitesimal distance along the curve.

Substituting this expression into the work integral, we get: W = $\int_{C} \vec{F}·d\vec{r} = \int_{C} \vec{F}·\hat{\mathbf{T}}·ds$

Examples and Geometrical Interpretation

• Work Done by the Vector Field $\vec{F} = x\vec{i} + y\vec{j}$ along a Circular Path

Consider a circle C of radius “a”. The curve is traversed counterclocwise. The vector field is given by $\vec{F} = x\vec{i} + y\vec{j}$. We want to compute the work done by the force as we move along the circle C.

Geometric interpretation

Geometrically, the work done by the vector field $\vec{F}$ along the curve C can be intuitively understood by considering the relationship between the force vector and the direction of the motion and is zero because the force is perpendicular (point radially outward from the origin) to the motion (tangent to the circle) at every point on the circle (Figure B) ↭ $\vec{F} ⊥\hat{\mathbf{T}}⇒ \vec{F} · \hat{\mathbf{T}} = 0 ⇒\int_{C} \vec{F}·\hat{\mathbf{T}}·ds = 0$.

Detailed Calculation.

1. Parameterize the circle. The circle can be parameterized using: x = acos(θ), y = asin(θ), where 0 ≤ θ ≤ 2π. The differentials are: dx = −asin(θ)dθ, dy = acos(θ)dθ.
2. Compute the work integral. W = $\int_{C} \vec{F}·d\vec{r} = \int_{C} Mdx + Ndy =\int_{C} xdx +ydy =[\text{Substituting the parameterization}] \int_{C} (acos(θ))(-asin(θ)dθ) + asin(θ)(acos(θ)dθ) =[\text{Simplify}] \int_{0}^{2π}a^2(-cos(θ)sin(θ)+sin(θ)cos(θ))dθ = \int_{0}^{2π} 0dθ = 0.$

• Work done by the vector field $\vec{F} = -y\vec{i} + x\vec{j}$ along a circular path

Now consider the same circle C of radius “a” centered at the origin, traversed counterclockwise. The vector field is $\vec{F} = -y\vec{i} + x\vec{j}$ (Figure C). We are to compute the work done by the force as we move along the circle C.

Geometric interpretation

Geometrically, we can see that the force (it is a field that is tangential to the circle) is parallel to the motion ↭ $\vec{F} || \hat{\mathbf{T}} ⇒ \vec{F}·\hat{\mathbf{T}} = |\vec{F}| ⇒ \int_{C} \vec{F}·\hat{\mathbf{T}}·ds = \int_{C} |\vec{F}|·ds = \int_{C} a·ds = a\int_{C}ds = a·length(C) = a·2πa = 2πa^2$. Hence, the work done by the force field is not zero.

Detailed Calculation.

1. Parameterize the Circle: The circle can be parameterized using: x = acos(θ), y = asin(θ), where 0 ≤ θ ≤ 2π. The differentials are: dx = −asin(θ)dθ, dy = acos(θ)dθ.
2. Compute the work integral. W = $\int_{C} \vec{F}·d\vec{r} = \int_{C} Mdx + Ndy =\int_{C} -ydx +xdy =[\text{Substituting the parameterization:}] \int_{C} -(asin(θ))(-asin(θ)dθ)+(acos(θ))(acos(θ))dθ =[\text{Simplifying:}] \int_{C} a^2(sin^2(θ)+cos^2(θ)) = \int_{0}^{2π} a^2dθ = a^2\int_{0}^{2π} dθ = a^2·2π$

• Work done by the vector field $\vec{F}= y\vec{i}+x\vec{j}$ along a path composed of three segments.

Consider the vector field $\vec{F}= y\vec{i}+x\vec{j}$ = ⟨y, x⟩ and we want to calculate the work done by the force as we move along a path C consisting of three segments C₁, C₂, and C₃ in the unit circle, C = C₁+ C₂ + C₃, covering the sector 0 ≤ θ ≤ ^π⁄₄ (Figure D)

The work done by the force $\vec{F}$ is given by: W = $\int_{C} \vec{F}·d\vec{r} = \int_{C_1} \vec{F}·d\vec{r} + \int_{C_2} \vec{F}·d\vec{r} + \int_{C_3} \vec{F}·d\vec{r} = \int_{C_1} (ydx+xdy) + \int_{C_2} (ydx+xdy) + \int_{C_3} (ydx+xdy)$

1. Segment C₁: Along the x-axis from (0, 0) to (1,0), y = 0, hence dy = 0. So, the integral simplifies to: $\int_{C_1} (ydx+xdy) = \int_{C_1} (0·dx+x·0) = \int_{C_1} 0 = 0$. Geometrically, the vector field is perpendicular to the motion along the x-axis, resulting in zero work.
2. Segment C₂: Along the unit circle from (1, 0) to $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$. $\int_{C_2} (ydx+xdy) =$[Notice that C₂ is a portion of the unit circle, so polar coordinates are a good idea, x = cos(θ), y = sin(θ), 0 ≤ θ ≤ ^π⁄₄, dx = -sin(θ)dθ, dy = cos(θ)dθ] =$\int_{C_2} (sin(θ)(-sin(θ)dθ))+ cos(θ)·(cos(θ)dθ) = \int_{0}^{\frac{π}{4}} (cos^2(θ)-sin^2(θ))dθ = \int_{0}^{\frac{π}{4}} cos(2θ)dθ = \frac{1}{2}sin(2θ)\bigg|_{0}^{\frac{π}{4}} = \frac{1}{2}sin(\frac{π}{2})-\frac{1}{2}sin(0) = \frac{1}{2}·1-0 = \frac{1}{2}$
3. Segment C₃: Along the line from $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$ to (0, 0). $\int_{C_3} (ydx+xdy) =$ [For C₃, parametrize the line. A first approach would be $x = \frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}t, y = \frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}t$, 0 ≤ t ≤ 1, but a simpler, better idea is x = t, y = t, 0 ≤ t ≤ $\frac{1}{\sqrt{2}}$ and we will get the work backwards ⇒ dx = dy = dt]. Therefore,

$\int_{-C_3} (ydx+xdy) = \int_{0}^{\frac{1}{\sqrt{2}}} tdt+tdt = \int_{0}^{\frac{1}{\sqrt{2}}} 2tdt = t^2\bigg|_{0}^{\frac{1}{\sqrt{2}}} = \frac{1}{2}$.

Since we are integrating backwards, we take the negative: $\int_{C_3} (ydx+xdy) = -\frac{1}{2}$.

Combining the results from the three segments (work done on each segment): W (Total Work Done) = $\int_{C_1} (ydx+xdy) + \int_{C_2} (ydx+xdy) + \int_{C_3} (ydx+xdy) = 0 + \frac{1}{2} -\frac{1}{2} = 0.$

• Work done by the vector field $\vec{F}$ = ⟨xy², 2xy⟩ where the path C is the smaller arc of x² + y² from (1, 0) to (0, -1).

The path C is part of the circle x² + y² = 1. The points (1, 0) and (0, −1) are on this circle. The smaller arc of this circle that goes from (1, 0) to (0, −1) corresponds to moving clockwise.

For a circle of radius 1, a general parametrization is: x(t) = cos(t), y(t) = sin(t), where t ranges over the interval corresponding to the points on the arc.

1. Parametrized the smaller arc. Since we are moving clockwise from (1, 0) to (0, −1), the correct parametrization that reflects the clockwise motion is: x = cos(t), y = -sin(t) where 0 ≤ t ≤ $\frac{π}{2},$ i.e., t ranges from t = 0 (corresponding to (1, 0)) to t = $\frac{π}{2}$ (corresponding to (0, -1)) (Refer to Figure ii for a visual representation and aid in understanding it). The differentials are dx = -sin(t)dt, dy = -cos(t)dt.

2. Set up the work integral. The work done by the vector field $\vec{F} = ⟨M, N⟩=⟨xy^2, 2xy⟩$ along the path C is given by W = $\int_{C} \vec{F}·d\vec{r} = \int_{C} Mdx + Ndy =\int_{C} xy^2dx +2xydy$.

Substituting the parameterization: $\int_{0}^{\frac{π}{2}} cos(t)sin^2(t)(-sin(t)dt)+2cos(t)(-sin(t))(-cos(t)dt)=[\text{Simplifying:}] \int_{0}^{\frac{π}{2}} (-sin^3(t)cos(t)+2cos^2(t)sin(t))dt = \int_{0}^{\frac{π}{2}} -sin^3(t)cos(t)dt + \int_{0}^{\frac{π}{2}} 2cos^2(t)sin(t)dt$

Let’s use substitution (first integral): u = sin(t), du = cos(t)dt & (second part) v = cos(t), dv = -sin(t)dt

$\int_{0}^{\frac{π}{2}} -u^3dt \int_{0}^{\frac{π}{2}} - 2v^2dv = -\frac{u^4}{4}-\frac{2v^3}{3}\bigg|_{0}^{\frac{π}{2}}$

$-\frac{1}{4}sin^4(t)-\frac{2}{3}cos^3(t)\bigg|_{0}^{\frac{π}{2}} = \frac{-1}{4}-(-\frac{2}{3}) = \frac{-1}{4}+\frac{2}{3} = \frac{-3}{12}+\frac{8}{12} = \frac{5}{12}$

• Calculate the work done by the field $\vec{F} = ⟨xsin(y), y⟩$ along the curve C defined as y = x² from (-1, 1) to (2, 4).

Parametrize the curve C. x = t, y = t² where t ranges from -1 to 2. When t = −1, x = −1 and y = 1, so the point is (−1, 1). When t = 2, x = 2 and y = 4, so the point is (2, 4). The differentials are dx = dt, dy = 2tdt.

Set up the work integral. The work done by the vector field $\vec{F} = ⟨M, N⟩=⟨xy^2, 2xy⟩$ along the path C is given by W = $\int_{C} \vec{F}·d\vec{r} = \int_{C} Mdx + Ndy =\int_{C} xsin(y)dx+ydy$.

Substituting the parameterization. $\int_{-1}^{2} (tsin(t^2))dt +t^2(2tdt) =[\text{Simplify}] \int_{-1}^{2} (tsin(t^2)+2t^3)dt$

First integral: $\int_{-1}^{2} tsin(t^2)dt =[u = t^2, du = 2tdt] \frac{1}{2}\int_{-1}^{2} sin(u)du = \frac{-1}{2}cos(u) = \frac{-1}{2}cos(t^2)\bigg|_{-1}^{2} = \frac{-1}{2}[cos(4)-cos(1)]$

Second integral: $\int_{-1}^{2} 2t^3dt = 2\frac{t^4}{4} = \frac{t^4}{2}\bigg|_{-1}^{2} = \frac{16}{2}-\frac{1}{2} = \frac{15}{2}$

W = $\int_{C} \vec{F}·d\vec{r} = \int_{C} Mdx + Ndy =\int_{C} xsin(y)dx+ydy = \frac{-1}{2}cos(4)+\frac{1}{2}cos(1)+\frac{15}{2}$

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