Quantity has a quality of its own, Joseph Stalin.
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.
We are going to illustrate the following vector fields:
Arrows point away from the origin, growing longer as they move further away, indicating that the strength of the field increases with distance.
These illustrations help visualize how vector fields vary across space and provide insights into the behavior of different types of fields.
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 t_{1} to t_{2} is the integral of the dot product of the force vector and the velocity vector over that time interval.
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:
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 = x^{2} 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 = t^{2}] = ⟨-t^{2}, t⟩. Besides, $\frac{dx}{dt}=1,\frac{dy}{dt}= 2t, \frac{d\vec{r}}{dt}= ⟨\frac{dx}{dt}, \frac{dy}{dt}⟩ = ⟨1,2t⟩$
W = $\int_{0}^{1} (\vec{F}·\frac{d\vec{r}}{dt}dt) =\int_{0}^{1} ⟨-t^2, t⟩ · ⟨1, 2t⟩ dt = \int_{0}^{1} (-t^2+2t^2)dt = \int_{0}^{1} t^2dt = \frac{t^3}{3}\bigg|_{0}^{1} = \frac{1}{3}$
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 = $\int_{C} \vec{F}·d\vec{r} = \int_{C} Mdx + Ndy$
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 = t^{2}, 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 = sin^{2}(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.
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 = $\int_{C} \vec{F}·d\vec{r} = \int_{C} \vec{F}·\hat{\mathbf{T}}·ds$
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.
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.
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.
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.
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 C_{1}, C_{2}, and C_{3} in the unit circle, C = C_{1}+ C_{2} + C_{3}, covering the sector 0 ≤ θ ≤ ^{π}⁄_{4} (Figure D)
The work done by the force $\vec{F}$ is given by: W = $\int_{C} \vec{F}·d\vec{r} = \int_{C_1} \vec{F}·d\vec{r} + \int_{C_2} \vec{F}·d\vec{r} + \int_{C_3} \vec{F}·d\vec{r} = \int_{C_1} (ydx+xdy) + \int_{C_2} (ydx+xdy) + \int_{C_3} (ydx+xdy)$
$\int_{-C_3} (ydx+xdy) = \int_{0}^{\frac{1}{\sqrt{2}}} tdt+tdt = \int_{0}^{\frac{1}{\sqrt{2}}} 2tdt = t^2\bigg|_{0}^{\frac{1}{\sqrt{2}}} = \frac{1}{2}$.
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) = $\int_{C_1} (ydx+xdy) + \int_{C_2} (ydx+xdy) + \int_{C_3} (ydx+xdy) = 0 + \frac{1}{2} -\frac{1}{2} = 0.$
The path C is part of the circle x^{2} + y^{2} = 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.
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.
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
$\int_{0}^{\frac{π}{2}} -u^3dt \int_{0}^{\frac{π}{2}} - 2v^2dv = -\frac{u^4}{4}-\frac{2v^3}{3}\bigg|_{0}^{\frac{π}{2}}$
$-\frac{1}{4}sin^4(t)-\frac{2}{3}cos^3(t)\bigg|_{0}^{\frac{π}{2}} = \frac{-1}{4}-(-\frac{2}{3}) = \frac{-1}{4}+\frac{2}{3} = \frac{-3}{12}+\frac{8}{12} = \frac{5}{12}$
Parametrize the curve C. x = t, y = t^{2} 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 = $\int_{C} \vec{F}·d\vec{r} = \int_{C} Mdx + Ndy =\int_{C} xsin(y)dx+ydy = \frac{-1}{2}cos(4)+\frac{1}{2}cos(1)+\frac{15}{2}$