No pressure, no diamonds, Thomas Carlyle.
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 P_{0} and P_{1} are the initial and final points of the curve C, respectively.
A line integral in space allows us to integrate a function along a curve. This is particularly useful in physics and engineering, where we often need to calculate quantities such as:
Consider a vector field $\vec{F}$ in three dimensional space. A vector field can be expressed as: $\vec{F} = P\hat{\mathbf{i}} + Q\hat{\mathbf{j}}+R\hat{\mathbf{k}}$ where P, Q, and R are functions of the coordinates x, y, and z respectively. Each of these functions represents the components of the vector field in the $\hat{\mathbf{i}}$ (x-axis), $\hat{\mathbf{j}}$ (y-axis), and $\hat{\mathbf{k}}$ (z-axis) directions, respectively.
The vector field $\vec{F}$ can represent various physical phenomena where a vector quantity is defined at every point in space, such as:
To perform a line integral, we first need to describe the curve C along which we are integrating. A curve in three dimensions can be parametrized by a parameter t, which varies over an interval [a, b]. The curve is represented as: $\vec{r}(t)=⟨x(t), y(t), z(t)⟩$ where a ≤ t ≤ b. The functions x(t), y(t), and z(t) are the coordinates that describe the positions of points on the curve C as the parameter t varies from a to b.
The line integral of the vector field $\vec{F}$ along the curve C is defined as: $\int_{C} \vec{F}d\vec{r}$ where $d\vec{r}$ is an infinitesimal vector tangent to the curve at each point. The vector $d\vec{r}$ can be expressed as: $d\vec{r} = ⟨dx, dy, dz⟩$.
In the context of physics, when the vector field represents a force, the line integral of the vector field along a curve C gives the work done by the force as it moves a particle along the curve.
To compute the line integral, we express $d\vec{r}$ in terms of the parameter t. Given the parametrization $\vec{r}(t)=⟨x(t), y(t), z(t)⟩$, the differential vector $d\vec{r}$ can be written as: $d\vec{r} = \frac{d\vec{r}}{dt}dt = ⟨\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}⟩dt$.
Now, substituting $d\vec{r}$ into the line integral expression:
$\int_{C} \vec{F}d\vec{r} = \int_{a}^{b} \vec{F}(\vec{r}(t))\frac{d\vec{r}}{dt}dt$
Given the vector field $\vec{F} = P\hat{\mathbf{i}} + Q\hat{\mathbf{j}}+R\hat{\mathbf{k}}$, we can express it as: $\vec{F}(\vec{r}(t)) = ⟨P(x(t), y(t), z(t)), Q(x(t), y(t), z(t)), R(x(t), y(t), z(t))⟩$
Thus, the dot product $\vec{F}d\vec{r}$ becomes, $\vec{F}d\vec{r} = ⟨P, Q, R⟩· \frac{d\vec{r}}{dt}dt = ⟨P, Q, R⟩·⟨\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}⟩dt = P\frac{dx}{dt}dt + Q\frac{dy}{dt}dt + R\frac{dz}{dt}dt$
Substituting this into the line integral gives: $\int_{C} \vec{F}d\vec{r} = \int_{a}^{b} (P\frac{dx}{dt} + Q\frac{dy}{dt} + R\frac{dz}{dt})dt = \int_{a}^{b} (P(x(t), y(t), z(t))\frac{dx}{dt} + Q(x(t), y(t), z(t))\frac{dy}{dt} + R(x(t), y(t), z(t))\frac{dz}{dt})dt$
Conclusion: Given a vector field $\vec{F} = ⟨P(x, y, z), Q(x, y, z), R(x, y, z)⟩$ and a curve C parametrized by $\vec{r}(t)=⟨x(t), y(t), z(t)⟩$, the line integral is defined as: $\int_{C} \vec{F}d\vec{r} = \int_{a}^{b} (P\frac{dx}{dt} + Q\frac{dy}{dt} + R\frac{dz}{dt})dt = \int_{a}^{b} (P(x(t), y(t), z(t))\frac{dx}{dt} + Q(x(t), y(t), z(t))\frac{dy}{dt} + R(x(t), y(t), z(t))\frac{dz}{dt})dt$
Let’s work through exercises to illustrate these concepts.
Work done by a force field along a path is calculated using the line integral of the vector field along the curve. If $\vec{F} =⟨P(x,y),Q(x,y)⟩$ is the vector field and C is the curve, the work done is given by the line integral: $\int_{C} \vec{F}·d\vec{r}$ where $d\vec{r}$ is the differential displacement vector along the curve C. The vector field $\vec{F}·d\vec{r}$ represents the dot product of the force field with the tangent to the curve.
Understanding the Curve. (-1, 1) to (2, 4) is not a line segment, they lie on the parabola y = x^{2}.
To find the work done along this curve, we can parametrize the curve using a parameter t. A natural choice for this parameterization is: x = t, y = x^{2}=t^{2}.
Parameter Range. Since x = t ranges from -1 to 2, the parameter t also varies from -1 to 2, -1 ≤ t ≤ 2.
Compute the Differentials, dx = $\frac{d}{dt}(t)dt$ = dt, dy = $\frac{d}{dt}(t^2)dt = 2tdt$
Substitute Parametrization into the vector field $\vec{F}$: P = xsin(y) = tsin(t^{2}), Q = y = t^{2}.
Evaluate the Line Integral
$\int_{C} \vec{F}d\vec{r} = \int_{C} [P\frac{dx}{dt}+Q\frac{dy}{dt}+R\frac{dz}{dt}]dt$ =[Substitute P, Q, R, and their differentials:] $\int_{-1}^{2} [tsin(t^2)·1+t^2·2t]dt = \int_{-1}^{2} [tsin(t^2)+2t^3]dt$
Integrate each term separately
$\int_{-1}^{2} tsin(t^2)dt=$[To solve this integral, we use a substitution method. Let u = t^{2}. Then, du = 2tdt ↭ ^{1}⁄_{2}du = tdt] $\frac{1}{2}\int_{-1}^{2} sin(u)du = -\frac{1}{2}cos(u)+C =$[Substitute back u = t^{2}] $\frac{-1}{2}cos(t^2)+C$
$\int_{-1}^{2} 2t^3dt = 2\frac{t^4}{4} = \frac{t^4}{2}$
Combine the results and evaluate the definite integral:
$\frac{-1}{2}cos(t^2) +\frac{t^4}{2}\bigg|_{-1}^{2} =$
$[\frac{-1}{2}cos(4) +\frac{16}{2}]-[\frac{-1}{2}cos(1) +\frac{1}{2}] = \frac{-1}{2}cos(4) + 8 + \frac{1}{2}cos(1)-\frac{1}{2} = \frac{-1}{2}cos(4) + \frac{1}{2}cos(1) +\frac{15}{2}.$
Conclusion: The work done by the vector field $\vec{F} = ⟨xsin(y), y⟩$ along the curve y = x^{2} from (-1, 1) to (2, 4) is $\int_{C} \vec{F}d\vec{r} =\frac{-1}{2}cos(4) + \frac{1}{2}cos(1) +\frac{15}{2}.$
Given a vector field $\vec{F} = ⟨P(x, y, z), Q(x, y, z), R(x, y, z)⟩$ and a curve C parametrized by $\vec{r}(t)=⟨x(t), y(t), z(t)⟩$, the line integral is defined as: $\int_{C} \vec{F}d\vec{r} = \int_{a}^{b} (P\frac{dx}{dt} + Q\frac{dy}{dt} + R\frac{dz}{dt})dt = \int_{a}^{b} (P(x(t), y(t), z(t))\frac{dx}{dt} + Q(x(t), y(t), z(t))\frac{dy}{dt} + R(x(t), y(t), z(t))\frac{dz}{dt})dt$
Compute the Differentials, dx = $\frac{d}{dt}(cos(t))dt$ = -sin(t)dt, dy = cos(t)dt, dz = 2tdt
Substitute Parametrization into the vector field $\vec{F}$: P = y = sin(t), Q = x = cos(t), R = z = t^{2}.
Evaluate the Line Integral
$\int_{C} \vec{F}d\vec{r} = \int_{0}^{2π} [P\frac{dx}{dt}+Q\frac{dy}{dt}+R\frac{dz}{dt}]dt$ =[Substitute P, Q, R, and their differentials:] $\int_{0}^{2π}[sin(t)(-sin(t))+cos(t)cos(t)+t^2(2t)]dt =[\text{Simplifying inside the integral:}] \int_{0}^{2π} [-sin^2(t)dt+cos^2(t)+2t^3]dt$
Integrate Each Term Separately.
We can use the identities, $sin^2(t)=\frac{1-cos(2t)}{2}, cos^2(t)=\frac{1+cos(2t)}{2}$
$\int_{0}^{2π} -sin^2(t)dt = -\frac{1}{2}\int_{0}^{2π}(1-cos(2t))dt = -\frac{1}{2}(t-\frac{1}{2}sin(2t))\bigg|_{0}^{2π} = -\frac{1}{2}(2π-0) = -π$
$\int_{0}^{2π} cos^2(t)dt = \frac{1}{2}\int_{0}^{2π}(1+cos(2t))dt = \frac{1}{2}(t+\frac{1}{2}sin(2t))\bigg|_{0}^{2π} = \frac{1}{2}(2π-0) = π$
$\int_{0}^{2π} 2t^3dt = \frac{1}{2}t^4\bigg|_{0}^{2π} = \frac{(2π)^4}{2}-\frac{0^4}{2} = 8π^4$
Combining everything:
$-π + π + 8π^4 = 8π^4$
Conclusion: $\int_{C} \vec{F}d\vec{r} = 8π^4$
Given a vector field $\vec{F} = ⟨P(x, y, z), Q(x, y, z), R(x, y, z)⟩$ and a curve C parametrized by $\vec{r}(t)=⟨x(t), y(t), z(t)⟩$, the line integral is defined as: $\int_{C} \vec{F}d\vec{r} = \int_{a}^{b} (P\frac{dx}{dt} + Q\frac{dy}{dt} + R\frac{dz}{dt})dt = \int_{a}^{b} (P(x(t), y(t), z(t))\frac{dx}{dt} + Q(x(t), y(t), z(t))\frac{dy}{dt} + R(x(t), y(t), z(t))\frac{dz}{dt})dt$
Compute the Differentials, dx = $\frac{d}{dt}(t^3)dt$ = 3t^{2}dt, dy = 2tdt, dz = dt
Substitute Parametrization into the vector field $\vec{F}$: P = yz = t^{2}·t = t^{3}, Q = zx = t^{2}·t = t^{4}, R = xy = t^{3}t^{2} = t^{5}
Evaluate the Line Integral
$\int_{C} \vec{F}d\vec{r} = \int_{0}^{1} [P\frac{dx}{dt}+Q\frac{dy}{dt}+R\frac{dz}{dt}]dt $ =[Substitute P, Q, R, and their differentials:]$ \int_{0}^{1} [t^3·3t^2+t^4·2t+ t^5·1] dt =[\text{Simplifying inside the integral:}] \int_{0}^{1} 6t^5dt = t^6\bigg|_{0}^{1} = 1.$
Conclusion. The work done by the vector field $\vec{F} = ⟨yz, xz, xy⟩$, along the curve C parametrized by x = t^{3}, y = t^{2}, z = t, 0 ≤ t ≤ 1 is: $\int_{C} \vec{F}d\vec{r} = 1$.
The goal is to evaluate the line integral $\int_{C’} \vec{F}d\vec{r}$ along the piecewise curve.
The line integral of the vector field $\vec{F}$ along the curve C’ is:
$\int_{C’} \vec{F}d\vec{r} = \int_{C’} [P\frac{dx}{dt}+Q\frac{dy}{dt}+R\frac{dz}{dt}]dt = \int_{C’} [yzdx +xzdy + xydz]dt$
Since C′ is a piecewise curve consisting of three segments, the line integral can be broken down into three separate integrals, one for each segment:
$\int_{C’} \vec{F}d\vec{r} = \int_{C_1} [P\frac{dx}{dt}+Q\frac{dy}{dt}+R\frac{dz}{dt}]dt + \int_{C_2} [P\frac{dx}{dt}+Q\frac{dy}{dt}+R\frac{dz}{dt}]dt + \int_{C_3} [P\frac{dx}{dt}+Q\frac{dy}{dt}+R\frac{dz}{dt}]dt$
C_{1}: The curve C_{1} can be parametrized as $\vec{r}(t)=⟨t, 0, 0⟩$ where t ranges from 0 to 1. Compute differentials: dx = 1·dt, dy = 0, dz = 0. Integrand components: P = yz = 0, Q = xz = 0, R = xy = 0. Integral: $\int_{C_1} [P\frac{dx}{dt}+Q\frac{dy}{dt}+R\frac{dz}{dt}]dt = 0$
C_{2}: The curve C_{2} can be parametrized as $\vec{r}(t)=⟨1, t, 0⟩$ where t ranges from 0 to 1. Compute differentials: dx = 0, dy = 1·dt, dz = 0. Integrand components: P = yz = 0, Q = xz = 0, R = xy = t. Integral: $\int_{C_2} [P\frac{dx}{dt}+Q\frac{dy}{dt}+R\frac{dz}{dt}]dt = 0$
C_{3}: The curve C_{3} can be parametrized as $\vec{r}(t)=⟨1, 1, t⟩$ where t ranges from 0 to 1. Compute differentials: dx = 0, dy = 0, dz = 1·dt. Integrand components: P = yz = t, Q = xz = t, R = xy = 1. Integral: $\int_{C_3} [P\frac{dx}{dt}+Q\frac{dy}{dt}+R\frac{dz}{dt}]dt = \int_{0}^{1} (t⋅0+t⋅0+1·1)dt = \int_{0}^{1} 1·dt = \int_{0}^{1} dt = 1.$
Adding up the contributions from each segment:
$\int_{C’} \vec{F}d\vec{r} = 0 + 0 + 1 = 1$
In fact, we get the same result because $\vec{F}$ is conservative because it can be expressed as the gradient of a scalar potential function: $\vec{F} = ∇f = ∇(x·y·z) = ⟨yz, xz, xy⟩$ where f(x, y, z) = xyz.
For conservative vector fields, the line integral between two points depends only on the endpoints and not on the path taken. This means that for any two curves C and C′ connecting the same points, the line integrals will be the same: $\int_{C} \vec{F}d\vec{r} = \int_{C’} \vec{F}d\vec{r}$
In this case, the points are P_{0}(0, 0, 0) and P_{1}(1, 1, 1). Using the Fundamental Theorem of Calculus for line integrals: $\int_{C} \vec{F}d\vec{r} = \int_{C} ∇f d\vec{r} = f(P_1)-f(P_0) = f(1, 1, 1) - f(0, 0, 0) = 1 - 0 = 1$.
In calculus, a vector field $\vec{F} = ⟨P, Q, R⟩$ is considered conservative if it can be expressed as the gradient of some scalar potential function f(x, y, z). In other words, there exist a scalar function f such that: $\vec{F} = ∇f = ⟨\frac{∂f}{∂x}, \frac{∂f}{∂y}, \frac{∂f}{∂z}⟩$.
If such a function f exists, it means that the vector field $\vec{F}$ is a gradient field and has a few important properties, such as the fact that the line integral of $\vec{F}$ between two points depends only on the endpoints, not on the path taken.
To determine if a vector field $\vec{F}= ⟨P, Q, R⟩$ is conservative, you can check the following criteria:
These conditions arise from the fact that if $\vec{F}= ∇f$, then the mixed partial derivatives of f should be equal (by Clairaut’s theorem on the equality of mixed partials).
When given a vector field $\vec{F} = ⟨P, Q, R⟩$ follow these steps to test if it is conservative:
The vector field $\vec{F} = ⟨P, Q, R⟩$ is conservative if there exist a scalar function f such that: $\vec{F} = Δf = ⟨\frac{∂f}{∂x},\frac{∂f}{∂y}, \frac{∂f}{∂z}⟩$. In other words, the vector field is conservative if it can be written as the gradient of some scalar function f.
Integrate P = yz = $\frac{∂f}{∂x}$ with respect to x. f(x, y, z) = $ = \int yzdx = xyz +g(y, z)$ where g(y, z) is a function of y and z because any terms that do not depend on x could be part of this function.
Differentiate f(x, y, z) = xyz + g(y, z) with respect to y and compare it to Q(x, y, z) = xz: $\frac{∂f}{∂y} = xz + \frac{∂g}{∂y} = Q = xz ⇒ \frac{∂g}{∂y} = 0$⇒[Therefore, g(y, z) does not depend on y, so g(y, z) = h(z), where h(z) is a function of z alone.] g(y, z) = h(z).
Differentiate f = xyz + h(z) with respect to z and compare it to R(x, y, z) = xy:
$\frac{∂f}{∂z} = xy + h’(z) = R = zy ⇒ h’(z) = 0 ⇒ h(z) = C ⇒f = xyz + C.$
So, the potential function is: f(x, y, z) = xyz + C. Therefore, $\vec{F} = ⟨yz,xz,xy⟩$ is indeed conservative, and its potential function is f(x, y, z) = xyz (ignoring the constant C since it does not affect the gradient).
An exact differential form means that it can be written as the differential of some scalar potential function f(x, y, z), i.e., $df = \frac{∂f}{∂x}dx + \frac{∂f}{∂y}dy + \frac{∂f}{∂z}dz = Pdx + Qdy + Rdz$. This implies the vector field $\vec{F}$ is conservative.
$\vec{F} = ⟨P, Q, R⟩$ where P = axy, Q = x^{2} +z^{3}, R = byz^{2}-4z^{3}.
To check if the differential form is exact, we need to verify the equality of mixed partial derivatives:
P_{y} = ax = 2x = Q_{x} ⇒ a = 2. P_{z} = 0 = 0 = R_{x}. Finally, Q_{z} = 3z^{2} = bz^{2} = R_{y} ⇒ b = 3 ⇒Thus, the values of a and b such that the differential form is exact are: a = 2,b = 3. The exact differential form becomes 2xydx + (x^{2}+z^{3})dy +(3yz^{2} -4z^{3})dz
To find the potential function f(x, y, z), we have two approaches:
First approach. Using line integrals. We integrate $\vec{F}$ along a well-chosen curve C, $f(x_1, y_1, z_1) = \int_{C} \vec{F}d\vec{r} + constant$ where C would typically be (for easy calculations’ sake) a curve from the origin (0, 0, 0) to (x, y, z) similar to Figure B.
C_{1}: From (0, 0, 0) to (x, 0, 0) along the x-axis. On C_{1}, y = 0 and z = 0, then dy = dz = 0. $\int_{C_1} \vec{F}d\vec{r} = \int_{0}^{x} Pdx = \int_{0}^{x} 2xydx = \int_{0}^{x} 0dx = 0.$
C_{2}: From (x, 0, 0) to (x, y, 0) along the y-axis. On C_{2}, x is constant and z = 0, then dx = dz = 0. $\int_{C_2} \vec{F}d\vec{r} = \int_{0}^{y} Qdy = \int_{0}^{y} (x^2+z^3)dy = \int_{0}^{y} x^2dy = x^2y\bigg|_{0}^{y} = x^2y$
C_{3}: From (x, y, 0) to (x, y, z) along the z-axis. On C_{3}, x and y are constant, then dx = dy = 0. $\int_{C_3} \vec{F}d\vec{r} = \int_{0}^{z} Rdz = \int_{0}^{z} (3yz^2-4z^3)dz = yz^3-z^4\bigg|_{0}^{z} = yz^3-z^4$
Combining all parts, the potential function is: f(x, y, z) = $0 + x^2y + yz^3-z^4 = x^2y + yz^3-z^4.$
Approach B. Using the gradient. We find a function f(x, y, z) such that $\vec{F} = ∇f$.
$\int_{C} \vec{F}d\vec{r} = \int_{C} ∇f d\vec{r} = \int_{C} ⟨\frac{∂f}{∂x},\frac{∂f}{∂y},\frac{∂f}{∂z}⟩ d\vec{r} = \int_{C} f_xdx +f_ydy + f_zdz$ where $\vec{F}$ is the vector field, f is the scalar potential function, ∇f is the gradient of f, $d\vec{r}$ is the differential displacement vector along the curve C, and f_{x}, f_{y} and f_{z} are the partial derivatives of f with respect to x, y, and z respectively.
The second approach involves integrating the partial derivatives of f:
f_{x} = P = 2xy ⇒[Integrate f_{x} with respect to x:] $f(x, y, z) = \int (2xy)dx$ = x^{2}y + g(y, z) (*) where g(y, z) is a function of y and z that arises because it is independent of x.
f_{y} = Q = x^{2} + z^{3}. Using the previous result (*) f = x^{2}y + g(y, z). $f_y = \frac{∂f}{∂y} = $ x^{2} +g_{y}⇒ x^{2} + z^{3} = x^{2} +g_{y} ⇒ g_{y} = z^{3} ⇒[By integrating with respect to y, $\int g_ydy = \int z^3dy$] g = yz^{3} + h(z)
f_{z} = R = 3yz^{2} -4z^{3} =[Considering f = x^{2}y +g = x^{2}y +yz^{3} + h(z), and differentiating this result with respect to z] 3yz^{2}+h’(z) ⇒ h’(z) = -4z^{3}⇒[Integrate with respect to z, $\int h’(z)dz = \int -4z^3dz$] h = -z^{4} + c.
Combining all parts, the potential function is: f = x^{2}y + g(y, z) = x^{2}y + yz^{3} + h(z) = x^{2}y + yz^{3} -z^{4} + c. Both approaches yield the same potential function for the vector field $\vec{F} = ⟨P, Q, R⟩ = ⟨2xy, x^2+z^3, 3yz^2-4z^3⟩$
$\vec{F} = ∇f = ⟨\frac{∂f}{∂x},\frac{∂f}{∂y},\frac{∂f}{∂z}⟩$
$\frac{∂f}{∂x} = P = ycos(x)+y$⇒[Integrate with respect to x:] $\int \frac{∂f}{∂x}dx = \int (ycos(x)+y)dx ⇒ f = ysin(x) + yx + g(y, z)$ where g(y, z) is an arbitrary function of y and z (since the derivative with respect to x was taken, this part could still depend on y and z).
$\frac{∂f}{∂y} = sin(x)+x+\frac{∂g(y, z)}{∂y} = Q = sin(x)+x ⇒ \frac{∂g(y, z)}{∂y} = 0$ ⇒ g is a constant with respect to y, g(y, z) is a function of z alone, g = g(z), hence f = ysin(x) + yx + g(z)
$\frac{∂f}{∂z} = 0 + 0 + g’(z) = R = 1 ⇒ g’(z) = 1 ⇒[\text{Integrate with respect to z}] g(z) = z + c⇒ f = ysin(x) + yx + z + c$
Evaluate the line integral. $\int_{C} \vec{F}d\vec{r} =$[Fundamental Theorem of Calculus for line integrals] f(π, 1, 3) -f(0, 2, 1).
f(π, 1, 3) = 1⋅sin(π) +1⋅π+ 3 + c = π + 3 + c. f(0, 2, 1) = 2⋅sin(0) + 2⋅0 +1 +c = 1 + c.
$\int_{C} \vec{F}d\vec{r} =$ f(π, 1, 3) -f(0, 2, 1) = $\int_{C} \vec{F}d\vec{r} =$ f(π, 1, 3) -f(0, 2, 1) = (π +3 +c) − (1 + c) = π + 2.