Success is not final, failure is not fatal: It is the courage to continue that counts, Winston Churchill
We shall go on to the end. We shall fight in France, we shall fight on the seas and oceans, we shall fight with growing confidence and growing strength in the air, we shall defend our island, whatever the cost may be. We shall fight on the beaches, we shall fight on the landing grounds, we shall fight in the fields and in the streets, we shall fight in the hills; we shall never surrender, Winston Churchill
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.
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 $\int_{C} \vec{F}·d\vec{r} = 0$.
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.
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: $\int_{C} \vec{F}·d\vec{r} = f(P_1)-f(P_0).$
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.
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.
A vector field $\vec{F}$ is conservative if it can be expressed as the gradient of some scalar potential function f(x, y). In mathematical terms, this means:$\vec{F} = ∇f = (\frac{∂f}{∂x}, \frac{∂f}{∂y})$.
To find a potential function for a conservative vector field, we first need to ensure that the vector field is conservative. A vector field $\vec{F}$ is conservative 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.
For the vector field $\vec{F}= (4x^2+8xy)\vec{i}+(3y^2+4x^2)\vec{j}$, let’s check if F is conservative.
Since these conditions are met, $\vec{F}$ is indeed conservative, meaning there exists a potential function f(x,y) such that $\vec{F}$ = ∇f.
We will find the potential function f(x, y) by computing a line integral along a path or trajectory C from the origin (0, 0) to a point (x_{1}, y_{1}) (see Figure F.a in the image below). The potential function is defined as:
$f(x_1, y_1) = \int_{C} \vec{F}·d\vec{r} + f(0, 0)$
where f(0, 0) is an arbitrary constant.
To simplify the calculation, we break the path C into two segments:
Compute the line integral along C, $\vec{F}=⟨4x^2+8xy, 3y^2+4x^2⟩$, we have: $\int_{C} \vec{F}·d\vec{r} = \int_{C} (4x^2+8xy)dx + (3y^2+4x^2)dy$
This can be broken down as:
$\int_{C_1} (4x^2+8xy)dx + (3y^2+4x^2)dy + \int_{C_2} (4x^2+8xy)dx + (3y^2+4x^2)dy$
$\int_{C_1} (4x^2+8xy)dx + (3y^2+4x^2)dy$ = [x varies from 0 to x_{1}, y = 0 ⇒ dy = 0] $\int_{0}^{x_1} 4x^2dx = \frac{4}{3}x^3\bigg|_{0}^{x_1} \frac{4}{3} = x_1^3$
$\int_{C_2} (4x^2+8xy)dx + (3y^2+4x^2)dy$ = [y varies from 0 to y_{1}, x = x_{1} ⇒ dx = 0] $\int_{0}^{y_1} (3y^2+4x_1^2)dy = (y^3+4x_1^2y)\bigg|_{0}^{y_1} = y_1^3+4x_1^2y_1$
Combine the results: $f(x_1, y_1) = \int_{C} \vec{F}·d\vec{r} + f(0, 0) ⇒ f(x_1, y_1) = \int_{C} \vec{F}·d\vec{r} + c ↭ f(x_1, y_1) = \frac{4}{3}x_1^3 + y_1^3+4x_1^2y_1 + c ↭[\text{Therefore,}] f(x, y) = \int_{C} \vec{F}·d\vec{r} + c = \frac{4}{3}x^3 + y^3+4x^2 + c$ where c is a constant.
For the same vector field, we know that $\vec{F} = ∇f$ where F = ⟨M, N⟩ = $⟨\frac{∂f}{∂x}, \frac{∂f}{∂y}⟩ = ⟨f_x, f_y⟩$ where f_{x} = 4x^{2} +8xy (i), f_{y} = 3y^{2} +4x^{2} (ii).
Starting with the partial derivative f_{x} (i), we integrate with respect to x:
f(x, y) = $\int (4x^2+8xy)dx = \frac{4}{3}x^3+4x^2y+g(y)$. Here, g(y) is a function of y that acts as the “constant” of integration when integrating with respect to x.
To determine g(y), we differentiate f with respect to y and match it with the given expression for f_{y}:
Differentiate f with respect to y: f_{y} = 4x^{2} + g’(y).
Matching this with (ii) f_{y} = 3y^{2} +4x^{2}, we get: 4x^{2} + g’(y) = 3y^{2} +4x^{2} ⇒ g’(y) = 3y^{2}
Integrate g′(y) with respect to y: g(y) = $\int g’(y)dy = \int 3y^2dy = y^3 + c$ and c is a real constant.
Substituting g(y) =y^{3} +c into the expression for f(x, y), we get: f(x, y) = $\frac{4}{3}x^3+4x^2y+g(y) = \frac{4}{3}x^3+4x^2y+ y^3 +c$
Step 1. Check if $\vec{F}$ is a conservative vector field.
To find a potential function for a conservative vector field, we first need to ensure that the vector field is conservative. A vector field $\vec{F}$ is conservative 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.
Given: M = $ye^x+sin(y)$, N = $e^x+xcos(y)$
Since these conditions are met, $\vec{F}$ is indeed conservative, meaning there exists a potential function f(x,y) such that $\vec{F}$ = ∇f.
Step 2. Finding the potential function f using Antiderivatives
For the same vector field, we know that $\vec{F} = ∇f$ where F = ⟨M, N⟩ = $⟨\frac{∂f}{∂x}, \frac{∂f}{∂y}⟩ = ⟨f_x, f_y⟩$ where f_{x}(x, y) = $ye^x+sin(y)$ (i), f_{y}(x, y) = $e^x+xcos(y)$ (ii).
From (i), we integrate with respect to x:
f(x, y) = $\int (ye^x+sin(y))dx = ye^x+xsin(y)+g(y)$
where g(y) is a function of y (acting as the constant of integration with respect to x, but not necessarily for y), i.e., the integration constant can depend on y.
To determine g(y), we differentiate f with respect to y:
f_{y}(x, y) = $\frac{∂f}{∂y} = \frac{∂}{∂y}(ye^x+xsin(y)+g(y))$ = e^{x} + xcos(y) + g’(y). Matching this with (ii) f_{y}(x, y) = $e^x+xcos(y)$, we get: $e^x+xcos(y) = e^x+cos(y)+g’(y)$ ⇒ g’(y) =0
Integrating g′(y) with respect to y: g(y) = $\int 0dy = c$ and c is a real constant.
Therefore, the potential function f is: f(x, y) = ye^{x} +xsin(y) + c. This function represents the scalar potential associated with the conservative vector field $\vec{F}$.
Given: p = yz, q = xz, r = (xy+2z).
Step 0: Check if $\vec{F}$ is conservative. The vector field is in three dimensions, so we will follow a similar process as before but adapted for three variables.
Since these conditions are met, $\vec{F}$ is indeed conservative, meaning there exists a potential function f(x, y, z) such that $\vec{F}$ = ∇f.
Step 2: Find the Potential Function f(x, y, z)
We have f_{x}(x, y, z) = yz (i), f_{y}(x, y, z) = xz (ii), f_{z}(x, y, z) = xy + 2z (iii).
From (i) integrate f_{x} = yz with respect to x:
f(x, y, z) = $\int (yz)dx = yzx + g(y, z)$ where g(y, z) is a function of y and z because it acts as the “constant” of integration with respect to x, but not necessarily for y and z.
To determine g(y), we differentiate f with respect to y and z:
Differentiate with Respect to y: f_{y}(x, y, z) = $\frac{∂}{∂y}(yzx + g(y, z))$ = xz + g_{y}(y, z). Matching this with (ii), f_{y}(x, y, z) = xz, we get: xz = xz + g_{y}(y, z) ⇒ g_{y}(y, z) = 0. Integrating g_{y}(y, z) with respect to y, we get, g(y, z) = h(z), so g(y, z) is a function of z alone.
Differentiate with Respect to z: f_{z}(x, y, z) = $\frac{∂}{∂z}(yzx + h(z))$ = yx + h’(z). Matching this with (iii), f_{z}(x, y, z) = xy + 2z we get: yx + h’(z) = xy + 2z⇒ h’(z) = 2z. Integrating h(z) with respect to z, we get, h(z) = z^{2} + k
Therefore, the potential function f(x, y, z) = xyz + g(y, z) = xyz + h(z) = xyz + z^{2} + k.
Step 3. Evaluating the line integral $\int_{C} \vec{F}·d\vec{r}$
The line integral of a conservative vector field $\vec{F}$ over a curve C from point A to point B can be evaluated using the potential function f: $\int_{C} \vec{F}·d\vec{r} = f(B) - f(A)$
Here, C is the line segment from A = (1, 0, −2) to B = (4, 6, 3).
$\int_{C} \vec{F}·d\vec{r} = \int_{C} ∇f·d\vec{r} = f(4, 6, 3) -f(1, 0, -2) = 4·6·3+3^2+c-(1·0·(-2)+(-2)^2-c)=72+9-4 = 81 -4 = 77.$
Step 0: Check if $\vec{F}$ is conservative.
Since all the conditions are met, $\vec{F}$ is indeed a conservative vector field.
Step 1. Calculate the potential function
f_{x} = ycos(x)+y^{2}. Let’s integrate with respect to x, $f = \int f_xdx = \int ycos(x)+y^2dx = ysin(x) + y^2x + g(y)$. Here, g(y) is a function of y that acts as the “constant” of integration when integrating with respect to x.
f_{y} = sin(x) + 2xy -2y. Let’s derivative f with respect to y, $f_y = \frac{∂f}{∂y} = \frac{∂}{∂y}(ysin(x) + y^2x + g(y)) = sin(x) +2yx +g’(y) = sin(x) + 2xy -2y ⇒g’(y) = -2y ⇒[\text{Let’s integrate g with respect to y}] g = \int -2ydy = -y^2 + k$.
Combining these results together: $f = ysin(x) + y^2x + g(y)= ysin(x) + y^2x + -y^2 + k$
Step 2. Evaluating work or the line integral $\int_{C} \vec{F}·d\vec{r}$
The line integral of a conservative vector field $\vec{F}$ over a curve C from point A to point B can be evaluated using the potential function f: Work = $\int_{C} \vec{F}·d\vec{r} = f(\frac{π}{2}, -1) - f(-π, 2)$.
$f(\frac{π}{2}, -1) = (-1)sin(\frac{π}{2})+\frac{π}{2}(-1)^2-(-1)^2+k = -1 + \frac{π}{2}-1+ k = \frac{π}{2}-2+k.$
$f(-π, 2) = 2sin(-π) +2^2(-π)-4+k = 0-4π-4+C$
Work = $\int_{C} \vec{F}·d\vec{r} = f(\frac{π}{2}, -1) - f(-π, 2) = \frac{π}{2}-2+k +4π+4-k = \frac{9π}{2}+2$
Step 0: Check if $\vec{F}$ is conservative.
Since all the conditions are met, $\vec{F}$ is indeed a conservative vector field.
Step 1. Calculate the potential function
f_{x} = e^{2y}. Let’s integrate with respect to x, $f = \int f_xdx = \int e^{2y}dx = e^{2y}x + g(y)$. Here, g(y) is a function of y that acts as the “constant” of integration when integrating with respect to x.
Similarly f_{y} = 1 +2xe^{2y}. Let’s integrate with respect to y, $f = \int f_ydy = \int 1 +2xe^{2y}dy = y + xe^{2y} + h(x)$.
Combining these results together: g(y) = y, h(x) = 0, f(x) = $y + xe^{2y}$
Step 2. Evaluating work or the line integral $\int_{C} \vec{F}·d\vec{r}$
The line integral of a conservative vector field $\vec{F}$ over a curve C from point A to point B can be evaluated using the potential function f: Work = $\int_{C} \vec{F}·d\vec{r} =[\text{t=1→B =(e,2),t=0→A = (0,1)}] f(e, 2) - f(0, 1) = (2+e·e^4)-(0+1) = e^5+1$
Step 1: Express the Vector Field
The given integral can be interpreted as a line integral of a vector field. We define the vector field $\vec{F}(x, y)$ as $\vec{F}(x, y) = ⟨M, N⟩ = M\vec{i}+ N\vec{j} = 2xe^{-y}\vec{i}+ (2y-x^2e^{-y})\vec{j}$
Step 2: Ensure the vector field $\vec{F}$ is Defined Everywhere
We can confirm that $\vec{F}$ is defined everywhere in R^{2}. The components M(x, y) and N(x,y) involve polynomials and exponential functions, which are continuous (we are being very “economical” with the language, what we really mean is that the condition: continuous first partial derivatives is satisfied) and defined for all real values of x and y. Therefore, the vector field $\vec{F}$ is defined everywhere in ℝ^{2} and since ℝ^{2} which is a open and simply-connected domain, we proceed to check if $\vec{F}$ is conservative.
Step 3. Check if $\vec{F}$ is a conservative vector field.
To determine if the vector field $\vec{F} = ⟨M, N⟩$ is conservative, we check if it satisfies the condition N_{x} = M_{y}. This means that the mixed partial derivatives of the potential function must be equal.
Given: M = $2xe^{-y}$, N = $2y-x^2e^{-y}$
Calculate the partial derivatives:
$\frac{∂M}{∂y} = \frac{∂}{∂y}(2xe^{-y}) = -2xe^{-y}, \frac{∂N}{∂x} = \frac{∂}{∂x}(2y-x^2e^{-y}) = -2xe^{-y}$. M_{y} = N_{x} ⇒both conditions are satisfied, $\vec{F}$ is a conservative vector field, and therefore, the line integral over any path from (1, 0) to (2, 1) is path-independence.
Step 4. Calculate the potential function
For the same vector field, we know that $\vec{F} = ∇f$ where F = ⟨M, N⟩ = $⟨\frac{∂f}{∂x}, \frac{∂f}{∂y}⟩ = ⟨f_x, f_y⟩$
Given: M = f_{x}(x, y) = $2xe^{-y}$ (i), N = f_{y}(x, y) = $2y-x^2e^{-y}$ (ii)
M = f_{x}(x, y) = $2xe^{-y}$ ⇒[To find f, we integrate f_{x} with respect to x] f(x, y) = $\int f_x(x, y)dx = \int 2xe^{-y}dx = 2e^{-y}\int xdx = 2e^{-y}\frac{x^2}{2}+g(y) = x^2e^{-y} + g(y)$
Here, g(y) is a function of y acting as the constant of integration with respect to x.
To determine g(y), we differentiate $f(x, y) = x^2e^{-y} + g(y)$ with respect to y:
f_{y}(x, y) = $\frac{∂}{∂y}(x^2e^{-y} + g(y)) = -x^2e^{-y}+g’(y)$
Matching this with (ii) f_{y}(x, y) = $2y-x^2e^{-y}$, we get $2y-x^2e^{-y} = -x^2e^{-y}+g’(y)⇒ g’(y)=2y ⇒$[Let’s integrate with respect to y] $g(y)= y^2 +c$
Therefore, the potential function f is: $f(x, y) = x^2e^{-y} + g(y) = x^2e^{-y} + y^2 +c$
Step 4. Evaluating the line integral $\int_{C} \vec{F}·d\vec{r}$
The line integral of a conservative vector field $\vec{F}$ over a curve C from point A to point B can be evaluated using the potential function f: $\int_{C} \vec{F}·d\vec{r} = f(B) - f(A)$ Here, C is any path from (1, 0) to (2, 1).
$\int_{C} \vec{F}·d\vec{r} = \int_{C} ∇f·d\vec{r} = \int_{(1, 0)}^{(2, 1)} (x^2e^{-y} + y^2 +c)·d\vec{r} = f(2, 1)-f(1, 0)= (4e^{-1} + 1^2 +c)-(1^2e^{0} + 0^2 +c) = \frac{4}{e}+1+c-1-c = \frac{4}{e}.$