Question

If \(\mathbf{u}=\operatorname{grad} F\) in a domain \(D\), then show that a) \(\int_{\left(x_{1}, y_{1}, z_{1}\right)}^{\left(x_{2}, z_{2}\right)} u_{T} d s=F\left(x_{2}, y_{2}, z_{2}\right)-F\left(x_{1}, y_{1}, z_{1}\right)\), where the integral is along any path in \(D\) joining the two points; b) \(\int_{C} u_{T} d s=0\) on any closed path in \(D\).

Short answer

Question: Prove the following results based on the fact that vector field 𝐮 is the gradient of a scalar function 𝐹: a) the line integral along any path in the domain 𝐷 connecting the points (𝑥1,𝑦1,𝑧1) and (𝑥2,𝑦2,𝑧2) is equal to the difference of the values of 𝐹 at these points, b) the line integral along any closed path in the domain 𝐷 is zero. Answer: The first result can be proved by expressing the line integral using the gradient of 𝐹, parametrizing the path, and applying the fundamental theorem of calculus in multiple variables. This gives us the expression: $$ \int_C \mathbf{u}\cdot d\boldsymbol{\ell}=F(x_2, y_2, z_2)-F(x_1, y_1, z_1) $$ For the second result, we can show that the integral along a closed path is zero since the starting and ending points are equal, resulting in: $$ \int_C \mathbf{u}\cdot d\boldsymbol{\ell} = F(x_2, y_2, z_2)-F(x_1, y_1, z_1)=0 $$

Step-by-step solution

Definition of gradient and line integral

Given that \(\mathbf{u}=\operatorname{grad} F\), we have \(\mathbf{u} =\frac{\partial F}{\partial x}\hat{\textbf{i}}+\frac{\partial F}{\partial y}\hat{\textbf{j}}+\frac{\partial F}{\partial z}\hat{\textbf{k}}\). The line integral along any path \(C\) is defined as \(\int_C \mathbf{u}\cdot d\boldsymbol{\ell}=\int_C u_T ds\), where \(u_T\) is the tangential component of the vector field \(\mathbf{u}\) along the path.

Calculate the line integral

Let \(\boldsymbol{\ell}(t)=(x(t),y(t),z(t))\) be a parametrization of the path \(C\) connecting the points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\). The differential path vector can be expressed as \(d\boldsymbol{\ell}=dx\hat{\textbf i}+dy\hat{\textbf j}+dz\hat{\textbf k} =\frac{d\mathbf{\ell}}{dt}dt\), where \(dx=\frac{dx}{dt}dt\), \(dy=\frac{dy}{dt}dt\), \(dz=\frac{dz}{dt}dt\). Now we can compute the line integral: $$ \int_C \mathbf{u}\cdot d\boldsymbol{\ell}=\int_C \left(\frac{\partial F}{\partial x}\hat{\textbf{i}}+\frac{\partial F}{\partial y}\hat{\textbf{j}}+\frac{\partial F}{\partial z}\hat{\textbf{k}}\right) \cdot \left( \frac{d\mathbf{\ell}}{dt} dt\right) = \int_a^b \left(\frac{\partial F}{\partial x}\frac{dx}{dt} + \frac{\partial F}{\partial y}\frac{dy}{dt}+\frac{\partial F}{\partial z}\frac{dz}{dt} \right) dt $$ where \(a\) and \(b\) are the limits of the parameter \(t\) corresponding to the points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) respectively.

Fundamental theorem of calculus in multiple variables

The integral above can be written as: $$ \int_C \mathbf{u}\cdot d\boldsymbol{\ell}=\int_C \left( \frac{dF}{dt}\right) dt $$ According to the fundamental theorem of calculus in multiple variables, the integral of a gradient along any path in the domain is equal to the difference of the scalar function at the endpoints of the path: $$ \int_C \mathbf{u}\cdot d\boldsymbol{\ell}=F(x_2, y_2, z_2)-F(x_1, y_1, z_1) $$ This proves the first result.

Closed path integral

In the case of a closed path, the starting and ending points are the same. This means that \((x_1, y_1, z_1)=(x_2, y_2, z_2)\). Using the result we just proved above, the line integral along a closed path becomes: $$ \int_C \mathbf{u}\cdot d\boldsymbol{\ell} = F(x_2, y_2, z_2)-F(x_1, y_1, z_1)=0 $$ This proves the second result.

Key concepts

Gradient Vector Field

A gradient vector field is the collection of vectors pointing in the direction of the steepest ascent for a scalar function. Mathematically, if we have a scalar function F(x, y, z), the gradient of F, denoted as grad F or ∇F, forms a vector field. This vector field is given by

\[\begin{equation}\mathbf{u} = abla F = \frac{\partial F}{\partial x}\hat{\textbf{i}} + \frac{\partial F}{\partial y}\hat{\textbf{j}} + \frac{\partial F}{\partial z}\hat{\textbf{k}}\end{equation}\]where \(\hat{\textbf{i}}\), \(\hat{\textbf{j}}\), and \(\hat{\textbf{k}}\) are the unit vectors in the direction of the x, y, and z coordinates, respectively. It's important to highlight that the gradient vector field is conservative, meaning a line integral's result solely depends on the initial and final points and not on the path taken.

Fundamental Theorem of Calculus for Line Integrals

The Fundamental Theorem of Calculus for Line Integrals establishes a profound connection between a gradient field and the scalar function from which it is derived. If \(\mathbf{u} = abla F\) is a gradient of F, then the line integral of \(\mathbf{u}\) along a curve C from a point A to a point B is given by F(B) - F(A). This theorem mirrors the fundamental theorem of calculus for single-variable functions and allows us to evaluate line integrals easily.

It simplifies the computation of work done by a force field or the change in energy within a field, often represented as a line integral, to a calculation of the difference in potential at two points.

Closed Path Integral

A closed path integral is a line integral where the start and end points are identical, forming a loop. In the context of gradient vector fields, this integral is especially significant because, for any conservative field (like a gradient field), the closed path integral is always zero. This is a result of the path-independence of conservative fields. In practical terms, it means that the net work done by a conservative force around any closed loop is zero, highlighting a type of 'energy conservation' within the field. The mathematics behind it utilizes the fact that returning to the initial position leaves no net change in the scalar function—that is the potential function of the force field.

This property has profound implications in physics, particularly in the study of conservative forces such as gravity and electrostatic forces.

Vector Calculus

The field of vector calculus extends the rules and techniques of calculus to vector fields and functions of several variables. It's a powerful toolset that applies to many physical phenomena, from electromagnetism to fluid dynamics. In vector calculus, operations such as gradient, divergence, and curl, alongside line, surface, and volume integrals, allow for the analysis of vector fields and the flow of substances or forces within these fields.

Understanding line integrals, as deployed for assessing the work done by a force field, is pivotal in vector calculus. They help quantify how a vector field interacts with a curve or path in space. This area of mathematics is essential for students who are pursuing engineering, physics, or any applied sciences, as it provides the language and tools needed to mathematically model and solve real-world problems.