Problem 36 - Advanced Engineering Mathematics

Question

In an inverse square force field \(\mathbf{F}=c \mathbf{r} /\|\mathbf{r}\|^{3}\), where \(c\) is a constant and \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k},^{*}\) find the work done in moving a particle along the line from \((1,1,1)\) to \((3,3,3)\).

Step-by-Step Solution

Answer

The work done is \(\frac{2c}{3\sqrt{3}}\)."

Step 1: Understand the Work Done Formula

The work done by a force over a path can be found using the line integral of the force field along the path. For a vector field \(\mathbf{F}\) and a path from point \(\mathbf{r}_1\) to \(\mathbf{r}_2\), the work done is \( W = \int_{\mathbf{r}_1}^{\mathbf{r}_2} \mathbf{F} \cdot d\mathbf{r} \).

Step 2: Set Up the Parametrization of the Path

We need to parameterize the path from \((1,1,1)\) to \((3,3,3)\). A simple way to parametrize a straight line is \( \mathbf{r}(t) = (1+2t, 1+2t, 1+2t) \) where \( t \) ranges from 0 to 1.

Step 3: Differentiate the Parametrization

Find the differential \( d\mathbf{r} = \frac{d\mathbf{r}}{dt} dt \). Differentiating the parametrization with respect to \(t\), we get \( \frac{d\mathbf{r}}{dt} = (2, 2, 2) \). Thus, \( d\mathbf{r} = (2, 2, 2) dt \).

Step 4: Write the Force Field in Terms of Parameter \(t\)

Substitute \( \mathbf{r}(t) \) into the inverse square field \( \mathbf{F} = c\frac{\mathbf{r}}{\|\mathbf{r}\|^3} \). We have \( \mathbf{F}(t) = c\frac{(1+2t, 1+2t, 1+2t)}{((1+2t)^2 + (1+2t)^2 + (1+2t)^2)^{3/2}} \).

Step 5: Simplify the Magnitude

The magnitude \( \|\mathbf{r}\| \) is found by simplifying \( ((1+2t)^2 + (1+2t)^2 + (1+2t)^2)^{1/2} = \sqrt{3(1+2t)^2} = \sqrt{3}(1+2t) \). Thus, \( \mathbf{F} = c \frac{(1+2t, 1+2t, 1+2t)}{3\sqrt{3}(1+2t)^3} \).

Step 6: Find the Dot Product for the Integral

Calculate the dot product \( \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} \). We have: \( c \frac{(1+2t, 1+2t, 1+2t)}{3\sqrt{3}(1+2t)^3} \cdot (2, 2, 2) = c \frac{3\times 2(1+2t)}{3\sqrt{3}(1+2t)^3} = \frac{2c}{\sqrt{3}(1+2t)^2} \).

Step 7: Evaluate the Integral

The work done \( W = \int_{0}^{1} \frac{2c}{\sqrt{3}(1+2t)^2} dt \). This integral evaluates to \(-\frac{2c}{\sqrt{3}} \left[ \frac{1}{1+2t} \right]_{0}^{1} \). Substituting the limits gives \(-\frac{2c}{\sqrt{3}} \left( \frac{1}{3} - 1 \right) = \frac{2c}{3\sqrt{3}} \).

Key Concepts

Inverse Square Force Field

In physics and vector calculus, an inverse square force field is a vector field where the force exerted by the field diminishes with the square of the distance from a source point. This means that as you move away from the source, the intensity of the force decreases sharply.

Mathematically, it's expressed as:
\[ \mathbf{F} = c \frac{\mathbf{r}}{\|\mathbf{r}\|^3} \]
where \(c\) is a constant, and \(\mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k}\) is the position vector. The \(\|\mathbf{r}\|^3\) in the denominator indicates that the force decreases with the cube of the distance, maintaining the inverse square relationship due to vector calculus's divergence properties.

• The gravitational force is a classic example of an inverse square law in action.
• Another instance is the electric force between two point charges.

Parametrization of Path

Parametrizing a path is a crucial step in computing line integrals, especially when dealing with vector fields. It involves breaking down a path into a simpler form using a parameter, typically \(t\). For a straight line from point \((1,1,1)\) to point \((3,3,3)\), we use the parametrization:

\[ \mathbf{r}(t) = (1+2t, 1+2t, 1+2t) \]
where \(t\) varies from 0 to 1, representing the start and end points of the path.

This method allows you to traverse the path smoothly, compute differentials easily, and ultimately evaluate integrals along that path. By differentiating \(\mathbf{r}(t)\) with respect to \(t\), we find the change in the path per unit parameter change:
\[ d\mathbf{r} = \frac{d\mathbf{r}}{dt} dt = (2, 2, 2) dt \]

• Provides a standardized method to simplify complex path evaluations.
• Ensures consistency in calculations within vector fields.

Line Integrals

Line integrals extend the concept of integration to vector fields beyond just single-variable functions. They compute the cumulative effect of a field, like force, along a curve or path.

For a vector field \(\mathbf{F}\) and a parametrized curve \(\mathbf{r}(t)\), the line integral is determined by:
\[ W = \int_{\mathbf{r}_1}^{\mathbf{r}_2} \mathbf{F} \cdot d\mathbf{r} \]
In the exercise above, we computed the line integral to find the work done along a path in an inverse square force field.

By finding the dot product \(\mathbf{F}(t) \cdot d\mathbf{r}\), the integral simplifies to calculating how much force is applied along each segment of the path.

• Measures the "weighted" effect of the field over distance.
• Essential for understanding dynamics in electromagnetism and fluid dynamics.

Vector Calculus

Vector calculus is the branch of mathematics focusing on vector fields and differential calculus concepts. It leverages the power of vectors to model and analyze physical systems like fluid flow, electromagnetic fields, and other forces in three-dimensional space.

Key operations in vector calculus include differentiation and integration of vector fields, as well as applications of divergence, curl, and gradient. In the context of the exercise, we dealt primarily with:

• Vector Differentiation: Calculating the change in vector fields along paths.
• Line Integrals: Summing the effects of fields over a path, revealing crucial insights into the nature of forces and work.

Understanding vector calculus is vital for effectively solving complex problems such as computing work done in diverse force fields. It bridges the gap between mathematical abstraction and real-world physical phenomena.