Problem 24 - Calculus: Early Transcendental Functions

Question

Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}.\) \(\mathbf{F}(x, y, z)=\frac{\langle x, y, z\rangle}{x^{2}+y^{2}+z^{2}}, C\) runs from \((2,0,0)\) to \((0,1,-1)\)

Step-by-Step Solution

Answer

The evaluated line integral \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) is -2.

Step 1: Parameterizing the Path

Firstly, the straight line path C from point A \((2,0,0)\) to point B \((0,1,-1)\) must be parameterized. This can be done in terms of a new variable, let's say \(t\), where \(t\) varies from 0 to 1. The parameterized points \(\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle\) become \(x(t) = 2(1-t)\), \(y(t) = t\), and \(z(t) = -t\). Also, compute \(d\mathbf{r} = \frac{d \mathbf{r}}{dt} dt = \langle -2, 1, -1 \rangle dt \).

Step 2: Performing the Dot Product

Substitute the parameterized points into the vector field \(\mathbf{F}(x, y, z)\) given as \(\frac{\langle x, y, z\rangle}{x^{2}+y^{2}+z^{2}}\) and also replace \(x, y, z\) in the integrand with \(x(t), y(t)\) and \(z(t)\) respectively. The result after substitution is \(\mathbf{F}(x(t), y(t), z(t)) \cdot d \mathbf{r} = \langle \frac{2(1-t)}{(2-2t)^{2} + t^{2} - t^{2}}, \frac{t}{(2-2t)^{2} + t^{2} - t^{2}}, - \frac{t}{(2-2t)^{2} + t^{2} - t^{2}} \rangle \cdot \langle -2, 1, -1 \rangle dt\). Then, perform the dot product, yielding \((-4+4t)dt\).

Step 3: Performing the Integral

Now, the integral has been converted to a form that can be easily solved. Integrate \((-4+4t)\) with respect to \(t\) from 0 to 1. The integral of this function is \(-4t + 2t^{2}\) when evaluated from 0 to 1, which simplifies to -2.

Key Concepts

Vector Fields

In the realm of mathematics and physics, vector fields are a fundamental concept that describes a spatial distribution of vectors. Imagine each point in space having a vector attached to it. This vector, which could represent any quantity that has direction and magnitude (like force or velocity), can vary from one point to another. Vector fields are widely used in physics to model magnetic and gravitational fields, among numerous other applications. Let's dive deeper into what defines a vector field and how it applies to the problem at hand.

In the given exercise, the vector field \( \mathbf{F}(x, y, z) = \frac{\langle x, y, z\rangle}{x^{2}+y^{2}+z^{2}} \) is dependent on the coordinates \((x, y, z)\), with each component of the vector \(\langle x, y, z \rangle\) divided by the square of their sums: \(x^2 + y^2 + z^2\). This kind of vector field plays a crucial role in understanding the problem as it determines how the vector behaves in space as we move along the curve, which in this case is a line from \((2,0,0)\) to \((0,1,-1)\).

This kind of vector function helps us in evaluating line integrals which are used to determine the work done by a force field in moving an object along a path. It greatly simplifies calculations as it provides us with a direct formula based on the position coordinates.

Parameterization of Curves

Parameterization is a valuable technique in calculus, especially when evaluating line integrals. It allows us to express a curve in terms of a single variable, usually denoted as \(t\). This variable moves through the curve from one point to another, providing a way to systematically evaluate functions along the path.

In the exercise step 1, we are tasked with parameterizing the straight line path \(C\) from point \((2,0,0)\) to point \((0,1,-1)\). This is achieved by expressing the line in terms of \(t\), where \(t\) ranges from 0 to 1.

Here’s how the parameterization is done:

• x-coordinate: \(x(t) = 2(1-t)\)
• y-coordinate: \(y(t) = t\)
• z-coordinate: \(z(t) = -t\)

Differentiation gives us \(d\mathbf{r} = \langle -2, 1, -1 \rangle dt.\)

This parameterization redefines our line integral in terms of \(t\), enabling us to systematically take into account every point along the path, which is crucial for evaluating the line integral.

Integral Evaluation

Evaluating an integral, particularly a line integral, involves summing up values along a curve. In line integrals, instead of a standard function, we work with vector fields and parameterized paths to integrate over. This involves steps that translate the vector field into an expression in terms of our parameterized path, allowing for efficient computation.

In step 2, after substituting the parameterization \(\mathbf{r}(t)\) into the vector field \(\mathbf{F}(x, y, z),\) we compute the dot product with the differential \(d\mathbf{r}.\) The outcome for this specific problem yields \((-4 + 4t)dt.\)

Finally, in step 3, the actual integral evaluation is performed:

• We integrate \((-4 + 4t)\) with respect to \(t\) over the interval from 0 to 1.
• This integration results in the function \(-4t + 2t^2.\)
• Evaluating this expression from 0 to 1, we find the solution to be \(-2.\)

This result represents the total work done moving along the vector field's path \(C.\) Line integral evaluation thus allows us to calculate cumulative values like work along a field efficiently and is crucial in physics for solving practical problems relating to energy and force fields.