Problem 23 - Calculus: Early Transcendental Functions

Question

Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}.\) \(\mathbf{F}(x, y, z)=\frac{\langle x, y, z\rangle}{\sqrt{x^{2}+y^{2}+z^{2}}}, C\) runs from \((1,3,2)\) to \((2,1,5)\)

Step-by-Step Solution

Answer

Since the integral in this expression cannot be solved using standard analytical methods, you will need to use a numerical method to evaluate it. The general steps for solving this problem involve parametrizing the curve, substituting this into the field vector and then evaluating the line integral.

Step 1: Parametrize the Curve

The curve \(C\) runs from \(P(1,3,2)\) to \(Q(2,1,5)\). A common way to parametrize this type of curve is to take a straight line from \(P\) to \(Q\). This parametrization is given by \(\boldsymbol{r}(t)=\boldsymbol{P}+t(\boldsymbol{Q}-\boldsymbol{P})\) where \(0 \leq t \leq 1\). Plugging in the coordinates for \(P\) and \(Q\) gives us \(\boldsymbol{r}(t)=\langle 1, 3, 2 \rangle+t(\langle 2, 1, 5 \rangle-\langle 1, 3, 2 \rangle)=\langle 1, 3, 2 \rangle+t\langle 1, -2, 3 \rangle\), which simplifies to \(\boldsymbol{r}(t)=\langle 1+t, 3-2t, 2+3t \rangle\).

Step 2: Compute the Derivative of the Parametrization

The derivative \(d\mathbf{r}/dt\) is needed to calculate the line integral. We find that \(d\mathbf{r}/dt = \langle 1, -2, 3 \rangle\).

Step 3: Substituting the Parametrization into the Vector Field

We substitute the parametrization of the curve into the vector field \(\mathbf{F}\), we get: \(\mathbf{F}(r(t))=\dfrac{\langle 1+t, 3-2t, 2+3t \rangle}{\sqrt{(1+t)^{2}+(3-2t)^{2}+(2+3t)^{2}}}\).

Step 4: Compute the Dot Product of \(\mathbf{F}(r(t)) \) and \(d\mathbf{r}/dt \)

\(\mathbf{F}(r(t)) \cdot \frac{d\mathbf{r}}{dt}=\dfrac{\langle 1+t, 3-2t, 2+3t \rangle \cdot \langle 1, -2, 3 \rangle}{\sqrt{(1+t)^{2}+(3-2t)^{2}+(2+3t)^{2}}}\).

Step 5: Evaluate the Line Integral

The last step is the evaluation of the line integral: \(\int_{0}^{1}\mathbf{F}(r(t)) \cdot \frac{d\mathbf{r}}{dt} dt = \int_{0}^{1}\dfrac{\langle 1+t, 3-2t, 2+3t \rangle \cdot \langle 1, -2, 3 \rangle}{\sqrt{(1+t)^{2}+(3-2t)^{2}+(2+3t)^{2}}} dt\). You will likely need to use numerical methods to evaluate this integral.

Key Concepts

Vector Field

A vector field is a function that assigns a vector to every point in space. Imagine a field of arrows, where each arrow represents a vector, indicating both direction and magnitude at each point. This concept is fundamental in understanding forces, fluid dynamics, and more. When we have a vector field in three-dimensional space, it can be expressed as \(\mathbf{F}(x, y, z)\). In our exercise, the vector field given is \(\mathbf{F}(x, y, z) = \frac{\langle x, y, z \rangle}{\sqrt{x^{2} + y^{2} + z^{2}}}\).
The function divides each component of the position vector by the magnitude of that vector. This operation gives a unit vector, preserving direction but normalizing magnitude to 1. Analyzing vector fields helps us understand physical situations by studying how vectors vary over space. Often, these fields represent physical quantities like gravitational or electromagnetic forces.

Parametrization

Parametrization is the process of defining a curve by a parameter, usually denoted by \(t\). It simplifies complex geometries into manageable mathematical expressions. The exercise we are working on involves a straight line from point \(P(1,3,2)\) to point \(Q(2,1,5)\). To parametrize the line between these two points, we use the following process:

• Identify the starting and ending points, \(P\) and \(Q\).
• Define the vector from \(P\) to \(Q\), given by \(\langle Q - P \rangle = \langle 2 - 1, 1 - 3, 5 - 2 \rangle = \langle 1, -2, 3 \rangle\).
• The line can be described by \(\boldsymbol{r}(t) = \boldsymbol{P} + t(\boldsymbol{Q} - \boldsymbol{P})\), which translates to \(\langle 1+t, 3-2t, 2+3t \rangle\) with \(0 \leq t \leq 1\).

This creates a neat formula representing the position along the line as a function of \(t\). Such parametrization is crucial for defining paths along which we calculate integrals, ensuring precise control over the trajectory.

Dot Product

The dot product, also known as the scalar product, is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. In the realm of vector calculus, the dot product is crucial for finding how much one vector aligns with another. Here's how it works:

• Given two vectors, \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\) and \(\mathbf{b} = \langle b_1, b_2, b_3 \rangle\), their dot product is calculated as \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \).
• It results in a scalar value.
• Geometry-wise, it equals the product of their magnitudes and the cosine of the angle between them.

In our line integral problem, the dot product between \(\mathbf{F}(r(t))\) and \(\frac{d\mathbf{r}}{dt}\) represents the projection of the field along the curve. Evaluating this dot product tells us how strongly and in what manner the vector field aligns with the path. This is a vital part in computing the line integral, as it reduces the problem to integrating scalar values over the curve.