Problem 58 - Thomas Calculus

Question

In Exercises \(55-58, \mathbf{F}\) is the velocity field of a fluid flowing through a region in space. Find the flow along the given curve in the direction of increasing \(t .\) $$\begin{array}{l}{\mathbf{F}=-y \mathbf{i}+x \mathbf{j}+2 \mathbf{k}} \\\ {\mathbf{r}(t)=(-2 \cos t) \mathbf{i}+(2 \sin t) \mathbf{j}+2 t \mathbf{k}, \quad 0 \leq t \leq 2 \pi}\end{array}$$

Step-by-Step Solution

Answer

The flow along the curve is \( 8\pi \).

Step 1: Understand the Flow Integral

The flow of the fluid along a curve is given by the line integral \( \int_C \mathbf{F} \cdot d\mathbf{r} \), where \( \mathbf{F} \) is the velocity field and \( d\mathbf{r} \) is the differential element of the curve. In this problem, \( \mathbf{r}(t) \) defines the curve parameterized by \( t \).

Step 2: Compute the Derivative of \( \mathbf{r}(t) \)

To find \( d\mathbf{r} \), compute \( \frac{d\mathbf{r}}{dt} \). Given \( \mathbf{r}(t)=(-2 \cos t) \mathbf{i}+(2 \sin t) \mathbf{j}+2t \mathbf{k} \), we find: \( \frac{d\mathbf{r}}{dt} = (2 \sin t) \mathbf{i} + (2 \cos t) \mathbf{j} + 2 \mathbf{k} \).

Step 3: Set Up the Line Integral

Express the dot product \( \mathbf{F} \cdot \frac{d\mathbf{r}}{dt} \). Substituting \( \mathbf{F}=-y \mathbf{i}+x \mathbf{j}+2 \mathbf{k} \) and \( \frac{d\mathbf{r}}{dt} = (2 \sin t) \mathbf{i} + (2 \cos t) \mathbf{j} + 2 \mathbf{k} \), it simplifies to \( (-2\sin t)(2\sin t) + (2 \cos t)(2 \cos t) + 2(2) \).

Step 4: Simplify the Dot Product

Calculate \( \mathbf{F} \cdot \frac{d\mathbf{r}}{dt} \). This becomes: \( -4 \sin^2 t + 4 \cos^2 t + 4 \). Simplifying gives: \( 4(\cos^2 t - \sin^2 t) + 4 \).

Step 5: Evaluate the Integral

Integrate \( \int_0^{2\pi} \left( 4(\cos^2 t - \sin^2 t) + 4 \right) \, dt \). Use the identity \( \cos^2 t - \sin^2 t = \cos(2t) \), so the integral becomes \( \int_0^{2\pi} (4\cos(2t) + 4) \, dt \).

Step 6: Integrate and Solve

Separate the integral: \( 4\int_0^{2\pi} \cos(2t) \, dt + 4\int_0^{2\pi} 1 \, dt \). The first integral evaluates to zero over the interval \( 0 \) to \( 2\pi \) since \( \int_0^{2\pi} \cos(2t) \, dt = 0 \). The second integral evaluates to \( 8\pi \). Thus, the total flow is \( 8\pi \).

Key Concepts

Velocity Field

In any study of fluid dynamics, a velocity field is crucial. It represents how the velocity of fluid particles varies in a region of space. A velocity field is a vector field where each vector gives the velocity of the fluid at that point in space. This is typically represented by an equation, like \( \mathbf{F} = -y \mathbf{i} + x \mathbf{j} + 2 \mathbf{k} \), given in the exercise.

The beauty of a velocity field lies in its ability to describe complex behaviors of moving fluids. It tells us not just the speed of the fluid, but also the direction it is moving in. This information is vital for predicting how a fluid will behave under various conditions. Understanding and applying the concept of a velocity field helps us solve problems related to fluid flow and can be extended to various fields such as meteorology and oceanography.

• The vector \( \mathbf{i} \) represents the flow in the x-direction.
• The vector \( \mathbf{j} \) represents the flow in the y-direction.
• The vector \( \mathbf{k} \) represents the flow in the z-direction.

These directions and rates of flow are necessary to compute the line integral along a path, helping quantify the fluid's behavior along a specified curve.

Parameterization of Curves

The parameterization of a curve involves describing a curve using a set of equations, typically involving a parameter \( t \). These equations map out the curve in a coordinate space. For instance, in the exercise, the curve is parameterized by \( \mathbf{r}(t) = (-2 \cos t) \mathbf{i} + (2 \sin t) \mathbf{j} + 2t \mathbf{k} \).

Each value of \( t \) denotes a specific point on the curve, which in this case spans from \( t = 0 \) to \( t = 2\pi \), effectively tracing out a helix or spiral over this interval. Parameterization simplifies the process of calculating integrals along curves by converting them into functions of a single variable.

• Allows for easier computation of derivatives, like \( \frac{d \mathbf{r}}{dt} \).
• Enables straightforward calculation of vector fields along curves.
• Important for translating complex shapes into manageable mathematical forms.

This method is essential when dealing with vector fields and velocity, providing a way to break down a curve into simpler, more tackleable segments.

Definite Integral

The definite integral is a powerful mathematical tool used to find the total accumulation of quantities, which, in this context, helps determine the total flow of fluid along a curve. By setting up a line integral, \( \int_C \mathbf{F} \cdot d\mathbf{r} \), we can calculate the cumulative effect of the fluid's velocity field on a parameterized path.

In the given exercise, once the derivative \( \frac{d\mathbf{r}}{dt} \) is computed, the integral of the function defined by the dot product \( \mathbf{F} \cdot \frac{d\mathbf{r}}{dt} \) is evaluated from \( t = 0 \) to \( t = 2\pi \). This definite integral calculation provides the necessary measure of the flow through the region in question.

• Provides the method to aggregate the net flow along the defined path.
• Easily computed using the fundamental theorem of calculus when evaluated over specific intervals.
• Helps in interpreting physical phenomena related to accumulations and changes, such as pressure and fluid flow.

Understanding how to set up and evaluate definite integrals gives you the ability to quantify and make predictions about fluid movement through spatial regions.

Trigonometric Identities

Trigonometric identities are vital when simplifying expressions, especially when dealing with problems in calculus involving sine and cosine functions. These identities, like the Pythagorean identity or the angle sum formulas, allow us to rewrite complex expressions into simpler forms.

For instance, in step 5 of the solution, the identity \( \cos^2 t - \sin^2 t = \cos(2t) \) was used to simplify the function \( 4(\cos^2 t - \sin^2 t) + 4 \) before integrating. This simplification turned the integral into \( \int_0^{2\pi} (4\cos(2t) + 4) \, dt \), making it much more manageable.

• Trigonometric identities help streamline mathematical calculations.
• They are critical for converting complex trigonometric expressions.
• Essential for evaluating integrals in calculus related tasks.

By leveraging these identities, we can simplify integral calculations and solve problems efficiently. They highlight the importance of mathematical theorems in practical problem-solving scenarios.