Problem 56 - 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 \(t .\) $$\begin{array}{l}{\mathbf{F}=x^{2} \mathbf{i}+y z \mathbf{j}+y^{2} \mathbf{k}} \\ {\mathbf{r}(t)=3 t \mathbf{j}+4 t \mathbf{k}, \quad 0 \leq t \leq 1}\end{array}$$

Step-by-Step Solution

Answer

The flow along the curve is 24.

Step 1: Understand the Problem

The exercise requires us to find the flow of a fluid's velocity field \( \mathbf{F} \) along a curve in a particular direction. The field is described by \( \mathbf{F}=x^2 \mathbf{i} + yz \mathbf{j} + y^2 \mathbf{k} \), and the path is given by the vector function \( \mathbf{r}(t) = 3t \mathbf{j} + 4t \mathbf{k} \) for \( 0 \leq t \leq 1 \). We want to find the line integral of \( \mathbf{F} \) along this path.

Step 2: Calculate \( \mathbf{F}(\mathbf{r}(t)) \)

Substitute \( \mathbf{r}(t) = 3t \mathbf{j} + 4t \mathbf{k} \) into the velocity field \( \mathbf{F} \). Since \( x = 0 \), \( y = 3t \), and \( z = 4t \), substitute these into \( \mathbf{F} \): \( \mathbf{F}(\mathbf{r}(t)) = 0 \cdot \mathbf{i} + (3t)(4t) \cdot \mathbf{j} + (3t)^2 \cdot \mathbf{k} \). Simplifying, we get \( \mathbf{F}(\mathbf{r}(t)) = 12t^2 \mathbf{j} + 9t^2 \mathbf{k} \).

Step 3: Find \( \frac{d\mathbf{r}}{dt} \)

Differentiate \( \mathbf{r}(t) = 3t \mathbf{j} + 4t \mathbf{k} \) with respect to \( t \) to find \( \frac{d\mathbf{r}}{dt} \). This gives us \( \frac{d\mathbf{r}}{dt} = 3 \mathbf{j} + 4 \mathbf{k} \).

Step 4: Compute the Dot Product

Now find the dot product of \( \mathbf{F}(\mathbf{r}(t)) \) and \( \frac{d\mathbf{r}}{dt} \): \( (12t^2 \mathbf{j} + 9t^2 \mathbf{k}) \cdot (3 \mathbf{j} + 4 \mathbf{k}) = 36t^2 + 36t^2 = 72t^2 \).

Step 5: Evaluate the Line Integral

The flow along the curve is given by the line integral \( \int_{0}^{1} 72t^2 \, dt \). Compute this integral: \[ \int_{0}^{1} 72t^2 \, dt = 72 \int_{0}^{1} t^2 \, dt = 72 \left[ \frac{t^3}{3} \right]_{0}^{1} = 72 \times \frac{1}{3} = 24. \]

Step 6: Conclude the Solution

The flow of the velocity field \( \mathbf{F} \) along the curve \( \mathbf{r}(t) \) from \( t = 0 \) to \( t = 1 \) is 24.

Key Concepts

Velocity Field

A velocity field is a vector field that represents the velocity of a fluid at different points in space. Imagine a vast ocean where at every point in the water, there is a small arrow pointing in the direction the water is moving, and the length of the arrow indicates the speed of the flow. This is the essence of a velocity field.

In the given exercise, the velocity field is expressed as \( \mathbf{F} = x^2 \mathbf{i} + yz \mathbf{j} + y^2 \mathbf{k} \), where each component represents the velocity in the respective coordinate direction. Understanding the representation of these vectors is key to calculating how the fluid flows along a specific path.

Vector Calculus

Vector calculus is the branch of mathematics that deals with vector fields and operations like differentiation and integration on them. It provides the tools necessary to analyze and solve problems involving velocity fields.

• In this problem, we used differentiation to find \( \frac{d\mathbf{r}}{dt} \), which represents the tangent vector to the path described by the function \( \mathbf{r}(t) \).
• The dot product operation helped us determine how the velocity field interacts with the movement along the path, giving us a scalar value that's critical for integration.

Through these operations, vector calculus lets us understand the behavior of fields along curves and surfaces, making it indispensable in the study of fluid dynamics.

Fluid Dynamics

Fluid dynamics involves studying the movement of fluid substances like liquids and gases. It encompasses concepts like velocity fields which are crucial for understanding how fluids move through space.

In solving the exercise, our goal was to determine the flow of the fluid along a defined path. This path is given by \( \mathbf{r}(t) = 3t \mathbf{j} + 4t \mathbf{k} \), which signifies a movement solely in the y and z directions. The fluid dynamics approach links the velocity field to this path, allowing us to calculate the flow rate, which tells us how much fluid passes through any point along the path over time.

Path Integration

Path integration is an integral calculus operation used to compute the accumulation of a field along a path or curve. This concept is central to finding the total work done by a force field, or in our case, the flow of a fluid along a curve.

• The line integral \( \int_{0}^{1} 72t^2 \, dt \) represents the sum of the field component along the curve for each infinitesimal piece of the path from \( t = 0 \) to \( t = 1 \).
• It effectively integrates the influence of the vector field along the trajectory of the curve.

In the exercise, computing this integral provided us with a flow value of 24, which quantifies the total movement of the fluid along the specified path.