Problem 48 - Thomas Calculus

Question

\(\mathbf{F}\) 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 \(t .\) $$ \begin{array}{l}{\mathbf{F}=-4 x y \mathbf{i}+8 y \mathbf{j}+2 \mathbf{k}} \\\ {\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+\mathbf{k}, \quad 0 \leq t \leq 2}\end{array} $$

Step-by-Step Solution

Answer

The flow along the curve is 48 units.

Step 1: Understand the Concept of Flow

The flow of a fluid along a curve is calculated using a line integral of the vector field \(\mathbf{F}\) along the curve defined by the parametric equation \(\mathbf{r}(t)\). This can be expressed as \(\int_C \mathbf{F} \cdot d\mathbf{r}\) where \(C\) is the curve along which we are integrating.

Step 2: Determine the Differential Element \(d\mathbf{r}\)

The differential element for the curve is found by differentiating \(\mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + \mathbf{k}\) with respect to \(t\). This results in \(d\mathbf{r} = (1 \mathbf{i} + 2t \mathbf{j} + 0 \mathbf{k}) \, dt = \mathbf{i} \, dt + 2t \mathbf{j} \, dt\).

Step 3: Set up the Line Integral

The line integral of the vector field \(\mathbf{F}\) along the curve is given by \(\int_0^2 \mathbf{F}(\mathbf{r}(t)) \cdot d\mathbf{r}\). This simplifies to \(\int_0^2 (-4xy \mathbf{i} + 8y \mathbf{j} + 2 \mathbf{k}) \cdot (\mathbf{i} \, dt + 2t \mathbf{j} \, dt)\).

Step 4: Substitute the Parameterization into \(\mathbf{F}\)

Substitute \(x = t, y = t^2\) into \(\mathbf{F}\): \(-4xy = -4t^3\), \(8y = 8t^2\), and \(2 = 2\). Thus, \(\mathbf{F}(\mathbf{r}(t)) = -4t^3 \mathbf{i} + 8t^2 \mathbf{j} + 2 \mathbf{k}\).

Step 5: Evaluate the Dot Product

Calculate the dot product \(\mathbf{F}(\mathbf{r}(t)) \cdot d\mathbf{r} = (-4t^3 \mathbf{i} + 8t^2 \mathbf{j} + 2 \mathbf{k}) \cdot (\mathbf{i} \, dt + 2t \mathbf{j} \, dt)\). This results in \(-4t^3 \cdot 1 + 8t^2 \cdot 2t = -4t^3 + 16t^3 = 12t^3\). Since the \(\mathbf{k}\) term has no \(dt\) component, it contributes nothing.

Step 6: Integrate with Respect to \(t\)

Integrate \(12t^3\) with respect to \(t\) from 0 to 2: \(\int_0^2 12t^3 \, dt = [3t^4]_0^2 = 3(2)^4 - 3(0)^4 = 48\).

Step 7: Interpret the Result

The flow of the fluid along the curve from \(t = 0\) to \(t = 2\) is 48 units in the direction of increasing \(t\).

Key Concepts

Velocity Field

A velocity field is a vector field that represents the speed and direction of a point in a fluid's flow through space. Consider it as a map that shows how quickly and in which direction the fluid moves at every point.

• In our exercise, the velocity field is given by \( \mathbf{F} = -4xy \mathbf{i} + 8y \mathbf{j} + 2 \mathbf{k} \).
• This vector field allows us to understand how the fluid particles are moving within the specified space.
• The components of the field can change with varying positions \((x, y, z)\) in space, affecting the overall flow pattern.

By analyzing this field, we can determine specific behaviors of the fluid such as acceleration and vorticity.

Line Integral

A line integral is like a sum across a curve, taking into account the vector field's effect along that path. Think of it as measuring the total accumulation of a vector field's effect over the path of integration.

• In this problem, the line integral \( \int_C \mathbf{F} \cdot d \mathbf{r} \) evaluates the sum of the fluid's flow along a curve \( C \).
• It involves the dot product of the vectors \( \mathbf{F} \) and the differential element \( d \mathbf{r} \), which translates to detecting the component of \( \mathbf{F} \) along the curve.
• By evaluating this integral, you calculate the total flow across the curve, gaining insight into the fluid's behavior between two points.

Parameterization

Parameterization is a technique used to define a curve using a parameter \( t \). This simplifies calculations by expressing points on the curve as functions of \( t \).

• Our given curve is represented as \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + \mathbf{k} \), where \( t \) is between 0 and 2.
• Each point on the curve corresponds to a specific value of \( t \). For instance, at \( t = 0 \), the point is \( (0, 0, 1) \).
• This method simplifies substitution into the vector field equation, making it easier to find \( \mathbf{F}(\mathbf{r}(t)) \).

Through parameterization, complex spatial paths become more manageable via simpler variable expressions in terms of \( t \).

Differential Elements

Differential elements are small vector segments of a curve. They essentially break down a curve into infinitely small pieces for analysis, aiding in precise integrations.

• In the exercise, the differential element is derived as \( d\mathbf{r} = \mathbf{i} \, dt + 2t \mathbf{j} \, dt \).
• This is obtained by differentiating the parameterized curve \( \mathbf{r}(t) \) with respect to \( t \), representing a tiny vector of change along the curve.
• When working with line integrals, these differential vectors align with the vector field components to evaluate subtle interactions over the curve.

Differential elements serve as the foundational blocks, allowing complex integrals to transform into accumulation of these tiny pieces over a given curve.