Problem 50 - 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 \(t .\) $$ \begin{array}{l}{\mathbf{F}=-y \mathbf{i}+x \mathbf{j}+2 \mathbf{k}} \\\ {\mathbf{r}(t)=(-2 \cos t) \mathbf{i}+(2 \sin t) \mathbf{j}+2 t \mathbf{k}, \quad 0 \leq t \leq 2 \pi}\end{array} $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

The problem requires finding the flow of a fluid along a given parametric curve \( \mathbf{r}(t) \) under a velocity field \( \mathbf{F} \). The flow is given by the line integral of \( \mathbf{F} \cdot d\mathbf{r} \) along the curve.

Step 2: Calculate \( d\mathbf{r} \)

The differential \( d\mathbf{r} \) is the derivative of \( \mathbf{r}(t) \) with respect to \( t \), represented as \( d\mathbf{r} = \frac{d \mathbf{r}(t)}{dt}\, dt \). Calculate: \[ \frac{d\mathbf{r}(t)}{dt} = \frac{d}{dt}[-2 \cos t, 2 \sin t, 2t] = [2 \sin t, 2 \cos t, 2] \]

Step 3: Set Up the Dot Product \( \mathbf{F} \cdot d\mathbf{r} \)

Substitute \( \mathbf{r}(t) \) into \( \mathbf{F} \): \( \mathbf{F}(\mathbf{r}(t)) = -2\sin t \mathbf{i} + 2\cos t \mathbf{j} + 2 \mathbf{k} \).Then, compute the dot product:\[ \mathbf{F} \cdot d\mathbf{r} = (-2\sin t)(2\sin t) + (2\cos t)(2\cos t) + 2(2) \]

Step 4: Simplify the Dot Product

Simplifying the expression obtained from the dot product:\[= -4\sin^2 t + 4\cos^2 t + 4 \]\[= 4(\cos^2 t - \sin^2 t) + 4 \]

Step 5: Evaluate the Line Integral

The integral becomes: \[ \int_{0}^{2\pi} (4\cos^2 t - 4\sin^2 t + 4) \, dt \]This simplifies to:\[ \int_{0}^{2\pi} 4(\cos 2t + 1) \, dt \]Now evaluate the integral:\[ = 4 \left[ \int_{0}^{2\pi} \cos 2t \, dt + \int_{0}^{2\pi} 1 \, dt \right] = 4[0 + 2\pi] = 8\pi \]

Step 6: Conclusion

Based on our calculations, the flow along the curve from \( t = 0 \) to \( t = 2\pi \) is \( 8\pi \).

Key Concepts

Velocity Field

Imagine a field where every point in space has a vector that represents a velocity. This is known as a velocity field. In the context of fluid dynamics, it represents the speed and direction of fluid flow at every point.

Each vector in the field can indicate how fast and in which direction the fluid is moving locally. This is crucial for understanding the behavior of the fluid as it passes through different regions.

• The velocity field usually depends on the spatial location. In many cases, it's a function of coordinates (e.g., \(x\), \(y\), \(z\)).
• In our problem, the velocity field \( extbf{F} = -y \textbf{i} + x \ extbf{j} + 2 \ extbf{k}\). This tells us how the \(x\) and \(y\) components change, while the \(z\) component is constant at 2.

Understanding the velocity field allows us to analyze how particles within the fluid move through space. This is especially useful when evaluating the line integral over a given curve.

Parametric Curve

A parametric curve is defined by parameterizing each coordinate as a function of a variable, typically \(t\). This allows us to describe complex shapes and paths in a clear and manageable form. In our problem, the curve \( extbf{r}(t) = (-2 \cos t) \textbf{i} + (2 \sin t) \textbf{j} + 2t \textbf{k}\) creates a spiral-like motion in space.

By parameterizing the curve with \(t\), we effectively trace the path of the particle moving through the velocity field over time:

• The \(i\) component, \(-2 \cos t\), corresponds to horizontal oscillation.
• The \(j\) component, \(2 \sin t\), adds vertical oscillation.
• The \(k\) component, \(2t\), causes upward, constant movement along the \(z\)-axis.

Understanding how these components interact allows us to map the trajectory accurately, which is crucial when calculating the velocity field's influence along the path.

Dot Product

The dot product is a mathematical operation that takes two equal-length sequences of numbers, typically coordinate vectors, and returns a single number. It's significant in the context of line integrals because it allows us to factor in both the direction and magnitude when analyzing vector functions.

To solve the integral of a velocity field over a curve, compute the dot product of the velocity vector \(\textbf{F}(\textbf{r}(t))\) and the differential path vector \(d\textbf{r}\). Here's how it was done in the provided problem:

• Substitute the parametric expressions for \(\textbf{r}(t)\) into \(\textbf{F}\) to get \(\textbf{F}(\textbf{r}(t))\).
• Calculate the dot product of \(-2 \sin t \textbf{i} + 2 \cos t \textbf{j} + 2 \textbf{k}\) with the derivative \[ \frac{d}{dt}[-2 \cos t, 2 \sin t, 2t] = [2 \sin t, 2 \cos t, 2] \].
• The result \[ \textbf{F} \cdot d\textbf{r} = -4 \sin^2 t + 4 \cos^2 t + 4 \] represents the flow along the curve, accounting for both magnitude and directional impact.

In essence, the dot product helps combine linear contributions of curves and vectors in a succinct, powerful expression. It's a central component in computing the line integral, revealing not just how much, but also in which direction, the flow occurs.