Problem 43 - Thomas Calculus

Question

Find the flow of the velocity field \(\mathbf{F}=\frac{x}{y+1} \mathbf{i}+\frac{y}{x+1} \mathbf{j}\) where velocity is measured in meters per second, over the curve \(\mathbf{r}(t)=t^{2} \mathbf{i}+t \mathbf{j}, 0 \leq t \leq 1.\)

Step-by-Step Solution

Answer

The flow is 1 + (3/2)ln(2) meters.

Step 1: Parameterize the Curve

We have been given the parametric vector function \( \mathbf{r}(t) = t^2 \mathbf{i} + t \mathbf{j} \). This describes the path of integration from \( t = 0 \) to \( t = 1 \).

Step 2: Compute the Derivative of the Curve

To find the flow of \( \mathbf{F} \) over this curve, we need \( \frac{d\mathbf{r}}{dt} \). Differentiating \( \mathbf{r}(t) = t^2 \mathbf{i} + t \mathbf{j} \), we get \( \frac{d\mathbf{r}}{dt} = 2t \mathbf{i} + \mathbf{j} \).

Step 3: Substitute into the Velocity Field

Substitute \( x = t^2 \) and \( y = t \) into \( \mathbf{F} = \frac{x}{y+1} \mathbf{i} + \frac{y}{x+1} \mathbf{j} \). This gives us \( \mathbf{F}(t) = \frac{t^2}{t+1} \mathbf{i} + \frac{t}{t^2+1} \mathbf{j} \).

Step 4: Calculate the Dot Product

The flow of the velocity field is computed using the dot product \( \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} \). Calculate this as: \[ \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} = \left( \frac{t^2}{t+1} \right) (2t) + \left( \frac{t}{t^2+1} \right) (1) = \frac{2t^3}{t+1} + \frac{t}{t^2+1}. \]

Step 5: Set Up the Integral

The flow of the velocity field over the curve is the integral \( \int_{0}^{1} \left( \frac{2t^3}{t+1} + \frac{t}{t^2+1} \right) \, dt \).

Step 6: Evaluate the Integral

Evaluate the integral \( \int_{0}^{1} \frac{2t^3}{t+1} \, dt + \int_{0}^{1} \frac{t}{t^2+1} \, dt \) separately. Use substitution if necessary to solve these integrals. The first requires long division and substitution, resulting in \( \int_{0}^{1} (2t^2 - 2t + \frac{2}{t+1}) \, dt \), which evaluates to \( 1 + \ln 2 \). The second integral evaluates directly using substitution, resulting in \( \ln(\sqrt{2}) \).

Step 7: Combine the Results

Combine the results from evaluating the two integrals to arrive at the total flow: \( 1 + \ln 2 + \ln(\sqrt{2}) = 1 + \frac{3}{2}\ln(2) \).

Key Concepts

Dot Product

The dot product is a mathematical operation that takes two equal-length sequences of numbers, usually coordinate vectors, and returns a single number. In the context of our problem, the dot product helps us find how much of one vector lies in the direction of another vector.

Here, we are looking at the velocity field vector, denoted as \( \mathbf{F}(t) \), and the derivative of the curve at each point, denoted as \( \frac{d\mathbf{r}}{dt} \). To compute the flow of the velocity field, we calculate the dot product of these two vectors to find their interaction over the curve's path.

• The vector \( \mathbf{F}(t) \) is substituted as \( \left(\frac{t^2}{t+1}\right) \mathbf{i} + \left(\frac{t}{t^2+1}\right) \mathbf{j} \).
• The derivative of the parametric curve is \( 2t \mathbf{i} + \mathbf{j} \).

Performing the dot product, we multiply the \( \mathbf{i} \) components and \( \mathbf{j} \) components separately before adding them together: \[ \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} = \left( \frac{t^2}{t+1} \right) (2t) + \left( \frac{t}{t^2+1} \right) (1) = \frac{2t^3}{t+1} + \frac{t}{t^2+1}. \]This resulting value helps in integrating over the curve to understand total flow.

Parametric Equations

Parametric equations represent a set of related quantities as explicit functions of a number of independent parameters. This is particularly useful for describing trajectories.

In the exercise, the trajectory of a curve is given by a parametric equation: \( \mathbf{r}(t) = t^2 \mathbf{i} + t \mathbf{j} \). This means we are using \( t\) as our parameter which varies from 0 to 1. As \(t\) changes, we trace out the curve in the \(xy\)-plane.

• \(x\) is defined as \(t^2\), meaning the horizontal displacement grows with the square of \(t\).
• \(y\) is defined as \(t\), implying a linear relationship between \(t\) and vertical displacement.

Examining curves through parametric equations allows for more flexible descriptions of paths, especially those that aren't easily describable using conventional \(y=f(x)\) relations.

Integration

Integration is a fundamental concept of calculus and involves finding the integral of a function. In applications like ours, it is used to determine the total flow along a path.

We need to integrate the dot product expression \( \int_{0}^{1} \left( \frac{2t^3}{t+1} + \frac{t}{t^2+1} \right) \, dt \) to capture the flow over the curve. This means summing tiny infinitesimal contributions along the interval from \(t=0\) to \(t=1\).

• The expression is split into two separate integrals for easier calculation: \( \int_{0}^{1} \frac{2t^3}{t+1} \, dt \) and \( \int_{0}^{1} \frac{t}{t^2+1} \, dt \).
• The first integral benefits from the substitution and partial fraction decomposition methods.

Evaluating each integral separately and combining the results gives the total flow, which integrates contributions from each parameter value.

Substitution Method

The substitution method in integration simplifies complex integrals by changing the variable of integration. This often makes it easier to find an antiderivative.

In the exercise provided, substitution plays a key role in evaluating the integrals that represent flow. For instance, tackling \( \int \frac{t}{t^2+1} \, dt \) involves a substitution that simplifies the denominator.

• Using the substitution \( u = t^2 + 1 \), we find \( du = 2t \, dt \), simplifying part of the integral structure.
• Rewriting the integral in terms of \( u \) often reduces complexity and makes the integral manageable.

The method enhances our ability to solve challenging integrals and aids in breaking down parts of the problem into easier computations, aiding us in determining the flow over a curve.