Problem 37 - Thomas Calculus

Question

Find the flow of the velocity field \(\mathbf{F}=y^{2} \mathbf{i}+2 x y \mathbf{j}\) along each of the following paths from \((0,0)\) to \((2,4)\) c. Use any path from \((0,0)\) to \((2,4)\) different from parts (a) and (b).

Step-by-Step Solution

Answer

The flow of the velocity field along the path \( y = 2x^2 \) from \((0,0)\) to \((2,4)\) is \( \frac{1152}{5} \).

Step 1: Understanding the Vector Field and Path

The vector field is given as \( \mathbf{F}=y^{2} \mathbf{i}+2 x y \mathbf{j} \). The flow of the velocity field along a path \( \gamma \) is computed by evaluating the line integral \( \int_{\gamma} \mathbf{F} \cdot d\mathbf{r} \). We need to select a path different from parts (a) and (b) to compute this integral. Let's choose a parabolic path \( y = 2x^2 \) from \((0,0)\) to \((2,4)\).

Step 2: Parameterizing the Path

The parabolic path \( y = 2x^2 \) can be parameterized as \( \mathbf{r}(t) = t \mathbf{i} + 2t^2 \mathbf{j} \), where \( t \) goes from 0 to 2. Here, the initial point \((0,0)\) corresponds to \( t = 0 \) and the final point \((2,4)\) corresponds to \( t = 2 \).

Step 3: Finding the Differential of the Path

Calculate the differential \( d\mathbf{r} \) along the path: \( d\mathbf{r} = \frac{d\mathbf{r}}{dt} dt = (1 \mathbf{i} + 4t \mathbf{j}) dt \).

Step 4: Compute \( \mathbf{F} \cdot d\mathbf{r} \)

Substitute \( \mathbf{r}(t) = t \mathbf{i} + 2t^2 \mathbf{j} \) into \( \mathbf{F} \):\[ \mathbf{F} = (2(2t^2)^2) \mathbf{i} + (4t^2) (2t) \mathbf{j} = 4t^4 \mathbf{i} + 8t^3 \mathbf{j}. \]Compute \( \mathbf{F} \cdot d\mathbf{r} \):\[ \mathbf{F} \cdot d\mathbf{r} = (4t^4 \mathbf{i} + 8t^3 \mathbf{j}) \cdot (1 \mathbf{i} + 4t \mathbf{j}) dt = (4t^4 + 32t^4) dt = 36t^4 dt. \]

Step 5: Evaluate the Integral

Integrate \( \int_{0}^{2} 36t^4 dt \):\[ \int_{0}^{2} 36t^4 dt = 36 \left[ \frac{t^5}{5} \right]_{0}^{2} = 36 \left( \frac{32}{5} - 0 \right) = \frac{1152}{5}. \]

Step 6: Conclusion

The flow of the velocity field \( \mathbf{F} \) along the parabolic path \( y = 2x^2 \) from \((0,0)\) to \((2,4)\) is \( \frac{1152}{5} \).

Key Concepts

Vector Field

In mathematics and physics, a vector field is a space where each point has a vector assigned to it. Imagine a weather map where each location has arrows showing the wind direction and speed; that's a simplified representation of a vector field. In our exercise, the vector field is defined by \( \mathbf{F}=y^{2} \mathbf{i}+2 x y \mathbf{j} \). These components, \( y^2 \mathbf{i} \) and \( 2xy \mathbf{j} \), determine how the vector changes with respect to \( x \) and \( y \). Understanding the nature of these vectors helps us understand the behavior of fields in real-life applications like fluid mechanics and electromagnetism.
Vector fields are essential in describing phenomena where direction and magnitude matter at every point in space. For example:

• Electromagnetic fields exert forces on charged particles.
• Gravitational fields attract masses.
• Temperature distribution can be represented as a gradient field.

By visualizing the arrow's direction and length at different points, we can predict how an object will move or change over time when subjected to the vector field.

Parameterization

Parameterization involves describing a curve or a path by assigning each point a unique parameter value, usually denoted by \( t \). This process simplifies the evaluation of integrals and calculations along paths in fields. In the exercise, the path from \((0,0)\) to \((2,4)\) is parameterized as \( \mathbf{r}(t) = t \mathbf{i} + 2t^2 \mathbf{j} \), which truly conveys the point of a parabolic path.What does this mean in simpler terms?

• It connects points on the plane through a mathematical expression based on time or a similar variable.
• Here, as \( t \) increases, \( x = t \) and \( y = 2t^2 \), effectively tracing the parabola \( y = 2x^2 \).

Parameterization transforms complex integrals over paths in vector fields into single-variable calculus problems. By using \( t \), we can simplify the mathematical description of the path, helping us evaluate complex expressions more efficiently.

Differential Calculus

Differential calculus helps understand how vectors change over a path. At its core, it deals with rates of change and slopes of curves, which are foundational concepts for analyzing real-world dynamics.In the exercise, we find the differential \( d\mathbf{r} \) of the path as we move over it. This involves taking the derivative of the parameterized path and is given by \( d\mathbf{r} = (1 \mathbf{i} + 4t \mathbf{j}) dt \). Here's what this step achieves:

• The calculation of \( \frac{d\mathbf{r}}{dt} \) shows the velocity at which you move along the curve at any given moment \( t \).
• This rate of change helps determine how the vector field's effect on a particle changes throughout the movement.

Differential calculus allows us to analyze the flow of a vector field along a path. This is particularly useful in engineering and physics, where understanding such flows is essential.

Velocity Field

A velocity field is a vector field representing the velocity of flowing matter, like air or water, at various points in space. In our exercise, \( \mathbf{F}=y^{2} \mathbf{i}+2 x y \mathbf{j} \) can be visualized as a velocity field where each vector represents the velocity at that spot in the field.
Understanding how to work with velocity fields is crucial for determining how particles will move through a medium:

• It combines direction and magnitude derived from components \( y^2 \) and \( 2xy \).
• The dot product \( \mathbf{F} \cdot d\mathbf{r} \) gives the flow across a path, essentially describing how much of the field lines the path penetrates.

By evaluating line integrals, as demonstrated, we can determine the cumulative effect of the field along a particular trajectory, essential in calculating work done by forces in mechanics. These insights have applications ranging from aerospace, predicting airflow around objects, to oceanography, modeling currents in vast water bodies.