Question

Flow curves in the plane \(\operatorname{Let} \mathbf{F}(x, y)=\langle f(x, y), g(x, y)\rangle b e\) defined on \(\mathbb{R}^{2}\). Find and graph the flow curves for the vector field \(\mathbf{F}=\langle x, x\rangle.\)

Short answer

Answer: The flow curve of the given vector field is defined by the parametric equations \(x(t) = \frac{1}{2}t^2 + C_1\) and \(y(t) = \frac{1}{6}t^3 + C_2\), where \(C_1\) and \(C_2\) are constants determined by initial conditions. The shape of the flow curve is parabolic, and as \(t\) increases, the particles move upwards and outwards along the path.

Step-by-step solution

Define the parametric equation for flow curves

The flow curves are the path taken by the particles in the given vector field \(\mathbf{F}\). The path of the particle can be represented by \(\mathbf{r} = \langle x(t), y(t) \rangle\) where \(\mathbf{r}\) is a position vector and \(t\) is a parameter representing time. To find the flow curves, we need to find \(\frac{d\mathbf{r}}{dt}\) and set it equal to the given vector field \(\mathbf{F}\).

Set the derivative of the position vector equal to the vector field

Our goal is to find the flow curves, so we need to set \(\frac{d\mathbf{r}}{dt}\) equal to the vector field \(\mathbf{F}=\langle x, x \rangle\). The derivative of the position vector is \(\frac{d\mathbf{r}}{dt} = \frac{d}{dt}\langle x(t), y(t) \rangle = \langle x'(t), y'(t) \rangle\). Setting the derivative equal to the vector field, we get: $$ \langle x'(t), y'(t) \rangle = \langle x, x \rangle $$ Now, we can solve for \(x(t)\) and \(y(t)\).

Solve the parametric equations

From the previous step, we have two differential equations: $$ x'(t) = x \\ y'(t) = x $$ Solving the first differential equation, we get: $$ \int\frac{dx}{dt}dt = \int x dt \\ x(t) = \frac{1}{2}t^2 + C_1 $$ Now, solving the second differential equation, we get: $$ \int\frac{dy}{dt}dt = \int x dt \\ y(t) = \frac{1}{6}t^3 + C_2 $$ So the parametric equations for the flow curves are: $$ x(t) = \frac{1}{2}t^2 + C_1 \\ y(t) = \frac{1}{6}t^3 + C_2 \\ $$

Graph the flow curves

To graph the flow curves, we can use the parametric equations obtained in step 3: $$ x(t) = \frac{1}{2}t^2 + C_1 \\ y(t) = \frac{1}{6}t^3 + C_2 \\ $$ These equations represent a family of flow curves for the given vector field, where \(C_1\) and \(C_2\) are constants determined by the initial conditions of the particle. The flow curves will have a parabolic shape, and as \(t\) increases, the particles will move upwards and outwards along the same shaped path. You can use any graphing software or plotting tool to graph these flow curves for different values of \(C_1\) and \(C_2\).

Key concepts

Flow Curves

In the study of vector fields, flow curves are crucial because they describe the path that a particle would take under the influence of a given vector field. Imagine a vector field as a map of wind directions; the flow curves would be the paths that leaves or lightweight particles take as they are swept along by the wind.

To find flow curves, we start by examining the vector field given, which in our exercise is \(\mathbf{F} = \langle x, x \rangle\). To identify the flow curves in this field, we need to determine how a particle moves through this field over time.

The path taken by those particles, depending on their starting positions, is represented in the form of parametric equations. The result is often a set of curves that show different potential paths through the vector field, which can be visually represented using a graph. In this particular example, the flow curves will assume a parabolic shape.

Parametric Equations

Parametric equations are instrumental in representing the path of objects or points as they move; these equations express the coordinates of the points as functions of a parameter, often representing time \(t\). In the context of vector fields, parametric equations allow us to denote the movement across the field as a function of time.

In our example, the position of a particle within the vector field \(\mathbf{F} = \langle x, x \rangle\) is defined through the parametric equations \(x(t) = \frac{1}{2}t^2 + C_1\) and \(y(t) = \frac{1}{6}t^3 + C_2\). These equations give us the x and y coordinates of the particle at any given time \(t\).

The constants \(C_1\) and \(C_2\) represent the initial conditions, meaning the starting position of the particle. By solving these equations, depending on different values of \(C_1\) and \(C_2\), we obtain different paths or flow curves, which each reveal potential trajectories through the vector field.

Differential Equations

Differential equations are equations that relate a function to its derivatives, providing a way to model various phenomena, particularly those involving change over time. In the study of flow curves, differential equations are used to derive the expressions for the path of particles in vector fields.

In the given problem, we approached the vector field \(\mathbf{F} = \langle x, x \rangle\) and set up a differential equation \(\frac{d\mathbf{r}}{dt} = \langle x', y' \rangle = \langle x, x \rangle\). Solving this results in two differential equations: \(x'(t) = x\) and \(y'(t) = x\).

By integrating each of these differential equations, we derived the parametric equations \(x(t)\) and \(y(t)\), which describe how the particle moves over time. Differential equations thus provide the framework that helps encapsulate the dynamic nature of vector fields, making it possible to determine the flow curves.