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Chapter 14: Problem 27

Gradient fields Find the gradient field \(\mathbf{F}=\nabla \varphi\) for the potential function \(\varphi .\) Sketch a few level curves of \(\varphi\) and a few vectors of \(\mathbf{F}\). $$\varphi(x, y)=x+y, \text { for }|x| \leq 2,|y| \leq 2$$

Short Answer

Expert verified
Answer: The gradient field of the potential function \(\varphi(x, y) = x + y\) is constant and given by \(\mathbf{F} = (1, 1)\). On a 2D plane in the domain \(|x| \leq 2, |y| \leq 2\), the level curves of the potential function are straight lines with a slope of -1, and the vectors of the gradient field have the same direction and magnitude everywhere in the domain.

Step by step solution

01

Find the gradient of the potential function

To find the gradient of the potential function, we will take partial derivatives with respect to both \(x\) and \(y\). Then we can represent the gradient field \(\mathbf{F}\) as a vector \(\nabla \varphi = (\frac{\partial \varphi}{\partial x}, \frac{\partial \varphi}{\partial y})\): $$\frac{\partial \varphi}{\partial x} = \frac{\partial}{\partial x}(x+y) = 1$$ $$\frac{\partial \varphi}{\partial y} = \frac{\partial}{\partial y}(x+y) = 1$$ So the gradient field is \(\mathbf{F} = \nabla \varphi = (1, 1)\).
02

Sketch level curves of the potential function

Level curves of a function are the set of points where the function has a constant value. In this case, we will consider a few constant values for \(\varphi\) and sketch the corresponding curves. The level curve for a constant value \(c\) is defined by the equation \(\varphi(x, y) = c\). Here, we have \(\varphi(x, y) = x + y = c\). We can rewrite this equation as \(y = c - x\). We can choose a few values for \(c\) and sketch the corresponding level curves in the domain \(|x| \leq 2, |y| \leq 2\). Let's consider \(c = -3, -1, 1, 3\) for illustration. - For \(c = -3\), we have the level curve \(y = -3 - x\). - For \(c = -1\), we have the level curve \(y = -1 - x\). - For \(c = 1\), we have the level curve \(y = 1 - x\). - For \(c = 3\), we have the level curve \(y = 3 - x\).
03

Sketch vectors of the gradient field

In the domain \(|x| \leq 2, |y| \leq 2\), we will draw a few vectors of the gradient field \(\mathbf{F} = (1, 1)\). Note that the gradient field \(\mathbf{F}\) is constant, and its components do not depend on the values of \(x\) and \(y\). This means that the vectors of \(\mathbf{F}\) will have the same direction and magnitude everywhere in the domain. Now we have the level curves and vectors for the gradient field \(\mathbf{F}\). To visualize this, you can sketch the level curves on a 2D plane and draw the vectors of \(\mathbf{F}\) at some points in the domain. This will help you understand the relationship between the potential function \(\varphi\) and its gradient field \(\mathbf{F}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Potential Function
A potential function is a scalar function, denoted by \( \varphi(x, y) \), from which a vector field is derived. In our exercise, the potential function is \( \varphi(x, y) = x + y \). The term "potential" indicates that the function has the capacity to create a vector field, analogous to how a potential energy field works in physics.

To explore the potential function, you determine its gradient, which results in a vector field. This specific family of vector fields is called a gradient field. The gradient field represents the rate of change and direction in which \( \varphi \) increases most rapidly. Thus, understanding the potential function is crucial for describing the underlying vector field behavior.
Partial Derivatives
Partial derivatives play a key role in identifying how a multivariable function like \( \varphi(x, y) = x + y \) changes with respect to each of its variables independently. In simpler terms, a partial derivative examines the change in the potential function \( \varphi \) along one axis of the graph while keeping the other variable constant.

Take, for instance, the partial derivative of \( \varphi \) with respect to \( x \). It is represented as \( \frac{\partial \varphi}{\partial x} = 1 \). Similarly, for \( y \), the partial derivative is \( \frac{\partial \varphi}{\partial y} = 1 \). Here, both partial derivatives are constant, indicating that our function \( \varphi \) is linear, and the gradient field associated with it, \( \mathbf{F} = (1, 1) \), is uniform across the specified range.

Understanding partial derivatives helps discern the behavior of complicated multivariable functions in simpler, one-dimensional perspectives, which is essential in calculus and vector field analysis.
Level Curves
Level curves are a fascinating aspect of potential functions that can provide insight into the function's structure. For our function \( \varphi(x, y) = x + y \), level curves are lines of constant value \( c \). Imagine these curves as contour lines on a map, each representing a specific altitude. Here, each line on the 2D plane is where the function \( \varphi \) equals a particular constant.

In our example, the expression \( y = c - x \) represents various level curves. By choosing different values of \( c \) such as -3, -1, 1, and 3, different straight lines are drawn, each spaced evenly based on \( c \). These straight lines have a slope of -1, indicating steepness in the descent or ascent when moving across them.

Level curves are crucial for visualizing how a function varies across a plane. They also provide an easy reference to see how changes occur in relation to the potential function across different areas.

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Most popular questions from this chapter

Use Stokes' Theorem to write the circulation form of Green's Theorem in the \(y z\) -plane.

Find the general formula for the surface area of a cone with height \(h\) and base radius \(a\) (excluding the base).

Let \(R\) be a region in a plane that has a unit normal vector \(\mathbf{n}=\langle a, b, c\rangle\) and boundary \(C .\) Let \(\mathbf{F}=\langle b z, c x, a y\rangle\) a. Show that \(\nabla \times \mathbf{F}=\mathbf{n}\) b. Use Stokes' Theorem to show that $$\operatorname{area} \text { of } R=\oint_{C} \mathbf{F} \cdot d \mathbf{r}$$ c. Consider the curve \(C\) given by \(\mathbf{r}=\langle 5 \sin t, 13 \cos t, 12 \sin t\rangle\) for \(0 \leq t \leq 2 \pi .\) Prove that \(C\) lies in a plane by showing that \(\mathbf{r} \times \mathbf{r}^{\prime}\) is constant for all \(t\) d. Use part (b) to find the area of the region enclosed by \(C\) in part (c). (Hint: Find the unit normal vector that is consistent with the orientation of \(C\).)

Use the procedure in Exercise 57 to construct potential functions for the following fields. $$\mathbf{F}=\langle x, y\rangle$$

Consider the rotational velocity field \(\mathbf{v}=\mathbf{a} \times \mathbf{r},\) where \(\mathbf{a}\) is a nonzero constant vector and \(\mathbf{r}=\langle x, y, z\rangle .\) Use the fact that an object moving in a circular path of radius \(R\) with speed \(|\mathbf{v}|\) has an angular speed of \(\omega=|\mathbf{v}| / R\). a. Sketch a position vector a, which is the axis of rotation for the vector field, and a position vector \(\mathbf{r}\) of a point \(P\) in \(\mathbb{R}^{3}\). Let \(\theta\) be the angle between the two vectors. Show that the perpendicular distance from \(P\) to the axis of rotation is \(R=|\mathbf{r}| \sin \theta\). b. Show that the speed of a particle in the velocity field is \(|\mathbf{a} \times \mathbf{r}|\) and that the angular speed of the object is \(|\mathbf{a}|\). c. Conclude that \(\omega=\frac{1}{2}|\nabla \times \mathbf{v}|\).
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