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Chapter 14: Problem 32

Gradient fields Find the gradient field \(\mathbf{F}=\nabla \varphi\) for the following potential functions \(\varphi.\) $$\varphi(x, y)=\tan ^{-1}(y / x)$$

Short Answer

Expert verified
Answer: The gradient field of the potential function \(\varphi(x, y) = \tan^{-1}(y / x)\) is \(\mathbf{F} =\left<\frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2}\right>\).

Step by step solution

01

Identify the potential function

We are given the potential function \(\varphi(x, y)=\tan ^{-1}(y / x)\).
02

Calculate the partial derivatives of \(\varphi\)

To find the gradient field \(\mathbf{F}=\nabla \varphi\), we need to compute the partial derivatives \(\frac{\partial \varphi}{\partial x}\) and \(\frac{\partial \varphi}{\partial y}\): \(\frac{\partial \varphi}{\partial x} = \frac{\partial}{\partial x} \tan^{-1}(y / x) = \frac{-y}{x^2+y^2}\) (using the chain rule) \(\frac{\partial \varphi}{\partial y} = \frac{\partial}{\partial y} \tan^{-1}(y / x) = \frac{x}{x^2+y^2}\) (again using the chain rule)
03

Determine the gradient field \(\mathbf{F}\)

Now that we have the partial derivatives, we can write the gradient field \(\mathbf{F}\) as a vector: \(\mathbf{F}=\nabla \varphi = \left<\frac{\partial \varphi}{\partial x},\frac{\partial \varphi}{\partial y}\right> = \left<\frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2}\right>\) So the gradient field \(\mathbf{F}\) for the given potential function is: \(\boxed{\mathbf{F} = \left<\frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2}\right>}\)

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Key Concepts

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

Vector Calculus
Vector calculus is a branch of mathematics that focuses on the study of vector fields and vector-valued functions. This topic combines the principles of calculus with vectors and can operate in two, three, or more dimensions. In essence, vector calculus deals with differential and integral calculus of vector functions.

One of the key operations in vector calculus is finding the gradient of a scalar field, which is what this exercise explores. The gradient of a scalar field is a vector field that points in the direction of the greatest rate of increase of the scalar field, and its magnitude is the rate of that increase. Here, we used the gradient operation, denoted as abla, to compute the gradient field abla abla abla abla abla abla abla abla from a given potential function abla (x, y).abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla
Partial Derivatives
Partial derivatives are fundamental in multi-variable calculus. They represent the rate at which a function changes as one variable changes while all other variables are held constant. In the textbook exercise, we derive the gradient by finding the partial derivatives of the given potential function with respect to each variable
abla we're working in the xy-plane that means taking derivatives with respect to x and y. The abla derivative abla x is found by treating y as a constant (and vice versa for the y derivative), a method that allows us to see how abla changes along each axis individually.

Understanding the chain rule is crucial here as it enables us to handle composite functions like abla^{-1}(y / x)ablaif the inner function is y / xabla a two-variable functionabla the outer function is abla^{-1}, which is applied after the inner calculation. This technique allows us to break down complex calculations into more manageable steps, which is a vital skill for solving gradient fields.
Potential Function
A potential function, often denoted as abla, is a scalar field from which a vector field can be derived. If a vector field is the gradient of some scalar potential, it is said to be a conservative field, and the scalar potential is the function whose gradient gives us back that vector field.

In the given problem, abla(x, y) = abla^{-1}(y / x)ablaablaablaabla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla abla ablaablaablaablaablaablaablaablaablaablaablaablaablaablaablaablaablaablaablaablaablaablaablaablaablaablaablaablaablaabla abla abla abla ablaabla abla abla abla abla abla ablaabla abla abla abla abla ablaabla abla abla abla abla ablaabla abla abla abla abla ablaabla abla abla abla abla ablaabla abla abla abla abla ablaabla abla abla abla

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Most popular questions from this chapter

Let S be the disk enclosed by the curve \(C: \mathbf{r}(t)=\langle\cos \varphi \cos t, \sin t, \sin \varphi \cos t\rangle,\)for \(0 \leq t \leq 2 \pi,\) where \(0 \leq \varphi \leq \pi / 2\) is a fixed angle. What is the circulation on \(C\) of the vector field \(\mathbf{F}=\langle-y,-z, x\rangle\) as a function of \(\varphi ?\) For what value of \(\varphi\) is the circulation a maximum?

The cone \(z^{2}=x^{2}+y^{2},\) for \(z \geq 0,\) cuts the sphere \(x^{2}+y^{2}+z^{2}=16\) along a curve \(C\) a. Find the surface area of the sphere below \(C,\) for \(z \geq 0\). b. Find the surface area of the sphere above \(C\). c. Find the surface area of the cone below \(C,\) for \(z \geq 0\).

Suppose a solid object in \(\mathbb{R}^{3}\) has a temperature distribution given by \(T(x, y, z) .\) The heat flow vector field in the object is \(\mathbf{F}=-k \nabla T,\) where the conductivity \(k>0\) is a property of the material. Note that the heat flow vector points in the direction opposite that of the gradient, which is the direction of greatest temperature decrease. The divergence of the heat flow vector is \(\nabla \cdot \mathbf{F}=-k \nabla \cdot \nabla T=-k \nabla^{2} T\) (the Laplacian of \(T\)). Compute the heat flow vector field and its divergence for the following temperature distributions. $$T(x, y, z)=100 e^{-x^{2}+y^{2}+z^{2}}$$

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 \(\omega=\frac{1}{2}|\nabla \times \mathbf{v}|\).

Prove the following identities. Assume that \(\varphi\) is \(a\) differentiable scalar-valued function and \(\mathbf{F}\) and \(\mathbf{G}\) are differentiable vector fields, all defined on a region of \(\mathbb{R}^{3}\). $$\nabla(\mathbf{F} \cdot \mathbf{G})=(\mathbf{G} \cdot \nabla) \mathbf{F}+(\mathbf{F} \cdot \nabla) \mathbf{G}+\mathbf{G} \times(\nabla \times \mathbf{F})+\mathbf{F} \times(\nabla \times \mathbf{G})$$
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