/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 Gradient fields Find the gradien... [FREE SOLUTION] | 91Ó°ÊÓ

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Gradient fields Find the gradient field \(\mathbf{F}=\nabla \varphi\) for the following potential functions \(\varphi.\) $$\varphi(x, y)=x^{2} y-y^{2} x$$

Short Answer

Expert verified
Answer: The gradient field for the given potential function is F = ∇φ = 〈2xy - y^2, x^2 - 2yx〉.

Step by step solution

01

Calculate the partial derivative with respect to x

In order to find \(\frac{\partial\varphi}{\partial x}\), we differentiate the potential function with respect to \(x.\) $$ \frac{\partial\varphi}{\partial x} = \frac{\partial}{\partial x}(x^{2}y-y^{2}x) $$ Using the product and power rules, $$ \frac{\partial\varphi}{\partial x} = 2xy - y^2 $$
02

Calculate the partial derivative with respect to y

Similarly, we will differentiate the potential function with respect to \(y.\) $$ \frac{\partial\varphi}{\partial y} = \frac{\partial}{\partial y}(x^{2}y-y^{2}x) $$ Using the product and power rules, $$ \frac{\partial\varphi}{\partial y} = x^2 - 2yx $$
03

Write the gradient field as a vector

Now that we have the partial derivatives with respect to \(x\) and \(y\), we can write the gradient field as a vector. The gradient field \(\mathbf{F}\) can be denoted as: $$ \mathbf{F} = \nabla\varphi = \left\langle\frac{\partial\varphi}{\partial x},\frac{\partial\varphi}{\partial y}\right\rangle $$ Substituting the calculated partial derivatives, $$ \mathbf{F} = \left\langle 2xy - y^2, x^2 - 2yx \right\rangle $$ So, the gradient field for the given potential function is: $$ \mathbf{F} = \nabla\varphi = \left\langle 2xy - y^2, x^2 - 2yx \right\rangle $$

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Key Concepts

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

Partial Derivatives
When studying functions of multiple variables, understanding how these functions change with respect to each variable separately is crucial. Partial derivatives are the mathematical tools that provide this information. To find a partial derivative of a function, we differentiate with respect to one variable while keeping all other variables constant.

In the context of the given problem, we have a function \( \varphi(x, y) = x^{2} y - y^{2} x \). The partial derivative of \( \varphi \) with respect to \( x \) is denoted as \( \frac{\partial \varphi}{\partial x} \). This derivative tells us how the function \( \varphi \) changes as \( x \) changes, while \( y \) is kept constant. Similarly, \( \frac{\partial \varphi}{\partial y} \) tells us the rate of change of \( \varphi \) with respect to \( y \) when \( x \) is constant. These concepts build the basis for a gradient field which is a vector field representing the maximum rate of increase of the function at each point.

Understanding partial derivatives is fundamental not only in vector calculus but also in various applications such as optimization, physics, and engineering.
Potential Function
In vector calculus, a potential function is a scalar function whose gradient produces a vector field. This is analogous to gravitational or electrostatic fields in physics, where the field can be described as the gradient of a potential energy function. Given a potential function \( \varphi \) like in our exercise, \( \varphi(x, y) = x^{2} y - y^{2} x \), we can use it to generate a vector field.

The magic of potential functions lies in their ability to simplify many complex problems. In the context of conservative vector fields (where work done along a path is path-independent), every such field is the gradient of some potential function. This concept has a profound impact on how we understand physical phenomena and provides a powerful tool in the toolkit of scientists and engineers. The exercise provided is a classic example of using a potential function to find its associated vector field, thus illustrating a fundamental principle of physics and vector calculus.
Vector Calculus
Vector calculus is the branch of mathematics that deals with vector fields and operations on these fields. It generalizes the differentiation and integration of functions of several variables to vector fields, providing a language to describe and analyze physical phenomena like fluid flow, electromagnetic fields, and more.

In our exercise, we utilized vector calculus concepts to produce a gradient field, which is a significant type of vector field. By calculating the partial derivatives of the potential function, we have obtained the components of the gradient vector. The gradient vector at any point indicates the direction of the steepest ascent of the function at that point and the magnitude of this vector indicates how steep the ascent is.

Vector calculus is essential for understanding multivariable functions and is widely used in many fields, including engineering, physics, and computer graphics. Problems such as the one in the exercise can help students grasp how scalar fields and vector fields are closely related through operations like gradient, divergence, and curl.

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Most popular questions from this chapter

Find the work required to move an object in the following force fields along a line segment between the given points. Check to see whether the force is conservative. $$\mathbf{F}=e^{x+y}\langle 1,1, z\rangle \text { from } A(0,0,0) \text { to } B(-1,2,-4)$$

Consider the sphere \(x^{2}+y^{2}+z^{2}=4\) and the cylinder \((x-1)^{2}+y^{2}=1,\) for \(z \geq 0\) a. Find the surface area of the cylinder inside the sphere. b. Find the surface area of the sphere inside the cylinder.

The rotation of a three-dimensional velocity field \(\mathbf{V}=\langle u, v, w\rangle\) is measured by the vorticity \(\omega=\nabla \times \mathbf{V} .\) If \(\omega=\mathbf{0}\) at all points in the domain, the flow is irrotational. a. Which of the following velocity fields is irrotational: \(\mathbf{V}=\langle 2,-3 y, 5 z\rangle\) or \(\mathbf{V}=\langle y, x-z,-y\rangle ?\) b. Recall that for a two-dimensional source-free flow \(\mathbf{V}=(u, v, 0),\) a stream function \(\psi(x, y)\) may be defined such that \(u=\psi_{y}\) and \(v=-\psi_{x} .\) For such a two-dimensional flow, let \(\zeta=\mathbf{k} \cdot \nabla \times \mathbf{V}\) be the \(\mathbf{k}\) -component of the vorticity. Show that \(\nabla^{2} \psi=\nabla \cdot \nabla \psi=-\zeta\). c. Consider the stream function \(\psi(x, y)=\sin x \sin y\) on the square region \(R=\\{(x, y): 0 \leq x \leq \pi, 0 \leq y \leq \pi\\}\). Find the velocity components \(u\) and \(v\); then sketch the velocity field. d. For the stream function in part (c), find the vorticity function \(\zeta\) as defined in part (b). Plot several level curves of the vorticity function. Where on \(R\) is it a maximum? A minimum?

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. Use Stokes' Theorem and a surface integral to find the circulation on \(C\) of the vector field \(\mathbf{F}=\langle-y, x, 0\rangle\) as a function of \(\varphi .\) For what value of \(\varphi\) is the circulation a maximum?

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 \(\mathbb{R}^{3}\). $$\nabla \times(\varphi \mathbf{F})=(\nabla \varphi \times \mathbf{F})+(\varphi \nabla \times \mathbf{F}) \quad \text { (Product Rule) }$$

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