/*! 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 35 Find the gradient field \(\mathb... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Short Answer

Expert verified
Question: Determine the gradient field for the potential function \(\varphi(x, y) = x^2y - y^2x\). Answer: The gradient field for the potential function \(\varphi(x, y) = x^2y - y^2x\) is given by \(\mathbf{F} = \begin{pmatrix} 2xy - y^2 \\ x^2 - 2xy\end{pmatrix}\).

Step by step solution

01

Find the partial derivative of \(\varphi\) with respect to \(x\)

To find the partial derivative of \(\varphi\) with respect to \(x\), treat \(y\) as a constant and differentiate the function with respect to \(x\): $$\frac{\partial \varphi}{\partial x} = \frac{\partial}{\partial x}(x^2y - y^2x)$$ Applying the power rule, we get: $$\frac{\partial \varphi}{\partial x} = 2xy - y^2$$
02

Find the partial derivative of \(\varphi\) with respect to \(y\)

To find the partial derivative of \(\varphi\) with respect to \(y\), treat \(x\) as a constant and differentiate the function with respect to \(y\): $$\frac{\partial \varphi}{\partial y} = \frac{\partial}{\partial y}(x^2y - y^2x)$$ Applying the power rule, we get: $$\frac{\partial \varphi}{\partial y} = x^2 - 2xy$$
03

Write the gradient field as a vector field

Now that we have found the partial derivatives of \(\varphi\) with respect to both \(x\) and \(y\), we can represent the gradient field as a vector field \(\mathbf{F} = \nabla \varphi\) with components \(\frac{\partial \varphi}{\partial x}\) and \(\frac{\partial \varphi}{\partial y}\): $$\mathbf{F} = \begin{pmatrix} \frac{\partial \varphi}{\partial x}\\ \frac{\partial \varphi}{\partial y}\end{pmatrix} = \begin{pmatrix} 2xy - y^2 \\ x^2 - 2xy\end{pmatrix}$$ The gradient field \(\mathbf{F}\) for the given potential function \(\varphi(x, y) = x^2y - y^2x\) is: $$\mathbf{F} = \begin{pmatrix} 2xy - y^2 \\ x^2 - 2xy\end{pmatrix}$$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Partial Derivative
Understanding partial derivatives is crucial when working with multivariable functions. When we compute a partial derivative, we differentiate a function of multiple variables with respect to one variable, treating all other variables as constants.
Let's look at the potential function \( \varphi(x, y) = x^2 y - y^2 x \). This function involves two variables, \(x\) and \(y\). To find the partial derivative with respect to \(x\), we differentiate \( \varphi \) treating \(y\) as a constant:
  • Using the power rule, the derivative of \(x^2\) with respect to \(x\) is \(2x\), and since \(y\) acts as a constant, we multiply by \(y\), giving \(2xy\).
  • The derivative of \(-y^2 x\) with respect to \(x\) results in \(-y^2\), as \(y^2\) is treated as a constant factor.
Applying the same method to differentiate with respect to \(y\), treating \(x\) as a constant, allows us to systematically break down complex functions.
Potential Function
Potential functions are essential concepts in vector calculus. They are scalar fields where the gradient can give rise to a vector field. The function \( \varphi(x, y) = x^2 y - y^2 x \) is an example of a potential function. When we discuss potential functions, we aim to understand how they relate to vector fields through gradients.
The operation of taking the gradient of a potential function, noted as \( abla \varphi \), transforms it into a vector field. This is done using partial derivatives, as we've shown: calculating \( \frac{\partial \varphi}{\partial x} \) and \( \frac{\partial \varphi}{\partial y} \).
  • The gradient of \( \varphi \) leads to a vector that points in the direction of the greatest increase of the function.
  • If the vector field derived from a potential function is conservative, it implies that the line integral of the field over a closed path is zero.
This property makes working with potential functions very valuable in physics and engineering, especially in gravitational and electrostatic fields.
Vector Field
A vector field is a function that assigns a vector to every point in a space. With gradient fields specifically, such as \( \mathbf{F} = \begin{pmatrix} 2xy - y^2 \ x^2 - 2xy \end{pmatrix} \), these vectors represent the derivatives of the potential function \( \varphi \). The notion of vector fields is prevalent in areas like fluid dynamics, electromagnetism, and more.
Here's how a vector field operates:
  • The first component \( 2xy - y^2 \) represents the rate of change of the potential function in the x-direction, derived from the partial \( \frac{\partial \varphi}{\partial x} \).
  • The second component \( x^2 - 2xy \) represents the rate of change in the y-direction, derived from the partial \( \frac{\partial \varphi}{\partial y} \).
Visualizing a vector field often involves considering the direction and magnitude of vectors at each point in the field. These aspects signify the local direction and rate of fastest increase at any given point, painting a dynamic picture of the space in which they exist.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Linear and quadratic vector fields a. For what values of \(a, b, c,\) and \(d\) is the field \(\mathbf{F}=\langle a x+b y, c x+d y\rangle\) conservative? b. For what values of \(a, b,\) and \(c\) is the field \(\mathbf{F}=\left\langle a x^{2}-b y^{2}, c x y\right\rangle\) conservative?

Fourier's Law of heat transfer (or heat conduction ) states that the heat flow vector \(\mathbf{F}\) at a point is proportional to the negative gradient of the temperature; that is, \(\mathbf{F}=-k \nabla T,\) which means that heat energy flows from hot regions to cold regions. The constant \(k>0\) is called the conductivity, which has metric units of \(J /(m-s-K)\) A temperature function for a region \(D\) is given. Find the net outward heat flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d S=-k \iint_{S} \nabla T \cdot \mathbf{n} d S\) across the boundary S of \(D\) In some cases, it may be easier to use the Divergence Theorem and evaluate a triple integral. Assume \(k=1 .\) $$\begin{aligned} &T(x, y, z)=100+x^{2}+y^{2}+z^{2}\\\ &D=\\{(x, y, z): 0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1\\} \end{aligned}$$

Special case of surface integrals of scalar-valued functions Suppose a surface \(S\) is defined as \(z=g(x, y)\) on a region \(R\) Show that \(\mathbf{t}_{x} \times \mathbf{t}_{y}=\left\langle-z_{x},-z_{y}, 1\right\rangle\) and that \(\iint_{S} f(x, y, z) d S=\iint_{R} f(x, y, g(x, y)) \sqrt{z_{x}^{2}+z_{y}^{2}+1} d A\)

A beautiful flux integral Consider the potential function \(\varphi(x, y, z)=G(\rho),\) where \(G\) is any twice differentiable function and \(\rho=\sqrt{x^{2}+y^{2}+z^{2}} ;\) therefore, \(G\) depends only on the distance from the origin. a. Show that the gradient vector field associated with \(\varphi\) is \(\mathbf{F}=\nabla \varphi=G^{\prime}(\rho) \frac{\mathbf{r}}{\rho},\) where \(\mathbf{r}=\langle x, y, z\rangle\) and \(\rho=|\mathbf{r}|\)b. Let \(S\) be the sphere of radius \(a\) centered at the origin and let \(D\) be the region enclosed by \(S\). Show that the flux of \(\mathbf{F}\) across \(S\) is \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d S=4 \pi a^{2} G^{\prime}(a)\) c. Show that \(\nabla \cdot \mathbf{F}=\nabla \cdot \nabla \varphi=\frac{2 G^{\prime}(\rho)}{\rho}+G^{\prime \prime}(\rho)\) d. Use part (c) to show that the flux across \(S\) (as given in part (b)) is also obtained by the volume integral \(\iiint_{D} \nabla \cdot \mathbf{F} d V\) (Hint: Use spherical coordinates and integrate by parts.)

Green's Second Identity Prove Green's Second Identity for scalar-valued functions \(u\) and \(v\) defined on a region \(D:\) $$\iiint_{D}\left(u \nabla^{2} v-v \nabla^{2} u\right) d V=\iint_{S}(u \nabla v-v \nabla u) \cdot \mathbf{n} d S$$ (Hint: Reverse the roles of \(u\) and \(v\) in Green's First Identity.)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.