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Chapter 9: Problem 6

Determine whether the following vector fields are gradient fields. If so, find \(\phi\) such that \(\mathbf{F}=\nabla \phi\) : (a) \((x, y)\). (b) \(\left(y^{2}, x^{2}\right)\). (c) \((\sin x y, \cos x y)\). (d) \((x+y, x-y)\). (e) \(\left(y e^{x y}, x e^{y}\right)\). (f) \(\left(x^{2} y, y^{2} x\right)\).

Short Answer

Expert verified
Only the vector field \( (x+y, x-y) \) is a gradient field, with potential function \( \phi = \frac{1}{2} x^{2} + xy - y \).

Step by step solution

01

Title - Define a Gradient Field

A vector field \(abla \phi = \mathbf{F}\) is called a gradient field if there exists a scalar function \(abla \phi\) such that the vector field \(abla \phi \) equals the gradient of \(abla \phi\). This means that \(abla \phi = \left(\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}\right)\).
02

Title - Check if \(abla \phi = \mathbf{F}\) satisfies the Curl Condition

For a vector field \(abla \phi\) to be a gradient field, the curl of \(abla \phi\) must be zero. That is, \(abla \times \mathbf{F} = 0\). For \(2D\) vector fields \(abla \phi = (M, N)\), this is equivalent to checking if \(\frac{\partial N}{\partial x} = \frac{\partial M}{\partial y}\).
03

Title - Apply the Curl Condition to Each Vector Field

For each vector field, check the curl condition: \(\frac{\partial N}{\partial x} = \frac{\partial M}{\partial y}\).(a) \(M=x\) and \(N=y\): \(\frac{\partial y}{\partial x} = 0 \eq 1 = \frac{\partial x}{\partial y}\), so not a gradient field.(b) \(M=y^{2}\) and \(N=x^{2}\): \(\frac{\partial x^{2}}{\partial x} = 2x \eq 2y = \frac{\partial y^{2}}{\partial y}\), so not a gradient field.(c) \(M=\sin(xy)\) and \(N=\cos(xy)\): \(\frac{\partial \cos(xy)}{\partial x} = y\sin(xy) \eq -y\cos(xy) = \frac{\partial \sin(xy)}{\partial y}\), so not a gradient field.(d) \(M=x+y\) and \(N=x-y\): \(\frac{\partial (x-y)}{\partial x} = 1 \eq 1 = \frac{\partial (x+y)}{\partial y}\), so a gradient field.(e) \(M=y\e^{xy}\) and \(N=xe^{y}\): \(\frac{\partial x\e^{y}}{\partial x} = \e^{y} \eq y^{2}\e^{xy} = \frac{\partial y\e^{xy}}{\partial y}\), so not a gradient field.(f) \(M=x^{2}y\) and \(N=y^{2}x\): \(\frac{\partial y^{2}x}{\partial x} = y^{2} \eq 2xy = \frac{\partial x^{2}y}{\partial y}\), so not a gradient field.
04

Title - Find \(\phi\) for Gradient Field (\(x + y, x - y\))

To find the scalar function \(abla \phi\) for \(M = x + y\) and \(N = x - y\), we integrate each component:For \(M = x + y\), integrate with respect to \(x\): \(abla \phi = \int (x + y) \ dx = \frac{1}{2}x^{2} + xy + C(y)\).For \(N = x - y\), differentiate the found function with respect to \(y\) and equate to \(N\): \(\frac{\partial}{\partial y}\left(\frac{1}{2} x^{2} + xy + C(y)\right) = x - y\), which gives: \(x + C'(y) = x - 1 \Rightarrow C'(y) = -1 \Rightarrow C(y) = -y\).Thus, the potential function is: \(\phi = \frac{1}{2} x^{2} + xy - y\).

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Key Concepts

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

Vector Fields
In vector calculus, a vector field assigns a vector to every point in a subset of space.
Think of it as a collection of vectors that are functions of position.
For instance, in two dimensions, a vector field can be represented as \(\textbf{F}(x, y) = (M(x, y), N(x, y))\), where \(M\) and \(N\) are functions of \(x\) and \(y\).
These fields are visualized as arrows pointing in various directions over a region.
Common examples include the velocity fields of fluids and the gravitational fields around masses.
Gradient Field
A gradient field is a special type of vector field.
If there exists a scalar function \(\phi\) such that its gradient matches the vector field, then we have a gradient field.
Mathematically, this is written as \(\textbf{F} = \abla \phi\).
This means \(\textbf{F} = \left( \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y} \right)\).
Thus, we can reconstruct the vector field if we know the scalar potential \(\phi\).
For example, in the exercise, only part (d) \((x+y, x-y)\) was identified as a gradient field.
Curl Condition
The curl condition is crucial for determining if a vector field is a gradient field.
The curl of a vector field \(\textbf{F} = (M, N)\) in two dimensions is given by \(\partial N/\partial x - \partial M/\partial y\).
For the vector field to be a gradient field, this must be zero: \(\partial N/\partial x = \partial M/\partial y\).
In simple terms, the vector field should be irrotational.
This implies no tiny loops or rotations in the field.
For example, in the exercise, \((x+y, x-y)\) satisfied \(\partial N/\partial x = \partial M/\partial y\).
Scalar Function
A scalar function assigns a single value to every point in space.
In relation to gradient fields, it is the function from which the vector field can be derived through differentiation.
For example, \(\phi(x, y)\) could be a temperature distribution across a surface.
The gradient of \(\phi\) gives us a vector field representing the heat flow.
It's often denoted as \(\bigtriangledown \phi\).
Finding this scalar function from a known gradient vector field involves integration.
Potential Function
The potential function is another term for the scalar function \(\phi\) in the context of gradient fields.
It potential because it 'potentially' generates the vector field via its gradient.
For example, in exercise part (d), \((x+y, x-y)\) was identified as having a potential function \(\phi\).
By integrating \(x + y\) with respect to \(x\), we get \frac{1}{2} x^2 + xy + C(y)\.
Matching to \N = x - y\, we find \C(y) = -y\.
Thus, the potential function is \frac{1}{2} x^2 + xy - y\.

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Most popular questions from this chapter

For the following functions, sketch the surface corresponding to \(z=f(x, y)\) and the level curves in the \(x y\) plane: (a) \(f(x, y)=x^{2}+y^{2}\). (b) \(f(x, y)=-2 x^{2}-2 y^{2}\). (c) \(f(x, y)=\exp \frac{-\left(x^{2}+y^{2}\right)}{2}\) (d) \(f(x, y)=2 x+y\). (e) \(f(x, y)=x y .\) (f) \(f(x, y)=\sin x \cos y\).

Sketch the level curves described by the following equations. Give an equation for a surface that has these level curves. Sketch the vector field corresponding to \(\nabla f\) by using its property of orthogonality to level curves: (a) \(c=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\). (b) \(c=x^{2}+y\). (c) \(c=x-y\). (d) \(c=\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\). (e) \(c=\left(x^{2}+y^{2}\right)\).

Determine the boundary and initial conditions appropriate for the diffusion of salt in each of the following situations. (a) A hollow tube initially containing pure water connects two reservoirs whose salt concentrations are \(C_{1}\) and 0 respectively. (b) The tube is sealed at one end. Its other end is placed in a salt solution of fixed concentration \(C_{1}\). (c) The tube is sealed at both ends and initially has its greatest salt concentrations halfway along its length. Assume the initial distribution is a trigonometric function.

Solve the one-dimensional diffusion equation subject to the following conditions: (a) \(c(0, t)=c(1, t)=0\), $$ c(x, 0)=\sin \pi x . $$ (b) \(c(0, t)=c(L, t)=0\), $$ c(x, 0)=\sin \left(\frac{2 \pi}{L} x\right)+\sin \left(\frac{3 \pi}{L} x\right) $$

For the following vector fields, find \(\nabla \times \mathbf{F}, \nabla \cdot \mathbf{F}\) : (a) \((x, y, z)\). (b) \((y-z, z-x, x-y)\). (c) \(\left(x^{2}+2 y+z, y^{2}-x, z^{2}-y^{2}\right)\). (d) \(\left(\sin x y z, \cos x y z, e^{x q}\right)\). (e) \(\left(\frac{1}{x}, \frac{1}{y}, \frac{1}{z}\right)\). (f) \((x+y, y+z, z+x)\).
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