/*! 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 13 Testing for conservative vector ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Testing for conservative vector fields Determine whether the following vector fields are conservative (in \(\mathbb{R}^{2}\) or \(\mathbb{R}^{3}\) ). $$\mathbf{F}=\left\langle e^{-x} \cos y, e^{-x} \sin y\right\rangle$$

Short Answer

Expert verified
Based on the computed curl, which is \(2e^{-x} \cos y\), determine if the given vector field is conservative or not.

Step by step solution

01

Identify the components of the vector field

The given vector field \(\mathbf{F}\) has components: $$P(x,y) = e^{-x} \cos y \quad \text{and} \quad Q(x,y) = e^{-x} \sin y$$
02

Compute the partial derivatives

Calculate the partial derivatives of \(P\) and \(Q\) with respect to \(x\) and \(y\): $$\frac{\partial P}{\partial x} = -e^{-x} \cos y\quad \text{and} \quad \frac{\partial Q}{\partial y} = e^{-x} \cos y$$
03

Calculate the curl

The curl of a 2D vector field \(\mathbf{F}\) is given by: $$\operatorname{curl} \mathbf{F} = \frac{\partial Q}{\partial y} - \frac{\partial P}{\partial x}$$ Substitute the computed partial derivatives into the curl formula: $$\operatorname{curl} \mathbf{F} = e^{-x} \cos y - (-e^{-x} \cos y) = 2e^{-x} \cos y$$
04

Check if the curl is zero

The curl of the vector field \(\mathbf{F}\) is found to be: $$\operatorname{curl} \mathbf{F} = 2e^{-x} \cos y$$ As the curl is not equal to zero, the given vector field is not conservative.

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.

Curl of a Vector Field
Understanding the curl of a vector field is fundamental in vector calculus, particularly when assessing the behavior of fluid flow or magnetic fields. Consider it as a measure of the 'rotation' or 'twisting' within the field. In two dimensions, for a vector field \textbf{F}, with components P and Q, the curl is simply the difference \(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\).Break down in easy terms: the partial derivatives tell us how much Q changes when moving along the x-axis and how much P changes as we travel with the y-axis. If these changes are not equal, it implies there's a 'twisting force' at that point.In the exercise example, the nonzero curl of \(\textbf{F}\) indicates it's not just pushing everything in one direction but also inducing a rotational effect.
Partial Derivatives
When you step into the world of multivariable calculus, you'll meet partial derivatives. They are akin to the regular derivatives you'd find in single-variable calculus, but with a twist: you'll calculate the rate of change of a function with respect to one variable, keeping the others constant.Let's simplify it: imagine you're exploring a hilly terrain. A partial derivative would be like looking only in the north direction to see how steep the climb is, without caring about the east-west slope.In our exercise, the partial derivatives of the function components determine if there is any 'rotation' in the vector field, as they are the building blocks in computing the curl.
Vector Calculus
If you're diving into vector calculus, think of it as a toolkit filled with mathematical tools for analyzing fields that have both magnitude and direction (yes, vectors!). It's an extension of calculus into multiple dimensions and it's widely used for studying fluid dynamics, electromagnetism, and more.Picture this: you're playing with arrows that tell not just how fast you should go, but also the direction to travel. Vector calculus gives you concepts like gradient, divergence, and curl to explore how these arrows behave in space.In our exercise, we used tools from vector calculus, specifically the idea of curl, to investigate if our vector field is conservative—whether it has no 'whirlpool-like' feature at any point.
Multivariable Calculus
Now, broaden your horizons to multivariable calculus. It's all about dealing with functions that have more than one input. If single-variable calculus is a one-way road, multivariable calculus is the entire highway network. It allows us to model and solve problems in higher dimensions that involve surfaces, volumes, and much more.Simplify and visualize: while standing at an intersection in a city, you can move north, south, east, or west. Multivariable calculus helps you map the city, predicting where you’ll end up if you take a series of turns.The exercise falls under multivariable calculus as we look at a function with two inputs, x and y, and try to understand its behavior through concepts like the vector field and its curl.

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

Green's Theorem as a Fundamental Theorem of Calculus Show that if the flux form of Green's Theorem is applied to the vector field \(\left\langle\frac{f(x)}{c}, 0\right\rangle,\) where \(c>0\) and \(R=\\{(x, y): a \leq x \leq b, 0 \leq y \leq c\\},\) then the result is the Fundamental Theorem of Calculus, $$ \int_{a}^{b} \frac{d f}{d x} d x=f(b)-f(a) $$

A scalar-valued function \(\varphi\) is harmonic on a region \(D\) if \(\nabla^{2} \varphi=\nabla \cdot \nabla \varphi=0\) at all points of \(D\). Show that the potential function \(\varphi(x, y, z)=|\mathbf{r}|^{-p}\) is harmonic provided \(p=0\) or \(p=1,\) where \(\mathbf{r}=\langle x, y, z\rangle .\) To what vector fields do these potentials correspond?

Maximum surface integral Let \(S\) be the paraboloid \(z=a\left(1-x^{2}-y^{2}\right),\) for \(z \geq 0,\) where \(a>0\) is a real number. Let \(\mathbf{F}=\langle x-y, y+z, z-x\rangle .\) For what value(s) of \(a\) (if any) does \(\iint_{S}(\nabla \times \mathbf{F}) \cdot \mathbf{n} d S\) have its maximum value?

Surfaces of revolution Suppose \(y=f(x)\) is a continuous and positive function on \([a, b] .\) Let \(S\) be the surface generated when the graph of \(f\) on \([a, b]\) is revolved about the \(x\) -axis. a. Show that \(S\) is described parametrically by \(\mathbf{r}(u, v)=\langle u, f(u) \cos v, f(u) \sin v\rangle,\) for \(a \leq u \leq b\) \(0 \leq v \leq 2 \pi\) b. Find an integral that gives the surface area of \(S\) c. Apply the result of part (b) to the surface generated with \(f(x)=x^{3},\) for \(1 \leq x \leq 2\) d. Apply the result of part (b) to the surface generated with \(f(x)=\left(25-x^{2}\right)^{1 / 2},\) for \(3 \leq x \leq 4\)

The Navier-Stokes equation is the fundamental equation of fluid dynamics that models the flow in everything from bathtubs to oceans. In one of its many forms (incompressible, viscous flow), the equation is $$\rho\left(\frac{\partial \mathbf{V}}{\partial t}+(\mathbf{V} \cdot \nabla) \mathbf{v}\right)=-\nabla p+\mu(\nabla \cdot \nabla) \mathbf{V}$$ In this notation, \(\mathbf{V}=\langle u, v, w\rangle\) is the three-dimensional velocity field, \(p\) is the (scalar) pressure, \(\rho\) is the constant density of the fluid, and \(\mu\) is the constant viscosity. Write out the three component equations of this vector equation. (See Exercise 40 for an interpretation of the operations.)

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.