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Find a potential function for \(\mathbf{F}\) $$\mathbf{F}=\left(e^{x} \ln y\right) \mathbf{i}+\left(\frac{e^{x}}{y}+\sin z\right) \mathbf{j}+(y \cos z) \mathbf{k}$$

Short Answer

Expert verified
The potential function is \( f(x, y, z) = e^x \ln y + y \sin z + C \).

Step by step solution

01

Understand the Problem

We need to find a potential function \( f(x, y, z) \) such that \( \mathbf{F} = abla f \), where \( \mathbf{F} = \left(e^{x} \ln y\right) \mathbf{i}+ \left(\frac{e^{x}}{y}+\sin z\right) \mathbf{j}+(y \cos z) \mathbf{k} \).
02

Check and Use Gradient Matching

The components of \( abla f \) are derivative of \( f \) with respect to \( x, y, z \). So, \( f_x = e^x \ln y \), \( f_y = \frac{e^x}{y} + \sin z \), \( f_z = y \cos z \).
03

Integrate with respect to \( x \)

Integrate \( f_x = e^x \ln y \) with respect to \( x \) to get \( f(x, y, z) = e^x \ln y + g(y, z) \) where \( g(y, z) \) is the function of integration constants dependent on \( y \) and \( z \).
04

Use \( f_y \) to solve for \( g(y, z) \)

Differentiate \( f = e^x \ln y + g(y, z) \) with respect to \( y \). Set equal to \( f_y = \frac{e^x}{y} + \sin z \). We get: \( \frac{e^x}{y} + g_y(y, z) = \frac{e^x}{y} + \sin z \). So, \( g_y = \sin z \).
05

Integrate \( g_y \) with Respect to \( y \)

Integrate \( g_y = \sin z \) with respect to \( y \) to get \( g(y, z) = y \sin z + h(z) \), where \( h(z) \) is a function of \( z \) alone.
06

Use \( f_z \) to solve for \( h(z) \)

Differentiate \( f = e^x \ln y + y \sin z + h(z) \) with respect to \( z \). Set equal to \( f_z = y \cos z \). We get: \( y \cos z + h'(z) = y \cos z \). Thus, \( h'(z) = 0 \), so \( h(z) = C \) where \( C \) is a constant.
07

Compile the Potential Function

Since \( h(z) = C \), the potential function is \( f(x, y, z) = e^x \ln y + y \sin z + C \). The function satisfies the conditions, hence it is the potential function for \( \mathbf{F} \).

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Key Concepts

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

Gradient Field
In vector calculus, a gradient field is a special type of vector field. It represents the gradient of a function. Imagine a gradient field as a map of slopes.
Each point in the field has a direction and magnitude, stemming from changes in a function's value over space. The gradient points in the direction of greatest increase.
To better comprehend this, think of a hill. In a physical analogy, the gradient field would show you the steepest path up the hill at every point.
  • The notation, often written as \( abla f \), represents the gradient of a function \( f \).
  • In a function with variables \( x, y, \) and thus a vector field \( \textbf{F} = abla f \) equates to finding intense changes at each point \( (x, y) \).
Understanding gradient fields helps solve problems involving potential functions and explains how forces like gravity operate directionally.
Potential Function
A potential function offers a deeper view of a vector field. It's a function whose gradient is equal to the given vector field.
Think of it as the source function that generates the field. Finding a potential function is about retracing the steps to uncover this source.
The exercise given involves finding a potential function for the vector field \( \mathbf{F} = \left(e^{x} \ln y\right) \mathbf{i}+\left(\frac{e^{x}}{y}+\sin z\right) \mathbf{j}+(y \cos z) \mathbf{k} \). When we solve this, we look for a potential function \( f(x, y, z) \) such that the gradient, \( abla f \), perfectly matches \( \mathbf{F} \).
  • This requires integrating the components of \( \mathbf{F} \) to find corresponding pieces of \( f \).
  • Piece by piece, the potential function builds up to form a solution which is \( f(x, y, z) = e^x \ln y + y \sin z + C \), where \( C \) is a constant.
In sum, finding a potential function transforms vector field analysis into simpler, often solvable equations.
Vector Field
A vector field is a function that assigns a vector to each point in space. It can be visualized as a collection of arrows, where each arrow has a direction and magnitude depending on the position in the space.
Vector fields are crucial for modeling a variety of physical phenomena such as fluid flow, electromagnetic fields, and gravitational forces. For example, in the exercise, the vector field \( \mathbf{F} \) is represented as a combination of components along the \( x, y, \) and \( z \) axes.
Key aspects include:
  • The field \( \mathbf{F} \) in our exercise combines distinct functions like \( e^x \ln y, \frac{e^x}{y}, \) and \( y \cos z \).
  • Each component captures different variations, with \( i, j, \) and \( k \) denoting the unit vectors along the respective axes.
  • Exploring these components individually gives insight into how the field behaves at any given point in space.
Understanding vector fields helps in navigating many applied science areas. They allow us to predict movements and interactions within a spatial domain.

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Most popular questions from this chapter

Suppose that the parametrized curve \(C:(f(u), g(u))\) is revolved about the \(x\) -axis, where \(g(u)>0\) for \(a \leq u \leq b\) a. Show that $$\mathbf{r}(u, v)=f(u) \mathbf{i}+(g(u) \cos v) \mathbf{j}+(g(u) \sin v) \mathbf{k}$$ is a parametrization of the resulting surface of revolution, where \(0 \leq v \leq 2 \pi\) is the angle from the \(x y\) -plane to the point \(\mathbf{r}(u, v)\) on the surface. (See the accompanying figure.) Notice that \(f(u)\) measures distance along the axis of revolution and \(g(u)\) measures distance from the axis of revolution. b. Find a parametrization for the surface obtained by revolving the curve \(x=y^{2}, y \geq 0,\) about the \(x\) -axis.

If a simple closed curve \(C\) in the plane and the region \(R\) it encloses satisfy the hypotheses of Green's Theorem, the area of \(R\) is given by $$\text { Area of } R=\frac{1}{2} \oint_{C} x d y-y d x$$ The reason is that by Equation (4), run backward, $$\begin{aligned} \text { Area of } R &=\iint_{R} d y d x=\iint_{R}\left(\frac{1}{2}+\frac{1}{2}\right) d y d x \\ &=\oint_{C} \frac{1}{2} x d y-\frac{1}{2} y d x \end{aligned}.$$ Use the Green's Theorem area formula given above to find the areas of the regions enclosed by the curves. The circle \(\mathbf{r}(t)=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}, \quad 0 \leq t \leq 2 \pi\)

Use a parametrization to express the area of the surface as a double integral. Then evaluate the integral. (There are many correct ways to set up the integrals, so your integrals may not be the same as those in the back of the book. They should have the same values, however.) The portion of the sphere \(x^{2}+y^{2}+z^{2}=4\) between the planes \(z=-1\) and \(z=\sqrt{3}\)

Find the outward flux of the field \(\mathbf{F}=x z \mathbf{i}+y z \mathbf{j}+\mathbf{k}\) across the surface of the upper cap cut from the solid sphere \(x^{2}+y^{2}+z^{2} \leq 25\) by the plane \(z=3\).

Apply Green's Theorem to evaluate the integrals. \(\oint_{C}\left(y^{2} d x+x^{2} d y\right)\) \(C:\) The triangle bounded by \(x=0, x+y=1, y=0\)

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