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Determine whether the vector field F(x, y, z) is free of sources and sinks. If it is not, locate them. $$ \mathbf{F}(x, y, z)=(y+z) \mathbf{i}-x z^{3} \mathbf{j}+\left(x^{2} \sin y\right) \mathbf{k} $$

Short Answer

Expert verified
The vector field is free of sources and sinks.

Step by step solution

01

Understanding the Problem

To determine if the vector field \( \mathbf{F}(x, y, z) = (y+z) \mathbf{i} - x z^{3} \mathbf{j} + x^{2} \sin y \mathbf{k} \) is free of sources and sinks, we need to calculate its divergence. A vector field without sources or sinks has a divergence of zero everywhere.
02

Calculating the Divergence

The divergence of a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \) is given by \( abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \). For \( \mathbf{F}(x, y, z) = (y+z) \mathbf{i} - x z^{3} \mathbf{j} + x^{2} \sin y \mathbf{k} \), identify \( P = y+z \), \( Q = -xz^3 \), and \( R = x^2 \sin y \).
03

Partial Derivative of P with respect to x

Calculate \( \frac{\partial P}{\partial x} \). Here, \( P = y+z \) does not depend on \( x \), so \( \frac{\partial P}{\partial x} = 0 \).
04

Partial Derivative of Q with respect to y

Calculate \( \frac{\partial Q}{\partial y} \). Since \( Q = -xz^3 \) does not depend on \( y \), \( \frac{\partial Q}{\partial y} = 0 \).
05

Partial Derivative of R with respect to z

Calculate \( \frac{\partial R}{\partial z} \). Here, \( R = x^2 \sin y \) does not have a \( z \) term, so \( \frac{\partial R}{\partial z} = 0 \).
06

Summing the Derivatives

Combine the partial derivatives: \( abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} = 0 + 0 + 0 = 0 \).
07

Conclusion

Since the divergence of \( \mathbf{F} \) is zero everywhere, the vector field \( \mathbf{F}(x, y, z) \) is free of sources and sinks.

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Key Concepts

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

Sources and Sinks in Vector Fields
In the context of a vector field, sources and sinks represent points where the field behaves like a beginning or an end of flow. Imagine water flowing in a pond. If water is constantly added from a point, then that point is a **source** within the field. Alternatively, if water is being drawn from a point, then it's considered a **sink**. These concepts are crucial in understanding fluid dynamics and electromagnetism.
In calculus terms, sources and sinks relate directly to how the field behaves at certain points. Specifically, they can be identified mathematically by checking the divergence of a vector field. If its divergence is zero at all points, no matter where you look, there are neither sources nor sinks present. This means the flow neither begins nor terminates anywhere within the field.
Understanding Divergence in a Vector Field
Divergence is a measure of how much a vector field spreads out from a given point. To visualize it, imagine a single point where vectors are either emanating outward or converging inward. Divergence helps to quantify this behavior.
Mathematically, if a vector field is given by \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \), its divergence is found using the equation:
  • \( abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \).
These partial derivatives measure how each vector component changes with respect to its respective coordinate direction. To check for sources or sinks, we determine if this sum equals zero throughout the field. If so, the field is said to be divergence-free, meaning it has balanced inflow and outflow.
Utilizing Partial Derivatives for Calculating Divergence
Partial derivatives allow us to examine how a function changes as one variable changes, keeping the other variables constant. For a vector field, partial derivatives are essential tools for calculating its divergence. This involves differentiating each component of the field with respect to its respective variable. Here’s how it's done:
  • Let's break down our vector field components: for \( P = y+z \), since it doesn't change with \( x \), \( \frac{\partial P}{\partial x} = 0 \).
  • For \( Q = -xz^3 \), as it remains constant with changes in \( y \), \( \frac{\partial Q}{\partial y} = 0 \).
  • Regarding \( R = x^2 \sin y \), because \( z \) has no role here, \( \frac{\partial R}{\partial z} = 0 \).
These partial derivatives together determine the divergence of the field, allowing us to see whether any sources or sinks exist.
The Role of Calculus in Vector Field Analysis
Calculus is fundamental in understanding and analyzing vector fields. It provides the mathematical framework for exploring how these fields change and interact. Central to this is the concept of differentiation, which helps us understand how various field components behave under changes in space.
With the aid of calculus, particularly through vector calculus operations like divergence, curl, and gradient, we can deduce critical properties of a vector field. For example, divergence—a result of calculating partial derivatives—is crucial in determining the presence of sources or sinks within a field.
Thus, calculus offers a powerful set of tools not only to calculate but to also visually interpret the behavior of complex systems represented by vector fields, ensuring an accurate and profound comprehension of fluid flows, electromagnetic fields, and other physical phenomena.

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Most popular questions from this chapter

Evaluate the integral \(\iint_{\sigma} f(x, y, z) d S\) over the surface \(\sigma\) represented by the vector-valued function \(\mathbf{r}(u, v) .\) $$ \begin{array}{l}{f(x, y, z)=\frac{1}{\sqrt{1+4 x^{2}+4 y^{2}}}} \\\ {\mathbf{r}(u, v)=u \cos v \mathbf{i}+u \sin v \mathbf{j}+u^{2} \mathbf{k}} \\\ {(0 \leq u \leq \sin v, 0 \leq v \leq \pi)}\end{array} $$

Evaluate the surface integral $$ \iint_{\sigma} f(x, y, z) d S $$ \(f(x, y, z)=x y ; \sigma\) is the portion of the plane \(x+y+z=1\) lying in the first octant.

Use Stokes' Theorem to evaluate \(\oint_{C} \mathbf{F} \cdot d \mathbf{r}\) $$ \begin{array}{l}{\mathbf{F}(x, y, z)=(z+\sin x) \mathbf{i}+\left(x+y^{2}\right) \mathbf{j}+\left(y+e^{z}\right) \mathbf{k} ; \quad C \text { is }} \\ {\text { the intersection of the sphere } x^{2}+y^{2}+z^{2}=1 \text { and the cone }} \\ {z=\sqrt{x^{2}+y^{2}} \text { with counterclockwise orientation looking }} \\ {\text { down the positive } z \text { -axis. }}\end{array} $$

Confirm that the force field \(\mathbf{F}\) is conservative in some open connected region containing the points \(P\) and \(Q,\) and then find the work done by the force field on a particle moving along an arbitrary smooth curve in the region from \(P\) to \(Q .\) $$ \mathbf{F}(x, y)=e^{-y} \cos x \mathbf{i}-e^{-y} \sin x \mathbf{j} ; \quad P(\pi / 2,1), Q(-\pi / 2,0) $$

Let \(\sigma\) be the cylindrical surface that is represented by the vector-valued function \(\mathbf{r}(u, v)=\cos v \mathbf{i}+\sin v \mathbf{j}+\) \(u \mathbf{k}\) with \(0 \leq u \leq 1\) and \(0 \leq v \leq 2 \pi\) (a) Find the unit normal \(\mathbf{n}=\mathbf{n}(u, v)\) that defines the positive orientation of \(\sigma .\) (b) Is the positive orientation inward or outward? Justify your answer.

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