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Use the Divergence Theorem to find all positive values of \(k\) such that $$ \mathbf{F}(\mathbf{r})=\frac{\mathbf{r}}{\|\mathbf{r}\|^{k}} $$ satisfies the condition \(\operatorname{div} \mathbf{F}=0\) when \(\mathbf{r} \neq \mathbf{0}\).

Short Answer

Expert verified
The positive value of \( k \) is 3.

Step by step solution

01

Calculate the Divergence of \( \mathbf{F} \)

The divergence of a vector field \( \mathbf{F}(x, y, z) = (F_1, F_2, F_3) \) is given by \( abla \cdot \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \). In this problem, \( \mathbf{F}(\mathbf{r}) = \frac{\mathbf{r}}{\|\mathbf{r}\|^k} = \left( \frac{x}{\|\mathbf{r}\|^k}, \frac{y}{\|\mathbf{r}\|^k}, \frac{z}{\|\mathbf{r}\|^k} \right) \). We denote \( \|\mathbf{r}\| = (x^2 + y^2 + z^2)^{1/2} \).
02

Apply the Product Rule for Divergence

For each component of \( \mathbf{F} \), the derivative can be computed using the product rule: \[ \frac{\partial}{\partial x} \left( \frac{x}{\|\mathbf{r}\|^k} \right) = \frac{1}{\|\mathbf{r}\|^k} - k \frac{x^2}{\|\mathbf{r}\|^{k+2}}. \] Similarly, for the other components, \( \frac{\partial}{\partial y} \left( \frac{y}{\|\mathbf{r}\|^k} \right) = \frac{1}{\|\mathbf{r}\|^k} - k \frac{y^2}{\|\mathbf{r}\|^{k+2}} \) and \( \frac{\partial}{\partial z} \left( \frac{z}{\|\mathbf{r}\|^k} \right) = \frac{1}{\|\mathbf{r}\|^k} - k \frac{z^2}{\|\mathbf{r}\|^{k+2}} \).
03

Sum the Partial Derivatives

Calculate \( abla \cdot \mathbf{F} \) by summing the partial derivatives: \[ abla \cdot \mathbf{F} = \left( \frac{1}{\|\mathbf{r}\|^k} - k \frac{x^2}{\|\mathbf{r}\|^{k+2}} \right) + \left( \frac{1}{\|\mathbf{r}\|^k} - k \frac{y^2}{\|\mathbf{r}\|^{k+2}} \right) + \left( \frac{1}{\|\mathbf{r}\|^k} - k \frac{z^2}{\|\mathbf{r}\|^{k+2}} \right). \] Simplify this expression: \[ abla \cdot \mathbf{F} = \frac{3}{\|\mathbf{r}\|^k} - k \frac{x^2 + y^2 + z^2}{\|\mathbf{r}\|^{k+2}}. \]
04

Factor and Simplify the Expression

Observe that \( x^2 + y^2 + z^2 = \|\mathbf{r}\|^2 \). Plug this back into the expression: \[ abla \cdot \mathbf{F} = \frac{3}{\|\mathbf{r}\|^k} - k \frac{\|\mathbf{r}\|^2}{\|\mathbf{r}\|^{k+2}} = \frac{3 - k}{\|\mathbf{r}\|^k}. \] The condition \( abla \cdot \mathbf{F} = 0 \) implies \( 3 - k = 0 \).
05

Solve for \( k \)

From \( 3 - k = 0 \), we find that \( k = 3 \). Since we are looking for positive values of \( k \), and \( k = 3 \) is positive, this is acceptable.

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Key Concepts

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

Vector Field Divergence
Understanding the divergence of a vector field is an essential concept in vector calculus. A vector field, like \( \mathbf{F}(x, y, z) = (F_1, F_2, F_3) \), assigns a vector to each point in space. Divergence measures how much the vector field is "spreading out" or "diverging" from a point. To find the divergence, we use the formula:
  • \( abla \cdot \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \)
This involves taking the partial derivatives of each component of the vector field with respect to its corresponding variable and summing them up. Divergence tells us about the behavior of a vector field in a specific region, helping us understand whether the field has sinks or sources. If the divergence is zero at all points, the field is said to be solenoidal, meaning there are no net sources or sinks present.
Partial Derivatives
Partial derivatives are used to analyze functions of multiple variables, like those in a vector field. In the context of our exercise, each component of the vector field \( \mathbf{F}(x, y, z) \) is a function that depends on three variables: \(x\), \(y\), and \(z\). A partial derivative of a function relative to one variable is found by treating all other variables as constants. This helps us understand how the function changes as just one variable changes, while the others are held constant. For example:
  • \( \frac{\partial}{\partial x} \left( \frac{x}{\|\mathbf{r}\|^k} \right) = \frac{1}{\|\mathbf{r}\|^k} - k \frac{x^2}{\|\mathbf{r}\|^{k+2}} \)
This formula results from applying the product rule, which is crucial when dealing with expressions containing both multiplication and division of variables. Partial derivatives provide valuable insights into the rate of change in a vector field, helping in solving complex mathematical problems in physics and engineering.
Vector Calculus
Vector calculus plays a vital role in understanding and analyzing fields in physics and engineering. It extends basic calculus into higher dimensions, considering functions that output vectors instead of only scalars. Concepts like divergence, curl, and gradient help comprehend how fields behave in space.In our exercise, vector calculus aids in determining conditions under which the vector field \( \mathbf{F}(x, y, z) = \frac{\mathbf{r}}{\|\mathbf{r}\|^k} \) has zero divergence. Part of vector calculus involves the use of the Divergence Theorem, which relates the flow (divergence) of a vector field through a closed surface to the behavior of the field inside the surface. This theorem can be stated as:
  • \( \int_V (abla \cdot \mathbf{F}) \, dV = \oint_S \mathbf{F} \cdot d\mathbf{S} \)
Where \(V\) is the volume enclosed by surface \(S\), \(dV\) is a volume element, and \(d\mathbf{S}\) is a surface element.By applying vector calculus concepts, one can solve problems that involve complex, multi-dimensional fields, understanding phenomena in three-dimensional space.

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Most popular questions from this chapter

Find the work done by the force field \(\mathbf{F}\) on a particle that moves along the curve \(C\). \(\mathbf{F}(x, y, z)=x y \mathbf{i}+y z \mathbf{j}+x z \mathbf{k}\) \(C: \mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+t^{3} \mathbf{k} \quad(0 \leq t \leq 1)\)

In 1831 the physicist Michael Faraday discovered that an electric current can be produced by varying the magnetic flux through a conducting loop. His experiments showed that the electromotive force \(\mathbf{E}\) is related to the magnetic induction \(\mathbf{B}\) by the equation $$ \oint_{C} \mathbf{E} \cdot d \mathbf{r}=-\iint_{\sigma} \frac{\partial \mathbf{B}}{\partial t} \cdot \mathbf{n} d S $$ Use this result to make a conjecture about the relationship between curl \(\mathbf{E}\) and \(\mathbf{B}\), and explain your reasoning.

In each part, find the work done by the three-dimensional inverse-square field $$ \mathbf{F}(\mathbf{r})=\frac{1}{\|\mathbf{r}\|^{3}} \mathbf{r} $$ on a particle that moves along the curve \(C\). (a) \(C\) is the line segment from \(P(1,1,2)\) to \(Q(3,2,1)\). (b) \(C\) is the curve $$ \mathbf{r}(t)=\left(2 t^{2}+1\right) \mathbf{i}+\left(t^{3}+1\right) \mathbf{j}+(2-\sqrt{t}) \mathbf{k} $$ where \(0 \leq t \leq 1\). (c) \(C\) is the circle in the \(x y\) -plane of radius 1 centered at \((2,0,0)\) traversed counterclockwise.

Evaluate the line integral along the curve \(C\).$$ \begin{aligned} &\int_{C} y z d x-x z d y+x y d z \\ &C: x=e^{t}, y=e^{3 t}, z=e^{-t} \quad(0 \leq t \leq 1) \end{aligned} $$

True-False Determine whether the statement is true or false. Explain your answer.If \(\mathbf{F}(x, y)=a y \mathbf{i}+b x \mathbf{j}\) is a conservative vector field, then \(a=b\)

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