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Chapter 6: Problem 242

Use \(r=|\mathbf{r}|\) and \(\mathbf{r}=(x, y, z)\). Let \(\mathbf{F}(x, y)=\frac{-y \mathbf{i}+x \mathbf{j}}{x^{2}+y^{2}}, \quad\) where \(\mathbf{F}\) is defined on \(\\{(x, y) \in \mathbb{R} \mid(x, y) \neq(0,0)\\} .\) Find \(\operatorname{curl} \mathbf{F}\)

Short Answer

Expert verified
The curl of \( \mathbf{F} \) is zero everywhere except at the origin.

Step by step solution

01

Understand the Components of the Vector Field

The given vector field is \( \mathbf{F}(x, y) = \frac{-y \mathbf{i} + x \mathbf{j}}{x^2 + y^2} \). This means the components of \( \mathbf{F} \) are \( M(x, y) = \frac{-y}{x^2 + y^2} \) and \( N(x, y) = \frac{x}{x^2 + y^2} \).
02

Recall the Curl in 2D

In two dimensions, the curl of a vector field \( \mathbf{F} = M \mathbf{i} + N \mathbf{j} \) is given by \( \operatorname{curl} \mathbf{F} = \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \). We need to calculate these partial derivatives for \( N \) and \( M \).
03

Compute \( \frac{\partial N}{\partial x} \)

Start by calculating \( \frac{\partial}{\partial x} \left( \frac{x}{x^2 + y^2} \right) \). Use the quotient rule: \( \frac{f}{g} \rightarrow \frac{g \cdot f' - f \cdot g'}{g^2} \), where \( f(x) = x \) and \( g(x) = x^2 + y^2 \). The derivative becomes:\[ \frac{(x^2 + y^2) \cdot 1 - x \cdot 2x}{(x^2 + y^2)^2} = \frac{(x^2 + y^2) - 2x^2}{(x^2 + y^2)^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2} \]
04

Compute \( \frac{\partial M}{\partial y} \)

Next, calculate \( \frac{\partial}{\partial y} \left( \frac{-y}{x^2 + y^2} \right) \). Again apply the quotient rule. Here, \( f(y) = -y \) and \( g(y) = x^2 + y^2 \), giving the derivative:\[ \frac{(x^2 + y^2)(-1) - (-y)\cdot 2y}{(x^2 + y^2)^2} = \frac{-(x^2 + y^2) + 2y^2}{(x^2 + y^2)^2} = \frac{2y^2 - x^2 - y^2}{(x^2 + y^2)^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2} \]
05

Calculate the Curl

Substitute the values from steps 3 and 4 into the formula for the curl: \( \operatorname{curl} \mathbf{F} = \frac{y^2 - x^2}{(x^2 + y^2)^2} - \frac{y^2 - x^2}{(x^2 + y^2)^2} \).This simplifies to zero, as both terms are identical and cancel each other out.

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Key Concepts

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

Vector Fields
Imagine a vector field as a map of invisible forces in a given space. These forces can influence objects within that space without physically touching them. In a 2D plane, a vector field assigns a vector to every point, which describes both the direction and strength of this force at that point.
  • For example, wind over a field can be represented as a vector field, where each vector shows the wind's speed and direction.
  • Mathematically, a vector field can be expressed in terms of coordinate functions; in two dimensions, it often looks like \(\mathbf{F}(x, y) = M(x, y)\mathbf{i} + N(x, y)\mathbf{j}\).
In our problem, the vector field \(\mathbf{F}(x, y)\) is given by \(\frac{-y \mathbf{i} + x \mathbf{j}}{x^2 + y^2}\), where the components \(M(x, y)\) and \(N(x, y)\) represent how the field behaves at each point \((x, y)\). Understanding these components helps us interpret the nature of the vector field, which is critical for analyzing its properties, like the curl.
Curl of a Vector Field
The curl of a vector field helps us understand how much the field rotates or "twists" around a point. This concept is crucial in physics when studying fields related to rotation, like magnetic fields.
  • In a 2D vector field, the curl is a scalar value that resembles the amount of rotation at each point.
  • The mathematical formula to find the curl in 2D is \( \text{curl} \mathbf{F} = \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\).
In the given exercise, to find the curl of \(\mathbf{F}\), we focus on computing the partial derivatives of its components. Subtracting these derivatives reveals whether and how much the vector field encapsulates circular motion. If the curl is zero, as in this problem, the field is irrotational, meaning it doesn鈥檛 exhibit any rotational behavior around any point.
Partial Derivatives
Partial derivatives are a cornerstone of vector calculus. When a function involves several variables, like \(x\) and \(y\) in our vector field \(\mathbf{F}\), a partial derivative measures how the function changes as one variable changes while others remain constant.
  • The partial derivative of \(N(x, y)\) with respect to \(x\) tells us the rate of change of \(N\) as \(x\) shifts.
  • Similarly, \(\frac{\partial M}{\partial y}\) explores how \(M(x, y)\) changes as \(y\) varies.
These partial derivatives are essential to calculating the curl of the field. They give us insight into how small changes impact the vector components locally, thus aiding in understanding the broader behavior of the vector field.
Quotient Rule
Handling derivatives of functions that are fractions is where the quotient rule comes into play. It's a technique used in differentiation when dealing with ratios, crucial for computing the derivatives in vector calculus.
  • The quotient rule is given by: if \(u(x)\) and \(v(x)\) are function components, then the derivative \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{v(x)^2}\).
  • In our exercise, it helps us differentiate the component functions \( \frac{x}{x^2 + y^2}\) and \( \frac{-y}{x^2 + y^2}\).
By applying the quotient rule, we accurately determine each partial derivative needed to find the curl. This rule simplifies the process by providing a structured method to tackle derivatives of complex fractions. Knowing when and how to use it is crucial for solving more advanced problems in vector calculus.

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Most popular questions from this chapter

For the following exercises, without using Stokes' theorem, calculate directly both the flux of \(\operatorname{curl} \mathbf{F} \cdot \mathbf{N}\) over the given surface and the circulation integral around its boundary, assuming all boundaries are oriented clockwise as viewed from above. \(\mathbf{F}(x, y, z)=z \mathbf{i}+2 x \mathbf{j}+3 y \mathbf{k} ; S\) is upper hemisphere \(z=\sqrt{9-x^{2}-y^{2}}\)

Let \(\quad \mathbf{F}=\left(x^{2}+y^{3}\right) \mathbf{i}+(2 x y) \mathbf{j} . \quad\) Calculate flux \(\mathbf{F}\) orientated counterclockwise across curve \(C: x^{2}+y^{2}=9\).

& \text { Evaluate } & \iint_{S} \mathbf{F} \cdot d \mathbf{S}\end{array}\( where \)\mathbf{F}(x, y, z)=b x y^{2} \mathbf{i}+b x^{2} y \mathbf{j}+\left(x^{2}+y^{2}\right) z^{2} \mathbf{k}\( and \)S\( is a closed surface bounding the region and consisting of solid cylinder \)x^{2}+y^{2} \leq a^{2}\( and \)0 \leq z \leq b$.

For the following exercises, use a CAS to evaluate the given line integrals. \([\mathrm{T}] \quad\) Evaluate \(\quad \mathbf{F}(x, y, z)=x^{2} z \mathbf{i}+6 y \mathbf{j}+y z^{2} \mathbf{k},\) where \(\quad C \quad\) is \(\quad\) represented \(\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+\ln t \mathbf{k}, 1 \leq t \leq 3\)

For the following exercises, without using Stokes' theorem, calculate directly both the flux of \(\operatorname{curl} \mathbf{F} \cdot \mathbf{N}\) over the given surface and the circulation integral around its boundary, assuming all boundaries are oriented clockwise as viewed from above. \(\mathbf{F}(x, y, z)=(x+2 z) \mathbf{i}+(y-x) \mathbf{j}+(z-y) \mathbf{k} ; S\) is a triangular region with vertices \((3,0,0),(0,3 / 2,0)\), and (0, 0,3)
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