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Find the divergence and curl of the given vector field. $$ \vec{F}=\left\langle x, y^{2}\right\rangle $$

Short Answer

Expert verified
Divergence: \(1 + 2y\); Curl: \(0\).

Step by step solution

01

Understand Divergence Formula

The divergence of a vector field \( \vec{F} = \langle P, Q \rangle \) in two dimensions is given by \( abla \cdot \vec{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \). Here, \( P = x \) and \( Q = y^2 \).
02

Calculate Partial Derivatives for Divergence

Calculate \( \frac{\partial P}{\partial x} \) and \( \frac{\partial Q}{\partial y} \). For \( P = x \), \( \frac{\partial x}{\partial x} = 1 \). For \( Q = y^2 \), \( \frac{\partial y^2}{\partial y} = 2y \).
03

Find the Divergence

Substitute the partial derivatives into the divergence formula: \( abla \cdot \vec{F} = 1 + 2y \). Thus, the divergence of \( \vec{F} \) is \( 1 + 2y \).
04

Understand Curl Formula

The curl of a vector field \( \vec{F} = \langle P, Q \rangle \) in two dimensions is expressed as \( abla \times \vec{F} = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \). Again, \( P = x \) and \( Q = y^2 \).
05

Calculate Partial Derivatives for Curl

Calculate \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \). For \( Q = y^2 \), \( \frac{\partial y^2}{\partial x} = 0 \). For \( P = x \), \( \frac{\partial x}{\partial y} = 0 \).
06

Find the Curl

Substitute the partial derivatives into the curl formula: \( abla \times \vec{F} = 0 - 0 = 0 \). Thus, the curl of \( \vec{F} \) is \( 0 \).

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Key Concepts

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

Divergence
Divergence measures how much a vector field spreads out from a point.
The concept is akin to how air flows out from a fan. In mathematics, for a two-dimensional vector field, like our example field \( \vec{F} = \langle x, y^2 \rangle \), the divergence is computed using the formula:
\[ abla \cdot \vec{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \] where \( P = x \) and \( Q = y^2 \). The divergence provides a scalar field that tells us about the volume density of the "outflow" at each point.
When computing:
  • \( \frac{\partial x}{\partial x} = 1 \)
  • \( \frac{\partial y^2}{\partial y} = 2y \)

Substituting these values gives us a divergence of \( 1 + 2y \). This indicates that at each point, the field diverges or spreads at a rate that depends on the \( y \)-coordinate.
This result can help understand how systems modeled by \( \vec{F} \) expand over an area.
Curl
Curl is a measure of the rotation of a vector field around a point.
Imagine the swirling of a whirlpool when visualizing curl. In two dimensions, curl is simpler and often just a scalar, calculated using the formula:
\[ abla \times \vec{F} = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \] In our case \( P = x \) and \( Q = y^2 \). The intuition behind curl is to find out how much "twisting" the vector field exhibits at a given point.
For our vector field \( \vec{F} \):
  • \( \frac{\partial y^2}{\partial x} = 0 \)
  • \( \frac{\partial x}{\partial y} = 0 \)

Inserting these into the formula gives a curl of \( 0 \), indicating there's no rotational behavior in this field.
This is why understanding curl is crucial while studying fluid dynamics or electromagnetic fields where rotation is significant.
Partial Derivatives
Partial derivatives help us understand how a function changes as we vary one variable, keeping others constant.
Imagine you're peeling an orange slice by slice, examining one section at a time. In vector fields, like\( \vec{F} = \langle x, y^2 \rangle \), partial derivatives are essential for calculating both divergence and curl.
They are denoted as \( \frac{\partial}{\partial x} \) or \( \frac{\partial}{\partial y} \), depending on the variable you change:
  • \( \frac{\partial P}{\partial x} \) in \( P = x \) results in \( 1 \), indicating how \( P \) changes as \( x \) changes.
  • \( \frac{\partial Q}{\partial y} \) in \( Q = y^2 \) gives \( 2y \), showing the change in \( Q \) with \( y \).
Partial derivatives allow us to focus on and analyze the behavior of a multi-variable function one dimension at a time.
Understanding these derivatives is foundational for exploring deeper concepts, like finding maxima or minima, or understanding growth rates in economics.

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Most popular questions from this chapter

Compute the flux of \(\vec{F}\) across \(\mathcal{S}\). (If \(\mathcal{S}\) is not a closed surface, choose \(\vec{n}\) so that it has a positive z-component, unless otherwise indicated.) \(\mathcal{S}\) is the square in space with corners at \((0,0,0),(1,0,0),\) (1,0,1) and (0,0,1) (choose \(\vec{n}\) such that it has a positive \(y\) -component \() ; \vec{F}=\langle 0,-z, y\rangle .\)

Exercises \(21-24\) are designed to challenge your understanding and require no computation. Let \(\mathcal{S}\) be any closed surface enclosing a domain \(D\). Consider \(\vec{F}_{1}=\langle x, 0,0\rangle\) and \(\vec{F}_{2}=\left\langle y, y^{2}, z-2 y z\right\rangle\) These fields are clearly very different. Why is it that the total outward flux of each field across \(\mathcal{S}\) is the same?

In Exercises \(13-16,\) a closed surface \(S\) and a vector field \(\vec{F}\) are given. Find the outward flux of \(\vec{F}\) over \(S\) either through direct computation or through the Divergence Theorem. $$ \begin{aligned} &\mathcal{S} \text { is the surface formed by the intersections of the planes }\\\ &z=\frac{1}{2}(3-x), x=1, y=0, y=2 \text { and } z=0 ; \vec{F}=\left\langle x, y^{2}, z\right\rangle \end{aligned} $$

A vector field \(\vec{F}\) and a curve \(C\) are given. Evaluate \(\int_{C} \vec{F} \cdot \vec{n} d s,\) the flux of \(\vec{F}\) over \(C\). \(\vec{F}=\langle y, 0\rangle ; C\) is the line segment from (0,0) to \((1,1) .\)

In Exercises \(17-20\), a closed curve \(C\) that is the boundary of a surface \(S\) is given along with a vector field \(\vec{F}\). Find the circulation of \(\vec{F}\) around \(C\) either through direct computation or through Stokes' Theorem. \(C\) is the curve whose \(x\) - and \(y\) -values are given by \(\vec{r}(t)=\) \(\langle 2 \cos t, 2 \sin t\rangle\) and the \(z\) -values are determined by the function \(z=x^{2}+y^{3}-3 y+1 ; \vec{F}=\langle-y, x, z\rangle .\)

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