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Under "reasonable conditions," if curl \(\vec{F}=\overrightarrow{0},\) what can we conclude about the vector field \(\vec{F} ?\)

Short Answer

Expert verified
The vector field \( \vec{F} \) is conservative and can be expressed as the gradient of a scalar potential function.

Step by step solution

01

Understanding the Problem

The problem asks us to determine what we can conclude about a vector field \( \vec{F} \) when its curl is zero, i.e., \( \text{curl } \vec{F} = \overrightarrow{0} \). Essentially, we need to recognize the implications of a curl being zero in the context of vector calculus.
02

Recall the Definition of Curl

Curl is a vector operation that describes the infinitesimal rotation of a 3-dimensional vector field. Mathematically, for a vector field \( \vec{F} = (F_1, F_2, F_3) \), curl is given by:\[ \text{curl } \vec{F} = abla \times \vec{F} = \left( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}, \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right) \] When \( \text{curl } \vec{F} = \overrightarrow{0} \), it means each component of the curl is zero.
03

Implication of Zero Curl

A vector field with zero curl is called irrotational. Under 'reasonable conditions' generally refers to the vector field being defined on a simply-connected domain. In such cases, the Poincaré lemma implies that the vector field \( \vec{F} \) can be expressed as the gradient of some scalar potential function \( \phi \), i.e., \( \vec{F} = abla \phi \).
04

Conclusion

Thus, if \( \text{curl } \vec{F} = \overrightarrow{0} \), the vector field \( \vec{F} \) is irrotational and can be represented as the gradient of a scalar potential function \( \phi \) in a simply-connected domain, making \( \vec{F} \) a conservative vector field.

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Key Concepts

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

Curl
In vector calculus, the concept of curl involves measuring the rotation at a point in a 3D vector field. Imagine putting a tiny paddle wheel inside the vector field—if the wheel imparts a rotational movement, that's where the curl comes into play. Mathematically, the curl of a vector field \( \vec{F} \) is defined by the operation \( abla \times \vec{F} \), resulting in another vector field. This operation measures how the field "circulates" or "rotates" around a particular point. It effectively captures the infinitesimal rotation at each point within the vector field.
  1. Representation: In a vector field \( \vec{F} = (F_1, F_2, F_3) \), the curl is calculated using the formula:\[ abla \times \vec{F} = \left( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}, \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right) \]
  2. Zero Curl: When the curl of \( \vec{F} \) is zero, i.e., \( abla \times \vec{F} = \overrightarrow{0} \), it signifies that there is no rotation at that point—in other words, the vector field is irrotational at that location.
Irrotational Vector Field
An irrotational vector field is a very special type of vector field where the curl is zero everywhere. This implies that if you place a tiny paddle wheel anywhere in the field, it will not rotate. Simple as that!These fields have no "vortex" or rotational movement within the field, which is a sign of striking things happening.
  • Characterized by zero curl: An irrotational vector field satisfies \( abla \times \vec{F} = \overrightarrow{0} \).
  • Smoothness: Typically, these fields are smooth and undisturbed over the area they cover, presenting an even flow that lacks any swirling action.
  • Potential: In geographic terms, irrotational fields often correspond to areas where energy or some other quantities, can be moved with minimal resistance.
If you encounter an irrotational field on test problems, it's already quite solved in terms of rotations—it means smooth sailing when calculating the curl!
Conservative Vector Field
In the world of vector fields, conservative fields are the kind that conserve energy in physical terms. This means that moving along a path in a conservative field will store or release energy with no net loss or gain.

Properties of Conservative Fields

  • Irrotational by Nature: If a vector field is irrotational in a simply-connected domain, it's conservative too! Remember, this includes having a zero curl \( abla \times \vec{F} = \overrightarrow{0} \).
  • Gradient of a Scalar Potential: A key upshot for conservative fields is that they are gradients of a scalar potential function \( \phi \). This means \( \vec{F} = abla \phi \), allowing us to move from the complex world of vectors to the simpler world of scalars.
  • Path Independence: Moving between two points in this field doesn't depend on the path taken. So if you go from point A to B using any route, the net work done is the same.
This field is useful in electrostatics and gravity where forces play out conservatively. If you're asked to find whether a vector field is conservative, always check for zero curl and if the domain is simply-connected. That's your big clue!

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Most popular questions from this chapter

In Exercises \(17-20\), a conservative vector field \(\vec{F}\) and a curve C are given. 1\. Find a potential function \(f\) for \(\vec{F}\) 2\. Compute curl \(\vec{F}\). 3\. Evaluate \(\int_{C} \vec{F} \cdot d \vec{r}\) directly, i.e., using Key Idea 14.3.1. 4\. Evaluate \(\int_{C} \vec{F} \cdot d \vec{r}\) using the Fundamental Theorem of Line Integrals. $$ \vec{F}=\langle y+1, x\rangle, C \text { is the line segment from }(0,1) \text { to }(1,0) . $$

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. $$ \begin{aligned} &\mathcal{S} \text { is the surface formed by the intersections of the planes }\\\ &z=\frac{1}{2}(3-x), x=1, y=0, y=2 \text { and } z=0 ; \vec{F}=\left\langle x, y^{2}, z\right\rangle \end{aligned} $$

If \(\mathcal{S}\) is a plane, and \(\vec{F}\) is always parallel to \(\mathcal{S},\) then the flux of \(\vec{F}\) across \(\mathcal{S}\) will be ________.

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 disk in the \(y-z\) plane with radius 1 , centered at (0,1,1) (choose \(\vec{n}\) such that it has a positive \(x\) -component); \(\overrightarrow{\vec{F}}=\langle y, z, x\rangle\)

In Exercises \(7-12,\) a vector field \(\vec{F}\) and a curve \(C\) are given. Evaluate \(\int_{c} \vec{F} \cdot d \vec{r} .\) \(\vec{F}=\langle y, x\rangle ; C\) is the top half of the unit circle, beginning at (1,0) and ending at (-1,0) .

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