Problem 37
Question
If \(\mathbf{F}\) and \(\mathbf{G}\) are vector fields, prove that $$ \nabla \cdot(\mathbf{F} \times \mathbf{G})=\mathbf{G} \cdot(\nabla \times \mathbf{F})-\mathbf{F} \cdot(\nabla \times \mathbf{G}) $$
Step-by-Step Solution
Verified Answer
The given identity is proven by using the BAC-CAB rule, which yields \( \nabla \cdot(\mathbf{F} \times \mathbf{G}) = \mathbf{F} \cdot(\nabla \times \mathbf{G}) - \mathbf{G} \cdot(\nabla \times \mathbf{F})\). This can be rearranged to \( \nabla \cdot(\mathbf{F} \times \mathbf{G}) = \mathbf{G} \cdot(\nabla \times \mathbf{F}) - \mathbf{F} \cdot(\nabla \times \mathbf{G})\), thereby proving the given identity.
1Step 1: Write down known Vectors and Operators
We have two vector fields \(\mathbf{F}\) and \(\mathbf{G}\), and we have to work with the divergence operator \(\nabla \cdot\) and the curl operator \(\nabla \times\). We can generally express this as \( \nabla \cdot(\mathbf{F} \times \mathbf{G}) = \mathbf{G} \cdot(\nabla \times \mathbf{F}) - \mathbf{F} \cdot(\nabla \times \mathbf{G})\)
2Step 2: Applying the BAC-CAB rule
The BAC-CAB rule is a mnemonic for a result in vector calculus. The result is expressed in the following way: For three vectors \(\mathbf{A}\), \(\mathbf{B}\), and \(\mathbf{C}\), \(\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = \mathbf{B} \cdot (\mathbf{A} \times \mathbf{C}) - \mathbf{C} \cdot (\mathbf{A} \times \mathbf{B})\). In this problem, we can identify the vectors as \(\mathbf{A} = \nabla\), \(\mathbf{B} = \mathbf{F}\), and \(\mathbf{C} = \mathbf{G}\). So, we get that \( \nabla \cdot(\mathbf{F} \times \mathbf{G}) = \mathbf{F} \cdot(\nabla \times \mathbf{G}) - \mathbf{G} \cdot(\nabla \times \mathbf{F})\)
3Step 3: Rearranging
From step 2 we have the following equation \( \nabla \cdot(\mathbf{F} \times \mathbf{G}) = \mathbf{F} \cdot(\nabla \times \mathbf{G}) - \mathbf{G} \cdot(\nabla \times \mathbf{F})\). It can be rearranged to the following \( \nabla \cdot(\mathbf{F} \times \mathbf{G}) = \mathbf{G} \cdot(\nabla \times \mathbf{F}) - \mathbf{F} \cdot(\nabla \times \mathbf{G})\), proving the required identity.
Key Concepts
Divergence OperatorCurl OperatorBAC-CAB Rule
Divergence Operator
In vector calculus, the divergence operator is a vital concept often denoted as \( abla \cdot \). It is used to measure a vector field's tendency to originate from or converge at a point. In essence, divergence captures the "source" or "sink" magnitude of a vector field at a given point. It tells us how much a vector field like wind or fluid flow is expanding or compressing at a particular location.
The divergence of a vector field \( \mathbf{F} = (F_1, F_2, F_3) \) is calculated using the formula:
Divergence is particularly valuable in physics, especially in fluid dynamics and electromagnetism, where it's used to describe the behavior of fields. It's integral to understanding phenomena like
The divergence of a vector field \( \mathbf{F} = (F_1, F_2, F_3) \) is calculated using the formula:
- \( abla \cdot \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \)
Divergence is particularly valuable in physics, especially in fluid dynamics and electromagnetism, where it's used to describe the behavior of fields. It's integral to understanding phenomena like
- Expansion of gases
- Current flow in electrical circuits
- Behavior of magnetic and electric fields
Curl Operator
The curl operator is another fundamental tool in vector calculus, often symbolized by \( abla \times \). It helps us understand the rotational attributes of a vector field. Whereas divergence focuses on points spreading out or converging, curl analyzes how a vector field loops or swirls around a point.
The curl of a vector field \( \mathbf{F} = (F_1, F_2, F_3) \) is given as:
This concept is crucial in various physical fields, particularly where rotation and circulation are significant, such as:
The curl of a vector field \( \mathbf{F} = (F_1, F_2, F_3) \) is given as:
- \( abla \times \mathbf{F} = \left(\frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}, \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\right) \)
This concept is crucial in various physical fields, particularly where rotation and circulation are significant, such as:
- Fluid motion
- Electromagnetic fields
- Mechanics and dynamic systems
BAC-CAB Rule
The BAC-CAB rule is a handy mnemonic for remembering a specific identity in vector calculus concerning the cross product, often involving three vectors \( \mathbf{A} \), \( \mathbf{B} \), and \( \mathbf{C} \). This rule is particularly useful for simplifying expressions that involve multiple cross products by using known relationships.
Mathematically, this rule states:
Mathematically, this rule states:
- \( \mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = (\mathbf{A} \cdot \mathbf{C}) \mathbf{B} - (\mathbf{A} \cdot \mathbf{B}) \mathbf{C} \)
- Vector \( \mathbf{A} \) corresponds to \( abla \)
- Vector \( \mathbf{B} \) corresponds to \( \mathbf{F} \)
- Vector \( \mathbf{C} \) corresponds to \( \mathbf{G} \)
Other exercises in this chapter
Problem 37
Evaluate the flux integral \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d S\) \(\mathbf{F}=\langle x, y, z\rangle, S\) is the portion of \(z=4-x^{2}-y^{2}\) above th
View solution Problem 37
Find equations for the flow lines. $$\left\langle 2 y, 3 x^{2}\right\rangle$$
View solution Problem 37
Where is \(\mathbf{F}(x, y)=\left\langle\frac{2 x}{x^{2}+y^{2}}, \frac{2 y}{x^{2}+y^{2}}\right\rangle\) defined? Show that \(M_{y}=N_{x}\) everywhere the partia
View solution Problem 38
Evaluate the flux integral \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d S\) \(\mathbf{F}=\langle y,-x, 1\rangle, S\) is the portion of \(z=x^{2}+y^{2}\) below \(z=
View solution