Problem 36
Question
Let \(k\) be a constant, \(\mathbf{F}=\mathbf{F}(x, y, z), \mathbf{G}=\mathbf{G}(x, y, z),\) and \(\phi=\phi(x, y, z) .\) Prove the following identities, assuming that all derivatives involved exist and are continuous. $$ \operatorname{curl}(\phi \mathbf{F})=\phi \operatorname{curl} \mathbf{F}+\nabla \phi \times \mathbf{F} $$
Step-by-Step Solution
Verified Answer
The identity is proven: \(\operatorname{curl}(\phi \mathbf{F}) = \phi \operatorname{curl} \mathbf{F} + \nabla \phi \times \mathbf{F}\).
1Step 1: Understanding the Curl of a Vector Field
The curl of a vector field \(\mathbf{F} = (F_1, F_2, F_3)\) is given by the expression:\[ \operatorname{curl}(\mathbf{F}) = abla \times \mathbf{F} = \left( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}, \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right). \]This operation produces a new vector field from \(\mathbf{F}\).
2Step 2: Expanding the Expression \(\operatorname{curl}(\phi \mathbf{F})\)
We need to find the curl of \(\phi \mathbf{F}\). Using the product rule and the definition of the curl, the curl for this scalar multiplication gives:\[ \operatorname{curl}(\phi \mathbf{F}) = abla \times (\phi F_1, \phi F_2, \phi F_3). \]Applying the cross product of the gradient with this expression involves differentiating each term with respect to either \(x\), \(y\), or \(z\).
3Step 3: Applying the Product Rule for Derivatives
To expand \(\operatorname{curl}(\phi \mathbf{F})\), we use the product rule:\[ \operatorname{curl}(\phi \mathbf{F}) = \left( \frac{\partial (\phi F_3)}{\partial y} - \frac{\partial (\phi F_2)}{\partial z}, \frac{\partial (\phi F_1)}{\partial z} - \frac{\partial (\phi F_3)}{\partial x}, \frac{\partial (\phi F_2)}{\partial x} - \frac{\partial (\phi F_1)}{\partial y} \right). \]Using the product rule, we have:\[ \operatorname{curl}(\phi \mathbf{F}) = \phi abla \times \mathbf{F} + (abla \phi \times \mathbf{F}). \]
4Step 4: Verifying Each Component
Evaluate each component separately:\( \frac{\partial (\phi F_3)}{\partial y} = \phi \frac{\partial F_3}{\partial y} + \frac{\partial \phi}{\partial y}F_3 \) and \( \frac{\partial (\phi F_2)}{\partial z} = \phi \frac{\partial F_2}{\partial z} + \frac{\partial \phi}{\partial z}F_2 \). So the first component becomes \( \phi \left( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} \right) + F_3 abla \phi_y - F_2 abla \phi_z \), which is part of \(\phi \operatorname{curl} \mathbf{F} + abla \phi \times \mathbf{F}\).Repeat similar steps for the other two components, verifying that each part matches the components of the vector identity \(\phi \operatorname{curl} \mathbf{F} + abla \phi \times \mathbf{F}\).
5Step 5: Conclusion: Completing the Proof
After going through each component using the product rule and substituting back into the form of the curl, you arrive at:\[ \operatorname{curl}(\phi \mathbf{F}) = \phi \operatorname{curl} \mathbf{F} + abla \phi \times \mathbf{F}. \]This completes the proof, showing that the original identity holds true for the given conditions.
Key Concepts
Curl of a Vector FieldProduct Rule for DerivativesVector Field Identities
Curl of a Vector Field
To understand what the curl of a vector field is, imagine you have a vector field \( \mathbf{F} = (F_1, F_2, F_3) \) in a three-dimensional space. The curl operation, represented by \( abla \times \mathbf{F} \), takes this vector field and spins it around in a way. It's like examining how swirling or rotational the field is in space.
The formula for finding the curl of \( \mathbf{F} \) is given by:\[ \operatorname{curl}(\mathbf{F}) = abla \times \mathbf{F} = \left( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}, \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right)\]- The curl is a vector field that tells us how much the original field rotates around each point.- The greater the magnitude of the curl, the more intense the rotation or swirl.- A zero curl means the field is irrotational, like a smooth flow without eddies.Understanding the curl helps in visualizing fluid flow, electromagnetic fields, and more in vector calculus.
The formula for finding the curl of \( \mathbf{F} \) is given by:\[ \operatorname{curl}(\mathbf{F}) = abla \times \mathbf{F} = \left( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}, \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right)\]- The curl is a vector field that tells us how much the original field rotates around each point.- The greater the magnitude of the curl, the more intense the rotation or swirl.- A zero curl means the field is irrotational, like a smooth flow without eddies.Understanding the curl helps in visualizing fluid flow, electromagnetic fields, and more in vector calculus.
Product Rule for Derivatives
In calculus, the product rule is essential for finding derivatives when two functions are multiplied together. When extended to vector calculus, it's used to take derivatives of products involving vectors and scalars. Suppose you have a scalar function \( \phi \) and a vector field \( \mathbf{F} \). When you compute the curl of their product \( \phi \mathbf{F} \), the product rule aids in breaking it down.
This is illustrated by:\[ \operatorname{curl}(\phi \mathbf{F}) = \phi \operatorname{curl} \mathbf{F} + (abla \phi \times \mathbf{F})\]Here’s how the product rule works in this context:- Differentiate the scalar \( \phi \) separately from the vector \( \mathbf{F} \).- The first term \( \phi \operatorname{curl} \mathbf{F} \) keeps the original curl of \( \mathbf{F} \), weighted by \( \phi \).- The second term involves a cross-product between the gradient of \( \phi \) and \( \mathbf{F} \).The product rule is a powerful tool. It allows us to expand complex derivative expressions into simpler components.
This is illustrated by:\[ \operatorname{curl}(\phi \mathbf{F}) = \phi \operatorname{curl} \mathbf{F} + (abla \phi \times \mathbf{F})\]Here’s how the product rule works in this context:- Differentiate the scalar \( \phi \) separately from the vector \( \mathbf{F} \).- The first term \( \phi \operatorname{curl} \mathbf{F} \) keeps the original curl of \( \mathbf{F} \), weighted by \( \phi \).- The second term involves a cross-product between the gradient of \( \phi \) and \( \mathbf{F} \).The product rule is a powerful tool. It allows us to expand complex derivative expressions into simpler components.
Vector Field Identities
Vector identities are like shortcuts in vector calculus. They offer quick ways to compute various operations without redrawing all the derivations. In this article, we have explored one such identity: \[ \operatorname{curl}(\phi \mathbf{F}) = \phi \operatorname{curl} \mathbf{F} + abla \phi \times \mathbf{F}\]- This identity connects the operations of curl, gradient, and cross product.- It shows how a scalar field \( \phi \) interacting with a vector field \( \mathbf{F} \) transforms when considering curl.- Knowing this identity saves time by eliminating tedious computations whenever \( \phi \mathbf{F} \) appears.Vector field identities like these are fundamental in many areas such as electromagnetism, fluid dynamics, and engineering. They simplify calculations and enhance our conceptual understanding of how vector fields behave and interact.
Other exercises in this chapter
Problem 36
Evaluate the integral \(\iint_{\sigma} f(x, y, z) d S\) over the surface \(\sigma\) represented by the vector-valued function \(\mathbf{r}(u, v) .\) $$ \begin{a
View solution Problem 36
Determine whether the vector field F(x, y, z) is free of sources and sinks. If it is not, locate them. $$ \mathbf{F}(x, y, z)=\left(x^{3}-x\right) \mathbf{i}+\l
View solution Problem 36
Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) along the curve \(C\) $$ \begin{array}{l}{\mathbf{F}(x, y, z)=z \mathbf{i}+x \mathbf{j}+y \mathbf{k}} \\\ {C
View solution Problem 37
Evaluate the integral \(\iint_{\sigma} f(x, y, z) d S\) over the surface \(\sigma\) represented by the vector-valued function \(\mathbf{r}(u, v) .\) $$ \begin{a
View solution