Problem 35 - Calculus Early Transcendentals

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. $$ \operatorname{div}(\phi \mathbf{F})=\phi \operatorname{div} \mathbf{F}+\nabla \phi \cdot \mathbf{F} $$

Step-by-Step Solution

Answer

The identity is proved using the product rule for divergence: \(\operatorname{div}(\phi \mathbf{F}) = \phi \operatorname{div} \mathbf{F} + \nabla \phi \cdot \mathbf{F}\).

Step 1: Recall the Definition of Divergence

The divergence of a vector field \( \mathbf{F} = (F_1, F_2, F_3) \) is given by \( \operatorname{div} \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \). This is a scalar field derived from the vector field's rate of change.

Step 2: Apply the Product Rule for Divergence

Consider the divergence of the vector field \( \phi \mathbf{F} = (\phi F_1, \phi F_2, \phi F_3) \). The divergence is given by:\[\operatorname{div}(\phi \mathbf{F}) = \frac{\partial}{\partial x}(\phi F_1) + \frac{\partial}{\partial y}(\phi F_2) + \frac{\partial}{\partial z}(\phi F_3)\].Use the product rule to expand each term.

Step 3: Use Product Rule for Each Component

Apply the product rule to each component:\[ \frac{\partial}{\partial x}(\phi F_1) = \frac{\partial \phi}{\partial x}F_1 + \phi \frac{\partial F_1}{\partial x} \],\[ \frac{\partial}{\partial y}(\phi F_2) = \frac{\partial \phi}{\partial y}F_2 + \phi \frac{\partial F_2}{\partial y} \],\[ \frac{\partial}{\partial z}(\phi F_3) = \frac{\partial \phi}{\partial z}F_3 + \phi \frac{\partial F_3}{\partial z} \].Add these results together.

Step 4: Combine the Expanded Terms

Combine the expanded terms to form:\[\operatorname{div}(\phi \mathbf{F}) = \left( \phi \frac{\partial F_1}{\partial x} + \frac{\partial \phi}{\partial x}F_1 \right) + \left( \phi \frac{\partial F_2}{\partial y} + \frac{\partial \phi}{\partial y}F_2 \right) + \left( \phi \frac{\partial F_3}{\partial z} + \frac{\partial \phi}{\partial z}F_3 \right)\]\[= \phi \left(\frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \right) + \left(F_1 \frac{\partial \phi}{\partial x} + F_2 \frac{\partial \phi}{\partial y} + F_3 \frac{\partial \phi}{\partial z}\right).\]

Step 5: Interpret the Result

Recognize the resulting terms: \( \phi \operatorname{div} \mathbf{F} \) is the divergence of \( \mathbf{F} \) scaled by \( \phi \), and the expression \( abla \phi \cdot \mathbf{F} \) is the dot product of the gradient of \( \phi \) with \( \mathbf{F} \).Thus, the identity can be confirmed as:\[\operatorname{div}(\phi \mathbf{F}) = \phi \operatorname{div} \mathbf{F} + abla \phi \cdot \mathbf{F}.\]

Key Concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields and operations on them. It is essential in physics and engineering to describe physical phenomena like electromagnetic fields and fluid motion. In vector calculus:

• Vectors represent quantities with both magnitude and direction.
• Operations such as divergence, curl, and gradient are used to analyze vector fields.
• The Divergence Theorem is a fundamental principle connecting flux through a surface to the behavior inside a volume.

For this exercise, understanding divergence is key. Divergence gives you a scalar measuring how much a vector field is spreading out from a point. It's derived by summing partial derivatives of each field component. This concept helps analyze scenarios like where air is expanding or compressing in weather systems.
In this way, vector calculus provides tools and language to study and model real-world scenarios.

Product Rule

The product rule is a critical concept in calculus that extends to vector fields and is crucial in differentiation. It tells us how to differentiate the product of two functions or fields. Let's look at it closely:

• For scalar functions, if you have two functions, say, \( u(x) \) and \( v(x) \), the product rule states: \( \frac{d}{dx}[u(x) v(x)] = u'(x) v(x) + u(x) v'(x) \).

In vector calculus, this concept applies similarly but targets components. When dealing with divergence of products of scalar and vector fields, the product rule becomes:

• For a field \( \phi \mathbf{F} \), where \( \phi \) is a scalar and \( \mathbf{F} \) is a vector, the divergence can be expanded using the product rule: \( \operatorname{div}(abla \mathbf{F}) = \phi \operatorname{div} \mathbf{F} + abla \phi \cdot \mathbf{F} \).

This is exactly the identity you proved in the exercise. Recognizing how the product rule functions in such settings is necessary for solving complex problems across mathematics and physics.

Gradient

The gradient is a vector calculus tool providing crucial information about scalar fields. Essentially, it points in the direction of the greatest rate of increase of the scalar field and its length corresponds to the rate of increase at that point. You can think of it like this:

• If \( \phi \) is a scalar field, the gradient \( abla \phi \) is the vector field comprising partial derivatives of \( \phi \) with respect to all spatial variables.
• The gradient conveys both magnitude and direction of the steepest ascent, much like how steep a hill is and in which direction you would climb it fastest.

In the given problem, the gradient \( abla \phi \) connects with vector \( \mathbf{F} \) through the dot product, contributing to the terms on the right-hand side of the divergence identity: \( abla \phi \cdot \mathbf{F} \). Understanding gradients is beneficial for analyzing spatial variations and optimizing solutions in multidimensional space.