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Chapter 4: Problem 14

Prove that if \(f(\mathbf{r})\) and \(g(\mathbf{r})\) are any two scalar functions of \(\mathbf{r},\) then \(\nabla(f g)=f \nabla g+g \nabla f\)

Short Answer

Expert verified
The product rule for differentiation proves that \(\nabla(fg) = f \nabla g + g \nabla f\).

Step by step solution

01

Understand the Gradient of a Product

The task is to show that the gradient of a product of two scalar functions \(f(\mathbf{r})\) and \(g(\mathbf{r})\) follows the product rule. The gradient in vector calculus for a scalar field is the vector of its partial derivatives.
02

Apply the Gradient Operator to the Product

The gradient of the product \(fg\) is given by \(abla(fg) = \left(\frac{\partial}{\partial x}(fg), \frac{\partial}{\partial y}(fg), \frac{\partial}{\partial z}(fg)\right)\).
03

Apply Product Rule to Each Partial Derivative

Using the product rule for differentiation, compute the partial derivative of \(fg\) with respect to each variable:- \(\frac{\partial}{\partial x}(fg) = \frac{\partial f}{\partial x} g + f \frac{\partial g}{\partial x}\)- \(\frac{\partial}{\partial y}(fg) = \frac{\partial f}{\partial y} g + f \frac{\partial g}{\partial y}\)- \(\frac{\partial}{\partial z}(fg) = \frac{\partial f}{\partial z} g + f \frac{\partial g}{\partial z}\).
04

Reconstruct the Gradient Components

Substitute the results from Step 3 into the gradient formula:\[ abla(fg) = \left( \frac{\partial f}{\partial x}g + f\frac{\partial g}{\partial x}, \frac{\partial f}{\partial y}g + f\frac{\partial g}{\partial y}, \frac{\partial f}{\partial z}g + f\frac{\partial g}{\partial z} \right). \]
05

Factor and Rearrange Components

Factor out the components from the result:\[ abla(fg) = g(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}) + f(\frac{\partial g}{\partial x},\frac{\partial g}{\partial y},\frac{\partial g}{\partial z}) \]This simplifies to:\[ abla(fg) = gabla f + fabla g. \]
06

Conclusion: Verify Result Matches Desired Expression

The expression \(abla(fg) = f abla g + g abla f\) has been successfully derived using the product rule, thus proving the statement.

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Key Concepts

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

Product Rule in Calculus
The product rule in calculus is a fundamental principle used when differentiating the product of two functions. It can be stated simply: to differentiate the product of two functions, apply the derivative to each function in turn, keeping the other constant, and then sum the results. In mathematical notation, for functions \(u(x)\) and \(v(x)\):
\[ (uv)' = u'v + uv' . \]
This rule ensures that the contribution to the rate of change of the product is accounted for by both functions. In vector calculus, we extend this concept to scalar and vector fields, which is especially useful for functions of multiple variables.
Scalar Functions
Scalar functions are mathematical functions that map a vector space to a scalar (a single real number). Unlike vectors, they have no direction. Common examples include temperature, pressure, or density at a point in space.
For a scalar function of a vector \(\mathbf{r} = (x, y, z)\), say \(f(\mathbf{r})\), it assigns a single scalar value to each position \(\mathbf{r}\) in space.
  • They are straightforward to visualize because they only involve a scalar value.
  • Many physical phenomena can be described by scalar fields, such as sound intensity, gravitational potential, or electrical potential.
Understanding scalar functions is crucial for integrating concepts like the gradient, divergence, or Laplacian in vector calculus, each providing different insights into a scalar field's behavior.
Vector Calculus
Vector calculus is a branch of mathematics that deals with vector fields and the operations on these fields. It extends the basic principles of calculus to manipulations involving vectors, making it invaluable in physics and engineering. Some key operations in vector calculus include:
  • Gradient: It measures the rate and direction of change in a scalar field. The gradient of a scalar field is itself a vector, pointing in the direction of the greatest rate of increase.
  • Divergence: Represents the magnitude of a source or sink at a given point in a vector field.
  • Curl: Measures the rotation of a vector field around a point.
Each of these operations serves to analyze different aspects of vector fields, helping to solve real-world problems related to force fields, fluid dynamics, and electromagnetism. Understanding these core operations is essential for navigating topics like the gradient of a product, where vector calculus becomes a powerful tool for deeper analysis.

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Most popular questions from this chapter

(a) Consider a mass \(m\) in a uniform gravitational field \(\mathbf{g},\) so that the force on \(m\) is \(m \mathbf{g},\) where \(\mathbf{g}\) is a constant vector pointing vertically down. If the mass moves by an arbitrary path from point 1 to point \(2,\) show that the work done by gravity is \(W_{\mathrm{grav}}(1 \rightarrow 2)=-m g h\) where \(h\) is the vertical height gained between points 1 and 2. Use this result to prove that the force of gravity is conservative (at least in a region small enough so that \(\mathrm{g}\) can be considered constant). (b) Show that, if we choose axes with \(y\) measured vertically up, the gravitational potential energy is \(U=m g y\) (if we choose \(U=0\) at the origin).

If a particle's potential energy is \(U(\mathbf{r})=k\left(x^{2}+y^{2}+z^{2}\right),\) where \(k\) is a constant, what is the force on the particle?

Find the partial derivatives with respect to \(x, y,\) and \(z\) of the following functions: (a) \(f(x, y, z)=\) \(a x^{2}+b x y+c y^{2},(\mathbf{b}) g(x, y, z)=\sin \left(a x y z^{2}\right),(\mathbf{c}) h(x, y, z)=a e^{x y / z^{2}},\) where \(a, b,\) and \(c\) are constants. Remember that to evaluate \(\partial f / \partial x\) you differentiate with respect to \(x\) treating \(y\) and \(z\) as constants.

Both the Coulomb and gravitational forces lead to potential energies of the form \(U=\gamma / | \mathbf{r}_{1}-\) \(\mathbf{r}_{2} |,\) where \(\gamma\) denotes \(k q_{1} q_{2}\) in the case of the Coulomb force and \(-G m_{1} m_{2}\) for gravity, and \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\) are the positions of the two particles. Show in detail that \(-\nabla_{1} U\) is the force on particle 1 and \(-\nabla_{2} U\) that on particle 2.

The proof that the condition \(\nabla \times \mathbf{F}=0\) guarantees the path independence of the work \(\int_{1}^{2} \mathbf{F} \cdot d \mathbf{r}\) done by \(\mathbf{F}\) is unfortunately too lengthy to be included here. However, the following three exercises capture the main points: \(^{16}\) (a) Show that the path independence of \(\int_{1}^{2} \mathbf{F} \cdot d \mathbf{r}\) is equivalent to the statement that the integral \(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}\) around any closed path \(\Gamma\) is zero. (By tradition, the symbol \(\oint\) is used for integrals around a closed path \(-\) a path that starts and stops at the same point.) [Hint: For any two points 1 and 2 and any two paths from 1 to 2 , consider the work done by \(\mathbf{F}\) going from 1 to 2 along the first path and then back to 1 along the second in the reverse direction. \((\) b) Stokes's theorem asserts that \(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}=\int(\nabla \times \mathbf{F}) \cdot \hat{\mathbf{n}} d A,\) where the integral on the right is a surface integral over a surface for which the path \(\Gamma\) is the boundary, and \(\hat{\mathbf{n}}\) and \(d A\) are a unit normal to the surface and an element of area. Show that Stokes's theorem implies that if \(\nabla \times \mathbf{F}=0\) everywhere, then \(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}=0 .\) (c) While the general proof of Stokes's theorem is beyond our scope here, the following special case is quite easy to prove (and is an important step toward the general proof): Let \(\Gamma\) denote a rectangular closed path lying in a plane perpendicular to the \(z\) direction and bounded by the lines \(x=B, x=B+b, y=C\) and \(y=C+c .\) For this simple path (traced counterclockwise as seen from above), prove Strokes's theorem that \(\oint_{\Gamma} \mathbf{F} \cdot d \mathbf{r}=\int(\mathbf{\nabla} \times \mathbf{F}) \cdot \hat{\mathbf{n}} d A\) where \(\hat{\mathbf{n}}=\hat{\mathbf{z}}\) and the integral on the right runs over the flat, rectangular area inside \(\Gamma\). [Hint: The integral on the left contains four terms, two of which are integrals over \(x\) and two over \(y\). If you pair them in this way, you can combine each pair into a single integral with an integrand of the form \(F_{x}(x, C+c, z)-F_{x}(x, C, z)\) (or a similar term with the roles of \(x\) and \(y\) exchanged). You can rewrite this integrand as an integral over \(y\) of \(\partial F_{x}(x, y, z) / \partial y\) (and similarly with the other term), and you're home.]
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