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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

Which of the following forces is conservative? (a) \(\mathbf{F}=k(x, 2 y, 3 z)\) where \(k\) is a constant. (b) \(\mathbf{F}=k(y, x, 0) .\) (c) \(\mathbf{F}=k(-y, x, 0)\). For those which are conservative, find the corresponding potential energy \(U,\) and verify by direct differentiation that \(\mathbf{F}=-\nabla U\).

A particle of mass \(m_{1}\) and speed \(v_{1}\) collides with a second particle of mass \(m_{2}\) at rest. If the collision is perfectly inelastic (the two particles lock together and move off as one) what fraction of the kinetic energy is lost in the collision? Comment on your answer for the cases that \(m_{1} \ll m_{2}\) and that \(m_{2} \ll m_{1}\).

Calculate the gradient \(\nabla f\) of the following functions, \(f(x, y, z):\) (a) \(f=x^{2}+z^{3} .\) (b) \(f=k y\), where \(k\) is a constant. (c) \(f=r \equiv \sqrt{x^{2}+y^{2}+z^{2}} .\) [Hint: Use the chain rule.] (d) \(f=1 / r\).

Near to the point where I am standing on the surface of Planet \(X\), the gravitational force on a mass \(m\) is vertically down but has magnitude \(m \gamma y^{2}\) where \(\gamma\) is a constant and \(y\) is the mass's height above the horizontal ground. (a) Find the work done by gravity on a mass \(m\) moving from \(\mathbf{r}_{1}\) to \(\mathbf{r}_{2}\), and use your answer to show that gravity on Planet \(X,\) although most unusual, is still conservative. Find the corresponding potential energy. (b) Still on the same planet, I thread a bead on a curved, frictionless, rigid wire, which extends from ground level to a height \(h\) above the ground. Show clearly in a picture the forces on the bead when it is somewhere on the wire. (Just name the forces so it's clear what they are; don't worry about their magnitude.) Which of the forces are conservative and which are not? (c) If I release the bead from rest at a height \(h\), how fast will it be going when it reaches the ground?

(a) Consider an electron (charge \(-e\) and mass \(m\) ) in a circular orbit of radius \(r\) around a fixed proton (charge \(+e\) ). Remembering that the inward Coulomb force \(k e^{2} / r^{2}\) is what gives the electron its centripetal acceleration, prove that the electron's KE is equal to \(-\frac{1}{2}\) times its \(\mathrm{PE}\); that is, \(T=-\frac{1}{2} U\) and hence \(E=\frac{1}{2} U\). (This result is a consequence of the so-called virial theorem. See Problem 4.41.) Now consider the following inelastic collision of an electron with a hydrogen atom: Electron number 1 is in a circular orbit of radius \(r\) around a fixed proton. (This is the hydrogen atom.) Electron 2 approaches from afar with kinetic energy \(T_{2} .\) When the second electron hits the atom, the first electron is knocked free, and the second is captured in a circular orbit of radius \(r^{\prime} .\) (b) Write down an expression for the total energy of the three-particle system in general. (Your answer should contain five terms, three PEs but only two KEs, since the proton is considered fixed.) (c) Identify the values of all five terms and the total energy \(E\) long before the collision occurs, and again long after it is all over. What is the KE of the outgoing electron 1 once it is far away? Give your answers in terms of the variables \(T_{2}, r,\) and \(r^{\prime}\).

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