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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

For a system of \(N\) particles subject to a uniform gravitational field g acting vertically down, prove that the total gravitational potential energy is the same as if all the mass were concentrated at the center of mass of the system; that is, \(U=\sum_{\alpha} U_{\alpha}=M g Y\) where \(M=\sum m_{\alpha}\) is the total mass and \(\mathbf{R}=(X, Y, Z)\) is the position of the \(\mathrm{CM},\) with the \(y\) coordinate measured vertically up. [Hint: We know from Problem 4.5 that \(\left.U_{\alpha}=m_{\alpha} g y_{\alpha} .\right]\)

An interesting one-dimensional system is the simple pendulum, consisting of a point mass \(m\), fixed to the end of a massless rod (length \(l\) ), whose other end is pivoted from the ceiling to let it swing freely in a vertical plane, as shown in Figure \(4.26 .\) The pendulum's position can be specified by its angle \(\phi\) from the equilibrium position. (It could equally be specified by its distance \(s\) from equilibrium \(-\) indeed \(s=l \phi-\) but the angle is a little more convenient.) (a) Prove that the pendulum's potential energy (measured from the equilibrium level) is \(U(\phi)=m g l(1-\cos \phi)\). Write down the total energy \(E\) as a function of \(\phi\) and \(\dot{\phi}\). (b) Show that by differentiating your expression for \(E\) with respect to \(t\) you can get the equation of motion for \(\phi\) and that the equation of motion is just the familiar \(\Gamma=I \alpha\) (where \(\Gamma\) is the torque, \(I\) is the moment of inertia, and \(\alpha\) is the angular acceleration \(\ddot{\phi}\) ). (c) Assuming that the angle \(\phi\) remains small throughout the motion, solve for \(\phi(t)\) and show that the motion is periodic with period \(\tau_{\mathrm{o}}=2 \pi \sqrt{l / g}\).

(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).

Find the partial derivatives with respect to \(x, y,\) and \(z\) of the following functions: (a) \(f(x, y, z)=\) \(a y^{2}+2 b y z+c z^{2},(\mathbf{b}) g(x, y, z)=\cos \left(a x y^{2} z^{3}\right),(\mathbf{c}) h(x, y, z)=a r,\) where \(a, b,\) and \(c\) are constants and \(r=\sqrt{x^{2}+y^{2}+z^{2}} .\) Remember that to evaluate \(\partial f / \partial x\) you differentiate with respect to \(x\) treating \(y\) and \(z\) as constants.

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?

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