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Chapter 13: Problem 7

Find grad \(f=\left(f_{x}, f_{y}, f_{z}\right)\) for the functions from physics. \(\ln \left(x^{2}+y^{2}\right)\) (line source along \(z\) axis)

Short Answer

Expert verified
The gradient is \( \left( \frac{2x}{x^2+y^2}, \frac{2y}{x^2+y^2}, 0 \right) \)."

Step by step solution

01

Identify the Function

The given function, a scalar field, is given by \( f(x, y, z) = \ln(x^2 + y^2) \). It is a logarithmic function representing a potential caused by a line source along the \(z\)-axis.
02

Find Partial Derivative with respect to x

Compute \(f_x\), the partial derivative of \(f\) with respect to \(x\). From \(f(x, y, z) = \ln(x^2+y^2)\), use the chain rule: \( f_x = \frac{d}{dx}[\ln(x^2+y^2)] = \frac{1}{x^2+y^2} \cdot 2x = \frac{2x}{x^2+y^2} \).
03

Find Partial Derivative with respect to y

Compute \(f_y\), the partial derivative of \(f\) with respect to \(y\). Similarly, apply the chain rule: \( f_y = \frac{d}{dy}[\ln(x^2+y^2)] = \frac{1}{x^2+y^2} \cdot 2y = \frac{2y}{x^2+y^2} \).
04

Find Partial Derivative with respect to z

Since \(f(x, y, z) = \ln(x^2 + y^2)\) doesn't involve \(z\), the partial derivative \(f_z = \frac{d}{dz}[\ln(x^2+y^2)] = 0\).
05

Construct the Gradient Vector

The gradient of the function \(f\) is given by combining all the partial derivatives: \( abla f = \left( f_x, f_y, f_z \right) = \left( \frac{2x}{x^2+y^2}, \frac{2y}{x^2+y^2}, 0 \right) \).

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Key Concepts

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

Partial Derivative
A partial derivative represents the rate at which a function changes as one variable is altered, while keeping the others constant. When you have a multivariable function, like our function \( f(x, y, z) = \ln(x^2 + y^2) \), you can take the derivative with respect to one variable at a time. This process isolates the effect of changes in that variable alone.
  • Example: The partial derivative of \( f(x, y, z) \) with respect to \( x \) is labelled as \( f_x \).
  • This means differentiating \( \ln(x^2 + y^2) \) as though only \( x \) can change, not \( y \) or \( z \).
For this exercise, calculating \( f_x \) and \( f_y \) involves using the chain rule, yielding fractional expressions due to the \ln function.
Scalar Field
In mathematics and physics, a scalar field assigns a single value to every point in space. Think of it as a landscape where each point has a particular height, temperature, or potential. The function \( f(x, y, z) = \ln(x^2 + y^2) \) is a scalar field because it returns a single scalar value for any \( (x, y, z) \).
  • Interpretation: This scalar field can represent potential energy in a physical system.
  • It depends only on \( x \) and \( y \), meaning it's constant along the \( z \)-axis.
These fields are critical in physics, as they help map and calculate phenomena like temperature distribution or potential fields.
Line Source
A line source is a concept from physics where the source of a field, such as an electric or fluid potential, can be thought of as being spread out along a line rather than being located at a point. In our case, the line is along the \( z \)-axis, influencing the logarithmic potential \( f(x, y, z) = \ln(x^2 + y^2) \).
  • This particular form suggests that the influence diminishes as you move away from the line (line source), particularly in the \( x \) and \( y \) directions.
  • Understanding this helps in visualizing how the field interacts spatially.
Such representations are often used in fields like electrostatics and fluid dynamics to model infinite or long charge and flow distributions.
Chain Rule
The chain rule in calculus is a crucial tool for finding the derivative of composite functions. It explains how a change in one variable affects another when two functions are combined. In our exercise, it's used to differentiate \( \ln(x^2 + y^2) \) with respect to \( x \) and \( y \).
  • Usage: First, differentiate the outer function, \( ln(u) \), with respect to \( u \) giving \( \frac{1}{u} \).
  • Next, multiply by the derivative of the inner function \( u(x, y) = x^2 + y^2 \) with respect to \( x \) or \( y \).
This effectively links the rate of change in one part of the function to changes in its component functions, simplifying complex derivative calculations.

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Most popular questions from this chapter

(a) For \(f=x^{2}+y^{2}+z^{2}\) compute \(\partial f / \partial x\) (no subscript, \(x, y, z\) all independent). (b) When there is a further relation \(z=x^{2}+y^{2}\), use it to remove \(z\) and compute \((\partial f / \partial x)_{y}\). (c) Compute \(\\{\partial f / \partial x)_{y}\) using the chain rule \((\partial f / \partial x)+\) \((\partial f / \partial z)(\partial z / \partial x)\) (d) Why doesn't that chain rule contain \((\partial f / \partial y)(\partial y / \partial x) ?\)

Find lines along which \(f(x, y)\) is constant (these functions have \(f_{x x} f_{y y}=f_{x y}^{2}\) or \(\left.a c=b^{2}\right):\) (a) \(f=x^{2}-4 x y+4 y^{2}\) (b) \(f=e^{x} e^{y}\)

Find \(d f / d t\) from the chain rule (3). \(f=x^{2}+y^{2}, x=t, y=t^{2}\)

The complex equation \((x+i y)^{3}=1\) contains two real equations, \(x^{3}-3 x y^{2}=1\) from the real part and \(3 x^{2} y-y^{3}=0\) from the imaginary part. Search by computer for the basins of attraction of the three solutions \((1,0),(-1 / 2, \sqrt{3} / 2),\) and \((-1 / 2,-\sqrt{3} / 2)-\) which give the cube roots of \(1 .\)

Develop the theory of Lagrange multipliers. (No maximum) Find a point on the line \(g=x+y=1\) where \(f(x, y)=2 x+y\) is greater tban 100 (or 1000 ). Write out grad \(f=\lambda\) grad \(g\) to see that there is no solution.
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