Chapter 14: Problem 52
If \(f\) is a scalar function, \(\mathbf{r}=\langle x, y\rangle\) and \(r=\|\mathbf{r}\|,\) show that $$\nabla^{2} f(r)=f^{\prime \prime}(r)+\frac{1}{r} f^{\prime}(r)$$
Short Answer
Expert verified
The Laplacian (\(\nabla^{2} \)) of the scalar function \( f \) with respect to \( r \) is indeed equal to \( f''(r) + \frac{1}{r}f'(r) \).
Step by step solution
01
Definitions and Preparation
Let's start by preparing for the task. Let \( f \) be a function of \( r \), where \( r \) is the magnitude of the position vector \( \mathbf{r} = \langle x, y \rangle \), that is, \( r = \|\mathbf{r}\| = \sqrt{x^2 + y^2} \). This exercise requires demonstrating that the Laplace operator applied to \( f \) gives \( \nabla^{2} f(r)=f^{\prime \prime}(r)+\frac{1}{r} f^{\prime}(r) \).
02
Derivative of \( f \) with respect to \( r \)
The derivative of \( f \) with respect to \( r \) can be written as \( f^{\prime}(r) \). Let's denote this as \( g(r) \). Now, the derivative of \( g \) with respect to \( r \) would hence be \( g'(r) \), or written in terms of \( f \), this is \( f''(r) \).
03
Gradient of \( f \)
Next, let's calculating the gradient of \( f \) in polar coordinates. The gradient vector in 2D polar coordinates can be represented as \( \nabla f = f'(r) \hat{r} \), where \( \hat{r} \) is the unit vector in the radial direction.
04
Applying the Laplacian operator
We know that the Laplacian operator applied to a scalar function \( f \) in polar coordinates can be computed as \( \nabla^{2} f(r) = \frac{1}{r} \frac{\partial}{\partial r}\left( r \frac{\partial f(r)}{\partial r} \right) \). We can substitute \( f'(r) = g(r) \) and then apply the derivative rules. Doing this, yields \( \nabla^{2} f(r) = \frac{1}{r}[rg'(r) + g(r)] = f''(r) + \frac{1}{r}f'(r) \). Thus, we have verified the given equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Gradient and Its Role
The gradient is a vector that describes the rate and direction of change in a scalar field. In simpler terms, it tells you the way in which a function increases or decreases most quickly and how steep that change is.
In our context, we're dealing with a scalar function, specifically in polar coordinates, which means the function depends on the radial distance, \( r \), from the origin.
This scalar field, \( f(r) \), has its gradient expressed in 2D polar coordinates as \( abla f = f'(r) \hat{r} \).
Here, \( \hat{r} \) is the unit vector pointing outwards in the radial direction.
This essentially encapsulates the rate of change of the function \( f \) along the direction of \( r \).
In our context, we're dealing with a scalar function, specifically in polar coordinates, which means the function depends on the radial distance, \( r \), from the origin.
This scalar field, \( f(r) \), has its gradient expressed in 2D polar coordinates as \( abla f = f'(r) \hat{r} \).
Here, \( \hat{r} \) is the unit vector pointing outwards in the radial direction.
This essentially encapsulates the rate of change of the function \( f \) along the direction of \( r \).
- The gradient provides a local direction at each point of \( f \).
- It shows how steeply \( f \) rises or falls at each point.
- In polar coordinates, it simplifies handling radial symmetry.
Deciphering Scalar Function
A scalar function assigns a single value to every point in space. Think of it as a temperature distribution across a region, where each point has one temperature value.
This exercise presents \( f \) as a scalar function of \( r \), indicating that \( f \) depends only on the distance from the origin, making it spherically symmetric.
Such functions are crucial in physical scenarios where only the distance from a reference point (the origin, in this case) determines values like potential energy in a field.
This exercise presents \( f \) as a scalar function of \( r \), indicating that \( f \) depends only on the distance from the origin, making it spherically symmetric.
Such functions are crucial in physical scenarios where only the distance from a reference point (the origin, in this case) determines values like potential energy in a field.
- Scalar functions are easier to work with due to their single-dimensional output.
- They often describe properties distributed in space (like temperature or pressure).
- In polar coordinates, representation focuses on radial distances, simplifying calculations.
Exploring Radial Direction
The radial direction refers to a line or path that extends from the center point (or origin) outwards in a circular pattern. In polar coordinates, this becomes particularly meaningful.
The radial direction in our exercise is crucial since \( f \) changes its value only by varying \( r \), the distance from the origin.
Each point of \( \mathbf{r} = \langle x, y \rangle \) maps to a specific \( r \) in this direction, reaffirming the concept of radial symmetry.
The radial direction in our exercise is crucial since \( f \) changes its value only by varying \( r \), the distance from the origin.
Each point of \( \mathbf{r} = \langle x, y \rangle \) maps to a specific \( r \) in this direction, reaffirming the concept of radial symmetry.
- The radial direction streamlines calculations in circular or spherical domains.
- It gives insight into how functions vary with distance from a central point.
- This is highly relevant to disciplines like electromagnetism and gravitation fields.
Diving into Derivative Concepts
A derivative represents how a function changes as its input changes. It's a fundamental concept in calculus that shows the rate of change at any given point.
In this exercise, we focus on derivatives of the scalar function \( f \) with respect to \( r \). The notation \( f'(r) \) represents the first derivative, indicating how \( f \) changes per unit change in \( r \).
\( f''(r) \) would be the second derivative, showing how the rate of change itself changes. **This is very insightful in understanding acceleration in physical terms.**
In this exercise, we focus on derivatives of the scalar function \( f \) with respect to \( r \). The notation \( f'(r) \) represents the first derivative, indicating how \( f \) changes per unit change in \( r \).
\( f''(r) \) would be the second derivative, showing how the rate of change itself changes. **This is very insightful in understanding acceleration in physical terms.**
- Derivatives measure the local rate of change in functions, like slope for lines.
- The second derivative reveals the curvature or acceleration of a function's graph.
- In polar coordinates, it significantly aids in revealing symmetry properties.