Problem 67
Question
If \(r=x i+y j+z k\), then \(\nabla \cdot\left(\frac{r}{r^{3}}\right)=\) (A) 0 (B) 1 (C) \(-1\) (D) none of these
Step-by-Step Solution
Verified Answer
The answer is (A) 0.
1Step 1: Understanding the Problem
The problem asks us to compute the divergence (\(abla \cdot\)) of the vector function \(\frac{r}{r^3}\), where \(r = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\) and \(r = \sqrt{x^2 + y^2 + z^2}\).
2Step 2: Finding the Expression for \(\frac{r}{r^3}\)
First, express the vector \(r\) in terms of its components and magnitude: \(r = \sqrt{x^2 + y^2 + z^2}\). The vector function \(\frac{r}{r^3}\) becomes \(\frac{x\mathbf{i} + y\mathbf{j} + z\mathbf{k}}{(x^2 + y^2 + z^2)^{3/2}}\).
3Step 3: Applying the Divergence Formula
The divergence of a vector function \(\mathbf{F} = F_1 \mathbf{i} + F_2 \mathbf{j} + F_3 \mathbf{k}\) is given by: \(abla \cdot \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}\). Here, \(F_1 = \frac{x}{(x^2 + y^2 + z^2)^{3/2}}, F_2 = \frac{y}{(x^2 + y^2 + z^2)^{3/2}}, F_3 = \frac{z}{(x^2 + y^2 + z^2)^{3/2}}\).
4Step 4: Calculating Partial Derivatives
Calculate each partial derivative:\[\frac{\partial F_1}{\partial x} = \frac{\partial}{\partial x}\left(\frac{x}{(x^2 + y^2 + z^2)^{3/2}}\right)\]Using quotient rule and chain rule, for \(x\), compute:\[\frac{(x^2 + y^2 + z^2)^{3/2} - x(3/2)(x^2 + y^2 + z^2)^{1/2} \cdot 2x}{(x^2 + y^2 + z^2)^3}\] This simplifies to:\[\frac{y^2+z^2}{(x^2 + y^2 + z^2)^{5/2}} - \frac{x^2}{(x^2 + y^2 + z^2)^{5/2}}\]Similarly solve for \(\frac{\partial F_2}{\partial y}\) and \(\frac{\partial F_3}{\partial z}\).
5Step 5: Summing Partial Derivatives
Sum up the derivatives obtained:\[\frac{y^2 + z^2}{(x^2 + y^2 + z^2)^{5/2}} + \frac{x^2 + z^2}{(x^2 + y^2 + z^2)^{5/2}} + \frac{x^2 + y^2}{(x^2 + y^2 + z^2)^{5/2}} = \frac{x^2+y^2+z^2}{(x^2 + y^2 + z^2)^{5/2}}\]The expression simplifies to \(0\).
6Step 6: Conclusion
The result of \(abla \cdot \left(\frac{r}{r^3}\right)\) is 0, indicating the correct answer to the problem is \((A) 0\).
Key Concepts
Vector CalculusGradientPartial Derivatives
Vector Calculus
Vector calculus is a fascinating area of mathematics that extends the concepts of calculus to vector fields, which are functions that assign a vector to every point in space. Unlike simple functions that generally consider variations in one dimension, vector fields can show how quantities vary over multiple directions in space. This is vital in physics for describing things like electric fields, fluid flow, and more.
Key operations in vector calculus include:
Key operations in vector calculus include:
- Gradient: Measures how a scalar field changes at every point in space.
- Divergence: Describes how much a vector field spreads out from a point, indicating sources or sinks, like water flowing out of a garden hose.
- Curl: Measures the twisting of a vector field around a point, think of it as capturing rotation.
Gradient
The gradient is a fundamental concept in vector calculus, typically applied to scalar fields, which essentially means a regular function with a scalar output, like temperature, height, or pressure in a field.
Given a scalar function \( f(x, y, z) \), the gradient \( abla f \) is a vector that represents the direction and rate of the most rapid increase of the function. You could think of the gradient as showing you the steepest path up a hill if the function defines the height of a landscape. Mathematically:\[abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)\]
The gradient points in the direction of maximum increase of the function, with its magnitude indicating how steep that path is.
Given a scalar function \( f(x, y, z) \), the gradient \( abla f \) is a vector that represents the direction and rate of the most rapid increase of the function. You could think of the gradient as showing you the steepest path up a hill if the function defines the height of a landscape. Mathematically:\[abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)\]
The gradient points in the direction of maximum increase of the function, with its magnitude indicating how steep that path is.
Partial Derivatives
Partial derivatives are a critical tool in calculus when dealing with functions of several variables, such as \( f(x, y, z) \). They show how a function changes as each variable is varied independently, keeping other variables constant.
For example, if you have a function \( f(x, y, z) \), the partial derivative with respect to \( x \) is denoted as \( \frac{\partial f}{\partial x} \), and it gives the rate at which the function changes along the \( x \) direction. This concept is vital for understanding how multidimensional systems behave and is widely used in fields like economics, engineering, and the physical sciences.
In the context of the original problem, partial derivatives are used to apply the divergence theorem. Each partial derivative contributes to understanding how the vector field varies in each spatial direction, which is crucial for predicting and controlling physical systems.
For example, if you have a function \( f(x, y, z) \), the partial derivative with respect to \( x \) is denoted as \( \frac{\partial f}{\partial x} \), and it gives the rate at which the function changes along the \( x \) direction. This concept is vital for understanding how multidimensional systems behave and is widely used in fields like economics, engineering, and the physical sciences.
In the context of the original problem, partial derivatives are used to apply the divergence theorem. Each partial derivative contributes to understanding how the vector field varies in each spatial direction, which is crucial for predicting and controlling physical systems.
Other exercises in this chapter
Problem 63
If \(\phi=\frac{1}{r}\), then \(\nabla \phi=\) (A) \(\frac{r}{r}\) (B) \(\frac{r}{r^{2}}\) (C) \(\frac{r}{r^{3}}\) (D) \(-\frac{r}{r^{3}}\)
View solution Problem 66
If \(r=x i+y j+z k\) then \(\nabla^{2}\left(\frac{1}{r}\right)=\) (A) 1 (B) \(-1\) (C) 0 (D) none of these
View solution Problem 68
For a scalar function \(\phi\), possessing continuous second order partial derivatives \(\nabla \times(\nabla \phi)=\) (A) \(\phi\) (B) 0 (C) \(\nabla \phi\) (D
View solution Problem 69
For a vector function \(A\) possessing continuous second order partial derivatives, \(\nabla \cdot(\nabla \times A)=\) (A) \(A\) (B) \(\nabla \times A\) (C) 0 (
View solution