Problem 66 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

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

Step-by-Step Solution

Answer

(C) 0

Step 1: Understand the Problem

The task is to find the Laplacian of the function \( \frac{1}{r} \), where \( r = \sqrt{x^2 + y^2 + z^2} \) and \( abla^2 \) denotes the Laplacian operator. The Laplacian in three dimensions is given by \( abla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2} \).

Step 2: Calculate the Gradient

First, compute the gradient \( abla \left( \frac{1}{r} \right) \). The gradient is given by \(-\frac{1}{r^2} \frac{\partial r}{\partial x} i - \frac{1}{r^2} \frac{\partial r}{\partial y} j - \frac{1}{r^2} \frac{\partial r}{\partial z} k \). Since \( r = \sqrt{x^2 + y^2 + z^2} \), \( \frac{\partial r}{\partial x} = \frac{x}{r} \), \( \frac{\partial r}{\partial y} = \frac{y}{r} \), \( \frac{\partial r}{\partial z} = \frac{z}{r} \).

Step 3: Compute the Divergence

The divergence of the gradient \( abla \cdot abla \left( \frac{1}{r} \right) \) gives the Laplacian. \( abla \cdot abla \left( \frac{1}{r} \right) = -abla \cdot \left( \frac{x}{r^3} i + \frac{y}{r^3} j + \frac{z}{r^3} k \right) \).

Step 4: Apply Divergence Formula

Use the divergence formula: \( abla \cdot (fA) = f abla \cdot A + A \cdot abla f \). Here, let \( A = \frac{x}{r^3} i + \frac{y}{r^3} j + \frac{z}{r^3} k \), then compute each term. For \( abla \cdot A \), calculate \( \frac{\partial}{\partial x} \left( \frac{x}{r^3} \right) + \frac{\partial}{\partial y} \left( \frac{y}{r^3} \right) + \frac{\partial}{\partial z} \left( \frac{z}{r^3} \right) \).

Step 5: Derivative Calculation

The derivative of each term is \( \frac{d}{dx} \left( \frac{x}{r^3} \right) = \frac{r^2 - 3x^2}{r^5} \), similarly for the terms with \(y\) and \(z\).

Step 6: Combine Results

Summing up the results from the previous step, we find: \[ \frac{r^2 - 3x^2}{r^5} + \frac{r^2 - 3y^2}{r^5} + \frac{r^2 - 3z^2}{r^5} = \frac{3r^2 - 3(x^2 + y^2 + z^2)}{r^5} = 0 \]. Since \(r^2 = x^2 + y^2 + z^2\), the expression simplifies to zero.

Key Concepts

Laplacian operator

The Laplacian operator, denoted as \(abla^2\), is a second-order differential operator in the realm of vector calculus. It essentially measures the rate at which a quantity diffuses over a region. In three dimensions, the Laplacian of a function \(f\) is calculated as \(abla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2}\). This operator is vital in physics and engineering, particularly in fields like electromagnetism, fluid dynamics, and quantum mechanics.

In the exercise, we looked at finding the Laplacian of \(\frac{1}{r}\), where \(r\) is the radial distance \(\sqrt{x^2 + y^2 + z^2}\). The key step involves calculating the gradient first, then using the divergence of the gradient as the Laplacian. This concept is especially useful when handling problems in spherical coordinates or dealing with potential fields.

gradient

The gradient is a vector operator denoted by \(abla\), and it indicates the direction of the steepest ascent of a scalar field. When applied to a scalar function \(f(x, y, z)\), the gradient \(abla f\) yields a vector comprising partial derivatives: \(abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)\). So, the gradient tells us in which direction the function increases most rapidly and how steep that incline is.

In the given exercise, we calculated the gradient of \(\frac{1}{r}\) as \(-\frac{1}{r^2} \left( \frac{x}{r} i + \frac{y}{r} j + \frac{z}{r} k \right)\). The negative sign indicates that \(\frac{1}{r}\) is decreasing in the direction of the gradient. Understanding the gradient is crucial for finding the Laplacian, as the next step involves computing the divergence of this gradient vector.

divergence

Divergence, represented by the symbol \(abla \cdot\), is a scalar operator that measures a vector field's tendency to originate from or converge at a given point. If \(A\) is a vector field \(A = (A_x, A_y, A_z)\), the divergence is given by \(abla \cdot A = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z}\). Essentially, it provides a scalar quantity that describes the rate at which "stuff" is expanding out of any point in the field.

In our problem, we found the divergence of the gradient \(abla \left( \frac{1}{r} \right)\), leading us to \(-abla \cdot \left( \frac{x}{r^3} i + \frac{y}{r^3} j + \frac{z}{r^3} k \right)\). When calculated, this divergence revealed that the function \(\frac{1}{r}\) resulted in a zero Laplacian, meaning there's no net "outflow" or "inflow" at any point.

vector calculus

Vector calculus is a branch of mathematics focused on differential and integral calculus involving vector fields. This area is crucial for fields like physics, engineering, and computer graphics, since it provides tools to describe electromagnetic fields, fluid flow, and more.

Within vector calculus, operators like the gradient, divergence, and Laplacian are foundational. They allow us to handle complex phenomena in multi-dimensional spaces beyond what simple calculus can offer.

• The gradient provides directions and rates of change for scalar fields.
• Divergence quantifies the magnitude of a source or sink within a vector field.
• The Laplacian combines these to express how a field fluctuates over a space.

By breaking down these concepts as in our exercise, students can better understand how mathematical tools are applied to solve real-world problems efficiently and accurately.