Problem 63 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\phi=\frac{1}{r}\), then \(\nabla \phi=\) (A) \(\frac{r}{r}\) (B) \(\frac{r}{r^{2}}\) (C) \(\frac{r}{r^{3}}\) (D) \(-\frac{r}{r^{3}}\)

Step-by-Step Solution

Answer

The answer is (D) \(-\frac{r}{r^3}\).

Step 1: Understand the Gradient

To find \( abla \phi \), we need to compute the gradient of the function \( \phi = \frac{1}{r} \). For a scalar field \( \phi(x, y, z) \), the gradient \( abla \phi \) is a vector consisting of partial derivatives with respect to \( x \), \( y \), and \( z \).

Step 2: Define \( r \)

In this context, \( r \) is typically the magnitude of the position vector \( \mathbf{r} = (x, y, z) \), so \( r = \sqrt{x^2 + y^2 + z^2} \).

Step 3: Chain Rule Application

We will use the chain rule to find the gradient. Express \( abla \phi = \frac{d}{dr} \left(\frac{1}{r}\right) \cdot abla r \), where \( abla r = \left( \frac{x}{r}, \frac{y}{r}, \frac{z}{r} \right) \).

Step 4: Differentiate \( \phi \) with Respect to \( r \)

Compute \( \frac{d}{dr} \left( \frac{1}{r} \right) = -\frac{1}{r^2} \).

Step 5: Compute the Gradient

Now, multiply \(-\frac{1}{r^2}\) by \( \left( \frac{x}{r}, \frac{y}{r}, \frac{z}{r} \right) \) to get \( abla \phi = -\frac{x}{r^3} \hat{i} - \frac{y}{r^3} \hat{j} - \frac{z}{r^3} \hat{k} \). This simplifies to \( abla \phi = -\frac{\mathbf{r}}{r^3} \).

Step 6: Compare with Options

The computed gradient is \( -\frac{\mathbf{r}}{r^3} \). Compare this with the given options. The correct choice is (D) \(-\frac{r}{r^3}\) which aligns with our solution.

Key Concepts

Partial Derivatives

Partial derivatives are fundamental in understanding how a function changes as each variable alters. In multivariable calculus, we often deal with functions that depend on more than one variable, like \( \phi(r) = \frac{1}{r} \). Such functions require a partial differentiation approach to comprehend how they behave when only one variable shifts, holding others constant.

This is particularly useful in physics and engineering. For example, while computing the gradient of a scalar field, we take partial derivatives. The gradient \( abla \phi \) is a vector composed of these partial derivatives, which shows the direction and rate of fastest increase of the scalar field. If our function is expressed in terms of variables \(x\), \(y\), and \(z\), we need the partial derivatives \( \frac{\partial \phi}{\partial x} \), \( \frac{\partial \phi}{\partial y} \), and \( \frac{\partial \phi}{\partial z} \).

• For \( \phi = \frac{1}{r} \), where \( r = \sqrt{x^2 + y^2 + z^2} \), the gradient vector is calculated by deriving the function with respect to each variable.

Understanding partial derivatives equips you with the ability to perform more complex calculations and dissect how small changes in input can influence the overall system behavior.

Vector Calculus

Vector calculus is an essential branch of mathematics used extensively in fields like physics and engineering to describe the dynamics of scalar and vector fields. When dealing with functions like \( \phi = \frac{1}{r} \), vector calculus offers tools to determine various properties, such as the direction of the field or the rate at which quantities change over space.

One key operation in vector calculus is computing the gradient of a scalar field. The gradient \( abla \phi \) of a scalar field \( \phi \) is a vector field, which points in the direction of the greatest rate of increase of the scalar function and whose magnitude is the rate of increase in that direction.

• In our example, to find \( abla \phi \), we compute the negative of \( \frac{1}{r^2} \) multiplied by the unit vector \( \left( \frac{x}{r}, \frac{y}{r}, \frac{z}{r} \right) \).

This blend of concepts from vector calculus allows us to understand forces and changes in physical systems, making it a powerful analytical tool.

Chain Rule

The chain rule is a fundamental mechanism in calculus for differentiating compositions of functions. It is often employed when a variable depends on one or more other variables, which in turn influence the function we are interested in differentiating.

In the context of our exercise, to compute \( abla \phi \) where \( \phi = \frac{1}{r} \), we apply the chain rule because the function is dependent on \( r \), which itself depends on \( x \), \( y \), and \( z \):

• We first differentiate the outer function \( \phi = \frac{1}{r} \) with respect to \( r \), obtaining \( \frac{d}{dr} \left( \frac{1}{r} \right) = -\frac{1}{r^2} \).
• Next, we find the gradient of \( r \) in the form \( abla r = \left( \frac{x}{r}, \frac{y}{r}, \frac{z}{r} \right) \).
• By multiplying these together, using the chain rule, we find that \( abla \phi = -\frac{\mathbf{r}}{r^3} \).

Mastering the chain rule allows deeper insight into how different variables interact within a system, fostering an adaptable skill set for solving complex problems.