Problem 46 - Calculus: Early Transcendental Functions

Question

Use the notation \(r=\langle x, y\rangle\) and \(r=\|\mathbf{r}\|=\sqrt{x^{2}+y^{2}}\) $$\text { Show that } \nabla\left(r^{2}\right)=2 \mathbf{r}$$

Step-by-Step Solution

Answer

\nabla r^{2} = 2 \mathbf{r}

Step 1: Express \(r^{2}\) in terms of x and y

We know that \( r = \sqrt{x^{2}+y^{2}} \), so \( r^{2} = x^{2}+y^{2} \). We will work with \( r^{2} = x^{2}+y^{2} \) since it's easier to deal with.

Step 2: Calculate the gradient of \(r^{2}\)

The gradient of a scalar field in Cartesian coordinates is given by \( \nabla F = \frac{\partial F}{\partial x} \mathbf{i} + \frac{\partial F}{\partial y} \mathbf{j} \). Apply this to \( r^{2} \), yielding \( \nabla r^{2} = 2x \mathbf{i} + 2y \mathbf{j} \).

Step 3: Express \(2 \mathbf{r}\) in terms of x and y

From the notation in the exercise, we know that \( \mathbf{r} = \langle x, y \rangle = x\mathbf{i} + y\mathbf{j} \), so multiplying by 2 will result in \( 2 \mathbf{r} = 2x\mathbf{i} + 2y\mathbf{j} \).

Step 4: Compare the results

We can see that the gradient of \( r^{2} \) computed in Step 2 is equal to \( 2 \mathbf{r} \) computed in Step 3.

Key Concepts

Vector Notation

Vectors are mathematical entities that have both magnitude and direction. In the context of this exercise, we use

• Vector notation: Often represented by bold characters like \( \mathbf{r} \).
• Component form: Shown as \( \langle x, y \rangle \), indicating the components along the x and y axes.

This notation allows us to easily perform operations, like addition, subtraction, or scaling with vectors, helping to simplify complex calculations.

Understanding how to manipulate vector notation is key in applications like physics and engineering where direction and magnitude are important.

Scalar Field

A scalar field is a function that assigns a scalar value to every point in space. It can be visualized as a landscape where the height of the landscape at each point represents the scalar value. Examples include temperature distribution in a room or air pressure in the atmosphere.

In this exercise, we're working with \( r^2 = x^2 + y^2 \), a scalar field that associates the sum of squares of coordinates (x and y) to a point in a plane. Calculating the gradient of such a field tells us how the field changes at different points.

Cartesian Coordinates

Cartesian coordinates are a system that uses two or three perpendicular axes to uniquely identify points in space. For our purposes, we focus on two-dimensional Cartesian coordinates \((x, y)\) which are ideal for describing locations on a flat plane.

This system is widely used in mathematics and science because it's intuitive and maps easily to real-world needs like plotting graphs or defining vector components. The exercise relies on Cartesian coordinates to express vectors and perform gradient calculations, essential in fields like engineering and computer graphics.

Gradient of a Scalar Field

The gradient is a vector that points in the direction of maximum increase of a scalar field. Imagine standing on a hill: the gradient tells you which way is uphill. It's calculated as the partial derivatives of the scalar field along each axis.

In the exercise, we computed this gradient for \( r^2 = x^2 + y^2 \) using

• Partial derivatives: \( \frac{\partial}{\partial x}(x^2+y^2) = 2x \) and \( \frac{\partial}{\partial y}(x^2+y^2) = 2y \).
• The result: The gradient \( abla r^2 = 2x\mathbf{i} + 2y\mathbf{j} \).

This matches \( 2 \mathbf{r} = 2x\mathbf{i} + 2y\mathbf{j} \), showing how gradients provide powerful insights into changes within scalar fields.