Problem 43 - Thomas Calculus

Question

Unit vectors pointing toward the origin Find a field \(\mathbf{F}=\) \(M(x, y) \mathbf{i}+N(x, y) \mathbf{j}\) in the \(x y-\) plane with the property that at each point \((x, y) \neq(0,0), \mathbf{F}\) is a unit vector pointing toward the origin. (The field is undefined at (0, 0).)

Step-by-Step Solution

Answer

The field is \(\mathbf{F} = \frac{-x}{\sqrt{x^2 + y^2}}\mathbf{i} + \frac{-y}{\sqrt{x^2 + y^2}}\mathbf{j}\).

Step 1: Define the Vector Field

To find a field \( \mathbf{F} = M(x, y) \mathbf{i} + N(x, y) \mathbf{j} \) such that it points towards the origin at any point \((x, y)\), we can start by considering a vector from point \((x, y)\) to the origin \((0,0)\). This vector can be expressed as \(-x\mathbf{i} - y\mathbf{j}\).

Step 2: Normalize the Field Vector

A unit vector has a magnitude of 1. To make \(\mathbf{F}\) a unit vector, we need to normalize the field vector \(-x\mathbf{i} - y\mathbf{j}\). The magnitude of this vector is \(\sqrt{x^2 + y^2}\). To normalize, divide each component by its magnitude: \[ \mathbf{F} = \frac{-x}{\sqrt{x^2 + y^2}}\mathbf{i} + \frac{-y}{\sqrt{x^2 + y^2}}\mathbf{j} \].

Step 3: Verify Field Properties

After equating \( \mathbf{F} \) to \(\frac{-x}{\sqrt{x^2 + y^2}}\mathbf{i} + \frac{-y}{\sqrt{x^2 + y^2}}\mathbf{j}\), we ensure it points towards the origin by examining its components, which are negative and directed oppositely to \((x, y)\). Its magnitude is 1, confirming that it's a unit vector.

Key Concepts

Unit Vectors

In geometry and physics, a unit vector is a vector with a magnitude of exactly one. It is often used to indicate direction without affecting magnitude. In simpler terms, unit vectors are the scaled down versions of larger vectors that only contain direction information. They don't grow in length, staying a steady size of one. This helps in calculations dealing with directions while keeping the problem simple.

• Unit vectors are commonly denoted with a hat, like \( \hat{i} \) or \( \hat{j} \).
• They serve as the building blocks for larger vectors.
• Any vector can be turned into a unit vector using a process called normalization.

Unit vectors are crucial in fields like physics, where only the direction of a force or motion is needed. These vectors also find application in simplifying complex calculations by focusing purely on direction.

Normalization

Normalization is a mathematical process used to transform any vector into a unit vector. This process ensures that the resulting vector has a magnitude of one, but maintains the original vector's direction. This is particularly useful in vector fields where a uniform direction is needed with consistent magnitude.
Given a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} \), normalization involves dividing each component of the vector by its magnitude. If the magnitude of the vector is \( |\mathbf{v}| \), the normalized vector \( \mathbf{u} \) is expressed as:

• \( \mathbf{u} = \frac{a}{|\mathbf{v}|} \mathbf{i} + \frac{b}{|\mathbf{v}|} \mathbf{j} \)

Here are the steps for vector normalization:

• Calculate the magnitude of the original vector.
• Divide each component of the vector by this magnitude.

The resultant vector is now a unit vector, retaining its original direction but with a magnitude of 1.

Magnitude of a Vector

The magnitude of a vector is a measure of its length. It can be thought of as the "size" of the vector. Calculating the magnitude is essential when normalizing a vector. In a 2D vector field, the magnitude \(|\mathbf{v}|\) of a vector \(\mathbf{v} = a \mathbf{i} + b \mathbf{j}\) is determined using the Pythagorean Theorem.The formula for the magnitude is:

• \(|\mathbf{v}| = \sqrt{a^2 + b^2} \)

This value tells us how long the vector is from the origin. When you see a vector expressed in terms of its magnitude, you're looking at a straightforward measure of length, kind of like measuring a straight line from one end to another.Understanding magnitude is vital for engineering, physics, and computer graphics, where precise vector sizing is crucial for function and simulation.