Problem 19 - Thomas Calculus

Question

Can anything be said about the speed of a particle whose acceleration is always orthogonal to its velocity? Give reasons for your answer.

Step-by-Step Solution

Answer

The speed of the particle is constant since acceleration is always orthogonal to velocity.

Step 1: Understanding Orthogonality

When two vectors are orthogonal, their dot product is zero. In this context, the acceleration vector \( extbf{a} \) is always orthogonal to the velocity vector \( extbf{v} \), i.e., \( extbf{a} \cdot extbf{v} = 0\).

Step 2: Defining Relationship between Acceleration and Velocity

The expression \( extbf{a} = \frac{d\textbf{v}}{dt} \) states that acceleration is the derivative of velocity with respect to time. Our condition becomes \( \frac{d\textbf{v}}{dt} \cdot extbf{v} = 0 \).

Step 3: Application of the Chain Rule

The dot product \( \frac{d\textbf{v}}{dt} \cdot extbf{v} \) can be seen as the derivative of \( \textbf{v}\cdot\textbf{v} \). Applying the chain rule, we have \( \frac{1}{2} \frac{d}{dt} (\textbf{v}\cdot\textbf{v}) = 0 \).

Step 4: Analyzing the Time Derivative

The expression \( \textbf{v}\cdot\textbf{v} = v^2 \) gives us the square of the speed. The derivative \( \frac{d}{dt}(v^2) = 0 \) implies that \( v^2 \) is constant.

Step 5: Concluding the Speed's Behavior

Since \( v^2 \) is constant over time, the speed \( v \) must also be constant. This means that although the direction of the velocity can change, its magnitude does not.

Key Concepts

Velocity

Velocity is a vector quantity describing the rate at which an object's position changes. It has both a magnitude, known as speed, and a direction.

• Magnitude is how fast the object is moving.
• Direction is the line along which the object is moving.

Understanding velocity is crucial because it describes how particles move through space. In our context, velocity is a key component because it interacts with acceleration. When dealing with vectors, it's important to consider both elements of velocity—how fast and where it goes. This dual characteristic of velocity is particularly relevant when we say that acceleration is orthogonally aligned with velocity, meaning it affects direction but not speed.
Thus, a constant speed implies that while velocity may constantly redirect, its rate does not change.

Acceleration

Acceleration is another vector quantity that indicates how quickly the velocity of an object changes over time.

• Speeding up, slowing down, and changing direction all constitute acceleration.
• It's calculated as the derivative of velocity in respect to time, denoted as \( \frac{d\textbf{v}}{dt} \).

In our problem, the acceleration vector is constantly orthogonal to the velocity vector, meaning the product of the two vectors is zero. This orthogonal relationship tells us that acceleration affects only the direction of velocity—turning it without changing its magnitude. That's why we find that acceleration doesn't alter the speed of the particle but does redirect its path.

Dot Product

The dot product is an important operation for vectors, calculating a scalar result from two vectors. This scalar can reveal the angle between the vectors. For two vectors \( \textbf{a} \) and \( \textbf{b} \), the dot product is calculated as \( \textbf{a} \cdot \textbf{b} = |\textbf{a}|\,|\textbf{b}|\cos(\theta) \).

• If the dot product is zero, it indicates that the two vectors are orthogonal (perpendicular to each other).

In our exercise, applying the dot product between acceleration and velocity being zero explains the orthogonal condition. Since the dot product of these vectors is zero, it results in orthogonality, where acceleration only changes the direction of velocity but not its speed. This underlines why while velocity direction changes, speed or magnitude stays constant.

Chain Rule

The chain rule is a principle in calculus used to differentiate compositions of functions. It provides a method to find the derivative of such compositions efficiently.

• If \( y = g(f(x)) \), then the derivative \( \frac{dy}{dx} = g'(f(x)) \cdot f'(x) \).

In vector calculus, this rule is paramount when dealing with changing vectors over time. Here, we used the chain rule to differentiate \( \textbf{v}\cdot\textbf{v} = v^2 \), ultimately leading to \( \frac{1}{2} \frac{d}{dt} (\textbf{v}\cdot\textbf{v}) = 0 \). This implies that \( v^2 \), the square of the speed, must be constant. Understanding the chain rule's application in this context reveals why the speed doesn't change even though the path or direction does, reinforcing the core principle of orthogonal vectors.