Problem 8 - Thomas Calculus

Question

Suppose that \(\mathbf{r}\) is the position vector of a particle moving along a plane curve and \(d A / d t\) is the rate at which the vector sweeps out area. Without introducing coordinates, and assuming the necessary derivatives exist, give a geometric argument based on increments and limits for the validity of the equation \begin{equation} \frac{d A}{d t}=\frac{1}{2}|\mathbf{r} \times \dot{\mathbf{r}}|. \end{equation}

Step-by-Step Solution

Answer

The equation \( \frac{d A}{d t} = \frac{1}{2}|\mathbf{r} \times \dot{\mathbf{r}}| \) is valid as it represents the area rate using the cross product.

Step 1: Understanding the setup

First, consider the position vector \( \mathbf{r}(t) \) representing the position of a particle as it moves along a plane curve. The velocity of the particle is given by the derivative of \( \mathbf{r}(t) \) with respect to time, \( \dot{\mathbf{r}}(t) \).

Step 2: Consider a small time interval

During a small time interval \( \Delta t \), the position vector changes from \( \mathbf{r}(t) \) to \( \mathbf{r}(t + \Delta t) \). The line segment connecting \( \mathbf{r}(t) \) and \( \mathbf{r}(t + \Delta t) \) forms a triangle with the origin.

Step 3: Calculate the area of the triangle

The area \( \Delta A \) of this triangle can be found using the cross-product formula for the area of a triangle. For two vectors \( \mathbf{a} \) and \( \mathbf{b} \), the area is \( \frac{1}{2}|\mathbf{a} \times \mathbf{b}| \). Here, \( \mathbf{a} = \mathbf{r}(t) \) and \( \mathbf{b} \approx \dot{\mathbf{r}}(t) \Delta t \).

Step 4: Express the area in terms of increments

In terms of increments, the area \( \Delta A \) can be expressed as: \( \Delta A \approx \frac{1}{2} | \mathbf{r}(t) \times [\dot{\mathbf{r}}(t) \Delta t] | = \frac{1}{2} | \mathbf{r}(t) \times \dot{\mathbf{r}}(t) | \Delta t \).

Step 5: Transition to the rate of change of area

The rate of change of area, \( \frac{dA}{dt} \), is the limit of \( \Delta A / \Delta t \) as \( \Delta t \to 0 \). Hence, \( \frac{dA}{dt} = \lim_{\Delta t \to 0} \frac{1}{2} | \mathbf{r} \times \dot{\mathbf{r}} | \Delta t / \Delta t = \frac{1}{2} | \mathbf{r} \times \dot{\mathbf{r}} | \).

Key Concepts

Cross Product

The cross product is a fundamental concept in vector calculus. It is a way to multiply two vectors in three-dimensional space to find another vector that is perpendicular to both. Imagine vectors like arrows pointing in space. The cross product is like finding a new arrow that stands straight up from the plane these two arrows make.

When you have two vectors, say \( \mathbf{a} \) and \( \mathbf{b} \), the cross product \( \mathbf{a} \times \mathbf{b} \) results in a vector that lies perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \). The magnitude, or length, of this resultant vector shows how much these two vectors "spread out" in their plane. It's found using the formula:

• \[ |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| |\mathbf{b}| \sin{\theta} \]

Here, \( \theta \) is the angle between the two vectors.

This concept is closely tied to finding areas. For instance, the area of the parallelogram formed by vectors \( \mathbf{a}\) and \( \mathbf{b} \) is given by the magnitude of their cross product. Similarly, the area of the triangle is half of that because a triangle is essentially half of a parallelogram.

Position Vector

A position vector is like a GPS pin on the map, marking where you are in a mathematical space. In the context of physics or mathematics, it points from the origin of your coordinate system to a particular location or particle.

• **Origin:** The starting point, typically \((0,0)\) in 2D or \((0,0,0)\) in 3D.
• **End Point:** The specific location the vector points to, usually given by coordinates \((x, y)\) or \((x, y, z)\).

Let's visualize it as a particle moving along a path. At any instant, the position vector \( \mathbf{r}(t) \) gives the exact location of the particle. It's like pressing pause on a moving object and finding exactly where it is frozen in time.

In our example, \( \mathbf{r}(t) \) changes with time as the particle moves, and the change tells us how the space is explored over an interval of time. The importance of position vectors lies in their ability to describe motion, geometry, and transformations within any given space.

Rate of Change

The rate of change concept is about measuring how one quantity changes concerning another. It's like watching how fast scenery passes as you're moving in a car.

• **Velocity:** For our problem, this is expressed by the velocity vector \( \dot{\mathbf{r}}(t) \), which tells how fast and in which direction the particle's position is changing.
• **Sweeping area:** The rate of change \( \frac{dA}{dt} \) explains how quickly the area is swept by the position vector as the particle moves.

This ties into calculus, where the derivative is a key component. Derivatives measure instantaneous rates of change. In the problem scenario, the derivative \( \frac{dA}{dt} \) is calculated as:

• \[ \frac{dA}{dt} = \frac{1}{2} | \mathbf{r} \times \dot{\mathbf{r}} | \]

This equation elegantly links the concept of velocity and space coverage. It demonstrates how much area the moving vector is sweeping over time, driven by the constant interaction of the vector and its velocity.