Problem 10 - 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 $$ \frac{d A}{d t}=\frac{1}{2}|\mathbf{r} \times \dot{\mathbf{r}}| $$

Step-by-Step Solution

Answer

The equation is valid because the cross product's magnitude represents twice the parallelogram's area, and half gives the triangle's area rate.

Step 1: Understanding the Vector Cross Product

The cross product \( \mathbf{r} \times \dot{\mathbf{r}} \) represents a vector that is perpendicular to both of the vectors \( \mathbf{r} \) and \( \dot{\mathbf{r}} \). The magnitude of this cross product represents twice the area of the parallelogram formed by the vectors \( \mathbf{r} \) and \( \dot{\mathbf{r}} \), as the area of a parallelogram is given by the base times the height, which is captured by the magnitude of the cross product.

Step 2: Relating the Area Swept Out to the Triangle Area

Consider the position vector \( \mathbf{r} \) at a point along its trajectory in the plane. As \( \mathbf{r} \) moves, it forms a triangle between its initial and later positions. The area \( A \) swept by \( \mathbf{r} \) as it moves through a small angle is approximately half the area of this parallelogram.

Step 3: Establishing the Rate of Change of Area

The area \( dA \) swept by the vector \( \mathbf{r} \) in an infinitesimally small time interval \( dt \) is given by the area of the triangle, which is \( \frac{1}{2} \) of the area of the parallelogram formed by \( \mathbf{r} \) and \( \dot{\mathbf{r}} \). Therefore, the rate at which the area is swept out is \( \frac{dA}{dt} = \frac{1}{2} | \mathbf{r} \times \dot{\mathbf{r}} | \).

Step 4: Conclusion: Equation Validity

The given equation \( \frac{dA}{dt} = \frac{1}{2} |\mathbf{r} \times \dot{\mathbf{r}}| \) is validated by the observation that the magnitude of the cross product represents twice the area of the parallelogram formed, and as described in the geometric context, half of this gives the rate of change of the triangular area swept out by the vector \( \mathbf{r} \).

Key Concepts

Cross Product

The cross product, also known as the vector product, is a fundamental operation in vector calculus, particularly when dealing with three-dimensional vectors. The cross product of two vectors, say \( \mathbf{a} \) and \( \mathbf{b} \), is denoted as \( \mathbf{a} \times \mathbf{b} \). This operation results in a new vector that is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \).

Some key properties of the cross product include:

• The magnitude of \( \mathbf{a} \times \mathbf{b} \) is equal to the area of the parallelogram formed by vectors \( \mathbf{a} \) and \( \mathbf{b} \).
• The direction of the cross product follows the right-hand rule, meaning if you point your index finger in the direction of \( \mathbf{a} \) and your middle finger in the direction of \( \mathbf{b} \), your thumb points in the direction of \( \mathbf{a} \times \mathbf{b} \).

Understanding these properties is crucial, particularly in contexts where you need to determine a perpendicular vector or compute areas related to vectors.

Area of a Parallelogram

The area of a parallelogram formed by vectors is directly tied to the notion of the cross product. When you have two vectors, say \( \mathbf{u} \) and \( \mathbf{v} \), positioned tail to tail, they can span a parallelogram. The formula for the area of this parallelogram is given by the magnitude of their cross product \( |\mathbf{u} \times \mathbf{v}| \).

This geometric interpretation links the algebraic operation of the cross product with a visual model:

• Imagine vectors \( \mathbf{u} \) and \( \mathbf{v} \) on a plane; they stretch to form the sides of the parallelogram.
• The cross product \( \mathbf{u} \times \mathbf{v} \) not only gives you a vector perpendicular to this plane but also the magnitude indicating twice the area of the parallelogram.

The concept of using cross products to find areas is useful in physics and engineering, particularly in calculations involving moments and forces.

Geometric Interpretation

Geometric interpretation helps in visualizing and understanding mathematical concepts. For the relationship between area and vectors, geometric interpretation involves imagining how a vector \( \mathbf{r} \) moves over time to sweep out area in a plane.

Consider the following:

• The vector \( \mathbf{r} \) represents a moving particle's position on a plane, tracing out a trajectory.
• As \( \mathbf{r} \) moves incrementally, it entails a region in the plane forming a sequential array of infinitesimal triangles.

By bisecting the parallelogram created by the initial and final positions of \( \mathbf{r} \), you understand that the total area swept is actually half of the larger parallelogram's area. This is a simple yet powerful way to connect algebraic cross product calculations to their geometric counterparts.

Rate of Change

In vector calculus, the rate of change is integral to understanding dynamic systems. It refers to how a certain quantity changes over time, often represented as a derivative.

When focusing on the rate at which area is swept out by a vector, \( \frac{dA}{dt} \) describes this change. The relationship:

• \( dA \) indicates the infinitesimal piece of area covered in an infinitesimal time \( dt \).
• The equation \( \frac{dA}{dt} = \frac{1}{2} |\mathbf{r} \times \dot{\mathbf{r}}| \) signifies that the rate of area change is half the magnitude of the cross product of \( \mathbf{r} \) and its derivative \( \dot{\mathbf{r}} \).

Here, the idea is to understand that velocities (via derivatives) are being mapped back to understand how they impact area, a principle quite essential for fields like physics.

Position Vector

A position vector describes the location of a point in space relative to an origin. It's a fundamental concept in both physics and vector calculus.

Key aspects of position vectors include:

• The vector \( \mathbf{r} \) is often used to depict position in space, with components describing each dimension, such as \( (x, y, z) \) in a three-dimensional space.
• A time-dependent position vector, \( \mathbf{r}(t) \), describes how a point moves over time, with \( \dot{\mathbf{r}}(t) \) representing its velocity.

This vector formulation is essential for mapping trajectories, understanding motion, and solving problems involving dynamic systems, where knowing how a position vector changes with time helps unravel the nature of motion.