Problem 38 - Thomas Calculus

Question

Motion along a circle Show that the vector-valued function \(\mathbf{r}(t)=(2 \mathbf{i}+2 \mathbf{j}+\mathbf{k})\) $$ +\cos t\left(\frac{1}{\sqrt{2}} \mathbf{i}-\frac{1}{\sqrt{2}} \mathbf{j}\right)+\sin t\left(\frac{1}{\sqrt{3}} \mathbf{i}+\frac{1}{\sqrt{3}} \mathbf{j}+\frac{1}{\sqrt{3}} \mathbf{k}\right) $$ describes the motion of a particle moving in the circle of radius 1 centered at the point \((2,2,1)\) and lying in the plane \(x+y-2 z=2\) .

Step-by-Step Solution

Answer

The vector function describes a circle of radius 1 centered at (2,2,1) in the plane x+y-2z=2.

Step 1: Identify the Fixed Part of the Vector Function

The constant vector part of the function is \((2 \mathbf{i} + 2 \mathbf{j} + \mathbf{k})\). This indicates that the circle is centered at the point \((2, 2, 1)\).

Step 2: Analyze the Circular Motion Part

The expression for the circular motion is given by:\[\cos t \left(\frac{1}{\sqrt{2}} \mathbf{i} - \frac{1}{\sqrt{2}} \mathbf{j}\right) + \sin t \left(\frac{1}{\sqrt{3}} \mathbf{i} + \frac{1}{\sqrt{3}} \mathbf{j} + \frac{1}{\sqrt{3}} \mathbf{k}\right)\]

Step 3: Consider the Coefficients

We see two orthogonal vectors forming the circular path as produced by the sine and cosine terms. Vector \(\left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0\right)\) and vector \(\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\) are orthogonal, ensuring a circular path.

Step 4: Verify the Radial Distance

To verify the radius, we calculate the magnitude of the vector comprising sine and cosine terms: \[\sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2} \]which simplifies to 1, confirming the radius of the circle is 1.

Step 5: Check Plane Condition

The plane equation is \(x + y - 2z = 2\). To check if the motion lies in this plane, consider the parametric position: \((x, y, z) = (2, 2, 1) + ...\). Substituting in the plane equation shows every point satisfies the plane equation hence the motion lies in this plane.

Key Concepts

Circular Motion

Circular motion refers to the movement of a particle along a circular path. In the context of vector-valued functions, this can be expressed with trigonometric functions like sine and cosine.
The function \( \mathbf{r}(t) = (2 \mathbf{i} + 2 \mathbf{j} + \mathbf{k}) + \cos t \left(\frac{1}{\sqrt{2}} \mathbf{i} - \frac{1}{\sqrt{2}} \mathbf{j}\right) + \sin t \left(\frac{1}{\sqrt{3}} \mathbf{i} + \frac{1}{\sqrt{3}} \mathbf{j} + \frac{1}{\sqrt{3}} \mathbf{k}\right) \) is designed to describe circular motion.

• The cosine and sine components determine the direction and distance from the center of the circle for each part of the function.
• The movement remains at a constant radius from the center, confirmed by the magnitude calculation \(\sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2} = 1 \).

The structure ensures that a circle with a radius of 1, centered at (2, 2, 1), guides the motion.

Orthogonal Vectors

Orthogonal vectors are vectors that are at right angles to each other. This is an important concept in defining the circular path.
To depict smooth circular motion, we look at two main vectors:

• \( \left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0\right) \) represents a horizontal component.
• \( \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \) stands for a radial component that includes a vertical aspect.

These vectors, found in the sine and cosine terms, are orthogonal.
This means when used together in the vector function, they form a perfect circle as their cross product results in zero, maintaining their orthognality.
Their magnitude contributes to defining the correct radius for the circular path.

Parametric Position

Parametric position refers to representing a particle's position using parameters, typically time. In our vector function,

• \( (2 \mathbf{i} + 2 \mathbf{j} + \mathbf{k}) \) defines the center point.
• The sine and cosine components define offsets from this center point, changing with time \( t\).

This time-dependent representation is useful for describing motion, as each \( t \) value gives a unique position.
These changes define the circle's path dynamically and continuously, ensuring the particle's path is correctly mapped.
Additionally, the parametric method maps this motion accurately to fit within the context of physics, such as for circular trajectories in space.

Plane Conditions

Plane conditions are the requirements for motion or points to lie within a certain plane. To verify the path of our vector function, we check against the plane equation \( x + y - 2z = 2 \).
The plane condition ensures that all measured points from the vector function adhere to this plane.
When substituting the parametric position into the plane equation, we confirm:

• The constant vector part \( (2, 2, 1) \) fits the plane.
• Trigonometric components, when broken down and checked, also satisfy the plane conditions due to repetitive sinusoidal paths.

Thus, the path indeed lies entirely in the given plane.
This verification gives confidence in the function's capability to describe not only circular motion but doing so within a constraint plane environment, critical in many engineering and physics applications.