Problem 26 - Thomas Calculus

Question

Motion along a cycloid A particle moves in the \(x y\) -plane in such a way that its position at time \(t\) is \begin{equation} \mathbf{r}(t)=(t-\sin t) \mathbf{i}+(1-\cos t) \mathbf{j} \end{equation} \(\begin{array}{l}{\text { a. Graph } \mathbf{r}(t) . \text { The resulting curve is a cycloid. }} \\ {\text { b. Find the maximum and minimum values of }|\mathbf{v}| \text { and }|\mathbf{a}| \text { . }} \\ {\text { (Hint: Find the extreme values of }|\mathbf{v}|^{2} \text { and }|\mathbf{a}|^{2} \text { first and take }} \\ {\text { square roots later.) }}\end{array}\)

Step-by-Step Solution

Answer

The graph is a cycloid. The maximum speed is 2 and the acceleration is constant at 1.

Step 1: Express the function

The position vector given for the particle, as a function of time \( t \), is \( \mathbf{r}(t) = (t - \sin t) \mathbf{i} + (1 - \cos t) \mathbf{j} \). Each component describes how the position changes with time in the \( x \) and \( y \) directions.

Step 2: Graph the Cycloid

To graph the cycloid, plot \( x(t) = t - \sin t \) and \( y(t) = 1 - \cos t \) over a range of \( t \) values. Generally, a cycloid starts at the origin, rises to a maximum point, and forms a series of arches as \( t \) increases.

Step 3: Find the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of \( \mathbf{r}(t) \) with respect to \( t \): \[ \mathbf{v}(t) = \left(1 - \cos t\right) \mathbf{i} + \sin t \mathbf{j}. \]

Step 4: Compute the Speed

The speed \( |\mathbf{v}(t)| \) is the magnitude of the velocity vector:\[ |\mathbf{v}(t)| = \sqrt{(1 - \cos t)^2 + (\sin t)^2}. \]Simplifying gives:\[ |\mathbf{v}(t)| = \sqrt{2(1 - \cos t)} = 2\sin(\frac{t}{2}). \]

Step 5: Derivative for Speed Extremes

To find the extremes of \( |\mathbf{v}(t)|^2 = 4\sin^2(\frac{t}{2}) \), note that it ranges from \(0\) to \(4\) as \(\sin(\frac{t}{2})\) varies from \(0\) to \(1\).

Step 6: Find the Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of \( \mathbf{v}(t) \):\[ \mathbf{a}(t) = \sin t \mathbf{i} + \cos t \mathbf{j}. \]

Step 7: Compute the Magnitude of Acceleration

The magnitude of acceleration \( |\mathbf{a}(t)| \) is:\[ |\mathbf{a}(t)| = \sqrt{(\sin t)^2 + (\cos t)^2} = 1. \]Since the magnitude of acceleration is constant, it has no maximum or minimum other than 1.

Key Concepts

Velocity Vector

When discussing the motion along a cycloid, understanding the velocity vector is key. The velocity vector, represented as \(\mathbf{v}(t)\), gives the rate of change of the position vector \(\mathbf{r}(t)\) with respect to time \(t\). This vector shows how fast and in which direction the point is moving at any point in time.
To find \(\mathbf{v}(t)\), we must derive the position vector, \(\mathbf{r}(t) = (t - \sin t) \mathbf{i} + (1 - \cos t) \mathbf{j}\). Taking the derivative with respect to \(t\) yields \(\mathbf{v}(t) = (1 - \cos t) \mathbf{i} + \sin t \mathbf{j}\). This means the velocity depends on sine and cosine functions, capturing the periodic nature of the cycloid motion.

Another important aspect is the speed of the object, which is determined by the magnitude of the velocity vector. For our cycloid, the speed \(|\mathbf{v}(t)| = \sqrt{(1 - \cos t)^2 + (\sin t)^2}\), which simplifies to \(|\mathbf{v}(t)| = 2\sin(\frac{t}{2})\). The speed varies from 0 to 2, illustrating the changing rate of motion along the cycloid path.

Acceleration Vector

In kinematics, the acceleration vector highlights the change in velocity as a function of time. For our cycloid example, the acceleration vector \(\mathbf{a}(t)\) is found by deriving the velocity vector \(\mathbf{v}(t)\). Thus, \(\mathbf{a}(t) = \sin t \mathbf{i} + \cos t \mathbf{j}\).
The acceleration vector reflects how the velocity is adjusted in response to the cycloid path. In this scenario, the expressions involve sine and cosine, indicating circular motion influences. Observing the magnitude \(|\mathbf{a}(t)|\), we derive it as \(\sqrt{(\sin t)^2 + (\cos t)^2} = 1\), which is constant. This illustrates that the acceleration, although always present, maintains a steady value throughout the motion.
This consistency in magnitude makes the motion easier to predict and analyze, as changes in direction occur smoothly without abrupt accelerations or decelerations.

Graphing Parametric Equations

Graphing the motion described by the parametric equations like a cycloid requires understanding the behavior of each component. Here, the equations \(x(t) = t - \sin t\) and \(y(t) = 1 - \cos t\) define the path as a series of loops or arches as \(t\) progresses.

- Start by varying \(t\) over a range, typically from 0 onwards, observing how the position \((x, y)\) evolves in the plane.
- The unique shape, the cycloid, emerges due to the combined sinusoidal effects in both components. At \(t = 0\), the curve starts at the origin \((0,0)\).

Each loop of the cycloid illustrates a complete cycle of motion where the particle rolls along the path, its height determined by the cosine terms and distance by sine. The cycloid is not just a mathematical curiosity but a practical path studied for optimizing motion in physics and engineering, recognized for its efficient traversal and minimal energy paths.