Problem 37 - Thomas Calculus

Question

Motion along a circle Each of the following equations in parts \((\mathrm{a})\) - (e) describes the motion of a particle having the same path, namely the unit circle \(x^{2}+y^{2}=1 .\) Although the path of each particle in parts \((a)-(e)\) is the same, the behavior, or "dynamics," of each particle is different. For each particle, answer the following questions. \(\begin{array}{l}{\text { i) Does the particle have constant speed? If so, what is its con- }} \\ {\text { stant speed? }} \\ {\text { ii) Is the particle's acceleration vector always orthogonal to its }} \\ {\text { velocity vector? }}\end{array}\) \(\begin{array}{l}{\text { iii) Does the particle move clockwise or counterclockwise around }}\\\ {\text { the circle? }} \\ {\text { iv) Does the particle begin at the point }(1,0) ?}\end{array}\) \(\begin{array}{l}{\text { a. } \mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}, \quad t \geq 0} \\ {\text { b. } \mathbf{r}(t)=\cos (2 t) \mathbf{i}+\sin (2 t) \mathbf{j}, \quad t \geq 0} \\ {\text { c. } \mathbf{r}(t)=\cos (t-\pi / 2) \mathbf{i}+\sin (t-\pi / 2) \mathbf{j}, \quad t \geq 0} \\ {\text { d. } \mathbf{r}(t)=(\cos t) \mathbf{i}-(\sin t) \mathbf{j}, \quad t \geq 0} \\ {\text { e. } \mathbf{r}(t)=\cos \left(t^{2}\right) \mathbf{i}+\sin \left(t^{2}\right) \mathbf{j}, \quad t \geq 0}\end{array}\)

Step-by-Step Solution

Answer

Particles a, b, and e have constant speeds of 1, 2, and varying, respectively; a, b, c, and d are orthogonal; a, b, and e initially move counterclockwise; c starts at (0,-1).

Step 1: Determine Velocity and Speed

For each function \(\mathbf{r}(t)\), the velocity \(\mathbf{v}(t)\) is the derivative of \(\mathbf{r}(t)\) with respect to time, \(t\). The speed is the magnitude of this velocity vector, \(||\mathbf{v}(t)||\). Derivatives are applied to get velocities: - **a.** \(\mathbf{v}(t) = \frac{d}{dt}((\cos t) \mathbf{i}+ (\sin t) \mathbf{j}) = -\sin t\, \mathbf{i} + \cos t\, \mathbf{j}\). Speed \(||\mathbf{v}(t)|| = \sqrt{(-\sin t)^2 + (\cos t)^2} = 1\) which means constant speed.- **b.** \(\mathbf{v}(t) = -2 \sin(2t) \mathbf{i} + 2 \cos(2t) \mathbf{j}\). Speed \(||\mathbf{v}(t)|| = 2\), constant.- **c.** \(\mathbf{v}(t) = \sin(t-\pi/2)\, \mathbf{i} + \cos(t-\pi/2)\, \mathbf{j}\) yielding a speed of \(1\), constant.- **d.** \(\mathbf{v}(t) = -\sin t\, \mathbf{i} - \cos t\, \mathbf{j}\). Speed \(||\mathbf{v}(t)|| = 1\), constant.- **e.** \(\mathbf{v}(t) = -2t \sin(t^2) \mathbf{i} + 2t \cos(t^2) \mathbf{j}\). Speed varies with \(||\mathbf{v}(t)|| = 2t\).

Step 2: Check Orthogonality of Acceleration and Velocity

To check if the acceleration is orthogonal to the velocity, compute the acceleration \(\mathbf{a}(t)\) which is the derivative of \(\mathbf{v}(t)\) with respect to \(t\), and then check \(\mathbf{v}(t) \cdot \mathbf{a}(t) = 0\):- **a.** \(\mathbf{a}(t) = -\cos t\, \mathbf{i} - \sin t\, \mathbf{j}\); \(\mathbf{v}(t) \cdot \mathbf{a}(t) = 0\), orthogonal.- **b.** \(\mathbf{a}(t) = -4\cos(2t) \mathbf{i} - 4\sin(2t) \mathbf{j}\); \(\mathbf{v}(t) \cdot \mathbf{a}(t) = 0\), orthogonal.- **c.** \(\mathbf{a}(t) = -\cos(t-\pi/2) \mathbf{i} - \sin(t-\pi/2) \mathbf{j}\); \(\mathbf{v}(t) \cdot \mathbf{a}(t) = 0\), orthogonal.- **d.** \(\mathbf{a}(t) = -\cos t \mathbf{i} + \sin t \mathbf{j}\); \(\mathbf{v}(t) \cdot \mathbf{a}(t) = 0\), orthogonal.- **e.** \(\mathbf{a}(t) = \left(-4t^2 \cos(t^2) - 2 \sin(t^2)\right) \mathbf{i} + \left(-4t^2 \sin(t^2) + 2 \cos(t^2)\right) \mathbf{j}\); Not always orthogonal unless at \(t=0\).

Step 3: Determine the Direction of Motion

To find whether the motion is clockwise or counterclockwise, determine the cross product direction or observe the rotation:- **a.** Vector \((\cos t, \sin t)\) moves counterclockwise.- **b.** Vector \((\cos(2t), \sin(2t))\) moves twice as fast counterclockwise.- **c.** With \(t - \frac{\pi}{2}\) phase shift, it remains counterclockwise.- **d.** Negating the sine function results in a clockwise rotation.- **e.** As \(t^2\) increases, direction is variable but moves counterclockwise initially as \(t^2\) increases like positive \(t\).

Step 4: Confirm Initial Position

To see if the particle starts at \((1,0)\), calculate \(\mathbf{r}(0)\):- **a.** \(\mathbf{r}(0) = (1,0)\): starts at \((1,0)\).- **b.** \(\mathbf{r}(0) = (1,0)\): starts at \((1,0)\).- **c.** \(\mathbf{r}(0) = (0, -1)\): does not start at \((1,0)\).- **d.** \(\mathbf{r}(0) = (1,0)\): starts at \((1,0)\).- **e.** \(\mathbf{r}(0) = (1,0)\): starts at \((1,0)\).

Key Concepts

Velocity Vector

Understanding the velocity vector is crucial in analyzing particle motion along a path. In mathematical terms, the velocity vector, \( \mathbf{v}(t) \), is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). Essentially, it indicates the direction and speed at which a particle moves at any given time.
For example, for a particle moving along the unit circle, let’s examine function (a): \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} \). The velocity vector can be calculated as:
\[ \mathbf{v}(t) = \frac{d}{dt} ((\cos t) \mathbf{i} + (\sin t) \mathbf{j}) = -\sin t \mathbf{i} + \cos t \mathbf{j} \]
Note:

• The result \(-\sin t \mathbf{i} + \cos t \mathbf{j}\) shows how the particle moves momentarily, indicating changes in the x and y direction.
• This vector not only gives us the instantaneous direction of the particle, but also hints at its speed when analyzed further.

Acceleration Vector

The acceleration vector represents the rate of change of the velocity vector over time. This vector, \( \mathbf{a}(t) \), offers insight into how the velocity changes, i.e., how the speed or direction of the particle is adjusted.
For instance, looking at example (a) again:
\[ \mathbf{a}(t) = \frac{d}{dt} (-\sin t \mathbf{i} + \cos t \mathbf{j}) = -\cos t \mathbf{i} - \sin t \mathbf{j} \]
This points out how much faster or in what new direction the velocity is shifting. Importantly:

• Orthogonality between velocity and acceleration (i.e., if \( \mathbf{v}(t) \cdot \mathbf{a}(t) = 0 \)) usually implies uniform circular motion. It verifies that the particle keeps to a predetermined path without spiraling outward or inward.

For particles moving in the unit circle, this condition of orthogonality typically holds, securing their constant circular path.

Unit Circle

The unit circle, defined by \( x^2 + y^2 = 1 \), plays a significant role in understanding particle motion. It is a circle of radius 1 centered at the origin of a coordinate system. When describing particle motion around the unit circle, trigonometric functions like sine and cosine often present themselves in parametric equations.
Consider a particle described by \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} \). This formulation ensures:

• The particle stays on the circle since substituting \( \cos t \) and \( \sin t \) into \( x^2 + y^2 = 1 \) renders this equation true for all \( t \).
• Positions around the circle are perfectly expressed by varying the parameter \( t \), meaning everything fits predictably into circular motion.

Having a firm grasp of the unit circle provides foundational understanding in topics like periodic motion and can aid in comprehending trigonometric relationships.

Constant Speed

In circular motion, the concept of constant speed implies that while the particle traverses around a path (here, the unit circle), its speed remains unchanged. However, the direction continually adjusts - a hallmark of rotational dynamics.
When determining constant speed mathematically, the magnitude of the velocity vector should be invariant with respect to time \( t \). Take \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} \) as an example:
\[ ||\mathbf{v}(t)|| = \sqrt{(-\sin t)^2 + (\cos t)^2} = 1 \]
This calculation gives a consistent speed of 1, which reinforces its constancy while the directional components alter.
Keep in mind:

• Constant speed isn't the same as constant velocity because velocity accounts for direction. In a circular path, speed can be constant while velocity isn't since direction changes.

This understanding is pivotal for navigating problems involving biomechanics, physics, and engineering where rotation and revolution are in play.