Problem 37 - Thomas Calculus

Question

Each of the following equations in parts (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. i. Does the particle have constant speed? If so, what is its constant speed? ii. Is the particle's acceleration vector always orthogonal to its velocity vector? iii. Does the particle move clockwise or counterclockwise around the circle? $$ \begin{array}{l}{\text { a. } \mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}, \quad t \geq 0} \\ {\text { b. } \mathbf{r}(t)=\cos (2 t) \mathbf{i}+\sin (2 t) \mathbf{j}, \quad t \geq 0} \\ {\text { c. } \mathbf{r}(t)=\cos (t-\pi / 2) \mathbf{i}+\sin (t-\pi / 2) \mathbf{j}, \quad t \geq 0} \\ {\text { d. } \mathbf{r}(t)=(\cos t) \mathbf{i}-(\sin t) \mathbf{j}, \quad t \geq 0} \\ {\text { e. } \mathbf{r}(t)=\cos \left(t^{2}\right) \mathbf{i}+\sin \left(t^{2}\right) \mathbf{j}, \quad t \geq 0}\end{array} $$

Step-by-Step Solution

Answer

(a) Constant speed: 1, Orthogonal: Yes, Direction: Counterclockwise. (b) Constant speed: 2, Orthogonal: Yes, Direction: Counterclockwise. (c) Constant speed: 1, Orthogonal: Yes, Direction: Counterclockwise. (d) Constant speed: 1, Orthogonal: Yes, Direction: Clockwise. (e) Variable speed, Not Orthogonal, Initial Direction: Counterclockwise.

Step 1: Find Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of \( \mathbf{r}(t) \). For each part, we calculate \( \mathbf{v}(t) \) as follows: For (a): \( \mathbf{v}(t) = (-\sin t)\mathbf{i} + (\cos t)\mathbf{j} \) For (b): \( \mathbf{v}(t) = -2\sin(2t)\mathbf{i} + 2\cos(2t)\mathbf{j} \) For (c): \( \mathbf{v}(t) = (-\sin(t-\pi/2))\mathbf{i} + \cos(t-\pi/2)\mathbf{j} \) For (d): \( \mathbf{v}(t) = (-\sin t)\mathbf{i} - (\cos t)\mathbf{j} \) For (e): \( \mathbf{v}(t) = (-2t\sin(t^2))\mathbf{i} + (2t\cos(t^2))\mathbf{j} \)

Step 2: Determine Speed

The speed is the magnitude of the velocity vector \( \|\mathbf{v}(t)\| \). For (a): \( \|\mathbf{v}(t)\| = \sqrt{(-\sin t)^2 + (\cos t)^2} = 1 \) hence constant speed.For (b): \( \|\mathbf{v}(t)\| = \sqrt{(-2\sin(2t))^2 + (2\cos(2t))^2} = 2 \) hence constant speed.For (c): \( \|\mathbf{v}(t)\| = \sqrt{(-\cos t)^2 + (\sin t)^2} = 1 \) hence constant speed.For (d): \( \|\mathbf{v}(t)\| = \sqrt{(-\sin t)^2 + (-\cos t)^2} = 1 \) hence constant speed.For (e): \( \|\mathbf{v}(t)\| = \sqrt{(-2t\sin(t^2))^2 + (2t\cos(t^2))^2} = 2t \) hence variable speed.

Step 3: Find Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of \( \mathbf{v}(t) \). For each part, calculate \( \mathbf{a}(t) \) as follows:For (a): \( \mathbf{a}(t) = (-\cos t)\mathbf{i} - (\sin t)\mathbf{j} \)For (b): \( \mathbf{a}(t) = (-4\cos(2t))\mathbf{i} - (4\sin(2t))\mathbf{j} \)For (c): \( \mathbf{a}(t) = (-\cos(t-\pi/2))\mathbf{i} - (\sin(t-\pi/2))\mathbf{j} \)For (d): \( \mathbf{a}(t) = -(\cos t)\mathbf{i} + (\sin t)\mathbf{j} \)For (e): \( \mathbf{a}(t) = (-4t^2\cos(t^2) - 2\sin(t^2))\mathbf{i} + (-4t^2\sin(t^2) + 2\cos(t^2))\mathbf{j} \)

Step 4: Check Orthogonality of Velocity and Acceleration

Two vectors are orthogonal if their dot product is zero. Find the dot product of \( \mathbf{v}(t) \) and \( \mathbf{a}(t) \):For (a): \( \mathbf{v}(t) \cdot \mathbf{a}(t) = (-\sin t)(-\cos t) + (\cos t)(-\sin t) = 0 \) (orthogonal).For (b): \( \mathbf{v}(t) \cdot \mathbf{a}(t) = (-2\sin(2t))(-4\cos(2t)) + (2\cos(2t))(-4\sin(2t)) = 0 \) (orthogonal).For (c): \( \mathbf{v}(t) \cdot \mathbf{a}(t) = (-\cos t)(-\cos t) + (\sin t)(-\sin t) = 0 \) (orthogonal).For (d): \( \mathbf{v}(t) \cdot \mathbf{a}(t) = (-\sin t)(-\cos t) + (-\cos t)(\sin t) = 0 \) (orthogonal).For (e): Examine individual components (non-zero terms can exist, indicating non-orthogonality).

Step 5: Determine Direction (Clockwise or Counterclockwise)

Check the sign of velocity components to determine rotation direction:For (a): \( (-\sin t, \cos t) \rightarrow \) counterclockwise.For (b): \( (-2\sin(2t), 2\cos(2t)) \rightarrow \) counterclockwise.For (c): shift by \( \pi/2 \) means starts at \((0,1)\) and spins counterclockwise.For (d): \( (-\sin t, -\cos t) \rightarrow \) clockwise.For (e): Non-constant components indicate varying behavior, but initial rotation is similar to counterclockwise.

Key Concepts

Constant Speed

When discussing the motion of a particle, one crucial concept is whether it has a constant speed. Speed is a measure of how fast a particle is moving regardless of its direction. It's calculated by finding the magnitude of the velocity vector.

• For the particles moving along the unit circle, a constant speed indicates that they move uniformly around the circle.
  
• From the solutions provided, we observe that in parts (a), (b), (c), and (d), the particles indeed travel at constant speeds — their speeds are 1 and 2 respectively.
• In part (e), however, the speed varies because it's dependent on time, expressed as \(2t\), meaning the particle's speed increases as it moves along the circle.

Understanding constant speed helps in predicting the time it will take for the particle to complete one revolution around the circle.

Acceleration Vector

The acceleration vector tells us how the velocity of a particle changes over time. For a particle moving in a circular path, acceleration is always directed towards the center.

• This concept is important for understanding how the particle is influenced by forces causing it to change its motion.
• In the given solutions, the acceleration vectors were derived by differentiating the velocity vectors for each part.
• The acceleration vectors display changes in direction but are at constant magnitude for each corresponding constant speed case.

Recognizing the acceleration helps in describing the dynamic behavior of particles, especially within circular motion.

Velocity Vector

Velocity is a vector quantity having both magnitude (speed) and direction. It's a derivative of the position vector and shows how position changes over time.

• For the unit circle, the velocity vectors for parts (a)-(d) were calculated and showed a consistent directional change.
• The direction of these vectors indicates the tangent to the path, pointing in the direction the particle is moving.

Grasping the nature of velocity vectors clarifies why particles in certain parts of the problem have velocities that exhibit uniform circular motion.

Orthogonality

Orthogonality is a condition where two vectors are perpendicular to each other. This occurs when their dot product equals zero.

• In circular motion, if the acceleration vector is orthogonal to the velocity vector, the motion is likely uniform with only direction changes and no speed alteration.
• From the solutions, parts (a)-(d) demonstrated orthogonal acceleration and velocity vectors, confirming constant speed.

Recognizing orthogonality in these vectors is key to understanding how particles maintain circular paths without speeding up or slowing down.

Clockwise or Counterclockwise

Determining the direction of a particle's motion is essential in analyzing circular dynamics. This can be observed by checking the sign of velocity components.

• When the velocity components show positive and negative value shifts in a specific order, it determines the rotational direction.
• For instance, parts (a), (b), and (c) move counterclockwise as indicated by their velocity components' arrangement.
• On the other hand, part (d) exhibits clockwise rotation due to reversed velocity vectors.

Understanding these directions is helpful for solving problems involving torque or angular velocity in circular motion scenarios.