Problem 13 - Thomas Calculus

Question

In Exercises \(13-18, \mathbf{r}(t)\) is the position of a particle in space at time \(t\) . Find the particle's velocity and acceleration vectors. Then find the particle's speed and direction of motion at the given value of \(t\) . Write the particle's velocity at that time as the product of its speed and direction. $$ \mathbf{r}(t)=(t+1) \mathbf{i}+\left(t^{2}-1\right) \mathbf{j}+2 t \mathbf{k}, \quad t=1 $$

Step-by-Step Solution

Answer

Velocity: \( \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \), Acceleration: \( 2\mathbf{j} \), Speed: 3, Direction: \( \frac{1}{3}(\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) \).

Step 1: Find the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). Calculate the derivative: \[ \mathbf{v}(t) = \frac{d}{dt}((t+1) \mathbf{i} + (t^2 - 1) \mathbf{j} + 2t \mathbf{k}) \] Computing each part: - \( \frac{d}{dt}(t+1) \mathbf{i} = 1 \mathbf{i} \)- \( \frac{d}{dt}(t^2 - 1) \mathbf{j} = 2t \mathbf{j} \)- \( \frac{d}{dt}(2t) \mathbf{k} = 2 \mathbf{k} \)Therefore, \[ \mathbf{v}(t) = \mathbf{i} + 2t \mathbf{j} + 2 \mathbf{k} \] At \( t = 1 \), the velocity vector is: \[ \mathbf{v}(1) = \mathbf{i} + 2 \mathbf{j} + 2 \mathbf{k} \]

Step 2: Find the Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \). Calculate the derivative: \[ \mathbf{a}(t) = \frac{d}{dt}(\mathbf{i} + 2t \mathbf{j} + 2 \mathbf{k}) \] Computing each part: - \( \frac{d}{dt}(\mathbf{i}) = 0 \mathbf{i} \)- \( \frac{d}{dt}(2t \mathbf{j}) = 2 \mathbf{j} \)- \( \frac{d}{dt}(2 \mathbf{k}) = 0 \mathbf{k} \)Therefore, \[ \mathbf{a}(t) = 2 \mathbf{j}\]

Step 3: Calculate the Speed

The speed of the particle is the magnitude of the velocity vector. Calculate the magnitude of \( \mathbf{v}(1) = \mathbf{i} + 2 \mathbf{j} + 2 \mathbf{k} \): \[ \text{Speed} = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3 \]

Step 4: Determine the Direction of Motion

The direction of motion is given by the unit vector of the velocity vector. Find the unit vector by dividing \( \mathbf{v}(1) \) by its magnitude (speed):\[ \text{Direction} = \frac{1}{3}(\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})\]

Step 5: Express Velocity as Product of Speed and Direction

Express the velocity vector at \( t=1 \) as the product of its speed and direction:\[ \mathbf{v}(1) = 3 \left(\frac{1}{3}(\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})\right) = 3 \times \frac{1}{3}(\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) \]This confirms \( \mathbf{v}(1) = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \).

Key Concepts

Velocity Vector

The velocity vector plays a crucial role in understanding particle motion. It's essentially the rate at which a particle's position changes over time. In mathematical terms, the velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). This derivative gives us the components of the velocity vector in the different coordinate directions (\( \mathbf{i}, \mathbf{j}, \mathbf{k} \)).

• The derivative of \( (t+1) \mathbf{i} \) with respect to \( t \) is \( \mathbf{i} \).
• The derivative of \( (t^2 - 1) \mathbf{j} \) is \( 2t \mathbf{j} \).
• The derivative of \( 2t \mathbf{k} \) is \( 2 \mathbf{k} \).

Therefore, the velocity vector becomes
\[ \mathbf{v}(t) = \mathbf{i} + 2t \mathbf{j} + 2 \mathbf{k} \].
When we substitute \( t = 1 \), we get
\[ \mathbf{v}(1) = \mathbf{i} + 2 \mathbf{j} + 2 \mathbf{k} \].
This vector indicates the direction and rate of movement of the particle at any moment.

Acceleration Vector

The acceleration vector is another fundamental concept in describing particle motion. It measures the rate at which the velocity of a particle changes with time. To find the acceleration vector \( \mathbf{a}(t) \), we take the derivative of the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \).

• The derivative of \( \mathbf{i} \) is \( 0 \mathbf{i} \), showing there's no change in the \( \mathbf{i} \)-direction.
• The derivative of \( 2t \mathbf{j} \) is \( 2 \mathbf{j} \), indicating a constant increase in the \( \mathbf{j} \)-direction.
• The derivative of \( 2 \mathbf{k} \) is \( 0 \mathbf{k} \), denoting no change in the \( \mathbf{k} \)-direction.

Thus, the acceleration vector simplifies to \( \mathbf{a}(t) = 2 \mathbf{j} \).
This tells us the only component of the particle's motion that is accelerating is along the \( \mathbf{j} \)-direction.

Speed

Speed is different from velocity. While velocity has both magnitude and direction, speed only concerns magnitude. It describes how fast a particle is moving at any given instant without considering its direction.
To determine the particle's speed, we calculate the magnitude of the velocity vector. For \( \mathbf{v}(1) = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \), the magnitude is found by computing:
\[ \text{Speed} = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3 \].
This tells us that at \( t = 1 \), the particle is moving at a speed of 3 units of distance per unit of time.

Direction of Motion

The direction of motion refers to the path that a particle is heading at a particular instant. Unlike speed, the direction of motion is independent of how fast the particle is moving.
To find the exact direction in which the particle is moving at \( t = 1 \), we calculate the unit vector of the velocity vector. A unit vector is found by dividing the velocity vector by its magnitude (or speed).
Given \( \mathbf{v}(1) = \mathbf{i} + 2 \mathbf{j} + 2 \mathbf{k} \), the unit vector representing the direction of motion is:
\[ \text{Direction} = \frac{1}{3} (\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) \].
This unit vector confirms that the particle is traveling in a direction that is determined by the components \( \mathbf{i}, \mathbf{j}, \) and \( \mathbf{k} \), appropriately scaled by the factor of \( \frac{1}{3} \). It tells us the particle's path but doesn't account for the speed of travel.