Problem 9 - Thomas Calculus

Question

In Exercises \(9-14, \mathbf{r}(t)\) 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. \begin{equation} \mathbf{r}(t)=(t+1) \mathbf{i}+\left(t^{2}-1\right) \mathbf{j}+2 t \mathbf{k}, \quad t=1 \end{equation}

Step-by-Step Solution

Answer

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

Step 1: Calculate the Velocity Vector

The velocity vector, denoted as \(\mathbf{v}(t)\), is the derivative of the position vector \(\mathbf{r}(t)\) with respect to time \(t\). We find the derivative: \[ \mathbf{v}(t) = \frac{d}{dt}[(t+1) \mathbf{i} + (t^2 - 1) \mathbf{j} + 2t \mathbf{k}] = \mathbf{i} + 2t \mathbf{j} + 2 \mathbf{k}. \] Substituting \(t=1\), we get \(\mathbf{v}(1) = \mathbf{i} + 2 \cdot 1 \mathbf{j} + 2 \mathbf{k} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}.\)

Step 2: Calculate the Acceleration Vector

The acceleration vector, denoted as \(\mathbf{a}(t)\), is the derivative of the velocity vector \(\mathbf{v}(t)\). We find the derivative: \[ \mathbf{a}(t) = \frac{d}{dt}[\mathbf{i} + 2t \mathbf{j} + 2 \mathbf{k}] = 2 \mathbf{j}. \] Thus, the acceleration vector is constant and equals \(2 \mathbf{j}\).

Step 3: Calculate the Particle's Speed

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

Step 4: Determine the Direction of Motion

The direction of motion is the unit vector in the direction of \(\mathbf{v}(1)\). We find it by dividing \(\mathbf{v}(1)\) by its magnitude: \[ \text{Direction} = \frac{\mathbf{v}(1)}{\|\mathbf{v}(1)\|} = \frac{\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}}{3} = \frac{1}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} + \frac{2}{3}\mathbf{k}. \]

Step 5: Express the Velocity as the Product of Speed and Direction

We express the velocity at \(t=1\) as the product of its speed and direction: \[ \mathbf{v}(1) = 3 \left( \frac{1}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} + \frac{2}{3}\mathbf{k} \right) = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}. \] Thus, the velocity \(\mathbf{v}(1)\) can be expressed in this form correctly.

Key Concepts

Velocity Vector

The velocity vector of a particle describes both how fast and in what direction the particle is moving at any given point in time. In our exercise, finding the velocity vector involved taking the derivative of the position vector \(\mathbf{r}(t)\). The position vector represents the particle's location in space as a function of time.
The velocity vector, \(\mathbf{v}(t)\), is derived as the rate at which this position changes. We achieve this by differentiating each component of the position vector with respect to time \(t\):

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

Hence, the velocity vector \(\mathbf{v}(t)\) becomes \( \mathbf{i} + 2t \mathbf{j} + 2 \mathbf{k} \). Evaluating this at \(t=1\), we find \(\mathbf{v}(1) = \mathbf{i} + 2 \mathbf{j} + 2 \mathbf{k}\). This tells us both the speed and the direction in which the particle is heading at that moment.

Acceleration Vector

Acceleration is about how the velocity of a particle changes over time; it's denoted as the acceleration vector \(\mathbf{a}(t)\). In simpler terms, while velocity tells you **how fast** and **where** you're going, acceleration tells you **how your speed or direction** is changing.
For each component of the velocity vector \(\mathbf{v}(t)\), we take another derivative. Here's what we did:

• The derivative of \(\mathbf{i}\) with respect to \(t\) is zero (since it's constant).
• The derivative of \(2t \mathbf{j}\) is \(2 \mathbf{j}\) (notice how \(2 \mathbf{j}\) remains).
• The derivative of \(2 \mathbf{k}\) is zero (also constant).

Thus, the full acceleration vector simplifies to \(2 \mathbf{j}\). This constant acceleration signifies that the particle's velocity is only changing along the \(\mathbf{j}\) axis, which could represent, for example, moving upwards in a three-dimensional space.

Speed and Direction

Speed, distinct from velocity, only tells us how fast a particle is moving regardless of direction. In our context, speed is the magnitude of the velocity vector.
To find out how quickly the particle moves at \(t=1\), we calculated the length (the magnitude) of the vector \(\mathbf{v}(1)\). This is computed using:\[\text{Speed} = \|\mathbf{v}(1)\| = \sqrt{1^2 + 2^2 + 2^2} = 3\]Thus, the particle's speed at this specific moment is 3 units per whatever time units you’re using.
The direction is essential to understand where the particle heads. We derive a unit vector by normalizing the velocity vector. This unit vector retains the direction, but scales the result to a length of 1, giving:\[\text{Direction} = \frac{\mathbf{v}(1)}{\|\mathbf{v}(1)\|} = \frac{1}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} + \frac{2}{3}\mathbf{k}.\]So the direction is a combination of these components, indicating the ratio of movement along each axis. Expressing the velocity as a product of its speed and direction emphasizes this relationship, showing \(\mathbf{v}(1)\) as a `stretched` unit direction vector by the speed factor.