Problem 11 - Essential Calculus Early Transcendentals

Question

Find the velocity and position vectors of a particle that has the given acceleration and the given initial velocity and position. $$\mathbf{a}(t)=\mathbf{i}+2 \mathbf{j}, \quad \mathbf{v}(0)=\mathbf{k}, \quad \mathbf{r}(0)=\mathbf{i} $$

Step-by-Step Solution

Answer

The velocity vector is \( \mathbf{v}(t) = t\mathbf{i} + 2t\mathbf{j} + \mathbf{k} \) and the position vector is \( \mathbf{r}(t) = \left(\frac{1}{2}t^2 + 1\right)\mathbf{i} + t^2\mathbf{j} + t\mathbf{k} \).

Step 1: Integrate acceleration to find velocity

The acceleration vector is \( \mathbf{a}(t) = \mathbf{i} + 2 \mathbf{j} \). Integrate this with respect to \( t \) to find the velocity vector. \[ \mathbf{v}(t) = \int \mathbf{a}(t) \, dt = \int (\mathbf{i} + 2 \mathbf{j}) \, dt = (t \mathbf{i} + 2t \mathbf{j} + \mathbf{C}) \] where \( \mathbf{C} \) is the integration constant representing the initial velocity.

Step 2: Apply initial condition for velocity

Use the given initial velocity \( \mathbf{v}(0) = \mathbf{k} \) to determine the constant \( \mathbf{C} \). Substitute \( t = 0 \) into the velocity equation:\[ \mathbf{v}(0) = (0 \mathbf{i} + 0 \mathbf{j} + \mathbf{C}) = \mathbf{k} \]Therefore, \( \mathbf{C} = \mathbf{k} \). The velocity vector becomes:\[ \mathbf{v}(t) = t \mathbf{i} + 2t \mathbf{j} + \mathbf{k} \]

Step 3: Integrate velocity to find position

Integrate the velocity vector \( \mathbf{v}(t) = t \mathbf{i} + 2t \mathbf{j} + \mathbf{k} \) to find the position vector. \[ \mathbf{r}(t) = \int \mathbf{v}(t) \, dt = \int (t \mathbf{i} + 2t \mathbf{j} + \mathbf{k}) \, dt = \left( \frac{1}{2} t^2 \mathbf{i} + t^2 \mathbf{j} + t \mathbf{k} + \mathbf{D} \right) \]where \( \mathbf{D} \) is the integration constant representing the initial position.

Step 4: Apply initial condition for position

Use the given initial position \( \mathbf{r}(0) = \mathbf{i} \) to determine the constant \( \mathbf{D} \). Substitute \( t = 0 \) into the position equation:\[ \mathbf{r}(0) = \left( 0 \mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k} + \mathbf{D} \right) = \mathbf{i} \]Therefore, \( \mathbf{D} = \mathbf{i} \). The position vector becomes:\[ \mathbf{r}(t) = \frac{1}{2} t^2 \mathbf{i} + t^2 \mathbf{j} + t \mathbf{k} + \mathbf{i} \]Simplified, this gives:\[ \mathbf{r}(t) = \left( \frac{1}{2} t^2 + 1 \right) \mathbf{i} + t^2 \mathbf{j} + t \mathbf{k} \]

Key Concepts

Velocity Vector

Velocity vectors play a key role in understanding the motion of particles. In vector calculus, the velocity vector describes the rate of change of the position of an object over time. It is derived from the integration of the acceleration vector.

To find the velocity vector, we begin with the acceleration vector provided in the exercise, \( \mathbf{a}(t) = \mathbf{i} + 2 \mathbf{j} \). By integrating this vector with respect to time \( t \), we can determine the velocity as a function of time.

• The integration of the unit vector \( \mathbf{i} \) with respect to \( t \) results in \( t \mathbf{i} \).
• The integration of \( 2 \mathbf{j} \) with respect to \( t \) results in \( 2t \mathbf{j} \).

This gives us the intermediate velocity vector \[ \mathbf{v}(t) = t \mathbf{i} + 2t \mathbf{j} + \mathbf{C} \] where \( \mathbf{C} \) is the integration constant. After considering the initial condition \( \mathbf{v}(0) = \mathbf{k} \), the constant \( \mathbf{C} \) is determined as \( \mathbf{k} \). Thus, the velocity vector becomes: \[ \mathbf{v}(t) = t \mathbf{i} + 2t \mathbf{j} + \mathbf{k} \] which captures how an object’s velocity evolves over time given the initial conditions.

Position Vector

The position vector provides a snapshot of the location of a particle at any given time and is derived by integrating the velocity vector. This vector is foundational in understanding an object's trajectory.

Starting with the velocity vector \( \mathbf{v}(t) = t \mathbf{i} + 2t \mathbf{j} + \mathbf{k} \), we integrate with respect to \( t \) to find the position. The integration process is as follows:

• Integrating \( t \mathbf{i} \) with respect to \( t \) gives \( \frac{1}{2}t^2 \mathbf{i} \).
• Integrating \( 2t \mathbf{j} \) gives \( t^2 \mathbf{j} \).
• Integrating the constant vector \( \mathbf{k} \) gives \( t \mathbf{k} \).

Including the constant \( \mathbf{D} \), the position vector is: \[ \mathbf{r}(t) = \frac{1}{2}t^2 \mathbf{i} + t^2 \mathbf{j} + t \mathbf{k} + \mathbf{D} \]. Applying the initial position \( \mathbf{r}(0) = \mathbf{i} \) helps us find that \( \mathbf{D} = \mathbf{i} \). Thus, the final position vector is:\[ \mathbf{r}(t) = \left( \frac{1}{2}t^2 + 1 \right) \mathbf{i} + t^2 \mathbf{j} + t \mathbf{k} \] which describes the path the particle follows.

Integration Constants

Integration constants are pivotal when determining both velocity and position vectors, serving to customize the general solutions to specific initial conditions.

When integrating a function to find its antiderivative, an arbitrary constant is introduced, known as the integration constant. In our exercise:

• The constant \( \mathbf{C} \) in the velocity vector results from integrating the acceleration vector. \( \mathbf{C} \) represents the initial velocity, which is corrected using the condition \( \mathbf{v}(0) = \mathbf{k} \).
• Similarly, the constant \( \mathbf{D} \) in the position vector comes from integrating the velocity vector. This constant aligns the equation with the initial position \( \mathbf{r}(0) = \mathbf{i} \).

Substituting these initial conditions refines the general formulas into precise and accurate mathematical representations. Without these adjustments, our solutions would not correctly illustrate the particle's motion based on its starting conditions. This highlights the fundamental importance of integrating constants in solving real-life calculus problems.