Problem 16 - Calculus Early Transcendentals

Question

15-16 Find the velocity and position vectors of a particle that has the given acceleration and the given initial velocity and position. $$\mathbf{a}(t)=2 \mathbf{i}+6 t \mathbf{j}+12 t^{2} \mathbf{k}, \quad \mathbf{v}(0)=\mathbf{i}, \quad \mathbf{r}(0)=\mathbf{j}-\mathbf{k}$$

Step-by-Step Solution

Answer

Velocity: \( \mathbf{v}(t) = (2t + 1) \mathbf{i} + 3t^2 \mathbf{j} + 4t^3 \mathbf{k} \) Position: \( \mathbf{r}(t) = (t^2 + t) \mathbf{i} + (t^3 + 1) \mathbf{j} + (t^4 - 1) \mathbf{k} \)

Step 1: Integrate the Acceleration Vector

To find the velocity vector \( \mathbf{v}(t) \), integrate the acceleration vector \( \mathbf{a}(t) = 2 \mathbf{i} + 6t \mathbf{j} + 12t^2 \mathbf{k} \) with respect to \( t \). The integral of the components gives:\[ \mathbf{v}(t) = \int(2 \mathbf{i} + 6t \mathbf{j} + 12t^2 \mathbf{k}) \, dt \]\( \int 2 \, dt = 2t + C_1 \) for the \( \mathbf{i} \) component, \( \int 6t \, dt = 3t^2 + C_2 \) for the \( \mathbf{j} \) component, and \( \int 12t^2 \, dt = 4t^3 + C_3 \) for the \( \mathbf{k} \) component. Hence:\[ \mathbf{v}(t) = (2t + C_1) \mathbf{i} + (3t^2 + C_2) \mathbf{j} + (4t^3 + C_3) \mathbf{k} \]

Step 2: Apply Initial Velocity Condition

Given \( \mathbf{v}(0) = \mathbf{i} \), substitute \( t = 0 \) into the velocity vector:\[ \mathbf{v}(0) = (2 \cdot 0 + C_1) \mathbf{i} + (3 \cdot 0^2 + C_2) \mathbf{j} + (4 \cdot 0^3 + C_3) \mathbf{k} = \mathbf{i} \]This means \( C_1 = 1 \), \( C_2 = 0 \), and \( C_3 = 0 \). Therefore:\[ \mathbf{v}(t) = (2t + 1) \mathbf{i} + 3t^2 \mathbf{j} + 4t^3 \mathbf{k} \]

Step 3: Integrate the Velocity Vector

To find the position vector \( \mathbf{r}(t) \), integrate the velocity vector \( \mathbf{v}(t) = (2t + 1) \mathbf{i} + 3t^2 \mathbf{j} + 4t^3 \mathbf{k} \):\[ \mathbf{r}(t) = \int ((2t + 1) \mathbf{i} + 3t^2 \mathbf{j} + 4t^3 \mathbf{k}) \, dt \]This results in \( \int (2t + 1) \mathbf{i} \, dt = (t^2 + t + C_4) \mathbf{i} \), \( \int 3t^2 \mathbf{j} \, dt = (t^3 + C_5) \mathbf{j} \), and \( \int 4t^3 \mathbf{k} \, dt = (t^4 + C_6) \mathbf{k} \). Hence:\[ \mathbf{r}(t) = (t^2 + t + C_4) \mathbf{i} + (t^3 + C_5) \mathbf{j} + (t^4 + C_6) \mathbf{k} \]

Step 4: Apply Initial Position Condition

Given \( \mathbf{r}(0) = \mathbf{j} - \mathbf{k} \), substitute \( t = 0 \) into the position vector:\[ \mathbf{r}(0) = (0^2 + 0 + C_4) \mathbf{i} + (0^3 + C_5) \mathbf{j} + (0^4 + C_6) \mathbf{k} = \mathbf{j} - \mathbf{k} \]This means \( C_4 = 0 \), \( C_5 = 1 \), and \( C_6 = -1 \). Therefore:\[ \mathbf{r}(t) = (t^2 + t) \mathbf{i} + (t^3 + 1) \mathbf{j} + (t^4 - 1) \mathbf{k} \]

Key Concepts

Particle Motion

Particle motion refers to the movement of a particle through space as it changes position over time. It's essential to understand that this movement is often described using vectors, which give both the direction and magnitude of the particle's motion. In classical physics, a particle's motion is determined by its initial conditions and the forces acting upon it. This motion can be analyzed by understanding several critical components such as velocity, acceleration, and position vectors.
The key is to follow the particle through its trajectory, predict future positions, and understand how various forces will impact its path. In mathematical terms, this involves calculating the derivatives and integrals of vector functions to move from one motion attribute, like acceleration, to another like velocity and position.

• Position describes where a particle is located at any given time.
• Velocity indicates the direction and speed of the particle.
• Acceleration details how the velocity changes over time.

Grasping these concepts is vital in mechanics and helps predict how a particle will behave under different conditions.

Velocity Vector

The velocity vector represents the rate of change of the position vector with respect to time. In simple terms, it shows how fast and in what direction a particle is moving. It's derived from the acceleration vector by integration over time.
For example, in the given exercise, the acceleration vector is provided as \( \mathbf{a}(t) = 2 \mathbf{i} + 6t \mathbf{j} + 12t^2 \mathbf{k} \). By integrating this vector, we find the velocity vector \( \mathbf{v}(t) \).

• To integrate the acceleration vector, take each component separately. For \( \mathbf{i} \), integrate a constant \( 2 \), which becomes \( 2t \).
• For the \( \mathbf{j} \) component, integrate \( 6t \), giving \( 3t^2 \).
• For the \( \mathbf{k} \) component, integrate \( 12t^2 \), which results in \( 4t^3 \).

This integration results in the velocity vector \( \mathbf{v}(t) = (2t + 1) \mathbf{i} + 3t^2 \mathbf{j} + 4t^3 \mathbf{k} \), with constants determined by initial conditions.

Position Vector

The position vector function gives the actual position of the particle at any given time. By knowing the initial position and integrating the velocity vector, students can find the position vector function.
In the exercise provided, the velocity vector, \( \mathbf{v}(t) = (2t + 1) \mathbf{i} + 3t^2 \mathbf{j} + 4t^3 \mathbf{k} \), is integrated to find the position vector \( \mathbf{r}(t) \).

• Integration of the \( \mathbf{i} \) component, \( 2t + 1 \), gives \( t^2 + t \).
• For the \( \mathbf{j} \) component, integrate \( 3t^2 \), which results in \( t^3 \).
• The \( \mathbf{k} \) component, \( 4t^3 \), integrates to \( t^4 \).

The full position vector is \( \mathbf{r}(t) = (t^2 + t) \mathbf{i} + (t^3 + 1) \mathbf{j} + (t^4 - 1) \mathbf{k} \), using the initial conditions to solve for constants.

Integration of Vectors

Integration of vectors plays a crucial role in understanding particle motion in classical physics. It allows us to go from understanding instantaneous changes (like acceleration) to understanding ongoing processes (like velocity and position).
To integrate a vector means to find a vector function that represents the accumulation of these changes over time. In practice, this involves finding antiderivatives for each component of a vector function separately.

• The integration process starts with the acceleration vector, yielding the velocity vector. Each element of the acceleration vector is treated separately and integrated with respect to time.
• The same process applies to transition from the velocity vector to the position vector.

It is essential to apply any initial conditions correctly after integration to find any unknown constants. This systematic approach convincingly connects the dots between a particle's acceleration and its trajectory. By solving these vector integrals, one gains a fuller picture of particle behavior in a given system.