Problem 12 - 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. $$\mathbf{a}(t)=2 \mathbf{i}+6 t \mathbf{j}+12 t^{2} \mathbf{k}, \quad \mathbf{v}(0)=\mathbf{i}, \quad \mathbf{r}(0)=\mathbf{j}-\mathbf{k}$$

Step-by-Step Solution

Answer

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

Step 1: Integrate Acceleration to Find Velocity

The given acceleration vector is \( \mathbf{a}(t) = 2 \mathbf{i} + 6t \mathbf{j} + 12t^2 \mathbf{k} \). To find the velocity vector \( \mathbf{v}(t) \), we integrate the acceleration vector with respect to time:\[ \mathbf{v}(t) = \int (2 \mathbf{i} + 6t \mathbf{j} + 12t^2 \mathbf{k}) \, dt = (2t + C_1) \mathbf{i} + (3t^2 + C_2) \mathbf{j} + (4t^3 + C_3) \mathbf{k} \]We will use the initial condition to solve for constants \( C_1, C_2, \) and \( C_3 \).

Step 2: Apply Initial Velocity to Determine Constants

Use the initial velocity \( \mathbf{v}(0) = \mathbf{i} \) to find the constants:\[ 2(0) + C_1 = 1, \3(0)^2 + C_2 = 0, \4(0)^3 + C_3 = 0 \]Solving these equations, we get \( C_1 = 1 \), \( C_2 = 0 \), and \( C_3 = 0 \). Thus, the velocity vector becomes:\[ \mathbf{v}(t) = (2t + 1) \mathbf{i} + 3t^2 \mathbf{j} + 4t^3 \mathbf{k} \]

Step 3: Integrate Velocity to Find Position

Next, integrate the velocity vector to find the position vector \( \mathbf{r}(t) \):\[ \mathbf{r}(t) = \int ((2t + 1) \mathbf{i} + 3t^2 \mathbf{j} + 4t^3 \mathbf{k}) \, dt = (t^2 + t + D_1) \mathbf{i} + (t^3 + D_2) \mathbf{j} + (t^4 + D_3) \mathbf{k} \]We will use the initial position to find the constants \( D_1, D_2, \) and \( D_3 \).

Step 4: Apply Initial Position to Determine Constants

Use the initial position \( \mathbf{r}(0) = \mathbf{j} - \mathbf{k} \) to find constants:\[ t^2 + t + D_1 = 0 \t^3 + D_2 = 1 \t^4 + D_3 = -1 \]Solving these equations, we get \( D_1 = 0 \), \( D_2 = 1 \), and \( D_3 = -1 \). The position vector becomes:\[ \mathbf{r}(t) = (t^2 + t) \mathbf{i} + (t^3 + 1) \mathbf{j} + (t^4 - 1) \mathbf{k} \]

Key Concepts

Velocity Vector

In vector calculus, understanding a velocity vector is crucial when analyzing the motion of particles. When you have the acceleration vector of a particle, you can find its velocity vector by integrating the acceleration over time. This integrates the rate of change of velocity over a specified time period to provide a vector that tells you the particle's speed and direction at any point in time.
For example, given the acceleration vector \( \mathbf{a}(t) = 2 \mathbf{i} + 6t \mathbf{j} + 12t^2 \mathbf{k} \), the velocity vector \( \mathbf{v}(t) \) is found by integrating each component of the acceleration:

• Integrate \( 2 \mathbf{i} \) to get \( 2t \mathbf{i} \)
• Integrate \( 6t \mathbf{j} \) to get \( 3t^2 \mathbf{j} \)
• Integrate \( 12t^2 \mathbf{k} \) to get \( 4t^3 \mathbf{k} \)

After finding these integrates components, an equation for constraints is utilized (such as initial conditions) to solve for any constants introduced during integration. Thus, you obtain an exact formula for the velocity vector where the constants are eliminated by substituting your known initial velocity conditions.

Position Vector

The position vector is another fundamental concept in vector calculus that describes the exact location of a particle in space over time. It is derived by integrating the velocity vector over time.
Given that we have determined the velocity vector \( \mathbf{v}(t) = (2t + 1) \mathbf{i} + 3t^2 \mathbf{j} + 4t^3 \mathbf{k} \), we can find the position vector \( \mathbf{r}(t) \) by integrating each of the components of \( \mathbf{v}(t) \):

• Integrating \( (2t + 1) \mathbf{i} \) results in \( t^2 + t \mathbf{i} \)
• Integrating \( 3t^2 \mathbf{j} \) results in \( t^3 \mathbf{j} \)
• Integrating \( 4t^3 \mathbf{k} \) results in \( t^4 \mathbf{k} \)

The initial position vector is then used to pinpoint any constants resulting from the integration process. By accurately determining these constants, you can represent the exact position of a moving particle at any given time during its journey.
Appropriate usage of the initial conditions, like the initial position \( \mathbf{r}(0) \), ensures the accuracy of the position vector trajectory, providing complete details about the particle's motion.

Integration of Vectors

Integration of vectors is the process used to determine quantities such as velocity and position vectors from the acceleration vector. Unlike scalar integration, vector integration considers the directional components, which often involve component-wise integration of the vector functions over a specific variable, typically time.
Each element of a vector is treated as an independent function, integrated separately. In the given exercise with acceleration vector \( \mathbf{a}(t) = 2 \mathbf{i} + 6t \mathbf{j} + 12t^2 \mathbf{k} \), each component \( 2 \), \( 6t \), and \( 12t^2 \) is integrated separately to find the respective components of the velocity vector.
The integration process introduces constants of integration, which are solved using initial conditions provided in the problem.

• Integration translates the acceleration components into velocity.
• Applying further integration to velocity gives the position vector.
• Initial conditions assure correctness by eliminating unknown constants.

Through understanding vector integration, students can predict the motion of particles in physical space, providing a fundamental tool in analyzing dynamic systems and understanding how a smoothly changing motion functions over space and time.