Problem 18 - Calculus Early Transcendentals

Question

\(17-18\) (a) Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. (b) Use a computer to graph the path of the particle. $$\mathbf{a}(t)=t \mathbf{i}+e^{t} \mathbf{j}+e^{-l} \mathbf{k}, \quad \mathbf{v}(0)=\mathbf{k}, \quad \mathbf{r}(0)=\mathbf{j}+\mathbf{k}$$

Step-by-Step Solution

Answer

Velocity: \( \mathbf{v}(t) = \frac{t^2}{2} \mathbf{i} + (e^t - 1) \mathbf{j} + (-e^{-t} + 2) \mathbf{k} \). Position: \( \mathbf{r}(t) = \frac{t^3}{6} \mathbf{i} + (e^t - t) \mathbf{j} + (e^{-t} + 2t) \mathbf{k} \).

Step 1: Integrate the Acceleration to Find Velocity

To find the velocity vector \( \mathbf{v}(t) \), integrate the acceleration vector \( \mathbf{a}(t) = t \mathbf{i} + e^t \mathbf{j} + e^{-t} \mathbf{k} \) with respect to \( t \). This gives: \[ \mathbf{v}(t) = \int (t \mathbf{i} + e^t \mathbf{j} + e^{-t} \mathbf{k}) \, dt = \left( \frac{t^2}{2} + C_1 \right) \mathbf{i} + (e^t + C_2) \mathbf{j} + (-e^{-t} + C_3) \mathbf{k} \]

Step 2: Apply Initial Velocity Condition

The initial condition \( \mathbf{v}(0) = \mathbf{k} \) helps determine constants. Substitute \( t = 0 \) in the velocity equation: \[ \mathbf{v}(0) = C_1 \mathbf{i} + (e^0 + C_2) \mathbf{j} + (-e^0 + C_3) \mathbf{k} = 0 \mathbf{i} + 0 \mathbf{j} + 1 \mathbf{k} \] Comparing components, we have: - \( C_1 = 0 \)- \( 1 + C_2 = 0 \) \( \Rightarrow C_2 = -1 \)- \( -1 + C_3 = 1 \) \( \Rightarrow C_3 = 2 \)

Step 3: Find the Velocity Vector

Substitute the constants back into \( \mathbf{v}(t) \):\[ \mathbf{v}(t) = \frac{t^2}{2} \mathbf{i} + (e^t - 1) \mathbf{j} + (-e^{-t} + 2) \mathbf{k} \]

Step 4: Integrate the Velocity to Find Position

To find the position vector \( \mathbf{r}(t) \), integrate the velocity vector:\[ \mathbf{r}(t) = \int \left( \frac{t^2}{2} \mathbf{i} + (e^t - 1) \mathbf{j} + (-e^{-t} + 2) \mathbf{k} \right) \, dt \]This results in:\[ \mathbf{r}(t) = \left( \frac{t^3}{6} + D_1 \right) \mathbf{i} + (e^t - t + D_2) \mathbf{j} + (e^{-t} + 2t + D_3) \mathbf{k} \]

Step 5: Apply Initial Position Condition

The initial condition \( \mathbf{r}(0) = \mathbf{j} + \mathbf{k} \) helps determine the other constants. Substitute \( t = 0 \) in the position equation:\[ \mathbf{r}(0) = D_1 \mathbf{i} + (1 + D_2) \mathbf{j} + (1 + D_3) \mathbf{k} = 0 \mathbf{i} + 1 \mathbf{j} + 1 \mathbf{k} \] Comparing components, we have: - \( D_1 = 0 \)- \( 1 + D_2 = 1 \) \( \Rightarrow D_2 = 0 \)- \( 1 + D_3 = 1 \) \( \Rightarrow D_3 = 0 \)

Step 6: Find the Position Vector

Substitute the constants back into \( \mathbf{r}(t) \):\[ \mathbf{r}(t) = \frac{t^3}{6} \mathbf{i} + (e^t - t) \mathbf{j} + (e^{-t} + 2t) \mathbf{k} \]

Step 7: Graph the Path of the Particle

Using a graphing software or tool, plot the parametric equations for \( x(t) = \frac{t^3}{6} \), \( y(t) = e^t - t \), and \( z(t) = e^{-t} + 2t \). Observe how the particle moves in three-dimensional space.

Key Concepts

Position Vector

In vector calculus, the **position vector** of a particle is a fundamental concept that describes the particle's location in three-dimensional space as a function of time. Typically denoted as \( \mathbf{r}(t) \), it maps the path taken by the particle as it moves through space.
In our exercise, we found the position vector by integrating the velocity function, which itself was derived from the given acceleration. The specific trajectory is expressed by the position vector:

• \( x(t) = \frac{t^3}{6} \)
• \( y(t) = e^t - t \)
• \( z(t) = e^{-t} + 2t \)

Here, each component function \( x(t) \), \( y(t) \), and \( z(t) \) gives the particle’s position along the respective axes. Hence, the position vector helps to visualize and understand the particle's path in the three-dimensional space.

Acceleration

**Acceleration** is a vector that represents the rate of change of velocity with respect to time. It tells us how the velocity of a particle changes as time progresses. In our problem, the acceleration given is:

• \( \mathbf{a}(t) = t \mathbf{i} + e^t \mathbf{j} + e^{-t} \mathbf{k} \)

This vector comprises three components, each expressing the acceleration in the respective \( x \), \( y \), and \( z \) directions. Because acceleration directly influences velocity, understanding it helps predict how motion changes over time. As shown, acceleration varies with both polynomial and exponential functions, indicating non-linear changes in velocity as time progresses.

Velocity

The **velocity** vector describes the speed and direction of the particle's motion in space. By integrating the acceleration vector, we derived the velocity vector \( \mathbf{v}(t) \). It was obtained as:

• \( \mathbf{v}(t) = \frac{t^2}{2} \mathbf{i} + (e^t - 1) \mathbf{j} + (-e^{-t} + 2) \mathbf{k} \)

Velocity provides critical insights into how fast and in what direction the particle is traveling at any given time.
To find the constants required for this equation, we used the **initial velocity condition**, where \( \mathbf{v}(0) = \mathbf{k} \), to solve for integration constants and accurately describe starting motion. This equation allows us to determine exactly how the velocity changes across different time points.

Integration

**Integration** is the process of finding the integral, which is essentially the inverse operation of differentiation. It's a key step in determining both velocity and position from acceleration. Through this process, we derived both the velocity and position vectors from the given functions.
For velocity, we integrated the acceleration vector, and for the position vector, we integrated the velocity vector. The integration process included an essential step: applying the specific initial conditions to determine the constant of integration. These constants were crucial to ensure the particle's motion adhered strictly to the initial stated conditions. Thus, integration underscores the dynamic relationship between acceleration, velocity, and position.

Initial Conditions

**Initial conditions** are the specific values provided at the start of observation, crucial for solving differential equations in vector calculus accurately. They ensure that derived functions accurately describe a particular scenario. In this exercise:

• The initial velocity was given as \( \mathbf{v}(0) = \mathbf{k} \).
• The initial position provided was \( \mathbf{r}(0) = \mathbf{j} + \mathbf{k} \).

By substituting these conditions into the integrated velocity and position equations, we solved for the constants in each vector expression. This process ensures that the motion of the particle starts from the correct state, depicting a realistic path and avoiding any arbitrary ambiguities in the mathematical model. Initial conditions, therefore, bind the mathematical solution of differential equations to practical, physical phenomena.