Problem 3 - Thomas Calculus

Question

\({r}(t)\) is the position of a particle in the \(x y\) -plane at time \(t .\) Find an equation in \(x\) and \(y\) whose graph is the path of the particle. Then find the particle's velocity and acceleration vectors at the given value of \(t .\) \begin{equation} \mathbf{r}(t)=e^{t} \mathbf{i}+\frac{2}{9} e^{2 t} \mathbf{j}, \quad t=\ln 3 \end{equation}

Step-by-Step Solution

Answer

Path is \( y = \frac{2}{9} x^2 \), velocity at \( t = \ln 3 \) is \( 3\mathbf{i} + 4\mathbf{j} \), acceleration is \( 3\mathbf{i} + 8\mathbf{j} \).

Step 1: Express the Parametric Equations

The position function \( \mathbf{r}(t) = e^{t} \mathbf{i} + \frac{2}{9} e^{2t} \mathbf{j} \) can be broken down into \( x(t) = e^{t} \) and \( y(t) = \frac{2}{9} e^{2t}. \)

Step 2: Eliminate Parameter to Find Path Equation

Using \( x = e^{t} \), we can express \( t \) as \( t = \ln(x) \). Substitute this into the equation for \( y \): \( y = \frac{2}{9} e^{2(\ln(x))} = \frac{2}{9} x^2 \), as \( e^{\ln(a)} = a \).

Step 3: Confirm Path Equation

The path of the particle is the graph of the equation \( y = \frac{2}{9} x^2 \). This confirms the path as a parabola opening upward.

Step 4: Find the Velocity Vector \( \mathbf{v}(t) \)

The velocity vector \( \mathbf{v}(t) \) is the derivative of \( \mathbf{r}(t) \) with respect to \( t \). Compute \( \frac{d}{dt}[e^{t}] \) to find \( v_x = e^{t} \) and \( \frac{d}{dt}[\frac{2}{9}e^{2t}] \) to find \( v_y = \frac{4}{9}e^{2t}. \) Thus, \( \mathbf{v}(t) = e^{t} \mathbf{i} + \frac{4}{9} e^{2t} \mathbf{j}. \)

Step 5: Evaluate Velocity at \( t = \ln 3 \)

Substituting \( t = \ln 3 \) into \( \mathbf{v}(t) \), compute the components: \( e^{\ln 3} = 3 \) leading to \( v_x = 3 \), and \( \frac{4}{9} e^{2(\ln 3)} = \frac{4}{9} \times 9 = 4 \), giving \( \mathbf{v}(\ln 3) = 3 \mathbf{i} + 4 \mathbf{j}. \)

Step 6: Find the Acceleration Vector \( \mathbf{a}(t) \)

The acceleration vector \( \mathbf{a}(t) \) is the derivative of \( \mathbf{v}(t) \). Compute \( \frac{d}{dt}[e^{t}] = e^{t} \) for \( a_x \) and \( \frac{d}{dt}[\frac{4}{9} e^{2t}] = \frac{8}{9} e^{2t} \) for \( a_y \). Thus, \( \mathbf{a}(t) = e^{t} \mathbf{i} + \frac{8}{9} e^{2t} \mathbf{j}. \)

Step 7: Evaluate Acceleration at \( t = \ln 3 \)

Substituting \( t = \ln 3 \) into \( \mathbf{a}(t) \), compute the components: \( a_x = 3 \), and \( \frac{8}{9}e^{2(\ln 3)} = \frac{8}{9} \times 9 = 8 \), giving \( \mathbf{a}(\ln 3) = 3 \mathbf{i} + 8 \mathbf{j}. \)

Key Concepts

Particle Path

The path of a particle in the plane can be understood using parametric equations. Here, these equations are given as \( x(t) = e^{t} \) and \( y(t) = \frac{2}{9} e^{2t} \). Each describes how one coordinate of the particle depends on time \( t \).
To find a single equation in terms of \( x \) and \( y \) that describes the particle's path, we eliminate the parameter \( t \). This can be done by expressing \( t \) from one equation and substituting into the other. In this problem, by substituting \( t = \ln(x) \) from \( x = e^{t} \) into \( y \), we obtain:

• \( y = \frac{2}{9} x^2 \)

This reveals the path as a parabola, specifically one that opens upward, illustrating a smooth and symmetrical curve.

Velocity Vector

The velocity vector provides critical information about the speed and direction of a particle along its path at any moment in time. For a position function \( \mathbf{r}(t) \), the velocity vector, \( \mathbf{v}(t) \), is its derivative.
In this example, differentiating \( \mathbf{r}(t) = e^{t} \mathbf{i} + \frac{2}{9} e^{2t} \mathbf{j} \) results in the velocity vector:

• \( \mathbf{v}(t) = e^{t} \mathbf{i} + \frac{4}{9} e^{2t} \mathbf{j} \)

Plugging \( t = \ln 3 \) into this vector, we find the velocity at that specific time:

• \( \mathbf{v}(\ln 3) = 3 \mathbf{i} + 4 \mathbf{j} \)

Thus, this velocity vector indicates that at \( t = \ln 3 \), the particle moves rightward with a speed of 3 units and upward with a speed of 4 units.

Acceleration Vector

The acceleration vector explains how the velocity of a particle changes over time. It is obtained by taking the derivative of the velocity vector \( \mathbf{v}(t) \). This second derivative of the position function \( \mathbf{r}(t) \) represents how the rate of movement itself is changing.
Here, we calculate the acceleration vector \( \mathbf{a}(t) \) from the velocity vector previously found:

• \( \mathbf{a}(t) = e^{t} \mathbf{i} + \frac{8}{9} e^{2t} \mathbf{j} \)

Evaluating this at \( t = \ln 3 \), we attain:

• \( \mathbf{a}(\ln 3) = 3 \mathbf{i} + 8 \mathbf{j} \)

This indicates an increasing velocity rightward and a more substantial increase upward at this particular moment in time.

Parabolic Motion

Parabolic motion is a type of projectile motion where the path or trajectory is a parabola. This occurs frequently in real-world scenarios, like the arc of a ball being tossed.
In this problem, we've found that \( y = \frac{2}{9} x^2 \) describes the path of the particle as a parabola opening upwards. The parabolic trajectory indicates that as the particle travels, it follows this specific curved pattern.
Characteristics of parabolic motion include:

• Symmetrical path about a vertical line
• Constant rate of velocity change linked to forces in motion
• Predictable acceleration due to a consistent rate of change in velocity

Understanding parabolic motion helps explain why many physical systems behave predictably, and offers insight into how forces like gravity act on particles in a plane.