Problem 3 - Thomas Calculus

Question

In Exercises \(1-4, \mathbf{r}(t)\) 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 \(t .\) $$ \mathbf{r}(t)=e^{t} \mathbf{i}+\frac{2}{9} e^{2 t} \mathbf{j}, \quad t=\ln 3 $$

Step-by-Step Solution

Answer

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

Step 1: Express Position in Terms of Components

The position vector \( \mathbf{r}(t) = e^{t} \mathbf{i} + \frac{2}{9} e^{2t} \mathbf{j} \) denotes the position of the particle with components in the \( x \) and \( y \) directions. Express this as: \[ x(t) = e^{t} \quad \text{and} \quad y(t) = \frac{2}{9} e^{2t}. \]

Step 2: Eliminate Parameter to Find Path Equation

To find an equation in terms of \( x \) and \( y \), eliminate the parameter \( t \). From \( x(t) = e^t \), we find \( t = \ln(x) \). Substitute \( t \) in the expression for \( y(t) \): \[ y = \frac{2}{9} e^{2 \ln(x)} = \frac{2}{9} x^{2}. \] The equation of the path is \( y = \frac{2}{9} x^2. \)

Step 3: Compute Velocity Vector

The velocity vector is the derivative of the position vector \( \mathbf{r}(t) \) with respect to \( t \). Compute \( \mathbf{v}(t) = \frac{d}{dt}(e^t \mathbf{i} + \frac{2}{9} e^{2t} \mathbf{j}) = e^t \mathbf{i} + \frac{4}{9} e^{2t} \mathbf{j}. \)

Step 4: Compute Acceleration Vector

The acceleration vector is the derivative of the velocity vector \( \mathbf{v}(t) \) with respect to \( t \). Compute \( \mathbf{a}(t) = \frac{d}{dt}(e^t \mathbf{i} + \frac{4}{9} e^{2t} \mathbf{j}) = e^t \mathbf{i} + \frac{8}{9} e^{2t} \mathbf{j}. \)

Step 5: Evaluate Velocity and Acceleration at Given Time

Substitute \( t = \ln(3) \) into the expressions for \( \mathbf{v}(t) \) and \( \mathbf{a}(t) \) obtained in Steps 3 and 4: - Velocity: \[ \mathbf{v}(\ln(3)) = 3 \mathbf{i} + \frac{4}{9}(3^2) \mathbf{j} = 3 \mathbf{i} + 4 \mathbf{j}. \]- Acceleration: \[ \mathbf{a}(\ln(3)) = 3 \mathbf{i} + \frac{8}{9}(3^2) \mathbf{j} = 3 \mathbf{i} + 8 \mathbf{j}. \]

Key Concepts

Particle Motion

In this exercise, we consider the motion of a particle in the plane described by the position vector \( \mathbf{r}(t) = e^{t} \mathbf{i} + \frac{2}{9} e^{2t} \mathbf{j} \). This vector gives us the position of the particle in terms of time \( t \). The concept of particle motion in the plane is crucial as it helps us understand how objects move in a two-dimensional space.
To find the path of the particle, we break down the position vector into its component functions: \( x(t) = e^t \) and \( y(t) = \frac{2}{9} e^{2t} \). These represent the particle's position along the \( x \)-axis and \( y \)-axis, respectively. The challenge here is to describe the motion not in terms of \( t \), but only using \( x \) and \( y \), which leads us to the step of eliminating the parameter \( t \).
Understanding particle motion enables us to predict the path, speed, and direction in which the particle will travel over time.

Velocity Vector

The velocity vector describes how fast and in what direction the particle is moving along its path. It is the first derivative of the position vector with respect to time \( t \).
In our example, the velocity vector \( \mathbf{v}(t) \) is calculated by differentiating each component of \( \mathbf{r}(t) \) with respect to \( t \). This gives us:

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

Thus, the velocity vector becomes \( \mathbf{v}(t) = e^t \mathbf{i} + \frac{4}{9} e^{2t} \mathbf{j} \). This vector provides instantaneous speed and direction at any given time \( t \). Calculating the velocity at a specific time, like \( t = \ln(3) \), can show us exactly how the particle is moving at that moment.

Acceleration Vector

The acceleration vector gives us a measure of how the velocity of a particle changes over time. This is particularly useful for understanding how the forces acting upon a particle will affect its motion.
To find the acceleration vector, we take the derivative of the velocity vector \( \mathbf{v}(t) \) with respect to \( t \). In our case, this involves:

• \( \frac{d}{dt}(e^t \mathbf{i}) = e^t \mathbf{i} \)
• \( \frac{d}{dt}(\frac{4}{9} e^{2t} \mathbf{j}) = \frac{8}{9} e^{2t} \mathbf{j} \)

Thus, the acceleration vector is \( \mathbf{a}(t) = e^t \mathbf{i} + \frac{8}{9} e^{2t} \mathbf{j} \). By evaluating this expression at \( t = \ln(3) \), we understand how quickly the particle's velocity is changing at that specific point in time.

Eliminating the Parameter

Eliminating the parameter \( t \) allows us to find a direct relationship between \( x \) and \( y \), which describes the trajectory of the particle independent of time. This is crucial for analyzing the path the particle takes.
In our exercise, we start with \( x(t) = e^t \) and invert it to find \( t = \ln(x) \). By substituting \( \ln(x) \) into \( y(t) \), we transform the \( y \)-component equation:\[ y(t) = \frac{2}{9} e^{2t} \] transforms into:\[ y = \frac{2}{9} e^{2 \ln(x)} = \frac{2}{9} x^2 \].This equation \( y = \frac{2}{9} x^2 \) represents the path of the particle in the \( x y \)-plane. By eliminating \( t \), we gain a clearer understanding of how the particle moves spatially, which is a fundamental aspect of studying motion.