Problem 6 - Calculus Early Transcendentals

Question

\(3-8\) Find the velocity, acceleration, and speed of a particle with the given position function. Sketch the path of the particle and draw the velocity and acceleration vectors for the specified value of \(t .\) $$\mathbf{r}(t)=e^{t} \mathbf{i}+e^{2 t} \mathbf{j}, \quad t=0$$

Step-by-Step Solution

Answer

Velocity: \(1 \mathbf{i} + 2 \mathbf{j}\); Acceleration: \(1 \mathbf{i} + 4 \mathbf{j}\); Speed: \(\sqrt{5}\).

Step 1: Find the Velocity

The velocity of a particle is the derivative of the position function with respect to time. Given \( \mathbf{r}(t) = e^t \mathbf{i} + e^{2t} \mathbf{j} \), differentiate each component with respect to \( t \):- Derivative of \( e^t \) is \( e^t \).- Derivative of \( e^{2t} \) is \( 2e^{2t} \).So, the velocity \( \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = e^t \mathbf{i} + 2e^{2t} \mathbf{j} \).

Step 2: Find the Acceleration

The acceleration of a particle is the derivative of the velocity function with respect to time. Differentiate \( \mathbf{v}(t) = e^t \mathbf{i} + 2e^{2t} \mathbf{j} \):- Derivative of \( e^t \) is \( e^t \).- Derivative of \( 2e^{2t} \) is \( 4e^{2t} \).Thus, the acceleration \( \mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = e^t \mathbf{i} + 4e^{2t} \mathbf{j} \).

Step 3: Calculate Velocity, Acceleration, and Speed at t = 0

Substitute \( t = 0 \) into the velocity and acceleration functions:- Velocity: \( \mathbf{v}(0) = e^0 \mathbf{i} + 2e^{0} \mathbf{j} = 1 \mathbf{i} + 2 \mathbf{j} \).- Acceleration: \( \mathbf{a}(0) = e^0 \mathbf{i} + 4e^{0} \mathbf{j} = 1 \mathbf{i} + 4 \mathbf{j} \).The speed of the particle is the magnitude of the velocity vector, calculated as:\[ \| \mathbf{v}(0) \| = \sqrt{1^2 + 2^2} = \sqrt{5}. \]

Step 4: Sketch the Path and Vectors

The path of the particle is described by the position function \( \mathbf{r}(t) = e^t \mathbf{i} + e^{2t} \mathbf{j} \) and sketches as an exponential curve in the plane, as both components grow exponentially. At \( t = 0 \), the position is \( (1, 1) \).The velocity vector \( \mathbf{v}(0) = (1, 2) \) can be drawn starting at point \( (1, 1) \), pointing in the direction of movement with length proportional to its magnitude.Similarly, the acceleration vector \( \mathbf{a}(0) = (1, 4) \) is drawn starting at the same point and is longer, reflecting higher magnitude due to the increase in acceleration.

Key Concepts

Velocity Vector

In physics, a **velocity vector** represents the rate of change of the position of a particle with respect to time. It gives both speed and direction. For a given position function \( \mathbf{r}(t) = e^t \mathbf{i} + e^{2t} \mathbf{j} \), the velocity vector \( \mathbf{v}(t) \) is obtained by taking its derivative with respect to time.

• Differentiate each component of the position function.
• The derivative of the term \( e^t \) is \( e^t \), and for \( e^{2t} \), it is \( 2e^{2t} \).

Thus, the velocity vector becomes:\[ \mathbf{v}(t) = e^t \mathbf{i} + 2e^{2t} \mathbf{j}. \]At \( t = 0 \), the velocity \( \mathbf{v}(0) = 1 \mathbf{i} + 2 \mathbf{j} \), indicating motion at a specific direction and rate at that exact moment. Velocity vectors are crucial for understanding how objects move and change over time.

Acceleration Vector

The **acceleration vector** signifies the rate of change of the velocity vector over time. It depicts how quickly a particle is speeding up or slowing down, as well as changes in direction. By taking the derivative of the velocity vector \( \mathbf{v}(t) = e^t \mathbf{i} + 2e^{2t} \mathbf{j} \), we obtain the acceleration vector:

• The derivative of \( e^t \) remains \( e^t \).
• The derivative of \( 2e^{2t} \) is \( 4e^{2t} \).

Therefore, the acceleration vector is:\[ \mathbf{a}(t) = e^t \mathbf{i} + 4e^{2t} \mathbf{j}. \]At \( t = 0 \), the acceleration is \( \mathbf{a}(0) = 1 \mathbf{i} + 4 \mathbf{j} \), suggesting that at this time, the particle's speed is increasing rapidly in the direction driven by its respective components. This vector provides insight into the dynamical nature of the particle's motion.

Exponential Functions

**Exponential functions** are mathematical functions of the form \( e^{kt} \), where \( e \) is Euler's number (approximately 2.718), and \( k \) is a constant. They are used to model growth or decay processes, extensively seen in both natural and applied sciences.

• In the position function \( \mathbf{r}(t) = e^t \mathbf{i} + e^{2t} \mathbf{j} \), the exponential components describe how the particle's position changes exponentially over time.
• The components \( e^t \) and \( e^{2t} \) mean the particle's position magnifies rather quickly as time progresses.

Exponential functions allow us to understand behaviors that involve continuous growth, like population growth or radioactive decay. In physics, they exemplify scenarios where rates of change are directly proportional to the existing amount, leading to rapid increases or decreases.

Position Function

A **position function** \( \mathbf{r}(t) \) provides the location of a particle at any time \( t \). It tells us where the particle is in space. For instance, \( \mathbf{r}(t) = e^t \mathbf{i} + e^{2t} \mathbf{j} \) describes a path in a 2-dimensional plane.

• The \( i \) and \( j \) components represent the horizontal and vertical directions respectively.
• The exponential terms user here reflect the changing position with time.

When time \( t = 0 \), the initial position is \( (1, 1) \), where \( e^0 = 1 \). The 'path of the particle' means tracing these changing positions over time, resulting in a curve on a graph. This function is crucial in analyzing and predicting the movement trajectories of particles in physics.