Problem 9 - Essential Calculus Early Transcendentals

Question

Find the velocity, acceleration, and speed of a particle with the given position function. $$\mathbf{r}(t)=\sqrt{2} t \mathbf{i}+e^{t} \mathbf{j}+e^{-t} \mathbf{k}$$

Step-by-Step Solution

Answer

Velocity: \( \mathbf{v}(t) = \sqrt{2} \mathbf{i} + e^{t} \mathbf{j} - e^{-t} \mathbf{k} \); Acceleration: \( \mathbf{a}(t) = e^{t} \mathbf{j} + e^{-t} \mathbf{k} \); Speed: \( \sqrt{2 + e^{2t} + e^{-2t}} \)."

Step 1: Find the Velocity

To find the velocity of a particle, we take the derivative of the position function \( \mathbf{r}(t) = \sqrt{2} t \mathbf{i} + e^{t} \mathbf{j} + e^{-t} \mathbf{k} \) with respect to time \( t \). The velocity \( \mathbf{v}(t) \) is given by:\[ \mathbf{v}(t) = \frac{d}{dt}(\mathbf{r}(t)) = \sqrt{2} \mathbf{i} + e^{t} \mathbf{j} - e^{-t} \mathbf{k} \]

Step 2: Find the Acceleration

Next, we find the acceleration by taking the derivative of the velocity function \( \mathbf{v}(t) \). The acceleration \( \mathbf{a}(t) \) is:\[ \mathbf{a}(t) = \frac{d}{dt}(\mathbf{v}(t)) = 0 \mathbf{i} + e^{t} \mathbf{j} + e^{-t} \mathbf{k} \]

Step 3: Calculate the Speed

The speed of the particle is the magnitude of the velocity vector \( \mathbf{v}(t) \). Calculate the magnitude as follows:\[ \text{Speed} = \| \mathbf{v}(t) \| = \sqrt{ (\sqrt{2})^2 + (e^{t})^2 + (-e^{-t})^2 } \]This simplifies to:\[ \text{Speed} = \sqrt{2 + e^{2t} + e^{-2t} } \]

Key Concepts

Position Function

The position function, denoted as \( \mathbf{r}(t) \), describes the location of a particle at any given time \( t \). It is a vector function that combines components in three-dimensional space, usually along the \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) unit vectors in Cartesian coordinates. For the given exercise, the position function is \( \mathbf{r}(t) = \sqrt{2} t \mathbf{i} + e^{t} \mathbf{j} + e^{-t} \mathbf{k} \).

This function represents motion where the particle moves linearly along the \( \mathbf{i} \) direction, and exponentially along the \( \mathbf{j} \) and \( \mathbf{k} \) axes. Understanding the position function is essential, as it is the starting point for finding velocity and acceleration by differentiation.

Velocity Vector

The velocity vector \( \mathbf{v}(t) \) measures how fast the position of the particle changes over time. It is the first derivative of the position function \( \mathbf{r}(t) \) with respect to \( t \). This derivative provides both the speed and direction of the particle's movement.

For the position function given, the velocity vector is calculated as \( \mathbf{v}(t) = \frac{d}{dt} (\mathbf{r}(t)) = \sqrt{2} \mathbf{i} + e^{t} \mathbf{j} - e^{-t} \mathbf{k} \). Each component of this vector tells us about the particle's velocity along that specific axis. Bullet points can help summarize key points:

• Constant velocity component in \( \mathbf{i} \): \( \sqrt{2} \)
• Exponential increasing velocity in \( \mathbf{j} \): \( e^{t} \)
• Exponential decreasing velocity in \( \mathbf{k} \): \( -e^{-t} \)

Acceleration Vector

Acceleration measures how quickly the velocity of an object changes over time. The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \). For the case at hand, it represents the second derivative of the position function.

Here, the acceleration vector is derived as follows: \( \mathbf{a}(t) = \frac{d}{dt} (\mathbf{v}(t)) = 0 \mathbf{i} + e^{t} \mathbf{j} + e^{-t} \mathbf{k} \). This tells us:

• No acceleration along \( \mathbf{i} \), indicated by zero.
• Positive, exponential acceleration in \( \mathbf{j} \).
• Positive, exponential acceleration in \( \mathbf{k} \).

Acceleration indicates how the velocity's direction and magnitude change.

Magnitude of Velocity

The magnitude of the velocity vector is commonly referred to as speed. It measures how fast a particle is moving irrespective of its direction. Mathematically, it is the length (norm) of the velocity vector.

To find the magnitude, apply the formula: \( \| \mathbf{v}(t) \| = \sqrt{ (\sqrt{2})^2 + (e^{t})^2 + (-e^{-t})^2 } \). Simplifying this, we obtain: \( \| \mathbf{v}(t) \| = \sqrt{2 + e^{2t} + e^{-2t} } \).

This expression represents the speed of the particle as a single value that combines the effects of all three velocity components.

Derivative in Calculus

The derivative is a fundamental concept in calculus, representing the rate at which a function changes. In vector calculus, the derivative of a vector function like the position function is crucial for finding other properties such as velocity and acceleration.

Here's how derivatives are applied in this context:

• Velocity \( \mathbf{v}(t) \) as the first derivative of \( \mathbf{r}(t) \): captures rate of displacement change.
• Acceleration \( \mathbf{a}(t) \) as the derivative of \( \mathbf{v}(t) \): measures rate of velocity change.

These derivatives tell us how the particle's motion evolves over time, allowing us to predict future positions and movements.