Problem 11 - Calculus Early Transcendentals

Question

\(9-14\) 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: \(\sqrt{2} \mathbf{i} + e^t \mathbf{j} - e^{-t} \mathbf{k}\); Acceleration: \(e^t \mathbf{j} + e^{-t} \mathbf{k}\); Speed: \(\sqrt{2 + e^{2t} + e^{-2t}}\).

Step 1: Differentiate Position for Velocity

To find the velocity, differentiate the position function \( \mathbf{r}(t) = \sqrt{2}t \mathbf{i} + e^t \mathbf{j} + e^{-t} \mathbf{k} \) with respect to \( t \). This gives: \[ \mathbf{v}(t) = \frac{d}{dt}(\sqrt{2}t) \mathbf{i} + \frac{d}{dt}(e^t) \mathbf{j} + \frac{d}{dt}(e^{-t}) \mathbf{k} = \sqrt{2} \mathbf{i} + e^t \mathbf{j} - e^{-t} \mathbf{k} \]

Step 2: Differentiate Velocity for Acceleration

Now, differentiate the velocity \( \mathbf{v}(t) = \sqrt{2} \mathbf{i} + e^t \mathbf{j} - e^{-t} \mathbf{k} \) to find the acceleration. This gives: \[ \mathbf{a}(t) = \frac{d}{dt}(\sqrt{2}) \mathbf{i} + \frac{d}{dt}(e^t) \mathbf{j} + \frac{d}{dt}(-e^{-t}) \mathbf{k} = 0 \mathbf{i} + e^t \mathbf{j} + e^{-t} \mathbf{k} \]

Step 3: Calculate Speed as Magnitude of Velocity

Speed is the magnitude of the velocity vector, \( \mathbf{v}(t) = \sqrt{2} \mathbf{i} + e^t \mathbf{j} - e^{-t} \mathbf{k} \). Calculate it as follows:\[ \text{Speed} = \| \mathbf{v}(t) \| = \sqrt{(\sqrt{2})^2 + (e^t)^2 + (-e^{-t})^2} = \sqrt{2 + e^{2t} + e^{-2t}} \]

Key Concepts

Velocity

Velocity is a key concept when studying motion along a curve. In simple terms, velocity refers to the rate at which the position of an object is changing over time. It is a vector quantity, which means it has both a magnitude and a direction.

To calculate velocity from a given position function like \( \mathbf{r}(t) = \sqrt{2} t \mathbf{i} + e^{t} \mathbf{j} + e^{-t} \mathbf{k} \), you need to differentiate it with respect to time \( t \).

The differentiation gives:

• \( \frac{d}{dt}(\sqrt{2}t) \mathbf{i} = \sqrt{2} \mathbf{i} \)
• \( \frac{d}{dt}(e^t) \mathbf{j} = e^t \mathbf{j} \)
• \( \frac{d}{dt}(e^{-t}) \mathbf{k} = -e^{-t} \mathbf{k} \)

By doing this, we find the velocity function \( \mathbf{v}(t) = \sqrt{2} \mathbf{i} + e^t \mathbf{j} - e^{-t} \mathbf{k} \).

This velocity vector indicates how the position changes in different directions (\( \mathbf{i}, \mathbf{j}, \mathbf{k} \)). It gives not just the speed but also the direction of the particle's motion.

Acceleration

Acceleration is the rate at which an object's velocity changes with time. Like velocity, it is a vector quantity, which means it has both magnitude and direction.

To calculate the acceleration from the velocity function \( \mathbf{v}(t) = \sqrt{2} \mathbf{i} + e^t \mathbf{j} - e^{-t} \mathbf{k} \), you need to differentiate it with respect to time \( t \).

The differentiation process involves:

• \( \frac{d}{dt}(\sqrt{2}) \mathbf{i} = 0 \mathbf{i} \) since \( \sqrt{2} \) is a constant
• \( \frac{d}{dt}(e^t) \mathbf{j} = e^t \mathbf{j} \)
• \( \frac{d}{dt}(-e^{-t}) \mathbf{k} = e^{-t} \mathbf{k} \)

The acceleration function thus becomes \( \mathbf{a}(t) = 0 \mathbf{i} + e^t \mathbf{j} + e^{-t} \mathbf{k} \).

This describes how the velocity in different directions is changing over time, showing not only how fast but also in which direction the acceleration occurs.

Vector Calculus

Vector calculus is a mathematical tool essential for exploring and analyzing motion along curves. It helps in understanding how particles move in space by dealing with vector functions and their derivatives.

When dealing with vectors like position, velocity, and acceleration, vector calculus provides you the methods to differentiate and integrate these vectors to understand motion characteristics.

Here are some fundamentals:

• Position Vectors: Point to the particle's position in space; for example, \( \mathbf{r}(t) = \sqrt{2} t \mathbf{i} + e^{t} \mathbf{j} + e^{-t} \mathbf{k} \).
• Velocity Vectors: Derived from differentiating position with time, giving the rate of change of position.
• Acceleration Vectors: Obtained by differentiating velocity, indicating the rate of change of velocity.

Vector calculus not only helps calculate these vectors but also provides insights into their geometric properties, like direction and magnitude, essential for fully understanding motion's nature. It applies in various fields, from physics and engineering to computer graphics and beyond.