Problem 10 - Calculus Early Transcendentals

Question

\(9-14\) Find the velocity, acceleration, and speed of a particle with the given position function. $$\mathbf{r}(t)=\langle 2 \cos t, 3 t, 2 \sin t\rangle$$

Step-by-Step Solution

Answer

Velocity: \( \mathbf{v}(t) = \langle -2\sin t, 3, 2\cos t \rangle \), Acceleration: \( \mathbf{a}(t) = \langle -2\cos t, 0, -2\sin t \rangle \), Speed: \( \sqrt{13} \).

Step 1: Find the Velocity Function

To find the velocity of the particle, we need to take the derivative of the position function \( \mathbf{r}(t) \) with respect to \( t \). This will give us the velocity function \( \mathbf{v}(t) = \frac{d \mathbf{r}}{dt} \).Given the position function: \[ \mathbf{r}(t) = \langle 2\cos t, 3t, 2\sin t \rangle \]Taking the derivative of each component:- Derivative of \(2 \cos t\) is \(-2 \sin t\),- Derivative of \(3t\) is \(3\),- Derivative of \(2 \sin t\) is \(2 \cos t\).Thus, the velocity function is:\[ \mathbf{v}(t) = \langle -2 \sin t, 3, 2 \cos t \rangle \]

Step 2: Find the Acceleration Function

To find the acceleration of the particle, we take the derivative of the velocity function \( \mathbf{v}(t) \) with respect to \( t \). This will give us the acceleration function \( \mathbf{a}(t) = \frac{d \mathbf{v}}{dt} \).The velocity function is:\[ \mathbf{v}(t) = \langle -2 \sin t, 3, 2 \cos t \rangle \]Taking the derivative of each component:- Derivative of \(-2 \sin t\) is \(-2 \cos t\),- Derivative of \(3\) is \(0\),- Derivative of \(2 \cos t\) is \(-2 \sin t\).Thus, the acceleration function is:\[ \mathbf{a}(t) = \langle -2 \cos t, 0, -2 \sin t \rangle \]

Step 3: Find the Speed

To find the speed of the particle, we need to calculate the magnitude of the velocity vector \( \mathbf{v}(t) \).The velocity function is:\[ \mathbf{v}(t) = \langle -2 \sin t, 3, 2 \cos t \rangle \]The magnitude of \( \mathbf{v}(t) \) is given by:\[\|\mathbf{v}(t)\| = \sqrt{(-2 \sin t)^2 + 3^2 + (2 \cos t)^2}\]Simplifying:\[\|\mathbf{v}(t)\| = \sqrt{4 \sin^2 t + 9 + 4 \cos^2 t}\]Notice that \(4(\sin^2 t + \cos^2 t) = 4\), because \(\sin^2 t + \cos^2 t = 1\). Therefore:\[\|\mathbf{v}(t)\| = \sqrt{4 + 9} = \sqrt{13}\]Thus, the speed of the particle is constant and equal to \( \sqrt{13} \).

Key Concepts

Velocity Function

The velocity function of a particle provides a measure of how the position of the particle changes with respect to time. It's an important concept in calculus because it describes the rate at which an object's position is changing. By taking the derivative of the position function, we obtain the velocity function.For a given position function \( \mathbf{r}(t) = \langle 2\cos t, 3t, 2\sin t \rangle \), which describes the path and position of a particle over time, the velocity function \( \mathbf{v}(t) \) is derived by differentiating each component of \( \mathbf{r}(t) \) with respect to \( t \):

• The derivative of \( 2 \cos t \) is \(-2 \sin t \).
• The derivative of \( 3t \) is \( 3 \).
• The derivative of \( 2 \sin t \) is \( 2 \cos t \).

Hence, the velocity function is expressed as \( \mathbf{v}(t) = \langle -2 \sin t, 3, 2 \cos t \rangle \). This indicates the speed and direction of the particle at any time \( t \).

Acceleration Function

The acceleration function indicates how an object's velocity changes over time. It's the derivative of the velocity function, just as the velocity is the derivative of the position function. This second derivative aspect in calculus helps us understand the changes a particle's velocity undergoes, which is crucial for analyzing the motion dynamics.For the velocity function \( \mathbf{v}(t) = \langle -2 \sin t, 3, 2 \cos t \rangle \), the acceleration function \( \mathbf{a}(t) \) is derived as follows:

• The derivative of \(-2 \sin t\) is \(-2 \cos t\).
• The derivative of \(3\) is \(0\) because it is constant.
• The derivative of \(2 \cos t\) is \(-2 \sin t\).

Therefore, the acceleration function becomes \( \mathbf{a}(t) = \langle -2 \cos t, 0, -2 \sin t \rangle \). This tells us about the change in velocity direction and magnitude as time progresses.

Position Function

The position function describes the location of a particle at any given time \( t \). It provides a clear snapshot of where the particle is located in a coordinate system. Understanding the position function is essential as it serves as the foundation for deriving other motion-related functions, such as velocity and acceleration.For example, the position function \( \mathbf{r}(t) = \langle 2 \cos t, 3t, 2 \sin t \rangle \) reveals the particle's position based on:

• The \( x \)-coordinate: \(2 \cos t\), indicating oscillatory motion.
• The \( y \)-coordinate: \(3t\), showing linear movement, indicating constant velocity in this direction.
• The \( z \)-coordinate: \(2 \sin t\), another oscillatory motion component.

By analyzing the position function, we gain a comprehensive understanding of the particle's path and motion characteristics over time. This understanding is critical when predicting future positions or understanding past motion behavior.