Problem 8 - Essential Calculus Early Transcendentals

Question

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

The velocity is \( \langle -2 \sin t, 3, 2 \cos t \rangle \), acceleration is \( \langle -2 \cos t, 0, -2 \sin t \rangle \), and speed is \( \sqrt{13} \).

Step 1: Differentiate the Position Function to Find Velocity

To find the velocity function, we need to differentiate the position function \( \mathbf{r}(t) = \langle 2 \cos t, 3t, 2 \sin t \rangle \) with respect to \( t \). \[\mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t) = \left\langle \frac{d}{dt}(2 \cos t), \frac{d}{dt}(3t), \frac{d}{dt}(2 \sin t) \right\rangle = \langle -2 \sin t, 3, 2 \cos t \rangle\]

Step 2: Differentiate the Velocity Function to Find Acceleration

To find the acceleration function, we differentiate the velocity function \( \mathbf{v}(t) = \langle -2 \sin t, 3, 2 \cos t \rangle \) with respect to \( t \). \[ \mathbf{a}(t) = \frac{d}{dt} \mathbf{v}(t) = \left\langle \frac{d}{dt}(-2 \sin t), \frac{d}{dt}(3), \frac{d}{dt}(2 \cos t) \right\rangle = \langle -2 \cos t, 0, -2 \sin t \rangle \]

Step 3: Calculate Speed from the Velocity Function

Speed is the magnitude of the velocity vector. To find the speed at any time \( t \), we calculate the magnitude of \( \mathbf{v}(t) = \langle -2 \sin t, 3, 2 \cos t \rangle \):\[\text{Speed} = ||\mathbf{v}(t)|| = \sqrt{(-2 \sin t)^2 + 3^2 + (2 \cos t)^2}\] Simplify the expression:\[\text{Speed} = \sqrt{4 \sin^2 t + 9 + 4 \cos^2 t } = \sqrt{4(\sin^2 t + \cos^2 t) + 9} = \sqrt{4(1) + 9} = \sqrt{13}\]

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the rate at which a function is changing at any given point. For a vector-valued position function like \( \mathbf{r}(t) = \langle 2 \cos t, 3t, 2 \sin t \rangle \), differentiation involves taking the derivative of each component separately. This process helps to determine how each coordinate of a particle is changing with respect to time.

• In the position function, each component is a function of time \( t \).
• The derivative with respect to \( t \) for each component tells us the velocity in that respective direction.

Think of differentiation as peeling back layers of a function to understand its behavior. It's like understanding how fast or slow something evolves at any precise moment.

Velocity

Velocity is not just about how fast something is moving. It is a vector, which means it has both magnitude and direction. To find the velocity from the position function, we need to differentiate each coordinate of the position function. For the given function \( \mathbf{r}(t) \), the velocity would be \( \mathbf{v}(t) = \langle -2 \sin t, 3, 2 \cos t \rangle \).

• The first component, \(-2 \sin t\), indicates how the x-position is changing.
• The second component, \(3\), is a constant, meaning the y-position increases linearly with time.
• The third component, \(2 \cos t\), describes the z-position change.

By analyzing the velocity vector, we can see how the particle's position is dynamically evolving over time in 3D space.

Acceleration

Acceleration is the rate at which velocity changes with time. Once we have the velocity vector, we differentiate it with respect to time to get the acceleration vector. For \( \mathbf{v}(t) = \langle -2 \sin t, 3, 2 \cos t \rangle \), the acceleration \( \mathbf{a}(t) \) is calculated as \( \langle -2 \cos t, 0, -2 \sin t \rangle \).

• The x-component \(-2 \cos t\) suggests periodic acceleration in the x-direction.
• The y-component is \(0\), indicating constant velocity here, with no acceleration.
• The z-component \(-2 \sin t\) reveals periodic changes in acceleration in the z-direction.

Understanding acceleration helps us predict how quick movements are changing directions or speeds, providing insight into the dynamics of motion.

Speed Calculation

Speed is a bit different from velocity because it is a scalar quantity. It doesn't have a direction, only magnitude. Speed is calculated as the magnitude of the velocity vector. For the velocity function \( \mathbf{v}(t) = \langle -2 \sin t, 3, 2 \cos t \rangle \), the speed is found by taking the square root of the sum of the squares of its components:\[\text{Speed} = \sqrt{(-2 \sin t)^2 + 3^2 + (2 \cos t)^2} = \sqrt{4(1) + 9} = \sqrt{13}\]

• Note how we used \( \sin^2 t + \cos^2 t = 1 \) from trigonometry, simplifying the expression.
• This constant value, \( \sqrt{13} \), indicates the particle moves with a consistent speed.

Calculating speed gives a complete view of how fast the particle moves, irrespective of its direction.