Problem 225 - Calculus Volume 3

Question

The position of a particle is given by \(\mathbf{r}(t)=\left\langle t^{2}, \ln (t), \sin (\pi t)\right\rangle, \quad\) where \(t \quad\) is measured in seconds and \(\mathbf{r}\) is measured in meters. Find the velocity, acceleration, and speed functions. What are the position, velocity, speed, and acceleration of the particle at \(1 \mathrm{sec} ?\)

Step-by-Step Solution

Answer

At 1 second: Position is \(\langle 1,0,0 \rangle\); Velocity is \(\langle 2,1,-\pi \rangle\); Speed is \(\sqrt{5+\pi^2}\); Acceleration is \(\langle 2,-1,0 \rangle\).

Step 1: Finding the Velocity Function

To determine the velocity function, we need to differentiate the position vector function \( \mathbf{r}(t) = \langle t^2, \ln(t), \sin(\pi t) \rangle \) with respect to time \( t \). This gives us the velocity vector \( \mathbf{v}(t) \).The differentiation is done component-wise:1. Differentiate \( t^2 \) to get \( 2t \).2. Differentiate \( \ln(t) \) to get \( \frac{1}{t} \).3. Differentiate \( \sin(\pi t) \) using the chain rule: \( \pi \cos(\pi t) \).So, the velocity function is:\[ \mathbf{v}(t) = \left\langle 2t, \frac{1}{t}, \pi \cos(\pi t) \right\rangle \]

Step 2: Finding the Acceleration Function

For the acceleration function, differentiate the velocity function \( \mathbf{v}(t) = \left\langle 2t, \frac{1}{t}, \pi \cos(\pi t) \right\rangle \) with respect to \( t \).Differentiating component-wise gives:1. Differentiate \( 2t \) to get \( 2 \).2. Differentiate \( \frac{1}{t} \) to get \( -\frac{1}{t^2} \).3. Differentiate \( \pi \cos(\pi t) \) using the chain rule: \(-\pi^2 \sin(\pi t) \).Thus, the acceleration function is:\[ \mathbf{a}(t) = \left\langle 2, -\frac{1}{t^2}, -\pi^2 \sin(\pi t) \right\rangle \]

Step 3: Finding the Speed Function

The speed of the particle is the magnitude of the velocity vector. Thus, we calculate it as follows:\[ \text{Speed} = \| \mathbf{v}(t) \| = \sqrt{(2t)^2 + \left(\frac{1}{t}\right)^2 + (\pi \cos(\pi t))^2} \]Essentially, simplify to:\[ \text{Speed} = \sqrt{4t^2 + \frac{1}{t^2} + \pi^2 \cos^2(\pi t)} \]

Step 4: Evaluating Functions at \(t=1\) second

Evaluate the position, velocity, acceleration, and speed at \( t = 1 \):1. Position: \[ \mathbf{r}(1) = \langle 1^2, \ln(1), \sin(\pi \times 1) \rangle = \langle 1, 0, 0 \rangle \]2. Velocity: \[ \mathbf{v}(1) = \langle 2 \times 1, \frac{1}{1}, \pi \cos(\pi \times 1) \rangle = \langle 2, 1, -\pi \rangle \]3. Acceleration: \[ \mathbf{a}(1) = \langle 2, -1, 0 \rangle \]4. Speed: \[ \| \mathbf{v}(1) \| = \sqrt{2^2 + 1^2 + (-\pi)^2} = \sqrt{4 + 1 + \pi^2} \]

Key Concepts

Particle Motion

When studying particle motion, we're observing how particles change position over time. This can be described by a position vector function, often denoted as \( \mathbf{r}(t) \). The function gives the coordinates of the particle in terms of time, \( t \). For instance, if a particle's position is given by \( \mathbf{r}(t) = \langle t^2, \ln(t), \sin(\pi t) \rangle \), each component of the vector function represents the particle's location in a different spatial dimension.
The first component, \( t^2 \), shows the particle's motion along the x-axis, the second component, \( \ln(t) \), along the y-axis, and the third, \( \sin(\pi t) \), along the z-axis. As time progresses, the particle traces out a path in three-dimensional space.

• The position of the particle at any specific time can be found by simply substituting the value of \( t \) into the position function.
• At \( t = 1 \) second, using the given function, the position vector is \( \langle 1, 0, 0 \rangle \), representing the particle's point in space.

Velocity Function

Velocity functions are crucial for understanding how quickly a particle's position changes over time. To find the velocity function \( \mathbf{v}(t) \) of a particle described by a position function \( \mathbf{r}(t) \), differentiate each component of the position vector with respect to time.
For our example position function \( \mathbf{r}(t) = \langle t^2, \ln(t), \sin(\pi t) \rangle \):

• Differentiating \( t^2 \) gives \( 2t \), illustrating the rate of change of the x-coordinate.
• Differentiating \( \ln(t) \) results in \( \frac{1}{t} \), indicating the rate of change in the y-coordinate.
• Applying the chain rule to \( \sin(\pi t) \), we get \( \pi \cos(\pi t) \), showing how quickly the z-coordinate changes.

Hence, the velocity function is \( \mathbf{v}(t) = \langle 2t, \frac{1}{t}, \pi \cos(\pi t) \rangle \).
At \( t = 1 \), we compute the velocity vector \( \langle 2, 1, -\pi \rangle \), which describes the particle’s speed and direction in each dimensional axis.

Acceleration Function

Acceleration functions detail how the velocity of a particle changes over time, offering insights into speeding up or slowing down motions. To determine the acceleration function \( \mathbf{a}(t) \), differentiate the velocity function \( \mathbf{v}(t) \) with respect to time.
Given the velocity function \( \mathbf{v}(t) = \langle 2t, \frac{1}{t}, \pi \cos(\pi t) \rangle \):

• The differentiation of \( 2t \) results in \( 2 \), representing constant acceleration along the x-axis.
• Differentiating \( \frac{1}{t} \) leads to \( -\frac{1}{t^2} \), showing how the y-coordinate’s rate of change decreases over time.
• Using the chain rule on \( \pi \cos(\pi t) \), we obtain \( -\pi^2 \sin(\pi t) \), indicating oscillatory acceleration along the z-axis.

Thus, the acceleration function is \( \mathbf{a}(t) = \langle 2, -\frac{1}{t^2}, -\pi^2 \sin(\pi t) \rangle \).
Evaluating this at \( t = 1 \) gives the acceleration \( \langle 2, -1, 0 \rangle \), reflecting the specific changes in velocity at that moment.

Speed Function

Speed is a scalar quantity that provides the magnitude of velocity without regard to direction. To find the speed of a particle, calculate the magnitude of its velocity vector \( \mathbf{v}(t) \). This involves taking the square root of the sum of the squares of its components.
For the velocity function \( \mathbf{v}(t) = \langle 2t, \frac{1}{t}, \pi \cos(\pi t) \rangle \), the speed is:

• \( \text{Speed} = \sqrt{(2t)^2 + \left(\frac{1}{t}\right)^2 + (\pi \cos(\pi t))^2} \)
• Simplifying yields \( \text{Speed} = \sqrt{4t^2 + \frac{1}{t^2} + \pi^2 \cos^2(\pi t)} \).

At \( t = 1 \), the speed computes to \( \sqrt{4 + 1 + \pi^2} \).
This tells us how fast the particle is moving in space, irrespective of its direction.