Problem 74 - Calculus III

Question

The position of a particle is given by \(\mathbf{r}(t)=\left\langle t^{2}, \ln (t), \sin (\pi t)\right\rangle\), where \(t\) 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 \( t=1 \): Position \( \langle 1, 0, 0 \rangle \), Velocity \( \langle 2, 1, -\pi \rangle \), Acceleration \( \langle 2, -1, 0 \rangle \), Speed \( \sqrt{5+\pi^2} \).

Step 1: Determine Velocity Function

To find the velocity, we need to take the derivative of the position function \( \mathbf{r}(t) \) with respect to time \( t \). The velocity \( \mathbf{v}(t) \) is the derivative \( \frac{d}{dt} \mathbf{r}(t) \).\[\mathbf{v}(t) = \left\langle \frac{d}{dt} t^2, \frac{d}{dt} \ln(t), \frac{d}{dt} \sin(\pi t) \right\rangle = \left\langle 2t, \frac{1}{t}, \pi \cos(\pi t) \right\rangle\]

Step 2: Determine Acceleration Function

The acceleration is the derivative of the velocity function with respect to time. Therefore, we take the derivative of \( \mathbf{v}(t) \).\[\mathbf{a}(t) = \left\langle \frac{d}{dt} (2t), \frac{d}{dt} \left(\frac{1}{t}\right), \frac{d}{dt}(\pi \cos(\pi t)) \right\rangle = \left\langle 2, -\frac{1}{t^2}, -\pi^2 \sin(\pi t) \right\rangle\]

Step 3: Determine Speed Function

The speed \( s(t) \) is the magnitude of the velocity vector \( \mathbf{v}(t) \). Calculate it by finding the square root of the sum of the squares of the components of \( \mathbf{v}(t) \).\[s(t) = \left\| \mathbf{v}(t) \right\| = \sqrt{(2t)^2 + \left(\frac{1}{t}\right)^2 + (\pi \cos(\pi t))^2}\]

Step 4: Evaluate Functions at \( t=1 \)

Now, we'll substitute \( t=1 \) into the position, velocity, acceleration, and speed functions to evaluate them at this specific time.- **Position at \( t=1 \):** \[ \mathbf{r}(1) = \left\langle 1^2, \ln(1), \sin(\pi \times 1) \right\rangle = \left\langle 1, 0, 0 \right\rangle \]- **Velocity at \( t=1 \):** \[ \mathbf{v}(1) = \left\langle 2 \times 1, \frac{1}{1}, \pi \cos(\pi \times 1) \right\rangle = \left\langle 2, 1, -\pi \right\rangle \]- **Acceleration at \( t=1 \):** \[ \mathbf{a}(1) = \left\langle 2, -\frac{1}{1^2}, -\pi^2 \sin(\pi \times 1) \right\rangle = \left\langle 2, -1, 0 \right\rangle \]- **Speed at \( t=1 \):** \[ s(1) = \sqrt{2^2 + 1^2 + (-\pi)^2} = \sqrt{4 + 1 + \pi^2} \]

Key Concepts

Velocity Function

In calculus, the velocity function is a critical component that helps us understand how the position of an object changes over time. To find the velocity function, we differentiate the position function with respect to time. The velocity vector contains components that describe the object's motion in different dimensions.

For our exercise, the position of a particle is described by the vector function \( \mathbf{r}(t) = \langle t^2, \ln(t), \sin(\pi t) \rangle \). The task is to find the derivative of each component of \( \mathbf{r}(t) \) to determine the velocity function \( \mathbf{v}(t) \).

This process involves applying basic differentiation rules:

• The derivative of \( t^2 \) is \( 2t \)
• The derivative of \( \ln(t) \) is \( \frac{1}{t} \)
• The derivative of \( \sin(\pi t) \) is \( \pi \cos(\pi t) \)

By combining these derivatives, the velocity function is \( \mathbf{v}(t) = \langle 2t, \frac{1}{t}, \pi \cos(\pi t) \rangle \).

This function tells us how fast and in what direction each component of the position vector is changing at any given time \( t \).

Acceleration Function

The acceleration function is the derivative of the velocity function, which illustrates how the velocity of an object changes over time. To find this, we need to take the derivative of each component of the velocity function.

From our earlier computation, we know the velocity function is \( \mathbf{v}(t) = \langle 2t, \frac{1}{t}, \pi \cos(\pi t) \rangle \). Now, we'll differentiate each component:

• The derivative of \( 2t \) is \( 2 \)
• The derivative of \( \frac{1}{t} \) is \( -\frac{1}{t^2} \)
• The derivative of \( \pi \cos(\pi t) \) is \(-\pi^2 \sin(\pi t)\)

This results in the acceleration function: \( \mathbf{a}(t) = \langle 2, -\frac{1}{t^2}, -\pi^2 \sin(\pi t) \rangle \).

In essence, acceleration gives us insights into how rapidly and in which direction the velocity of the particle is changing.

Speed Function

Speed is a scalar quantity representative of how fast an object is moving, irrespective of its direction. Unlike velocity, speed does not take direction into account, only magnitude. To find the speed of the particle, we calculate the magnitude of the velocity vector.

For our problem, the velocity vector is \( \mathbf{v}(t) = \langle 2t, \frac{1}{t}, \pi \cos(\pi t) \rangle \). To find the speed, we use the formula for the magnitude of a vector: The speed \( s(t) \) is determined by:\[s(t) = \| \mathbf{v}(t) \| = \sqrt{(2t)^2 + \left(\frac{1}{t}\right)^2 + (\pi \cos(\pi t))^2}\]

This calculation provides a scalar value indicating how fast the particle is moving at time \( t \), integrating the contributions of each individual component of velocity.