Problem 160 - Calculus Volume 3

Question

Find the velocity, acceleration, and speed of a particle with the given position function. \(\mathbf{r}(t)=\left\langle t^{2}-1, t\right\rangle\)

Step-by-Step Solution

Answer

Velocity: \( \langle 2t, 1 \rangle \), Acceleration: \( \langle 2, 0 \rangle \), Speed: \( \sqrt{4t^2 + 1} \).

Step 1: Understand the Position Function

The given position function of the particle is \[ \mathbf{r}(t) = \langle t^2 - 1, t \rangle \]where the particle's position is represented as a vector in the plane. Here, the x-coordinate is \( t^2 - 1 \), and the y-coordinate is \( t \).

Step 2: Find the Velocity Function

The velocity of the particle is the derivative of the position function with respect to time \( t \). Compute the derivatives of each component:\[ \mathbf{v}(t) = \frac{d}{dt}(t^2 - 1) \mathbf{i} + \frac{d}{dt}(t) \mathbf{j} \]This results in:\[ \mathbf{v}(t) = \langle 2t, 1 \rangle \]

Step 3: Find the Acceleration Function

The acceleration of the particle is the derivative of the velocity function with respect to time \( t \). Compute the derivatives of each component of the velocity vector:\[ \mathbf{a}(t) = \frac{d}{dt}(2t) \mathbf{i} + \frac{d}{dt}(1) \mathbf{j} \]This results in:\[ \mathbf{a}(t) = \langle 2, 0 \rangle \]

Step 4: Calculate the Speed of the Particle

The speed of the particle is the magnitude of the velocity vector. Calculate this by finding the magnitude, or Euclidean norm, of \( \mathbf{v}(t) = \langle 2t, 1 \rangle \):\[ \text{Speed} = \| \mathbf{v}(t) \| = \sqrt{(2t)^2 + 1^2} = \sqrt{4t^2 + 1} \]

Key Concepts

Velocity

Velocity describes how the position of a particle changes over time. It's a vector quantity, which means it has both magnitude and direction. For the given position function \( \mathbf{r}(t) = \langle t^2 - 1, t \rangle \), the velocity \( \mathbf{v}(t) \) is found by differentiating each component of the position vector with respect to time, \( t \). This results in:

• The x-component: \( \frac{d}{dt}(t^2 - 1) = 2t \)
• The y-component: \( \frac{d}{dt}(t) = 1 \)

Thus, the velocity vector is \( \mathbf{v}(t) = \langle 2t, 1 \rangle \).
This shows the rate of change along each axis, indicating how quickly and in what direction the particle is moving in the plane.
This is crucial because it helps us understand the instantaneous motion of the particle at any time \( t \).

Acceleration

Acceleration is the rate at which the velocity of a particle changes over time. Like velocity, it is also a vector quantity, meaning it has both magnitude and direction. To find the acceleration for the given position function, we need to compute the derivative of the velocity function \( \mathbf{v}(t) = \langle 2t, 1 \rangle \). The acceleration vector \( \mathbf{a}(t) \) is calculated as follows:

• Differentiate the x-component: \( \frac{d}{dt}(2t) = 2 \)
• Differentiate the y-component: \( \frac{d}{dt}(1) = 0 \)

Therefore, the acceleration vector is \( \mathbf{a}(t) = \langle 2, 0 \rangle \).
This tells us that the velocity is increasing by 2 units along the x-axis every second.
The lack of change in the y-component means there is no acceleration in the y-direction. This shows the particle is moving in a straight line parallel to the x-axis while maintaining constant velocity in the y-direction.

Position Function

The position function \( \mathbf{r}(t) = \langle t^2 - 1, t \rangle \) represents the location of a particle at any given time \( t \). The position is described as a vector in a plane, providing coordinates (x and y) to specify the exact spot of the particle at that moment. The components of this position function are:

• x-coordinate: \( t^2 - 1 \)
• y-coordinate: \( t \)

This means at time \( t \), the particle's x-position is \( t^2 - 1 \), and its y-position is \( t \).
This function is crucial because it describes how the location of the particle changes as time progresses.
By understanding the position function, we can determine where the particle starts and where it will be at any future time.

Speed

Speed refers to how fast a particle is moving, regardless of its direction. Unlike velocity, speed is a scalar quantity, meaning it only has magnitude. It is calculated as the magnitude, or Euclidean norm, of the velocity vector. For the velocity \( \mathbf{v}(t) = \langle 2t, 1 \rangle \), the speed is:

• Compute the square of each component: \( (2t)^2 \) and \( 1^2 \)
• Add these squares: \( 4t^2 + 1 \)
• Take the square root of the sum: \( \sqrt{4t^2 + 1} \)

Thus, the speed of the particle is \( \sqrt{4t^2 + 1} \).
This function tells us how much distance the particle covers in a small amount of time.
Remember, while velocity can be zero (if direction changes), as long as the particle is moving, its speed is positive.