Problem 6 - Calculus III

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: Differentiate to Find Velocity

To find the velocity of the particle, we take the derivative of the position function \( \mathbf{r}(t) \) with respect to \( t \), \( \mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t) \). The position function is \( \mathbf{r}(t) = \langle t^2 - 1, t \rangle \). Differentiate each component: \( \frac{d}{dt}(t^2 - 1) = 2t \) and \( \frac{d}{dt}(t) = 1 \). Thus, \( \mathbf{v}(t) = \langle 2t, 1 \rangle \).

Step 2: Differentiate to Find Acceleration

The acceleration of the particle is the derivative of the velocity function. Differentiate the velocity vector \( \mathbf{v}(t) = \langle 2t, 1 \rangle \) with respect to \( t \). Therefore, \( \mathbf{a}(t) = \frac{d}{dt} \mathbf{v}(t) = \langle 2, 0 \rangle \).

Step 3: Calculate Speed

Speed is the magnitude of the velocity vector. Compute this by finding \( \lVert \mathbf{v}(t) \rVert = \sqrt{(2t)^2 + 1^2} = \sqrt{4t^2 + 1}\). Thus, the speed is \( \sqrt{4t^2 + 1} \).

Key Concepts

Understanding Velocity in Vector Calculus

Velocity is a fundamental concept in vector calculus that describes how fast something is moving and in what direction. When dealing with a function like \( \mathbf{r}(t) = \langle t^2 - 1, t \rangle \), we find the velocity by differentiating the position function with respect to time \( t \). Each component of the position vector is differentiated separately:

• For \( t^2 - 1 \), the derivative is \( 2t \).
• For \( t \), the derivative is \( 1 \).

Thus, our velocity vector becomes \( \mathbf{v}(t) = \langle 2t, 1 \rangle \).
This vector illustrates both the speed and direction of the particle. However, velocity should not be confused with speed because speed doesn't include direction. With velocity, you can understand not just how fast the particle is moving but in which direction it's heading.

Exploring Acceleration: The Change in Velocity

In vector calculus, acceleration refers to the rate at which velocity changes with respect to time. It's the "second derivative" of the position function.
When we have the velocity function \( \mathbf{v}(t) = \langle 2t, 1 \rangle \), finding the acceleration involves taking the derivative of each component of this function:

• For \( 2t \), the derivative is \( 2 \).
• For \( 1 \), the derivative is \( 0 \).

This leads us to an acceleration vector \( \mathbf{a}(t) = \langle 2, 0 \rangle \).
This means the particle is accelerating at a constant rate along the x-axis while maintaining a constant velocity along the y-axis.
Understanding acceleration is crucial as it tells us how the velocity of the particle changes over time, providing insights into the forces acting on the particle.

Clarifying Speed: The Magnitude of Velocity

While velocity includes both speed and direction, speed alone is just a scalar quantity representing how fast an object is moving, regardless of direction. In vector calculus, we calculate speed as the magnitude of the velocity vector.
For our velocity vector \( \mathbf{v}(t) = \langle 2t, 1 \rangle \), the speed is calculated using the formula:
\[ \lVert \mathbf{v}(t) \rVert = \sqrt{(2t)^2 + 1^2} = \sqrt{4t^2 + 1}\]
This equation defines the speed of the particle at any point in time \( t \), providing a clear, numerical value of how quickly it moves.
Understanding speed in this context allows us to grasp the intensity of the particle's motion without getting into directional details.