Problem 7 - Essential Calculus Early Transcendentals

Question

Find the velocity, acceleration, and speed of a particle with the given position function. $$\mathbf{r}(t)=\left\langle t^{2}+t, t^{2}-t, t^{3}\right\rangle$$

Step-by-Step Solution

Answer

Velocity: \(\left\langle 2t + 1, 2t - 1, 3t^2 \right\rangle\); Acceleration: \(\left\langle 2, 2, 6t \right\rangle\); Speed: \(\sqrt{9t^4 + 8t^2 + 2}\)."

Step 1: Differentiate the Position Function to Find Velocity

The velocity of a particle is given by the derivative of its position function with respect to time. Given the position function \( \mathbf{r}(t) = \left\langle t^2 + t, t^2 - t, t^3 \right\rangle \), we differentiate each component separately. This gives us:\[\mathbf{v}(t) = \frac{d}{dt}\left\langle t^2 + t, t^2 - t, t^3 \right\rangle = \left\langle 2t + 1, 2t - 1, 3t^2 \right\rangle.\]

Step 2: Differentiate the Velocity Function to Find Acceleration

The acceleration of a particle is obtained by differentiating the velocity function with respect to time. Our velocity function is \( \mathbf{v}(t) = \left\langle 2t + 1, 2t - 1, 3t^2 \right\rangle \). Differentiating each component separately, we find:\[\mathbf{a}(t) = \frac{d}{dt}\left\langle 2t + 1, 2t - 1, 3t^2 \right\rangle = \left\langle 2, 2, 6t \right\rangle.\]

Step 3: Calculate Speed as the Magnitude of Velocity

The speed of a particle is the magnitude of the velocity vector. Given \( \mathbf{v}(t) = \left\langle 2t + 1, 2t - 1, 3t^2 \right\rangle \), we calculate the magnitude:\[|\mathbf{v}(t)| = \sqrt{(2t + 1)^2 + (2t - 1)^2 + (3t^2)^2}.\]Simplifying, we have:\[|\mathbf{v}(t)| = \sqrt{4t^2 + 4t + 1 + 4t^2 - 4t + 1 + 9t^4}= \sqrt{9t^4 + 8t^2 + 2}.\]

Key Concepts

Velocity

Velocity describes how much the position of an object changes over time. Imagine tracking the journey of a particle as it moves through space. If you want to know how quickly it moves and in which direction, you're looking for its velocity.
Different from mere speed, velocity has a direction, making it a vector quantity. This means it tells us not only how fast something moves, but also where it's headed. For our particle with position function \( \mathbf{r}(t) = \left\langle t^2 + t, t^2 - t, t^3 \right\rangle \),
the velocity function is found by calculating the derivative of each component:
\[ \mathbf{v}(t) = \frac{d}{dt}\left\langle t^2 + t, t^2 - t, t^3 \right\rangle = \left\langle 2t + 1, 2t - 1, 3t^2 \right\rangle. \]
Now, the velocity function \( \mathbf{v}(t) \) shows us that at any time \( t \),

• in the x-direction, the component is \( 2t + 1 \)
• in the y-direction, it is \( 2t - 1 \)
• for the z-direction, it's \( 3t^2 \)

With these equations, you can predict where the particle is going and at what rate.

Acceleration

Acceleration is all about change. It describes how the velocity of a particle changes over time. When we say a car accelerates, we really mean it's changing its speed or direction.
In our particle's adventure, we determine acceleration by taking the derivative of the velocity function. This gives us how quickly the velocity changes, or in math terms, the derivative of each velocity component.For our velocity function \( \mathbf{v}(t) = \left\langle 2t + 1, 2t - 1, 3t^2 \right\rangle \):
\[ \mathbf{a}(t) = \frac{d}{dt}\left\langle 2t + 1, 2t - 1, 3t^2 \right\rangle = \left\langle 2, 2, 6t \right\rangle. \]
This tells us that:

• The x-component of acceleration is constant at \( 2 \)
• The y-component is also constant at \( 2 \)
• The z-component, however, depends on time \( t \) and is \( 6t \)

The z-direction adds an interesting twist as it changes faster with increasing time, indicating some dynamic motion.

Speed

Speed gives us the bare-bones information of how fast our particle is moving, without concerning itself with direction. Unlike velocity, speed is a scalar, meaning it doesn't tell you which way the particle is headed, just how fast it's going anywhere.
We can find speed by calculating the magnitude of the velocity vector. For \( \mathbf{v}(t) = \left\langle 2t + 1, 2t - 1, 3t^2 \right\rangle \), the magnitude, or speed, is:
\[|\mathbf{v}(t)| = \sqrt{(2t + 1)^2 + (2t - 1)^2 + (3t^2)^2}.\]
When simplified, it becomes:
\[|\mathbf{v}(t)| = \sqrt{9t^4 + 8t^2 + 2}.\]
This value tells us how fast the particle is zooming through space at any given time \( t \), spelled out without worrying about which way it's zooming.