Problem 7 - Calculus III

Question

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

Step-by-Step Solution

Answer

Velocity: \(\langle e^t, -e^{-t} \rangle\); Acceleration: \(\langle e^t, e^{-t} \rangle\); Speed: \(\sqrt{e^{2t} + e^{-2t}}\).

Step 1: Find the Velocity

The velocity of a particle is the derivative of its position function with respect to time. Given that \( \mathbf{r}(t) = \langle e^t, e^{-t} \rangle \), we differentiate each component. The derivative of \( e^t \) is \( e^t \), and the derivative of \( e^{-t} \) is \( -e^{-t} \). Thus, the velocity function is:\[ \mathbf{v}(t) = \left\langle e^t, -e^{-t} \right\rangle \]

Step 2: Find the Acceleration

The acceleration of a particle is the derivative of its velocity function with respect to time. From the previous step, we have \( \mathbf{v}(t) = \langle e^t, -e^{-t} \rangle \). Differentiating each component:The derivative of \( e^t \) is \( e^t \) and the derivative of \( -e^{-t} \) is \( e^{-t} \). Therefore, the acceleration function is:\[ \mathbf{a}(t) = \left\langle e^t, e^{-t} \right\rangle \]

Step 3: Calculate the Speed

The speed of a particle is the magnitude of its velocity vector. From the velocity vector \( \mathbf{v}(t) = \langle e^t, -e^{-t} \rangle \), we calculate its magnitude:\[ \text{speed} = \sqrt{(e^t)^2 + (-e^{-t})^2} = \sqrt{e^{2t} + e^{-2t}} \]

Step 4: Final Result

We found that the velocity is \( \mathbf{v}(t) = \langle e^t, -e^{-t} \rangle \), the acceleration is \( \mathbf{a}(t) = \langle e^t, e^{-t} \rangle \), and the speed is \( \sqrt{e^{2t} + e^{-2t}} \).

Key Concepts

Velocity of a Particle

In vector calculus, the concept of velocity is critically important for understanding how a particle's position changes over time. Velocity is essentially a vector that indicates the direction and speed of a particle's movement. To find the velocity of a particle, you need to take the derivative of its position function with respect to time. This means you differentiate each component of the position vector separately.

For instance, if the position function is given by \( \mathbf{r}(t) = \langle e^t, e^{-t} \rangle \), then the velocity \( \mathbf{v}(t) \) of the particle is derived by:

• Taking the derivative of \( e^t \), which is \( e^t \)
• Taking the derivative of \( e^{-t} \), which is \( -e^{-t} \)

Thus, the velocity vector becomes \( \mathbf{v}(t) = \langle e^t, -e^{-t} \rangle \). It's important to remember that velocity has both magnitude and direction, and is measured in units of distance over time, such as meters per second (m/s).

This function tells you how the particle moves over time—faster or slower—in different directions.

Acceleration of a Particle

Acceleration is the rate at which the velocity of a particle changes with time. It is also a vector quantity, meaning it has both magnitude and direction. To find the acceleration, we differentiate the velocity function with respect to time. This gives us insight into how a particle's velocity varies, helping to understand its motion dynamics.

Taking our previous velocity function \( \mathbf{v}(t) = \langle e^t, -e^{-t} \rangle \):

• The derivative of \( e^t \) remains \( e^t \)
• The derivative of \( -e^{-t} \) becomes \( e^{-t} \)

This results in the acceleration function \( \mathbf{a}(t) = \langle e^t, e^{-t} \rangle \).

This tells us about the change in velocity. It means that as time progresses, the acceleration does not change in direction, and its components are the same as the original position function. So, the change in velocity over time is consistent, indicating a very stable motion path under these conditions.

Speed Calculation

Speed is a scalar quantity representing the magnitude of velocity, and it doesn’t account for direction—only how fast a particle is moving. While velocity tells you the speed and direction, speed gives the actual pace of movement. To find speed, you calculate the magnitude of the velocity vector.

Using the velocity vector \( \mathbf{v}(t) = \langle e^t, -e^{-t} \rangle \), the magnitude (or speed) is calculated as follows:

• Square each component: \( (e^t)^2 \) and \( (-e^{-t})^2 \)
• Add the squared components: \( e^{2t} + e^{-2t} \)
• Take the square root of the sum: \( \sqrt{e^{2t} + e^{-2t}} \)

This gives the particle's speed at any point in time \( t \).

Knowing the speed helps in understanding how quickly a particle moves, regardless of the direction. It is a crucial component for practical applications like predicting motion or planning trajectories.