Problem 161 - Calculus 3

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: Understand the position function

The position function is given as \( \mathbf{r}(t) = \langle e^{t}, e^{-t} \rangle \). This function describes the position of the particle at any time \( t \), represented as a vector in two-dimensional space.

Step 2: Find the velocity function

Velocity is the derivative of the position function \( \mathbf{r}(t) \) with respect to time \( t \). Calculate the derivative:\[\mathbf{v}(t) = \frac{d}{dt} \langle e^{t}, e^{-t} \rangle = \langle \frac{d}{dt}e^{t}, \frac{d}{dt}e^{-t} \rangle = \langle e^{t}, -e^{-t} \rangle.\]

Step 3: Determine acceleration function

Acceleration is the derivative of the velocity function \( \mathbf{v}(t) \). Differentiate the velocity function with respect to \( t \):\[\mathbf{a}(t) = \frac{d}{dt} \langle e^{t}, -e^{-t} \rangle = \langle \frac{d}{dt}e^{t}, \frac{d}{dt}(-e^{-t}) \rangle = \langle e^{t}, e^{-t} \rangle.\]

Step 4: Calculate the speed of the particle

Speed is the magnitude of the velocity vector \( \mathbf{v}(t) \). Calculate the magnitude:\[\text{Speed} = ||\mathbf{v}(t)|| = \sqrt{(e^{t})^2 + (-e^{-t})^2} = \sqrt{e^{2t} + e^{-2t}}.\]

Step 5: Summary

The velocity of the particle 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 Function

In vector calculus, understanding the velocity function is crucial to determining how fast a particle moves and in what direction. Velocity is defined as the derivative of a vector function that describes the position of a particle over time. By finding this derivative, you can obtain the velocity vector. For the given position function \( \mathbf{r}(t) = \langle e^{t}, e^{-t} \rangle \), the velocity function \( \mathbf{v}(t) \) is calculated as the derivative with respect to time \( t \). Steps to find the velocity:

• Differentiate \( e^{t} \) with respect to \( t \), which results in \( e^{t} \).
• Differentiate \( e^{-t} \) with respect to \( t \), giving \( -e^{-t} \). This step involves using the chain rule, noting that the derivative of \(-t\) is \(-1\).

Thus, the velocity function becomes \( \mathbf{v}(t) = \langle e^{t}, -e^{-t} \rangle \), indicating the rate and direction of the particle's movement in each respective component.

Acceleration Function

Acceleration in vector calculus represents how the velocity of a particle changes over time. To find the acceleration function, you differentiate the velocity function. This provides insights into whether the particle is speeding up, slowing down, or changing direction. From the earlier derived velocity \( \mathbf{v}(t) = \langle e^{t}, -e^{-t} \rangle \), the acceleration is the derivative of each component:

• Since the derivative of \( e^{t} \) is \( e^{t} \), it remains unchanged.
• For \( -e^{-t} \), the derivative is \( e^{-t} \). The negative sign in front is accounted for during differentiation.

Therefore, the acceleration function is \( \mathbf{a}(t) = \langle e^{t}, e^{-t} \rangle \), mirroring the position function and showing consistent growth in each component of velocity.

Speed Calculation

The speed of a particle is the magnitude of the velocity vector and indicates how fast the particle is traveling, regardless of direction. This scalar quantity is derived by finding the length of the velocity vector.Given the velocity \( \mathbf{v}(t) = \langle e^{t}, -e^{-t} \rangle \), the speed is calculated using the formula for the magnitude of a vector:\[ \text{Speed} = \sqrt{(e^{t})^2 + (-e^{-t})^2} \]Breaking it down:

• Square each component: \( (e^{t})^2 = e^{2t} \) and \( (-e^{-t})^2 = e^{-2t} \).
• Add them together to get \( e^{2t} + e^{-2t} \).
• Take the square root of this sum to find the magnitude or the speed.

This results in a speed of \( \sqrt{e^{2t} + e^{-2t}} \), capturing the total velocity magnitude without considering directional vectors.