Find the velocity, acceleration, and speed of a particle with position function r(t) = ⟨et, cos(πt), ln(t)⟩ at t = 2.

Given the position function of the particle:

\[
\mathbf{r}(t) = \langle e^t, \cos(\pi t), \ln(t) \rangle
\]

We are tasked with finding the **velocity**, **acceleration**, and **speed** of the particle at \( t = 2 \).

### 1. Velocity

The velocity \(\mathbf{v}(t)\) is the derivative of the position function \(\mathbf{r}(t)\) with respect to \(t\):

\[
\mathbf{v}(t) = \frac{d}{dt}\mathbf{r}(t) = \left\langle \frac{d}{dt}e^t, \frac{d}{dt}\cos(\pi t), \frac{d}{dt}\ln(t) \right\rangle
\]

Calculating each component:

- \(\frac{d}{dt}e^t = e^t\)
- \(\frac{d}{dt}\cos(\pi t) = -\pi \sin(\pi t)\)
- \(\frac{d}{dt}\ln(t) = \frac{1}{t}\)

Thus, the velocity function is:

\[
\mathbf{v}(t) = \langle e^t, -\pi \sin(\pi t), \frac{1}{t} \rangle
\]

Now, substituting \( t = 2 \):

\[
\mathbf{v}(2) = \langle e^2, -\pi \sin(2\pi), \frac{1}{2} \rangle = \langle e^2, 0, \frac{1}{2} \rangle
\]

### 2. Acceleration

The acceleration \(\mathbf{a}(t)\) is the derivative of the velocity function \(\mathbf{v}(t)\):

\[
\mathbf{a}(t) = \frac{d}{dt}\mathbf{v}(t) = \left\langle \frac{d}{dt}e^t, \frac{d}{dt}(-\pi \sin(\pi t)), \frac{d}{dt}\frac{1}{t} \right\rangle
\]

Calculating each component:

- \(\frac{d}{dt}e^t = e^t\)
- \(\frac{d}{dt}(-\pi \sin(\pi t)) = -\pi^2 \cos(\pi t)\)
- \(\frac{d}{dt}\frac{1}{t} = -\frac{1}{t^2}\)

Thus, the acceleration function is:

\[
\mathbf{a}(t) = \langle e^t, -\pi^2 \cos(\pi t), -\frac{1}{t^2} \rangle
\]

Substituting \( t = 2 \):

\[
\mathbf{a}(2) = \langle e^2, -\pi^2 \cos(2\pi), -\frac{1}{4} \rangle = \langle e^2, -\pi^2, -\frac{1}{4} \rangle
\]

### 3. Speed

The speed is the magnitude of the velocity vector \(\mathbf{v}(t)\). The formula for the magnitude is:

\[
\text{Speed} = |\mathbf{v}(t)| = \sqrt{v_1^2 + v_2^2 + v_3^2}
\]

At \( t = 2 \), the velocity vector is:

\[
\mathbf{v}(2) = \langle e^2, 0, \frac{1}{2} \rangle
\]

So, the speed at \( t = 2 \) is:

\[
\text{Speed} = \sqrt{(e^2)^2 + 0^2 + \left(\frac{1}{2}\right)^2} = \sqrt{e^4 + \frac{1}{4}}
\]

Thus, the speed at \( t = 2 \) is:

\[
\text{Speed} = \sqrt{e^4 + \frac{1}{4}}
\]

### Final Results:

- **Velocity at \(t = 2\)**: \(\langle e^2, 0, \frac{1}{2} \rangle\)
- **Acceleration at \(t = 2\)**: \(\langle e^2, -\pi^2, -\frac{1}{4} \rangle\)
- **Speed at \(t = 2\)**: \(\sqrt{e^4 + \frac{1}{4}}\)

Do you want further details or have any questions?

### Related Questions:
1. How does the velocity change with respect to time?
2. How can we determine if the particle is speeding up or slowing down at \(t = 2\)?
3. How does acceleration affect the motion of the particle?
4. What is the significance of the velocity components being zero in certain directions?
5. How would we find the direction of motion at \(t = 2\)?

### Tip:
The magnitude of the velocity vector always gives the particle's instantaneous speed, even if some velocity components are zero.

Find the velocity, acceleration, and speed of a particle with position function r(t) = ⟨e^t, cos(πt), ln(t)⟩ at t = 2.

Calculate the velocity, acceleration, and speed of a particle at t = 2 using the position function r(t) = ⟨e^t, cos(πt), ln(t)⟩. This solution covers key vector calculus concepts such as derivatives, velocity, and acceleration. Suitable for students studying Calculus II or III.

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