Find a tangent vector at the indicated value of *t*.

**r**(*t*) = 3*et* **i** + 4*e*−3*t* **j** + 2*e*4*t* **k**;  *t* = ln(2)

To find the tangent vector of the given vector function \( \mathbf{r}(t) = 3e^t \mathbf{i} + 4e^{-3t} \mathbf{j} + 2e^{4t} \mathbf{k} \) at \( t = \ln(2) \), we need to:

1. Differentiate \( \mathbf{r}(t) \) with respect to \( t \) to get the velocity (tangent) vector.
2. Evaluate the result at \( t = \ln(2) \).

### Step 1: Differentiate \( \mathbf{r}(t) \)

Given \( \mathbf{r}(t) = 3e^t \mathbf{i} + 4e^{-3t} \mathbf{j} + 2e^{4t} \mathbf{k} \), we differentiate each component with respect to \( t \).

\[
\frac{d}{dt} (3e^t) = 3e^t
\]
\[
\frac{d}{dt} (4e^{-3t}) = 4(-3)e^{-3t} = -12e^{-3t}
\]
\[
\frac{d}{dt} (2e^{4t}) = 2(4)e^{4t} = 8e^{4t}
\]

Thus, the derivative (tangent vector) is:

\[
\mathbf{r}'(t) = 3e^t \mathbf{i} - 12e^{-3t} \mathbf{j} + 8e^{4t} \mathbf{k}
\]

### Step 2: Evaluate at \( t = \ln(2) \)

Now substitute \( t = \ln(2) \) into \( \mathbf{r}'(t) \).

1. \( e^{\ln(2)} = 2 \)
2. \( e^{-3\ln(2)} = 2^{-3} = \frac{1}{8} \)
3. \( e^{4\ln(2)} = 2^4 = 16 \)

Substitute these into the components of \( \mathbf{r}'(t) \):

\[
\mathbf{r}'(\ln(2)) = 3(2) \mathbf{i} - 12\left(\frac{1}{8}\right) \mathbf{j} + 8(16) \mathbf{k}
\]

Simplifying:

\[
\mathbf{r}'(\ln(2)) = 6 \mathbf{i} - \frac{12}{8} \mathbf{j} + 128 \mathbf{k}
\]

\[
\mathbf{r}'(\ln(2)) = 6 \mathbf{i} - 1.5 \mathbf{j} + 128 \mathbf{k}
\]

### Final Answer:

The tangent vector at \( t = \ln(2) \) is:

\[
\mathbf{r}'(\ln(2)) = 6 \mathbf{i} - 1.5 \mathbf{j} + 128 \mathbf{k}
\]

Would you like more details or have any questions?

### Related Questions:
1. How do you find the unit tangent vector of a vector function?
2. What is the significance of the tangent vector in 3D space?
3. How do you compute the magnitude of a tangent vector?
4. How does the exponential function behave under differentiation?
5. What are the differences between velocity and acceleration vectors?

### Tip:
When differentiating vector functions, apply the derivative to each component separately, as if they were scalar functions.

Find a tangent vector at the indicated value of *t*. r(t) = 3e^t i + 4e^−3t j + 2e^4t k; t = ln(2)

This solution explains how to find the tangent vector of the vector function r(t) = 3e^t i + 4e^−3t j + 2e^4t k at t = ln(2). It involves differentiating each component of the vector and evaluating the result at t = ln(2). Suitable for students studying Calculus II or Multivariable Calculus.

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