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 to the vector-valued function \( \mathbf{r}(t) = 3e^t\, \mathbf{i} + 4e^{-3t}\, \mathbf{j} + 2e^{4t}\, \mathbf{k} \) at \( t = \ln(2) \), we need to follow these steps:

### Step 1: Find the derivative of \( \mathbf{r}(t) \)
The tangent vector is the derivative of the position vector \( \mathbf{r}(t) \), denoted \( \mathbf{r}'(t) \). We differentiate each component of the vector function.

\[
\mathbf{r}(t) = 3e^t\, \mathbf{i} + 4e^{-3t}\, \mathbf{j} + 2e^{4t}\, \mathbf{k}
\]

#### Component-wise derivatives:
1. For the \( \mathbf{i} \)-component: \( \frac{d}{dt}(3e^t) = 3e^t \)
2. For the \( \mathbf{j} \)-component: \( \frac{d}{dt}(4e^{-3t}) = 4(-3e^{-3t}) = -12e^{-3t} \)
3. For the \( \mathbf{k} \)-component: \( \frac{d}{dt}(2e^{4t}) = 2(4e^{4t}) = 8e^{4t} \)

Thus, the derivative of \( \mathbf{r}(t) \) is:
\[
\mathbf{r}'(t) = 3e^t\, \mathbf{i} - 12e^{-3t}\, \mathbf{j} + 8e^{4t}\, \mathbf{k}
\]

### Step 2: Evaluate the derivative at \( t = \ln(2) \)
Substitute \( t = \ln(2) \) into \( \mathbf{r}'(t) \).

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

Now, substitute these values into \( \mathbf{r}'(t) \):
\[
\mathbf{r}'(\ln(2)) = 3(2)\, \mathbf{i} - 12\left(\frac{1}{8}\right)\, \mathbf{j} + 8(16)\, \mathbf{k}
\]
\[
\mathbf{r}'(\ln(2)) = 6\, \mathbf{i} - \frac{12}{8}\, \mathbf{j} + 128\, \mathbf{k}
\]
\[
\mathbf{r}'(\ln(2)) = 6\, \mathbf{i} - 1.5\, \mathbf{j} + 128\, \mathbf{k}
\]

### 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 further details on this solution?

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Here are 5 related questions to deepen understanding:
1. How is the derivative of a vector function different from a scalar function?
2. What is the geometric interpretation of the tangent vector?
3. Can you find the magnitude of the tangent vector \( \mathbf{r}'(\ln(2)) \)?
4. What happens to the tangent vector if we change the function \( \mathbf{r}(t) \) slightly?
5. How do you compute the unit tangent vector from the tangent vector?

**Tip:** The derivative of a vector-valued function gives the direction and rate of change of the position vector, pointing in the direction of the tangent at any given point.

Find a tangent vector at the indicated value of t.

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

Learn how to calculate the tangent vector for the vector function r(t) = 3e^t i + 4e^−3t j + 2e^4t k at t = ln(2). This problem involves key vector calculus concepts, including differentiating vector-valued functions and applying the chain rule. Suitable for high school seniors or college-level calculus students.

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