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.