To find a parametrization of the tangent line to the vector-valued function $\mathbf{r}(t) = (\ln(t))\mathbf{i} + t^{-1}\mathbf{j} + 11t\mathbf{k}$ at the point $t = 1$, let's proceed with these steps:

Step 1: Calculate the Derivative $\mathbf{r}'(t)$

To determine the tangent line, we first need the derivative $\mathbf{r}'(t)$, which represents the velocity vector.

$\mathbf{r}(t) = \langle \ln(t), t^{-1}, 11t \rangle$

Now, differentiate each component with respect to $t$:

1. $x(t) = \ln(t)$
   $x'(t) = \frac{1}{t}$

2. $y(t) = t^{-1}$
   $y'(t) = -t^{-2}$

3. $z(t) = 11t$
   $z'(t) = 11$

Thus, the derivative vector is:

$\mathbf{r}'(t) = \left\langle \frac{1}{t}, -t^{-2}, 11 \right\rangle$

Step 2: Evaluate the Derivative at $t = 1$

To find the direction of the tangent line at $t = 1$, substitute $t = 1$ into $\mathbf{r}'(t)$:

$\mathbf{r}'(1) = \left\langle \frac{1}{1}, -1^{-2}, 11 \right\rangle = \langle 1, -1, 11 \rangle$

Step 3: Find the Point $\mathbf{r}(1)$

We also need the position vector at $t = 1$:

$\mathbf{r}(1) = \langle \ln(1), 1^{-1}, 11 \times 1 \rangle = \langle 0, 1, 11 \rangle$

Step 4: Parametrize the Tangent Line

The parametrization of the tangent line $\mathbf{L}(t)$ can be given by:

$\mathbf{L}(t) = \mathbf{r}(1) + t \mathbf{r}'(1)$

Substituting the values we have:

$\mathbf{L}(t) = \langle 0, 1, 11 \rangle + t \langle 1, -1, 11 \rangle$

Expanding this gives:

$\mathbf{L}(t) = \langle 0 + t \cdot 1, 1 + t \cdot (-1), 11 + t \cdot 11 \rangle$

$\mathbf{L}(t) = \langle t, 1 - t, 11 + 11t \rangle$

Final Answer

$\mathbf{L}(t) = \langle t, 1 - t, 11 + 11t \rangle$

Would you like further clarification on any part of the process? Here are some related questions to consider:

1. How do you find the derivative of a vector-valued function?
2. Why do we use the derivative to find the tangent line?
3. How do you parametrize lines in three-dimensional space?
4. Can we extend this method to curves in higher dimensions?
5. What is the geometric significance of the tangent vector at a given point?

Tip: When finding the tangent line to a curve, always evaluate both the position vector and the derivative at the point of interest to obtain the correct line parametrization.