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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.

Find a parametrization of the tangent line to r(t) = (ln(t))i + t^-1j + 11tk at the point t = 1.

This problem involves finding the parametrization of the tangent line to the vector-valued function r(t) = (ln(t))i + t^-1j + 11tk at t = 1. Learn how to compute the derivative of the vector function and apply it to obtain the tangent line parametrization. Suitable for students studying calculus at the university level.

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