https://file.aiquickdraw.com/mathbot/file/1725927773553g630v38v.jpg

The given vector function is:

\[
\mathbf{r}(t) = \sin(4t) \, \mathbf{i} + \ln(t) \, \mathbf{j} + 2te^{-9t} \, \mathbf{k}
\]

To find the second derivative, \(\mathbf{r}''(t)\), we need to first find the first derivative \(\mathbf{r}'(t)\) and then differentiate again to obtain \(\mathbf{r}''(t)\).

### Step 1: First derivative \(\mathbf{r}'(t)\)
Differentiate each component of \(\mathbf{r}(t)\) with respect to \(t\):

1. \(\sin(4t) \, \mathbf{i}\)
   \[
   \frac{d}{dt} [\sin(4t)] = 4\cos(4t) \, \mathbf{i}
   \]

2. \(\ln(t) \, \mathbf{j}\)
   \[
   \frac{d}{dt} [\ln(t)] = \frac{1}{t} \, \mathbf{j}
   \]

3. \(2te^{-9t} \, \mathbf{k}\)
   Using the product rule:
   \[
   \frac{d}{dt} [2te^{-9t}] = 2 \cdot \frac{d}{dt}[t] \cdot e^{-9t} + 2t \cdot \frac{d}{dt}[e^{-9t}]
   \]
   \[
   = 2e^{-9t} + 2t(-9e^{-9t}) = 2e^{-9t} - 18te^{-9t} \, \mathbf{k}
   \]

Thus, the first derivative is:
\[
\mathbf{r}'(t) = 4\cos(4t) \, \mathbf{i} + \frac{1}{t} \, \mathbf{j} + (2e^{-9t} - 18te^{-9t}) \, \mathbf{k}
\]

### Step 2: Second derivative \(\mathbf{r}''(t)\)
Now differentiate each component of \(\mathbf{r}'(t)\) again:

1. \(4\cos(4t) \, \mathbf{i}\)
   \[
   \frac{d}{dt} [4\cos(4t)] = -16\sin(4t) \, \mathbf{i}
   \]

2. \(\frac{1}{t} \, \mathbf{j}\)
   \[
   \frac{d}{dt} \left[\frac{1}{t}\right] = -\frac{1}{t^2} \, \mathbf{j}
   \]

3. \(2e^{-9t} - 18te^{-9t} \, \mathbf{k}\)
   Using the product rule on \(-18te^{-9t}\):
   \[
   \frac{d}{dt} [2e^{-9t}] = -18e^{-9t}
   \]
   \[
   \frac{d}{dt} [-18te^{-9t}] = -18e^{-9t} + 162te^{-9t}
   \]
   So, the derivative of \(2e^{-9t} - 18te^{-9t}\) is:
   \[
   (-18e^{-9t}) + (-18e^{-9t} + 162te^{-9t}) = -36e^{-9t} + 162te^{-9t} \, \mathbf{k}
   \]

Thus, the second derivative is:
\[
\mathbf{r}''(t) = -16\sin(4t) \, \mathbf{i} - \frac{1}{t^2} \, \mathbf{j} + (-36e^{-9t} + 162te^{-9t}) \, \mathbf{k}
\]

### Final Answer:
\[
\mathbf{r}''(t) = -16\sin(4t) \, \mathbf{i} - \frac{1}{t^2} \, \mathbf{j} + (-36e^{-9t} + 162te^{-9t}) \, \mathbf{k}
\]

Let me know if you need further clarification, or if you'd like more details on any step! 

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Here are 5 questions that build on this information:
1. How do we apply the product rule when differentiating vector functions?
2. What is the significance of the second derivative in the context of a position vector?
3. How does the exponential decay in the \(\mathbf{k}\)-component affect the behavior of the function as \(t\) increases?
4. What is the geometrical interpretation of \(\mathbf{r}(t)\) and \(\mathbf{r}''(t)\)?
5. How would the calculation change if \(\mathbf{r}(t)\) were defined in terms of \(e^{4t}\) instead of \(\ln(t)\)?

**Tip:** When differentiating vector functions, treat each component separately, applying the rules for differentiation as you would for scalar functions.

Given r(t) = sin(4t)i + ln(t)j + 2te^(-9t)k, find r''(t).

Learn how to find the second derivative of the vector function r(t) = sin(4t)i + ln(t)j + 2te^(-9t)k. This problem covers key calculus concepts, including the product rule, chain rule, and logarithmic differentiation, making it suitable for college-level students.

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