If 
r(t) = 

2e3t, 2e−3t, 3te3t

,
 find 
T(0),
 
r''(0),
 and 
r'(t) · r''(t).

Given the vector function \( r(t) = \begin{pmatrix} 2e^{3t} \\ 2e^{-3t} \\ 3te^{3t} \end{pmatrix} \), we need to find \( T(0) \), \( r''(0) \), and \( r'(t) \cdot r''(t) \).

### 1. Finding \( T(0) \)
\( T(t) \) represents the unit tangent vector at time \( t \), defined as:

\[
T(t) = \frac{r'(t)}{|r'(t)|}
\]

To find \( T(0) \), we first need to compute \( r'(t) \) and then evaluate it at \( t = 0 \).

### 1.1 Compute \( r'(t) \)
\( r(t) \) is given as:

\[
r(t) = \begin{pmatrix} 2e^{3t} \\ 2e^{-3t} \\ 3te^{3t} \end{pmatrix}
\]

Differentiate each component with respect to \( t \):

\[
r'(t) = \begin{pmatrix} 6e^{3t} \\ -6e^{-3t} \\ 3e^{3t} + 9te^{3t} \end{pmatrix}
\]

Now, evaluate \( r'(t) \) at \( t = 0 \):

\[
r'(0) = \begin{pmatrix} 6e^{0} \\ -6e^{0} \\ 3e^{0} + 9 \cdot 0 \cdot e^{0} \end{pmatrix} = \begin{pmatrix} 6 \\ -6 \\ 3 \end{pmatrix}
\]

### 1.2 Compute \( |r'(0)| \)

\[
|r'(0)| = \sqrt{6^2 + (-6)^2 + 3^2} = \sqrt{36 + 36 + 9} = \sqrt{81} = 9
\]

### 1.3 Compute \( T(0) \)

\[
T(0) = \frac{r'(0)}{|r'(0)|} = \frac{1}{9} \begin{pmatrix} 6 \\ -6 \\ 3 \end{pmatrix} = \begin{pmatrix} \frac{2}{3} \\ -\frac{2}{3} \\ \frac{1}{3} \end{pmatrix}
\]

### 2. Finding \( r''(0) \)
To find \( r''(0) \), differentiate \( r'(t) \) with respect to \( t \) and then evaluate it at \( t = 0 \).

\[
r''(t) = \frac{d}{dt} \begin{pmatrix} 6e^{3t} \\ -6e^{-3t} \\ 3e^{3t} + 9te^{3t} \end{pmatrix} = \begin{pmatrix} 18e^{3t} \\ 18e^{-3t} \\ 18te^{3t} + 18e^{3t} \end{pmatrix}
\]

Now, evaluate \( r''(t) \) at \( t = 0 \):

\[
r''(0) = \begin{pmatrix} 18e^{0} \\ 18e^{0} \\ 18 \cdot 0 \cdot e^{0} + 18e^{0} \end{pmatrix} = \begin{pmatrix} 18 \\ 18 \\ 18 \end{pmatrix}
\]

### 3. Finding \( r'(t) \cdot r''(t) \)
The dot product \( r'(t) \cdot r''(t) \) is given by:

\[
r'(t) \cdot r''(t) = \begin{pmatrix} 6e^{3t} \\ -6e^{-3t} \\ 3e^{3t} + 9te^{3t} \end{pmatrix} \cdot \begin{pmatrix} 18e^{3t} \\ 18e^{-3t} \\ 18te^{3t} + 18e^{3t} \end{pmatrix}
\]

This results in:

\[
r'(t) \cdot r''(t) = 6e^{3t} \cdot 18e^{3t} + (-6e^{-3t}) \cdot 18e^{-3t} + \left(3e^{3t} + 9te^{3t}\right) \cdot \left(18te^{3t} + 18e^{3t}\right)
\]

Simplify each term:

\[
r'(t) \cdot r''(t) = 108e^{6t} - 108e^{-6t} + \left(3e^{3t} \cdot 18e^{3t} + 3e^{3t} \cdot 18te^{3t} + 9te^{3t} \cdot 18e^{3t} + 9te^{3t} \cdot 18te^{3t}\right)
\]

Expanding and simplifying:

\[
r'(t) \cdot r''(t) = 108e^{6t} - 108e^{-6t} + 54e^{6t} + 54te^{6t} + 162te^{6t} + 162t^2e^{6t}
\]

Thus,

\[
r'(t) \cdot r''(t) = 162e^{6t} + 216te^{6t} + 162t^2e^{6t} - 108e^{-6t}
\]

### Summary of Answers:
1. \( T(0) = \begin{pmatrix} \frac{2}{3} \\ -\frac{2}{3} \\ \frac{1}{3} \end{pmatrix} \)
2. \( r''(0) = \begin{pmatrix} 18 \\ 18 \\ 18 \end{pmatrix} \)
3. \( r'(t) \cdot r''(t) = 162e^{6t} + 216te^{6t} + 162t^2e^{6t} - 108e^{-6t} \)

Would you like more details or have any questions?

Here are five related questions to explore:

1. How would you find the curvature of the curve \( r(t) \) at any point \( t \)?
2. What is the geometric interpretation of \( r'(t) \cdot r''(t) \)?
3. How does the unit normal vector \( N(t) \) relate to \( T(t) \) and \( r''(t) \)?
4. Can you find the arc length of \( r(t) \) from \( t = 0 \) to \( t = 1 \)?
5. How would you compute the torsion of the curve given by \( r(t) \)?

**Tip:** Always ensure that you differentiate correctly when dealing with vector functions. Each component should be treated independently, but remember to combine results appropriately.

Explore how to differentiate vector functions, compute unit tangent vectors, second derivatives, and dot products with a detailed example. This problem involves vector function differentiation, unit tangent vectors, and dot product calculations.

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