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.