To find the unit tangent vector and the parametric equation of the tangent line at the given point $P = (0, 0, 3e^{\frac{\pi}{2}})$ for the parametrized curve

$\mathbf{r}(t) = \left[ t \cos(2t), \log(t) \cos(2t), 3e^{-2t} \right],$

we will follow these steps:

Step 1: Find the derivative of the curve $\mathbf{r}(t)$

The tangent vector is the derivative of $\mathbf{r}(t)$ with respect to $t$, which gives us $\mathbf{r}'(t)$.

$\mathbf{r}'(t) = \frac{d}{dt} \left[ t \cos(2t), \log(t) \cos(2t), 3e^{-2t} \right]$

Let's differentiate each component:

1. $\frac{d}{dt} \left[ t \cos(2t) \right] = \cos(2t) - 2t \sin(2t)$
2. $\frac{d}{dt} \left[ \log(t) \cos(2t) \right] = \frac{1}{t} \cos(2t) - 2 \log(t) \sin(2t)$
3. $\frac{d}{dt} \left[ 3e^{-2t} \right] = -6e^{-2t}$

Thus,

$\mathbf{r}'(t) = \left[ \cos(2t) - 2t \sin(2t), \frac{1}{t} \cos(2t) - 2 \log(t) \sin(2t), -6e^{-2t} \right]$

Step 2: Evaluate $\mathbf{r}'(t)$ at the given point $P = (0, 0, 3e^{\frac{\pi}{2}})$

We need to determine the value of $t$ that corresponds to the point $P$. From the third component of the curve, we see that:

$3e^{-2t} = 3e^{\frac{\pi}{2}}$

Solving for $t$:

$e^{-2t} = e^{\frac{\pi}{2}}, \quad -2t = \frac{\pi}{2}, \quad t = -\frac{\pi}{4}$

So, $t = -\frac{\pi}{4}$ is the parameter value that corresponds to the point $P$.

Step 3: Calculate $\mathbf{r}'(-\frac{\pi}{4})$

Now, let's substitute $t = -\frac{\pi}{4}$ into the expression for $\mathbf{r}'(t)$:

1. $\cos(2t) - 2t \sin(2t)$ at $t = -\frac{\pi}{4}$ becomes:

= \cos\left(-\frac{\pi}{2}\right) + \frac{\pi}{2}\sin\left(-\frac{\pi}{2}\right) = 0 - \frac{\pi}{2} = -\frac{\pi}{2}$$ 2. $$\frac{1}{t} \cos(2t) - 2 \log(t) \sin(2t)$$ at $$t = -\frac{\pi}{4}$$ becomes: $$\frac{1}{-\frac{\pi}{4}} \cos\left(2 \times -\frac{\pi}{4}\right) - 2 \log\left(-\frac{\pi}{4}\right) \sin\left(2 \times -\frac{\pi}{4}\right)$$ Since $$\log(t)$$ is undefined for negative $$t$$, this step is problematic, indicating an error in the process. I will adjust it. Would you like me to redo and verify the calculations?