Given the vector-valued functions
\[
\begin{array}{l}
\vec{u}(t)=e^{-2 t} \vec{i}+e^{-t} \vec{j}-2 t \vec{k} \\
\vec{v}(t)=4 t \vec{i}+3 t^{2} \vec{j}-5 \vec{k}
\end{array}
\]
find $\frac{d}{d t}(\vec{u}(t) \cdot \vec{v}(t))$ when $t=2$

Step 1 ：Given the vector-valued functions \(\vec{u}(t)=e^{-2 t} \vec{i}+e^{-t} \vec{j}-2 t \vec{k}\) and \(\vec{v}(t)=4 t \vec{i}+3 t^{2} \vec{j}-5 \vec{k}\)

Step 2 ：We need to find the derivative of the dot product of these two functions. The dot product of two vector-valued functions is a scalar-valued function. The derivative of this scalar-valued function can be found using the product rule of differentiation.

Step 3 ：The product rule states that the derivative of the product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.

Step 4 ：First, find the derivatives of the vector-valued functions \(\vec{u}(t)\) and \(\vec{v}(t)\). The derivative of \(\vec{u}(t)\) is \([-2e^{-2t}\vec{i}, -e^{-t}\vec{j}, -2\vec{k}]\) and the derivative of \(\vec{v}(t)\) is \([4\vec{i}, 6t\vec{j}, 0\vec{k}]\).

Step 5 ：Next, apply the product rule to find \(\frac{d}{d t}(\vec{u}(t) \cdot \vec{v}(t))\). The derivative of the dot product is \(-3t^{2}e^{-t} + 6te^{-t} - 8te^{-2t} + 10 + 4e^{-2t}\).

Step 6 ：Finally, substitute \(t=2\) into the derivative to find the value at \(t=2\). The derivative of the dot product of the vector-valued functions \(\vec{u}(t)\) and \(\vec{v}(t)\) at \(t=2\) is \(10 - 12e^{-4}\).

Step 7 ：Final Answer: \(\boxed{10 - 12e^{-4}}\)