Answer
The distance is changing at a rate of
$\begin{equation}
21.01\frac{millimeters}{hour}.
\end{equation}$
Work Step by Step
let variables a represent the length of the minute hand; b to represent the length of the hour hand; and c to represent the length between their tips.
$\begin{equation}
a=8\\
b=4\\
\end{equation}$
and according to the law of cosines
$\begin{equation}
c^{2}=a^{2}+b^{2}-2ab\cosθ
\end{equation}$
where θ is the angle between the hands. so
$\begin{equation}
\theta=\frac{\pi}{6}\\
\end{equation}$
We know that the minute hand completes one rotation in one hour, and the hour hand completes one rotation in twelve hours, so
$\begin{equation}
\frac{d\theta_{minute hand}}{dt}=2\pi/hour\\
\frac{d\theta_{hour hand}}{dt}=\frac{2\pi}{12hours}=\frac{1}{6}\pi/hour\\
\end{equation}$
so the total change in angle is the difference between the two, thus
$\begin{equation}
\frac{d\theta}{dt}=2\pi-\frac{1}{6}\pi=\frac{11}{6}\pi
\end{equation}$
Now we take from where we left off with the law of cosines
$\begin{equation}
c=\sqrt{a^{2}+b^2-2ab\cosθ}\\
\frac{dc}{dt}=\frac{d}{dt}\left( \sqrt{a^{2}+b^2-2ab\cosθ} \right)
=\frac{d}{dt}\left(a^{2}+b^2-2ab\cosθ \right)^{\frac{1}{2}}\\
=\frac{1}{2}\left(a^{2}+b^2-2ab\cosθ \right)^{-\frac{1}{2}}\left[ \frac{d}{dt}\left(a^{2}+b^2-2ab\cosθ \right)\right]\\
=\frac{1}{2}\left(a^{2}+b^2-2ab\cosθ \right)^{-\frac{1}{2}}\left[2a+2b-2ab(-\sin\theta)\left(\frac{d\theta}{dt}\right)\right]
\end{equation}$
Now we plug in values and use a calculator to solve for the answer
$\begin{equation}
\frac{1}{2}\left(8^{2}+4^2-2(8)(4)\cos\frac{\pi}{6} \right)^{-\frac{1}{2}}\left[2(8)+2(4)+2(8)(4)(\sin\frac{\pi}{6})\left(\frac{11\pi}{6}\right)\right]
\approx21.01028518\frac{millimeters}{hour}
\end{equation}$
