An ice sculpture in the form of a sphere melts in such a way that it maintains its spherical shape. The volume of the sphere is decreasing at a constant rate of \(2\pi\) cubic meters per hour. At what rate, in square meters per hour, is the surface area of the sphere decreasing at the moment when the radius is 5 meters?

Note: For a sphere of radius \(r\), the surface area if \(4\pi r^2\) and the volume is \(\frac{4}{3}\pi r^3\).

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In the question, we need to find the change in the surface area. We are given the radius and the change in the volume.

We can start by deriving equations for the change in the surface area and the change in volume.

To obtain \(\frac{dA}{dt}\), we will need to find \(\frac{dr}{dt}\). Although this value is not given, we can substitute the value from our other equation.

Solving for \(\frac{dr}{dt}\)

$$ \frac{dV}{dt}=4\pi r^2 \frac{dr}{dt} $$ $$ \frac{dr}{dt}= \frac{\frac{dV}{dt}}{4\pi r^2} $$

Substituting this value into our change in surface area equation:

$$ \frac{dA}{dt}=8\pi r\frac{dr}{dt} $$ $$ \frac{dA}{dt}=8\pi r \left(\frac{\frac{dV}{dt}}{4\pi r^2} \right) $$ $$ \frac{dA}{dt}= \frac{2\frac{dV}{dt}}{r} $$

We are told that at the moment, the radius is 5 and the volume is decreasing at a rate of \(2\pi\).

$$ \frac{dA}{dt} = \frac{2(-2\pi)}{5} $$ $$ \frac{dA}{dt} = -\frac{4\pi}{5} $$

The surface area is decreasing at a rate of \(\boxed{\frac{4\pi}{5}}\)

Units are quite important for related rates problems. Generally, but not always, they can be ignored because the units given correspond to each other.

For this specific problem, all given rates and values used meters and hours.