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 2π 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 4πr2 and the volume is 34πr3.
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.
Surface Area
A=4πr2
dtdA=8πrdtdr
Volume
V=34πr3
dtdV=4πr2dtdr
To obtain dtdA, we will need to find dtdr. Although this value is not given, we can substitute the value from our other equation.
Solving for dtdr
dtdV=4πr2dtdr
dtdr=4πr2dtdV
Substituting this value into our change in surface area equation:
dtdA=8πrdtdr
dtdA=8πr(4πr2dtdV)
dtdA=r2dtdV
We are told that at the moment, the radius is 5 and the volume is decreasing at a rate of 2π.
dtdA=52(−2π)
dtdA=−54π
The surface area is decreasing at a rate of 54π
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.