Let's start by relating the rate of change of volume with the rate of change of radius.
The volume of a sphere is given by the formula: V = (4/3)πr^3.
We differentiate both sides of this equation with respect to time t:
dV/dt = d/dt [(4/3)πr^3].
Since the volume is decreasing at a constant rate of 2π cubic meters per hour, we have:
dV/dt = -2π.
Now, let's relate the rate of change of volume with the rate of change of surface area.
The surface area of a sphere is given by the formula: A = 4πr^2.
We differentiate both sides of this equation with respect to time t:
dA/dt = d/dt [4πr^2].
Using the chain rule, this becomes:
dA/dt = 8πr (dr/dt).
Since we are interested in finding the rate at which the surface area is decreasing, we take the negative value of dA/dt:
-dA/dt = -8πr (dr/dt).
We are given that the volume is decreasing at a rate of 2π cubic meters per hour, so dr/dt = -2/π.
Plugging in this value, along with the given radius r = 5, we can calculate the rate at which the surface area is decreasing:
-dA/dt = -8π(5)(-2/π) = 80 square meters per hour.
Therefore, the surface area of the sphere is decreasing at a rate of 80 square meters per hour when the radius is 5 meters.
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 2pi 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?
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