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?
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.
To find the rate at which the surface area of the sphere is decreasing, we need to use the formulas for the volume and surface area of a sphere.
The volume of a sphere is given by the formula:
V = (4/3) * π * r^3
The surface area of a sphere is given by the formula:
A = 4 * π * r^2
We are given that the volume of the sphere is decreasing at a constant rate of 2π cubic meters per hour. This rate is represented by dV/dt.
We need to find the rate at which the surface area is decreasing when the radius is 5 meters. This rate is represented by dA/dt.
To find dA/dt, we can use the chain rule of differentiation.
The chain rule states that if we have a function A = f(r), and r = g(t), then the derivative of A with respect to t is given by:
dA/dt = (df/dr) * (dr/dt)
In this case, A is the surface area of the sphere, which is a function of the radius r. So we have:
A = f(r) = 4 * π * r^2
We are given that r = 5 meters, so the derivative of A with respect to t is:
dA/dt = (dA/dr) * (dr/dt)
First, let's find (dA/dr), the derivative of the surface area with respect to the radius:
(dA/dr) = d/dt (4 * π * r^2)
= 8 * π * r
Next, let's find (dr/dt), the derivative of the radius with respect to time:
(dr/dt) = 0 (since the radius is not changing with time)
Now, substitute the values into the equation:
dA/dt = (dA/dr) * (dr/dt)
= (8 * π * r) * 0
= 0
Therefore, at the moment when the radius is 5 meters, the rate at which the surface area of the sphere is decreasing is 0 square meters per hour.