A spherical snowball is placed in the sun. The snowball melts so that it's surface area
decreases at a rate of 2 cm2
/min. Find the rate at which the diameter decreases when the
diameter is 8 cm
A = 4πr^2
dA/dt = 8πr dr/dt
when d = 8, r = 4, dA/dt = -2
-2 = 8π(4)dr/dt
dr/dt = -1/(16π)
so dd/dt = -1(8π) cm/min
To find the rate at which the diameter of the snowball decreases, we can use the relationship between the surface area and the diameter of a sphere.
The surface area, A, of a sphere with radius r is given by the formula:
A = 4πr²
Since the diameter, d, is equal to twice the radius (d = 2r), we can rewrite the formula for surface area in terms of the diameter:
A = 4π((d/2)²)
A = πd²/4
Now, we are given that the surface area decreases at a rate of 2 cm²/min. This means that dA/dt, the rate of change of surface area with respect to time, is -2 cm²/min (negative because the surface area is decreasing).
We want to find the rate at which the diameter, d, changes with time, or dd/dt. To do this, we can differentiate the formula for surface area with respect to time:
dA/dt = π(dd/dt)/2
Substituting the given rate of change of surface area, we get:
-2 = π(dd/dt)/2
Now, we can solve for dd/dt, the rate at which the diameter decreases:
dd/dt = -4/π cm/min
Therefore, when the diameter is 8 cm, the rate at which the diameter decreases is approximately -4/π cm/min.