The answer is C. 2.4 inches per minute.
How can you solve how fast the level in the pot rises when the coffee in the cone is 5 inches deep?
V c(t) = (1/3)πr²h c(t) = (1/3)π(3²)(5) = 25π cubic inches
The rate of change of the coffee volume in the pot is equal to the draining rate:
dV p(t)/dt = Rate = 8 cubic inches per minute
V p(t) = πr²h p(t)
The total volume of coffee (V c(t) + V p(t)) is constant:
V c(t) + V p(t) = V c(0) + V p(0) (initial volume)
V c(t) + πr²h p(t) = V c(0) + V p(0)
dV c(t)/dt + 2πr²h p(t)dh p(t)/dt = 0
8 + 2πr²h p(t)dh p(t)/dt = 0
dh p(t)/dt = -8 / (2πr²h p(t))
dh p(t)/dt = -8 / (2π(3²)(5)) ≈ -2.4 inches per minute
The negative sign indicates that the height of the coffee in the pot is decreasing, not rising. However, we need the absolute value of the rate of change. Therefore, the level in the pot is rising at a rate of:
|dh p(t)/dt| ≈ 2.4 inches per minute
Therefore, the answer is C. 2.4 inches per minute.