The County Roads Department is stockpiling salt for use next winter. The salt is being unloaded from

Question

railroad cars by a conveyor belt at a rate of 20 cubic feet per minute. The salt falls into a conical pile
whose diameter at the base is four times the height of the pile. At what rate is the diameter of the pile
changing when the pile is 12 feet high?

Show steps please!

Thank you!

Answer 1

let the height be h
then the diameter of the base is 4h, thus the radius is 2h

given: dV = 20 ft^3/min
find: d(2h)/dt , when h = 12

V = (1/3)π r^2 h
= (1/3) π (4h^2)(h)
= (4/3) π h^3

dV/dt = 4π h^2 dh/dt
20 = 4π((144) dh/dt
dh/dt = 5/(144π)
so
d(2h)/dt = 10/(144π)
= 5/(72π)
= appr .0221 ft/min

Answer 2

I was just reading the last solution and I was wondering why he had the 1/3 before the pir^2h.