Related Rate Volume Calculus help?

Question

At a sand and gravil plant, sand is falling off a conveyor belt and onto a conical pile at the rate of 10 cubic feet per min. The diameter of the base of the cone is approx three times the altitude. at what rate is the height of the pile changing when it is 15 ft high?

Answer 1

If the height is 15 ft, the diameter of the base is 3*15 = 45 ft. The volume of the cone is V = (1/3)pi(r^2)h. Now comes the tricky part. You can say that dV/dt = (dV/dh)(dh/dt), where those are all derivatives. You can use pretty simple calculus to see that dV/dh = (1/3)pi(r^2), which you can calculate because you know r (half of the diameter), and you've been given dV/dt = 10 ft^3/min. That means your only unknown is dh/dt, which you've been asked to find, and which you can now solve algebraically.

Answer 2

Volume of cone= V=(hpir^2)/3
DV/DT=10
H=15
D=45
R=45/2
H=3V/(pir^2)
DH/DT=(3/pir^2)DV/DT
120/2025=24/405=DH=.059DT