Gravel is being dumped from a conveyor belt at a rate of 10 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 14 feet high? Recall that the volume of a right circular cone with height h and radius of the base r is given by V=1/3πr2h
How do you solve this. I need an actual answer, and how to work it.
Category of question: Related Rates in Calculus
Let the radius of the cone's base be r ft
then we are told that h = 2r, where h is the height of the cone
V = (1/3)π r^2 h = (1/3)π r^2 (2r)
V = (2/3)π r^3
dV/dt = 2π r^2 dr/dt
when h = 14 ft
2r = 14 ft
r = 7 ft
now we can sub in values into
dV/dt = 2π r^2 dr/dt
10 = 2π (7^2) dr/dt
dr/dt = 10/98π = 5/ 49π ft/min
but you want dh/dt
h = 2r
dh/dt = 2 dr/dt
so dh/dt = 2(5/ 49π) ft/min
= 10/ 49π ft/min