[Related Rates]

MatthewLander

New member
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=1/3πr^2h

______ftminftmin

blamocur

Full Member
Have you tried to express the cone's radius (and thus its height too) as a function of time ?

Jomo

Elite Member
Gravel is being dumped from a conveyor belt at a rate of 10 cubic feet per minute.
Cubic feet is VOLUME. So dV/dT = 10ft^3/min

blamocur

Full Member
Gravel is being dumped from a conveyor belt at a rate of 10 cubic feet per minute.
Cubic feet is VOLUME. So dV/dT = 10ft^3/min
But the problems asks for dh/dT.

skeeter

Elite Member
But the problems asks for dh/dT.
yes, but one needs the value of dV/dt to determine dh/dt

equal base diameter and height implies r = h/2

[imath] V = \pi \left(\dfrac{h}{2}\right)^2 h [/imath]

simplify and find the time derivative

• Jomo

skeeter

Elite Member
forgot the “over 3” with pi …

[imath]V = \dfrac{\pi}{3} \left(\dfrac{h}{2}\right)^2 h[/imath]