A tank has the shape of an inverted circular cone with a base radius of 5 meters

Question

and a height of 20 meters. If water is being pumped into the tank at 2 cubic meters
per minute, find the rate at which the water level is rising when the water is 7
meters deep

WELL I HAVE THIS

v=1/3pi*r^2h
v=1/3pi*r2(4r)
v=4/3pi*r^3
dv/dt=4pi*r^2
2=4pi(7/4)^2
2=49pi/4 * dr/dt
dr/dt= 8/49pi = 0.052 meters per minute

but steve got
"I get dV/dt = pi/16 h^2 dh/dt
giving
2 = 49pi/16 dh/dt
or
dh/dt= 32/49pi
"

and reiny got
V = (1/3)πr^2 h = (1/3)π(h^2/16)(h)
= (1/48)π h^3
dV/dt = (1/16)π h^2 dh/dt
so when dV/dt = 2, h = 7
2 = (1/16)π(49) dh/dt
2(16)/(49π) = dh/dt
= 32/(49π)
= appr. .208 m/min

Answer 1

I answered in your other post on this

You have found the rate at which the radius is changing.

Unfortunately, your question asked for how fast the height is changing.

Did you notice that Steve and I both had the same answer?