Question

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

Answer 1

is the answer 8/49pi= 0.052 meters per minute

Answer 2

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

One of us has a factor of 2 out of place. Could be me ...

Answer 3

I did not get that

Let the height of the water be h m and the radius of the water level be r m
by ratios:
r/h = 5/20 = 1/4
r = h/4

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 4

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

Answer 5

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

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

Answer 6

should say:


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

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