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
6 answers
is the answer 8/49pi= 0.052 meters per minute
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 ...
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 ...
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
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
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
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
You have found the rate at which the rate is changing.
Unfortunately, your question asked for how fast the height is changing.
Unfortunately, your question asked for how fast the height is changing.
should say:
You have found the rate at which the radius is changing.
Unfortunately, your question asked for how fast the height is changing.
You have found the rate at which the radius is changing.
Unfortunately, your question asked for how fast the height is changing.