Question

Water is leaking out of an inverted conical tank at a rate of 700.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 10.0 meters and the diameter at the top is 3.5 meters. If the water level is rising at a rate of 26.0 centimeters per minute when the height of the water is 4.5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute

Answer 1

This looks fun

Answer 2

radius = 3.5/2 or 7/4
r/h = (7/4)/10
r = (7/40)h
plug this into volume formula
\[V = \pi r^{2}h = \pi \frac{49}{1600}h^{3}\]
rate that volume of water changes is the difference between water coming in and water leaking out
\[\frac{dV}{dt} = x-700\]
also remember
\[\frac{dV}{dt} = \frac{dh}{dt}*\frac{dV}{dh}\]
\[\frac{dV}{dh} = 3\pi \frac{49}{1600}h^{2}\]
Given dh/dt=26 when h=4.5, plug in 4.5 for h
\[x-700 = 26(3\pi \frac{49}{1600}4.5^{2})\]
solve for x
\[x = 851.9657\]

Answer 3

haha oops just realized i forgot the 1/3 in the volume formula :(

Answer 4

only thing that changes is dv/dh
\[\frac{dV}{dh} = \pi \frac{49}{1600}h^{2}\]
x = 750.655