Question

Water is leaking out of an inverted conical tank at a rate of 13300 cubic centimeters per minute at the same time that water is being pumped into the tank at a constant rate. The tank has height 10 meters and the diameter at the top is 4.5 meters. If the water level is rising at a rate of 24 centimeters per minute when the height of the water is 2.5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute. The answer must be in cubic centimeters per minute.

Answer 1

we need to write a function for the Volume of water based on its height from bottom of tank. (h)

So first we need to be able to write radius (r) in terms of h.

Answer 2

\[\frac rh={9/4 \over 10}\\ \implies r= {9 \over 40} h\]

Answer 3

\[V=\frac13 \pi r^2h\]

\[\therefore V(h) = \frac13 \pi \left( {9 \over 40}h \right)^2h\\ \ \ \ \ \ \ \ \ \ \ \ \ ={27 \over1600}\pi h^3\]

Answer 4

do you follow... so far?

Answer 5

Yes, I follow.

Answer 6

next from the chain rule we know:
\[{ d V \over dt} = {dV \over dh} \times {dh \over dt}\]

Answer 7

\[{dV \over dh} = V'(h)={81 \over 1600} \pi h^2\]

Answer 8

at \(h=2.5\ m = 250\ cm\), \(dh/dt = 24\) cm/min

since dV/dt is constant: \[{dV \over dt}=V'(250) \times 24\]

Answer 9

\[{dV \over dt}\\ ={81 \over 1600} \pi (250)^2 \times 24\\={151875 \over2} \pi\\ \approx 238565\ {cm^3 \over min}\]

Answer 10

\[{dV \over dt}=\text{rate at which water is being pumped in} - \text{rate at which water is leaking out}\]

\[\text{rate at which water is being pumped in}=238565+13300=\color{green}{251865\ {cm^3 \over min}}\]

Answer 11

Thank you so much!