Question

Calculus, I got some weird answer to this question:
Water is leaking out of an inverted conical tank at a rate of 11700 cubic centimeters per minute at the same time that water is being pumped into the tank at a constant rate. The tank has height 15 meters and the diameter at the top is 6.5 meters. If the water level is rising at a rate of 15 centimeters per minute when the height of the water is 5.0 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

Answer 1

h=15
r=3.25
r/h=3.25/15

Answer 2

Do you have the correct answer?

Answer 3

I do not. I was hoping to get help getting the answer.

Answer 4

I have some idea of how to solve it , but without the answer I'm not sure if I'm doing it ok

Answer 5

I thought I had some idea as well, but WebWork isn't accepting the answer. Pretty frustrating.

Answer 6

I found \[V' =\frac{ 3,25^{2} \pi }{9 }*0,15\]
then you have to subtract 0.011700m³ from it to get the V' he wants, That's what I think

Answer 7

subtract \[0.011700m ^{3}\] from it

Answer 8

I'll try it again when i get home

Answer 9

Let \(x\) be the rate at which water is leaking out, then
\[\frac{dV}{dt}=(11,700-x)\frac{\text{cm}^3}{\text{min}}\]
The volume of water in the tank is given by
\[V=\frac{1}{3}\pi r^2h\]
and differentiating with respect to time yields
\[\frac{dV}{dt}=\frac{1}{3}\pi\left(r^2\frac{dh}{dt}+2rh\frac{dr}{dt}\right)\]
Substituting all relevant information gives
\[(11,700-x)\frac{\text{cm}^3}{\text{min}}=\frac{\pi}{3}\left(r^2\frac{dh}{dt}\frac{\text{m}^3}{\text{min}}+2rh\left(\frac{3.25}{15}\frac{dh}{dt}\right)\frac{\text{m}^3}{\text{min}}\right)\]
Converting to cubic meters,
\[(11,700-x)\text{ cm}^3=\left[(\sqrt[3]{11,700-x})\text{ cm}\times\frac{1\text{ m}}{100\text{ cm}}\right]^3\frac{1}{\text{min}}=\frac{11,700-x}{100,000}\frac{\text{m}^3}{\text{min}}\]