Conical Tank Water Drainage: Finding Time to Empty with Differential Equations

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SUMMARY

The discussion focuses on solving a differential equation related to the drainage of water from an inverted conical tank. The equation derived is dy/dt = -k(y0/r0)²(1/πy), where 'k' is a constant. The tank's water level decreases from 16 feet to 9 feet in one hour, prompting the need to find the total time required to empty the tank completely. Participants emphasize the importance of integrating this differential equation to express 'y' as a function of time for further analysis.

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Homework Statement


Water drains out of an inverted conical tank at a rate proportional to the depth (y) of water in the tank. Write a diff EQ as a function of time.

This tank's water level has dropped from 16 feet deep to 9 feet deep in one hour. How long will it take before the tank is empty.

Homework Equations



V=1/3pir2y

dV/dt=pi(r0/y0)2y2dy/dt

The Attempt at a Solution


The solution to the first question is: dy/dt=-k(y0/r0)2(1/piy)

I don't really know how to go about finding the time it will take to empty the tank.

Please help!
 
Last edited:
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hey, welcome to physicsforums! your solution to the first question is correct. (I'm guessing you're using 'k' as a constant. this is good.) So now, you need to make use of this differential equation to get 'y' as a function of time. You have hopefully done this kind of integral before. It just needs a bit of rearranging to get a nice answer.
 

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