Not really homework, problem at my job right now.

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SUMMARY

The discussion revolves around solving a real-life ordinary differential equation (ODE) related to reservoir height and volume. The height of the reservoir as a function of volume is defined by the equation h(V) = 0.001V³ - 0.04V² + 1.7802V. The rate of change of height over time is given by h(t) = -0.2016t + 8.4213. To find the rate of discharge in gallons per minute when the reservoir height is at 15 feet, the user needs to calculate dV/dt using the chain rule, specifically dV/dt = dV/dh * dh/dt.

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Homework Statement



This is really a question that I can't figure out at work right now, so it is a real life example of (an ODE?) math problem.

reservoir height (ft) as a function of Volume (Million Gallons)

h(V) = 0.001V3-0.04V2+1.7802V

reservoir drain as a function of time (t=days) h = ft

h(t) = -0.2016t+8.4213 (based on a 5-day trend when reservoir is at 11ft-ish)

in other words, the important part of the equation is dh/dt= -0.2016ft/day

I want to find out what the rate of discharge is when the reservoir is at 15ft. i.e. gallons/min or million gallons / day when the height is at 15ft.

I believe it is a ODE? but I'm really stumped

The Attempt at a Solution



Lots of paper on my desk, and no answers. HAH!
 
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What you want to find is dV/dt at some point. You know that dh/dt = -0.2016.

You have h in some function as V, so you can get dh/dV.

Now dV/dt = dV/dh * dh/dt from your chain rule.
 

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