# Turning a real world problem into a calculus problem

In summary: Qe, Qh, Rh).In summary, the student is trying to solve a differential equation in order to find the height of the water in a swimming pool over a period of time. He has forgotten to include two other variables and needs help integrating the equation. He has asked around and learned that turning the volume of water into πr2h is necessary in order to find the height. He is also trying to start with an initial height of 0 and integrate to find the height at a later time.
Homework Statement
I have posted a picture of the HW problem below.
Relevant Equations
potentially could use:
Accumulation=Input-Output+Generation-Consumption

So I am a bit confused on how to get started. So far my thought process is we have water flowing in and water evaporating from the pool. The part that I think we are interested in is the leakage. The leakage has the rate it is flowing out per unit time. I will call it change in volume, or dV, per unit time, or dt.
My initial thought is:
dV/dt=QL from above equation and integrate both sides, but I don't have the variable "t" on the right hand side of the equation. If someone could walk me through a logical way of turning this word problem into a derivative that would be great.
Something else I would like to clarify: Are we trying to turn this into a differential equation? The course I am in uses DE's and I honestly don't know.

Thanks so much!

Last edited:
$${dV\over dt}=Q_L$$
Is a genuine differential equation. The fact that the righthand side does not depend on ##t## makes it a fairly easy one, but a DE nonetheless. However, you forget two other ##Q##.

And in
Accumulation=Input-Output-Generation+Consumption
I suspect a sign error: if someone swallows water, that does not increase he holdup of the pool ... and the sign for 'generation' likewise seems erroneous.

BvU said:
you forget two other Q.
Maybe not forgotten, just canceled out? Don't we know that , after filling, Qh=Qe?

A whole winter has gone by ... If ##Q_e = Q_{h, \text{last year}}## even getting it filled is a problem.

BvU said:
A whole winter has gone by ... If ##Q_e = Q_{h, \text{last year}}## even getting it filled is a problem.
If the same procedure is followed as in the first year, it is first filled at whatever rate will do then throttled back to match Qe.

haruspex said:
Maybe not forgotten, just canceled out?

BvU said:
Is a genuine differential equation. The fact that the righthand side does not depend on ##t## makes it a fairly easy one, but a DE nonetheless. However, you forget two other ##Q##.

And in I suspect a sign error: if someone swallows water, that does not increase he holdup of the pool ... and the sign for 'generation' likewise seems erroneous.

Hi BvU!
So I have spent some time looking at the problem and I now have a few more questions.
When setting up my D.E. I have learned that what I am doing with this problem is first: setting up boundaries. My boundaries are the swimming pool. I then want to know how the mass of the water is changing over a certain time period. I then choose my time period to be very close to zero so I can use differential calculus.
Let me show you with some work:

I will keep everything in units of mass. We are given flow rates of things (volumes), I will denote the density of liquid water ρl and vapor water ρv

ΔVρl=(Qhρl-Qeρv-Qlρl)Δt
I now let Δt-->0
dVρl/dt=Qhρl-Qeρv-Qlρl
now I think we have a separable D.E. so I can separate and integrate both sides.

I asked around and got some info from classmates that I am now trying to make sense of. They are saying that I should turn Vρl to the equivalent equation of πr2l. And my bounds of integration would be h=0 to h=? since we want an equation in terms of height.
Is the reason we need to turn our volume into πr2h so that we have the variable h, which is what the question is asking?
I was just going to leave it in terms of V...but haven't tried that yet.

BvU said:
No, I did forget/didn't understand where to put the other Q's at that point

Kudos for mentioning ##\rho_v## -- I had completely missed it and assumed ##Q_e## in m3/h water

So far you have $$\rho_L\, dV/dt= \rho_L\, Q_h - \rho_v \, Q_e - \rho_L \,Q_L$$ with ##Q_h## and ##Q_e## independent of ##V##, but ##Q_L ## not. Either you do ##V = \pi R^2 h## and ##Q_L = k_L\,2\pi Rh## to write an equation in terms of ##h##, or you leave ##V## and take ##Q_L = k_L\,2\pi R \displaystyle {V\over \pi R^2}\ ##.
Since the exercise asks for ##h(t)## the former is a good choice.
So is starting with ##h(0) = 0##.
And the integration goes from 0 to ##t## to get ##h## as a function of ##t##.

You consider ##Q_h## as input (but to get the thing filled before the summer is over, it should be bigger than the ##Q_h## after filling last year ) .

A bit strange that the exercise gives numerical values for a few of the variables, but not enough for a complete calculation (e.g. ##Q_e##).

BvU said:
but to get the thing filled before the summer is over, it should be bigger than
It is very confusing the way the description reuses Qh. It cannot be the same Qh as in the first year since that equalled Qe, and we are told the Qh in the second year is greater than that.
In the first year, it is first filled at some unknown but sufficient rate. After filling it is throttled back to match the evaporation rate. Were it at the lower rate through out it would not filled then either.

In the second year, it is unclear whether the same procedure is followed. I think I was wrong to assume so. Looks like we should take it as filling from empty with input constant and somewhat greater than the constant evaporation. For purposes of analysis, it is only the difference of the two that matters.
And it may well be that it never fills this time.

I get $$A_{top}\frac{dh}{dt}=Q_h-Q_e-k_L(2\pi R)h$$This is a first order linear ODE in h.

Chestermiller said:
I get $$A_{top}\frac{dh}{dt}=Q_h-Q_e-k_L(2\pi R)h$$This is a first order linear ODE in h.
There is the small matter of ##\ {\rho_v\over\rho_L} \ ## in front of ##Q_e##, but apart from that I agree (as in post #9)

BvU said:
There is the small matter of ##\ {\rho_v\over\rho_L} \ ## in front of ##Q_e##, but apart from that I agree (as in post #9)
To me, it is obvious that ##Q_e## is supposed to represent the rate at which liquid water is evaporating. In my judgment, for a problem presented to a student at this level, this is the only thing it could mean.

scottdave
Same here, but the wording

appears to state otherwise

BvU said:
Same here, but the wording

View attachment 256241

appears to state otherwise
Ooops. I missed that. But it makes no sense, unless it is at the interface temperature and equilibrium vapor pressure (which are not specified) and which the student at this level would not know how to determine anyway.

Forgive me if I’m wrong. But due to conservation of volume wouldn’t the evaporation rate of liquid water equal the volumetric gas flow rate? I.e. the “surroundings gain volume at the same rate the system loses volume”.

No. 1 kg of liquid water at 1 atm and 20 ##^\circ##C is 0.001 m3.
It evaporates to 57.8 m3 of water vapour.

And I never heard of a 'conservation of volume' law

BvU said:
No. 1 kg of liquid water at 1 atm and 20 ##^\circ##C is 0.001 m3.
It evaporates to 57.8 m3 of water vapour.

And I never heard of a 'conservation of volume' law

BvU said:
No. 1 kg of liquid water at 1 atm and 20 ##^\circ##C is 0.001 m3.
It evaporates to 57.8 m3 of water vapour.

And I never heard of a 'conservation of volume' law

So would ##\frac{\rho_g}{\rho_l} = \frac{m}{V_g} \frac{V_l}{m} = \frac{0.0001}{57.8} = 1.73 * 10^{-6}##??

For a final differential equation of (the h by the Q is a subscript)

##\frac{dh}{dt} = \frac{Q_h}{\pi R^2} - \left( 1.73 * 10^{-6} \right) k_e - k_l \left(\frac{2}{R} \right)h##which can then be solved using the method found here
http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx

##\frac{dy}{dt}+ p\left( t \right)y = g\left( t \right) ##

##y\left(t \right) = \frac{\int \mu \left( t \right) g\left( t \right) dt + c}{\mu \left(t \right)}##

where

## \mu \left( t \right) = e^{\int p\left(t\right) dt}##

If I'm spouting nonsense and detracting from the thread/hindering progress let me know and I will stop.

## 1. How can calculus be used to solve real world problems?

Calculus is a branch of mathematics that deals with rates of change and accumulation. By using derivatives and integrals, we can model and analyze real world problems such as motion, growth, and optimization. It provides a powerful tool for understanding and solving a wide range of problems in fields such as physics, economics, and engineering.

## 2. What are the steps involved in turning a real world problem into a calculus problem?

The first step is to identify the quantities involved and determine their relationship. Then, we use calculus concepts such as derivatives and integrals to create a mathematical model of the problem. This model can then be analyzed and used to find solutions to the problem.

## 3. Can any real world problem be transformed into a calculus problem?

While calculus can be applied to many real world problems, not all problems can be accurately modeled using this mathematical tool. Some problems may require other mathematical methods or may not have a precise solution. It is important to carefully consider the problem and determine if calculus is the appropriate approach.

## 4. How does turning a real world problem into a calculus problem benefit us?

By using calculus to solve real world problems, we can gain a deeper understanding of the problem and make predictions about its behavior. It allows us to optimize and improve systems, make informed decisions, and find efficient solutions. It also has practical applications in fields such as engineering, economics, and medicine.

## 5. Are there any tips for effectively using calculus to solve real world problems?

One tip is to carefully define and label all quantities involved in the problem. This will help in setting up the mathematical model. It is also important to understand the meaning and implications of the derivatives and integrals in the context of the problem. Practice and familiarization with different types of problems can also improve problem-solving skills using calculus.

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