Water Evaporation Rate Problem

[revisisland24]

Hey,

I'm having trouble starting with this one.

A tank is being slowly filled with drainage water, and simultaneously water is evaporating from the tank. Water is streaming at a rate of .1 cubic cm per minute, and 5% evaporates each minute.

a.) Find the differential equation for the volume of water in the tank, with respect to time.
b.) Solve the differential equation from part a, assuming that, at time t=0 minutes, the volume of water in the tank is 100 cubic centimeters.
c.) Assuming the reaction continues as described, find the limit of the volume of water in the tank as t approaches infinity.

So, i have V = pi*r^2*h and V' = 2pi*r*dh/dt but don't know how to go from there or if that's even the right setup. Thanks!

 

[MarkFL]

You seem to be assuming the tank is cylindrical, and for the purposes of this problem the geometry of the tank doesn't matter. What you want to do is (with linear measures in cm and time in minutes):

\(\displaystyle \text{time rate of change in total volume}=\dfrac{dV}{dt}=\text{flow in}-\text{evaporation out}\)

Can you identify expressions for the flow in and evaporation out?

 

[revisisland24]

You seem to be assuming the tank is cylindrical, and for the purposes of this problem the geometry of the tank doesn't matter. What you want to do is (with linear measures in cm and time in minutes):

\(\displaystyle \text{time rate of change in total volume}=\dfrac{dV}{dt}=\text{flow in}-\text{evaporation out}\)

Can you identify expressions for the flow in and evaporation out?

dV/dt = .1 - .05t

 

[MarkFL]

Yes, exactly! :cool:

What type of differential equation is this?

 

[revisisland24]

Yes, exactly! :cool:

What type of differential equation is this?

I think this is a related rates type problem

 

[Deleted member 4993]

dV/dt = .1 - .05t

dV = (0.1-0.5*t) dt

Now integrate both sides...

 

[revisisland24]

dV = (0.1-0.5*t) dt

Now integrate both sides...

would that be V = . 1- .2236t^2 + C

 

[MarkFL]

dV/dt = .1 - .05t

I'm sorry, in my haste, I saw what I wanted to see, this should be:

\(\displaystyle \frac{dV}{dt}=0.1-0.05V\)

Now, what type of ODE is this?

 

[revisisland24]

I'm sorry, in my haste, I saw what I wanted to see, this should be:

\(\displaystyle \frac{dV}{dt}=0.1-0.05V\)

Now, what type of ODE is this?

Would i find the anti-derivative for the next step?

 

[MarkFL]

No, this is a linear first order ODE, and so you need to compute the integrating factor. Are you familiar with this method?

 

[revisisland24]

No, this is a linear first order ODE, and so you need to compute the integrating factor. Are you familiar with this method?

we just learned it, so I'm a little unsure.

.05V dV/dt = .1 - then i take the anti-derivatives

.2236V^2 dV/dt = .1t + C and i'm unsure if this is the right step

 

[MarkFL]

First, you want to write the equation in standard linear form:

(1) \(\displaystyle \dfrac{dV}{dt}+0.05V=0.1\)

and so your integrating factor is:

\(\displaystyle \mu(t)=e^{0.05\int\,dt}=e^{0.05t}\)

Now, multiply (1) by the integrating factor, and you will find that the left hand side can be expressed as the product of a derivative. Can you give this a try?

 

[revisisland24]

First, you want to write the equation in standard linear form:

(1) \(\displaystyle \dfrac{dV}{dt}+0.05V=0.1\)

and so your integrating factor is:

\(\displaystyle \mu(t)=e^{0.05\int\,dt}=e^{0.05t}\)

Now, multiply (1) by the integrating factor, and you will find that the left hand side can be expressed as the product of a derivative. Can you give this a try?

Then, would dV/dt = e^.05t

 

[MarkFL]

No, what you want to do is multiply the ODE in standard linear form by \(\displaystyle \mu(t)\) to obtain:

\(\displaystyle e^{0.05t}\dfrac{dV}{dt}+0.05e^{0.05t}V=0.1e^{0.05t}\)

Now, observe that:

\(\displaystyle \dfrac{d}{dt}\left(e^{0.05t}V \right)=e^{0.05t}\dfrac{dV}{dt}+0.05e^{0.05t}V\)

which allows us to now write the ODE as:

\(\displaystyle \dfrac{d}{dt}\left(e^{0.05t}V \right)=0.1e^{0.05t}\)

This is why we use the integrating factor derived this way, it always allows us to write the left side as the derivative of the product of the integrating factor and the dependent variable. I recommend reviewing your textbook in the section on linear equations, as it should explain this in more detail.

Now, integrate both sides with respect to \(\displaystyle t\) to get:

\(\displaystyle \displaystyle \int\,d\left(e^{0.05t}V \right)=0.1\int e^{0.05t}\,dt\)

Can you proceed?

 

[Deleted member 4993]

\(\displaystyle \frac{dV}{dt}=0.1-0.05V\)

Another way

\(\displaystyle \frac{dV}{dt}=0.1-0.05V\)

\(\displaystyle \frac{dV}{0.1-0.05V} = dt\)

Integrating both sides, we get,

ln|0.1-0.05V| = 0.05*(-t + C₁)

0.1 - 0.05V = C₂*e^-t/20

V = 2 - C*e^-t/20

Now calculate C using initial condition.