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Amylia's question at Yahoo! Answers regarding an IVP

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MarkFL

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Feb 24, 2012
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Here is the question:

Question on differential equations?

"Water flows in a rectangular tank with base area A at a constant rate of n units of volume per unit time. Water flows out of the tank through a hole at the bottom at a rate which is proportional to the square root of depth of water in the tank. It is found that the when depth is h, the level of water remain constant. Initially the tank is filled to a depth of 4h. Obtain a differential equation for the depth z at time t."

this is what i did so far(im not exactly sure what im doing... T_T) , please guide me in getting the answers:

dV/dt = k√(z)

z(0) = 4h
z(1)=h

[1/√(z)](dV/dt) =k
[1/√(z)] dV =k dt
integrate both sides,
ln √(z) + C = kt + C

i dont know what happens after this.. help :(
I have posted a link there to this topic so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Hello Amylia,

We have water coming into the tank at a constant rate, increasing the volume, and water flowing out of the tank at a variable rate, decreasing the volume. So, using the given information, we may state:

\(\displaystyle \frac{dV}{dt}=n-k\sqrt{z}\)

where $0<k$ is the constant of proportionality.

We are told that \(\displaystyle z=h\implies \frac{dV}{dt}=0\,\therefore\,k=\frac{n}{\sqrt{h}}\)

Since the tank is a rectangular cuboid, we know the volume at time $t$ is given by:

\(\displaystyle V=Az\)

Differentiating with respect to $t$, we find:

\(\displaystyle \frac{dV}{dt}=A\frac{dz}{dt}\)

And thus, equating the two expressions for \(\displaystyle \frac{dV}{dt}\), we obtain the IVP:

\(\displaystyle A\frac{dz}{dt}=n\left(1-\sqrt{\frac{z}{h}} \right)\) where \(\displaystyle z(0)=4h\)

We are not instructed to solve the IVP, and in fact we cannot express $z$ as a function of $t$, however it is possible to express $t$ as a function of $z$.
 

Amylia

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Jul 20, 2013
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Hello Amylia,

We have water coming into the tank at a constant rate, increasing the volume, and water flowing out of the tank at a variable rate, decreasing the volume. So, using the given information, we may state:

\(\displaystyle \frac{dV}{dt}=n-k\sqrt{z}\)

where $0<k$ is the constant of proportionality.

We are told that \(\displaystyle z=h\implies \frac{dV}{dt}=0\,\therefore\,k=\frac{n}{\sqrt{h}}\)

Since the tank is a rectangular cuboid, we know the volume at time $t$ is given by:

\(\displaystyle V=Az\)

Differentiating with respect to $t$, we find:

\(\displaystyle \frac{dV}{dt}=A\frac{dz}{dt}\)

And thus, equating the two expressions for \(\displaystyle \frac{dV}{dt}\), we obtain the IVP:

\(\displaystyle A\frac{dz}{dt}=n\left(1-\sqrt{\frac{z}{h}} \right)\) where \(\displaystyle z(0)=4h\)

We are not instructed to solve the IVP, and in fact we cannot express $z$ as a function of $t$, however it is possible to express $t$ as a function of $z$.
okay, that is very different from the tutorials examples that i have. actually that was only half of the question. the other half is
by making substitution z=hu2, show that u satisfies the diffferential equation \(\displaystyle \frac{2Ah}{n}\)\(\displaystyle \frac{du}{dt}=\)\(\displaystyle \frac{u-1}{u}\)

Find the time at \(\displaystyle z=\frac{16h}{9}\) and describe how z varies with t.
so how do i go about this?
 
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MarkFL

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Feb 24, 2012
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Using the suggested substitution:

\(\displaystyle z=hu^2\)

we find by differentiating with respect to $t$:

\(\displaystyle \frac{dz}{dt}=2hu\frac{du}{dt}\)

and so the ODE becomes:

\(\displaystyle 2Ahu\frac{du}{dt}=n\left(1-\sqrt{\frac{hu^2}{h}} \right)\)

and so the IVP becomes:

\(\displaystyle \frac{2Ah}{n}\frac{du}{dt}=\frac{1-u}{u}\) where $u(0)=2$

This is the negative of what you gave as the result, and I will have to get back to you as I have to run for now.
 
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MarkFL

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After review, I am confident the IVP I gave is correct. So continuing, we need to solve the IVP in order to find \(\displaystyle t\left(\frac{16}{9}h \right)\).

Separating variables, and switching dummy variables of integration, and using the boundaries, we may write:

\(\displaystyle \frac{2Ah}{n}\int_2^u\frac{v}{1-v}\,dv=\int_0^t\,dw\)

\(\displaystyle \frac{2Ah}{n}\int_2^u\frac{1-(1-v)}{1-v}\,dv=\int_0^t\,dw\)

Applying the anti-derivative form of the FTOC, there results:

\(\displaystyle \frac{2Ah}{n}\left[-\ln|1-v|-v \right]_2^u=[w]_0^t\)

\(\displaystyle \frac{2Ah}{n}\left(-\ln|1-u|+2-u \right)=t\)

Now, letting \(\displaystyle z=\frac{16}{9}h\implies u=\frac{4}{3}\) we find:

\(\displaystyle t=\frac{2Ah}{n}\left(\ln(3)+\frac{2}{3} \right)\)

If we had used the ODE suggested by your textbook, we would have found a negative value for time $t$, since $z$ would be initially increasing, and this rate of increase would grow unbounded. And so in order for $z<4h$ we would need to look at a time before $t=0$.

To describe how $z$ varies with $t$, let's go back to:

\(\displaystyle A\frac{dz}{dt}=n\left(1-\sqrt{\frac{z}{h}} \right)\) where \(\displaystyle z(0)=4h\).

Initially, we have:

\(\displaystyle \frac{dz}{dt}=-\frac{n}{A}\)

As $z$ decreases, the rate of decrease of $z$ diminishes, so that $z$ approaches $h$ asymptotically, i.e.:

\(\displaystyle \lim_{t\to\infty}z(t)=h\).