Differential Equation Application

[james brug]

I'm having trouble understanding this post: https://www.physicsforums.com/showthread.php?t=81157

Specifically, part (a)

(1) \frac{dV}{dt} = k - \alpha \pi \left( \frac{3a}{\pi h} \right) ^{2/3} V ^{2/3}

Leaving out k, and considering a cone with no water coming in:

(2) \frac{dV}{dt} = - \alpha \pi \left( \frac{3a}{\pi h} \right) ^{2/3} V ^{2/3}

How would I solve this for an equation to determine the volume in terms of time t? How do I interpret V in terms of t?

(3) \frac{dV}{dt} = - \alpha \pi \left( \frac{3a}{\pi h} \right) ^{2/3} {\left(\frac{1}{3}\pi r^2 h\right)} ^{2/3}

I can't just solve eq. 3, that doesn't seem to make sense.

 

[gabbagabbahey]

Start by finding the relationship between r and h for the cone, this will allow you to express V and dV/dt in terms of a single time dependent variable (h or r, whichever you choose).

 

[james brug]

To make things a little clearer, say I change the h in thiago_j's posting to b, so that (3) looks like

(4) \frac{dV}{dt} = - \alpha \pi \left( \frac{3a}{\pi b} \right) ^{2/3} {\left(\frac{1}{3}\pi r^2 h\right)} ^{2/3}.


\frac{a}{b}=\frac{r}{h}, so r=\frac{h a}{b} ? Is this the relationship you are talking about? I don't understand.

 

[gabbagabbahey]

Yes, the opening angle of the cone \theta is constant, so if the full height of the cone is b, and the radius of the opening at the top is a, then tan(\theta /2)=a/b=r(t)/h(t) \Rightarrow r(t)= h(t)\frac{a}{b}.

So substitute this result into your equation 4 and into the equation for the Volume to obtain V(h). Then use the chain rule: \frac{dV}{dt}=\frac{dV}{dh} \frac{dh}{dt} to obtain a differential equation involving only h(t) and then solve.

 

[james brug]

\frac{dV}{dt} = - \alpha \pi \left( \frac{3a}{\pi b} \right) ^{2/3} {\frac{2}{3}\left(\frac{\pi}{3} h(t)^3 \left(\frac{a}{b}\right)^2 \right)} ^{-1/3} 3 h(t)^2 ?

I'm still not getting how this works.

 

[gabbagabbahey]

\frac{dV}{dt} = - \alpha \pi \left( \frac{3a}{\pi b} \right) ^{2/3} {\frac{2}{3}\left(\frac{\pi}{3} h(t)^3 \left(\frac{a}{b}\right)^2 \right)} ^{-1/3} 3 h(t)^2 ?

I'm still not getting how this works.

No, a straight substitution into your equation (4) gives:

<br /> \frac{dV}{dt} = - \alpha \pi \left( \frac{3a}{\pi b} \right) ^{2/3} {\left(\frac{1}{3}\pi h(t)^3(\frac{a}{b})^2 \right)} ^{2/3} <br />

But you can also express V as a function of h(t) because you know that the volume of the water at time t is V=\frac{1}{3} \pi r(t)^3 h(t)=\frac{1}{3} \pi (\frac{a}{b})^3 h(t)^4...use the chain rule on this last expression to find another expression for \frac{dV}{dt} in terms of \frac{dh}{dt}

 

[james brug]

But you can also express V as a function of h(t) because you know that the volume of the water at time t is V=\frac{1}{3} \pi r(t)^3 h(t)=\frac{1}{3} \pi (\frac{a}{b})^3 h(t)^4

How did you get r(t)^3?

...use the chain rule on this last expression to find another expression for \frac{dV}{dt} in terms of \frac{dh}{dt}

\frac{4 \pi}{3} \left(\frac{a}{b} \right)^3 h(t)^3 \frac{dh}{dt} ?

 

[gabbagabbahey]

How did you get r(t)^3?

\frac{4 \pi}{3} \left(\frac{a}{b} \right)^3 h(t)^3 \frac{dh}{dt} ?

Errg, sorry typo it should be r^2 and so V=\frac{1}{3} \pi r(t)^3\2 h(t)=\frac{1}{3} \pi (\frac{a}{b})^2 h(t)^3

And so the chain rule gives:

\frac{dV}{dt} =\pi \left(\frac{a}{b} \right)^2 h(t)^2 \frac{dh}{dt}

Now equate that with the other expresion for dV/dt and you will have a separable differential equation for h(t) which you can easily solve...After that, you can simply plug it into the expression for V annd obtain an expression for V(t)

 

[james brug]

Now equate that with the other expresion for dV/dt

I don't know what you mean by that.

 

[gabbagabbahey]

Well, you have 2 equations for dV/dt now:

\frac{dV}{dt} = - \alpha \pi \left( \frac{3a}{\pi b} \right) ^{2/3} {\left(\frac{1}{3}\pi h(t)^3(\frac{a}{b})^2 \right)} ^{2/3}

and

\frac{dV}{dt} =\pi \left(\frac{a}{b} \right)^2 h(t)^2 \frac{dh}{dt}

surely you can deduce that

\pi \left(\frac{a}{b} \right)^2 h(t)^2 \frac{dh}{dt}=- \alpha \pi \left( \frac{3a}{\pi b} \right) ^{2/3} {\left(\frac{1}{3}\pi h(t)^3(\frac{a}{b})^2 \right)} ^{2/3}

right?

 

[james brug]

It wasn't clear to me how those two equations were equivalent. Even then, with this, \frac{dh}{dt}=\frac{- \alpha \left( \frac{3a}{\pi b} \right) ^{2/3} {\left(\frac{1}{3}\pi h(t)^3\left(\frac{a}{b}\right)^2 \right)} ^{2/3} }{ \left(\frac{a}{b} \right)^2 h(t)^2}

Do I get h(t)=\alpha t +c ?

After that, you can simply plug it into the expression for V annd obtain an expression for V(t)

So, if V=\frac{1}{3} \pi r(t)^2 h(t), then V=\frac{1}{3} \pi \left({\frac{a}{b}}\right)^2 h(t)^3 , then V=\frac{1}{3} \pi \left({\frac{a }{b}}\right)^2 \left(\alpha t + c \right)^3 ?

 

[gabbagabbahey]

It wasn't clear to me how those two equations were equivalent. Even then, with this, \frac{dh}{dt}=\frac{- \alpha \left( \frac{3a}{\pi b} \right) ^{2/3} {\left(\frac{1}{3}\pi h(t)^3\left(\frac{a}{b}\right)^2 \right)} ^{2/3} }{ \left(\frac{a}{b} \right)^2 h(t)^2}

Do I get h(t)=\alpha t +c ?




So, if V=\frac{1}{3} \pi r(t)^2 h(t), then V=\frac{1}{3} \pi \left({\frac{a}{b}}\right)^2 h(t)^3 , then V=\frac{1}{3} \pi \left({\frac{a }{b}}\right)^2 \left(\alpha t + c \right)^3 ?

Close, you should h(t)=-\alpha t +c...in other words your missing a negative sign in front of your alpha...you can check that your solution (with -alpha) for V satisfies the DE in your first post (equation 2).

 

[james brug]

h(t)=\alpha t +c is what Maxima gives me for the equation

\frac{dh}{dt}=\frac{- \alpha \left( \frac{3a}{\pi b} \right) ^{2/3} {\left(\frac{1}{3}\pi h(t)^3\left(\frac{a}{b}\right)^2 \right)} ^{2/3} }{ \left(\frac{a}{b} \right)^2 h(t)^2}. But that equation simplifies to \frac{dh}{dt}=- \alpha which of course would give h(t)=-\alpha t +c. Perhaps Maxima was absorbing the minus sign into c?

...you can check that your solution (with -alpha) for V satisfies the DE in your first post (equation 2)

(5) \frac{dV}{dt}=-\alpha \pi \left(\frac{3a}{\pi b}\right)^{2/3}\left(\frac{\pi}{3} \left({\frac{a }{b}}\right)^2 \left(\alpha t + c \right)^3 \right)^{2/3}

So how does this bring me any closer to determining V at a time t?
I am told alpha is the proportionality constant, but I still don't understand its function. Couldn't a/b be a kind of proportionality constant? What do I do with alpha?

 

[gabbagabbahey]

You leave alpha as it is; it is just some constant which you may or may not know... a/b is also a constant which you can easily measure, since a is the radius of the conical depression, and b is its depth/height.

Given that h(t)=-\alpha t +c, V at tme t is just

V(t)=\frac{1}{3} \pi \left({\frac{a }{b}}\right)^2 \left(-\alpha t + c \right)^3

And that IS your solution!...you simply plug in the values of a,b and alpha and you can calculate V at a time t!

You can check to see that it is correct by differentiating it and making sure that it satisfies your original ODE:

\frac{dV}{dt} = - \alpha \pi \left( \frac{3a}{\pi h} \right) ^{2/3} V ^{2/3}

If it does (and it does!), then your solution is obviously correct.

 

[james brug]

How do I know where alpha comes from, from where its derived?


And I'm still not clear on how these two equations relate to each other,

(6) \frac{dV}{dt} = \underbrace{- \alpha \pi \left( \frac{3a}{\pi b} \right) ^{2/3}}_{u1} \underbrace{{\left(\frac{1}{3}\pi h(t)^3(\frac{a}{b})^2 \right)} ^{2/3}}_{u2}

(7) \frac{dV}{dt} =\underbrace{\pi \left(\frac{a}{b} \right)^2 h(t)^2}_{u3} \frac{dh}{dt}

So, u3 is dV/dh, correct? And u2 is V^(2/3) in terms of h(t).

What would u1 be?

 

[gabbagabbahey]

Yes, u_3 is dV/dh and u_2 is V^(2/3) in terms of h(t).

The quantity \pi \left( \frac{3a}{\pi b} \right) ^{2/3}\left(\frac{1}{3}\pi h(t)^3(\frac{a}{b})^2 \right) ^{2/3}} is the surface area of puddle at a time t

The proportionality constant \alpha is a quantity that can be experimentally measured, by measuring the height of the water at a few different times, or theoretically derived based on knowledge of how water evaporates.

 

[james brug]

Is there an easy way to find t, if V(0)=t_{0} and V(t)=t_0/2 ?

V(t)=\frac{1}{3} \pi \left({\frac{a }{b}}\right)^2 \left(-t + c \right)^3

c=\left(\frac{t_0}{\frac{\pi}{3} \left(\frac{a}{b}\right)^2}\right)^{1/3}

if t_0/2=\frac{1}{3} \pi \left({\frac{a }{b}}\right)^2 \left(-t + \left(\frac{t_0}{\frac{\pi}{3} \left(\frac{a}{b}\right)^2}\right)^{1/3} \right)^3 then t=\frac{-b^{2/9} \left(3t_0\right)^{1/9} \left( b^{4/9} \cdot 2^{2/3} \left( 3t_0 \right)^{2/9} -2a^{4/9} \cdot \pi^{2/9} \right) }{2a^{2/3} \cdot \pi^{1/3}}

but when I check it against real values, I don't get the same answers. I have the strong suspicion this is wrong.

 

[gabbagabbahey]

What happened to \alpha?

The easiest way is probably to just use:

\frac{V(t)}{V(0)}=\frac{\frac{t_0}{2}}{t_0}=\frac{1}{2}=\frac{(-\alpha t+c)^3}{c^3} \Rightarrow (\frac{-\alpha t}{c}+1)=\frac{1}{\sqrt[3]{2}}

 

[james brug]

What happened to \alpha?

It's not strictly necessary is it?

The easiest way is probably to just use:

\frac{V(t)}{V(0)}=\frac{\frac{t_0}{2}}{t_0}=\frac{ 1}{2}=\frac{(-\alpha t+c)^3}{c^3} \Rightarrow (\frac{-\alpha t}{c}+1)=\sqrt[3]{2}

So a and b don't matter?

 

[gabbagabbahey]

Yes, it is...is the proportionality constant so it determines how quiickly water evaproates after all, the original DE is essentially:

\frac{dV}{dt}=-\alpha( \text{Surface Area})

Which means setting \alpha=1 makes the puddle evaporate very quickly...but if you just want to check your solution for various values, I suppose you can set it equal to one. But your solution for t was incorrect regardless (although you got c right).

 

[gabbagabbahey]

It should have been \frac{1}{\sqrt[3]{2}}...I've edited that post.

And a and b do matter, but they are contained in your expression for c.

 

[james brug]

The easiest way is probably to just use:...

Is the way I was trying to do it hard?

I tried again and I get:
t=\frac{-b^{2/3} \left(2^{2/3}-2\right) \left( 3 t_0 \right) }{2a^{2/3} \cdot \pi^{1/3}}

 

[gabbagabbahey]

The way you are trying to do it is unnecessarily complicated, and you are making mistakes because of it.

I get (setting alpha=1):

(\frac{-t}{c}+1)=\frac{1}{\sqrt[3]{2}} \Rightarrow t=(1-\frac{1}{\sqrt[3]{2}})c=(1-\frac{1}{\sqrt[3]{2}})\left(\frac{t_0}{\frac{\pi}{3} \left(\frac{a}{b}\right)^2}\right)^{1/3}

 

[james brug]

Arrghh... if V(t)=t_0/2 , then what is the time period in between then and when V=0 ?

 

[gabbagabbahey]

V(T)=0 \Rightarrow \frac{1}{3} \pi \left({\frac{a }{b}}\right)^2 \left(-T + c \right)^3=0 \Rightarrow T=c=\left(\frac{t_0}{\frac{\pi}{3} \left(\frac{a}{b}\right)^2}\right)^{1/3}

And so computing the time period between t and T should be simple. It's just T-t=\frac{1}{\sqrt[3]{2}}\left(\frac{t_0}{\frac{\pi}{3} \left(\frac{a}{b}\right)^2}\right)^{1/3}.

 

[james brug]

Let's say I hadn't posted post #17 of this thread. Is the method you used in post #25 the most direct method? If I had only posted #24, is that what you would have said?

 

[gabbagabbahey]

If you hadn't posted #17, I would have had to use your method, but you would have to know what t was, because based only on #24 you would end up with T-t being a function of t (since c would be expressed in terms of t_0 and t)

 

[james brug]

Suppose that were the case--how would I do that? t is the number of units of time that have elapsed when half the water has evaporated, so I'm not quite sure what you mean.