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:

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

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.