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Homework Help: Differential Equation Application

  1. Nov 2, 2008 #1
    I'm having trouble understanding this post: https://www.physicsforums.com/showthread.php?t=81157

    Specifically, part (a)

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

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

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

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

    I can't just solve eq. 3, that doesn't seem to make sense.
  2. jcsd
  3. Nov 2, 2008 #2


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    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).
  4. Nov 2, 2008 #3
    To make things a little clearer, say I change the h in thiago_j's posting to b, so that (3) looks like

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

    [tex]\frac{a}{b}=\frac{r}{h}[/tex], so [tex]r=\frac{h a}{b}[/tex] ? Is this the relationship you are talking about? I don't understand.
  5. Nov 2, 2008 #4


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

    So substitute this result into your equation 4 and into the equation for the Volume to obtain V(h). Then use the chain rule: [tex]\frac{dV}{dt}=\frac{dV}{dh} \frac{dh}{dt}[/tex] to obtain a differential equation involving only h(t) and then solve.
  6. Nov 2, 2008 #5
    [tex]\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 [/tex] ?

    I'm still not getting how this works.
  7. Nov 2, 2008 #6


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    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 [itex]V[/itex] as a function of h(t) because you know that the volume of the water at time [itex]t[/itex] is [itex]V=\frac{1}{3} \pi r(t)^3 h(t)=\frac{1}{3} \pi (\frac{a}{b})^3 h(t)^4[/itex]....use the chain rule on this last expression to find another expression for [itex]\frac{dV}{dt}[/itex] in terms of [itex]\frac{dh}{dt}[/itex]
  8. Nov 2, 2008 #7
    How did you get [tex]r(t)^3[/tex]?
    [tex]\frac{4 \pi}{3} \left(\frac{a}{b} \right)^3 h(t)^3 \frac{dh}{dt}[/tex] ?
  9. Nov 2, 2008 #8


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

    And so the chain rule gives:

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

    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)
  10. Nov 2, 2008 #9

    I don't know what you mean by that.
  11. Nov 2, 2008 #10


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

    [tex]\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}[/tex]


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

    surely you can deduce that

    [tex]\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}[/tex]

  12. Nov 3, 2008 #11
    It wasn't clear to me how those two equations were equivalent. Even then, with this, [tex] \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}[/tex]

    Do I get [tex]h(t)=\alpha t +c[/tex] ?

    So, if [tex] V=\frac{1}{3} \pi r(t)^2 h(t)[/tex], then [tex] V=\frac{1}{3} \pi \left({\frac{a}{b}}\right)^2 h(t)^3[/tex] , then [tex] V=\frac{1}{3} \pi \left({\frac{a }{b}}\right)^2 \left(\alpha t + c \right)^3[/tex] ?
  13. Nov 3, 2008 #12


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    Close, you should [itex]h(t)=-\alpha t +c[/itex]....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).
  14. Nov 4, 2008 #13
    [tex]h(t)=\alpha t +c [/tex] is what Maxima gives me for the equation

    [tex]\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} [/tex]. But that equation simplifies to [tex]\frac{dh}{dt}=- \alpha [/tex] which of course would give [tex]h(t)=-\alpha t +c [/tex]. Perhaps Maxima was absorbing the minus sign into c?

    (5) [tex]\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}[/tex]

    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?
  15. Nov 4, 2008 #14


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    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 [itex]a[/itex] is the radius of the conical depression, and [itex]b[/itex] is its depth/height.

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

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

    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:

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

    If it does (and it does!), then your solution is obviously correct.
  16. Nov 4, 2008 #15
    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) [tex]\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} [/tex]

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

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

    What would u1 be?
  17. Nov 5, 2008 #16


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

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

    The proportionality constant [itex]\alpha[/itex] 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.
  18. Nov 7, 2008 #17
    Is there an easy way to find t, if [tex]V(0)=t_{0}[/tex] and [tex]V(t)=t_0/2[/tex] ?

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

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

    if [tex]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 [/tex] then [tex] 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}}[/tex]

    but when I check it against real values, I don't get the same answers. I have the strong suspicion this is wrong.
  19. Nov 7, 2008 #18


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    What happened to [itex]\alpha[/itex]?

    The easiest way is probably to just use:

    [tex]\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}}[/tex]
    Last edited: Nov 7, 2008
  20. Nov 7, 2008 #19
    It's not strictly necessary is it?

    So a and b don't matter?
    Last edited: Nov 7, 2008
  21. Nov 7, 2008 #20


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

    [tex]\frac{dV}{dt}=-\alpha( \text{Surface Area})[/tex]

    Which means setting [itex]\alpha=1[/itex] 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 [itex]t[/itex] was incorrect regardless (although you got [itex]c[/itex] right).
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