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    (Original post by RDKGames)
    Half the stuff he writes doesn't make sense, or is hard to follow where it comes from, or is unclear as to what he's trying to do with it so I give up after a while. I don't really need to explain anything.
    No - you probably just needed to give an hint and there's no need for rudeness because someone doesn't understand your method.
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    (Original post by RDKGames)
    \dfrac{dV}{dh} = 4000 since V=\pi R^2 h hence \dfrac{dV}{dh} = \pi R^2 = A, so that's fine.

    But I don't understand where you pull the last line from. The RHS of it is basically \dfrac{dV}{dt} = \pi R^2 \dfrac{dh}{dt} = \dfrac{dV}{dh} \dfrac{dh}{dt}. So the LHS needs to reflect that it is change in volume over time, and dV/dt is most certainly not 4000 as you have it.
    thats what you wrote earlier. I dont know how to use that with A = 4000

    either way, we have: \dfrac{dV}{dh} = 4000

    \dfrac{dV}{dh} = \pi R^2 = A

    where do I go from here?
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    (Original post by Maths&physics)
    dh/dv = 1/\pi r^2, dh/dv = 1/4000

    I get that.

    and you do: 1/\pi r^2 = 1/4000

    where do you go from here?
    The question tells us that liquid flows into the container at a rate of 1600, but is leaving at a rate proportional to the square root of the height.

    Therefore, the rate of change of volume can be expressed as:
    dv/dt = 1600 - C\sqrt h
    So now we can just substitute our differentials into our original equation for dh/dt = dh/dv\times dv/dt to get closer to our final answer.
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    (Original post by Maths&physics)
    thats what you wrote earlier.

    either way, we have: \dfrac{dV}{dh} = 4000

    \dfrac{dV}{dh} = \pi R^2 = A

    where do I go from here?
    I did not write 4000=(...) though which is what I've questioned.

    Anyway, yes we have \dfrac{dV}{dh} = 4000 and so \dfrac{dV}{dt} = 4000\dfrac{dh}{dt}

    Now what's dV/dt using the info about inflow and outflow of liquid?
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    (Original post by DisneylandChina)
    The question tells us that liquid flows into the container at a rate of 1600, but is leaving at a rate proportional to the square root of the height.

    Therefore, the rate of change of volume can be expressed as:
    dv/dt = 1600 - C\sqrt h
    So now we can just substitute our differentials into our original equation for dh/dt = dh/dv\times dv/dt to get closer to our final answer.
    ah, ok.

    when we're dealing with proportions, do we always have a constant (in this case: c)?.

    ok, I said that earlier, but I write: dv/dt = 1600 - \sqrt h
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    (Original post by Maths&physics)
    ah, ok.

    when we're dealing with proportions, do we always have a constant (in this case: c)?.

    ok, I said that earlier, but I write: dv/dt = 1600 - \sqrt h
    Yes, a constant is usually used. If they are proportional, it is very different from being equal.
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    (Original post by Maths&physics)
    all help is welcome.

    I thought: dV/dh = A or dV/dr = A

    but I'm unfamiliar with his approach although its probably the best. however, I still get stuck in the next part.


    4000 = A

    so, 4000 =\pi r^2

    or in the case of RDK: 4000 = \pi r^2 (dh/dt)
    Go back to the beginning.

    You said dh/dt = (dh/dV) x (dV/dt) which is correct

    then you seemd OK with the poster who said:




    Then you had:

    ,

    Now put that altogether:

    dh/dt = (1/4000) x (1600 - C h^1/2)

    Then expand.
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    (Original post by DisneylandChina)
    Yes, a constant is usually used. If they are proportional, it is very different from being equal.
    oh yeah, youre correct.
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    (Original post by Muttley79)
    Go back to the beginning.

    You said dh/dt = (dh/dV) x (dV/dt) which is correct

    then you seemd OK with the poster who said:




    Then you had:

    ,

    Now put that altogether:

    dh/dt = (1/4000) x (1600 - C h^1/2)

    Then expand.
    what happens to:
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    (Original post by Maths&physics)
    what happens to:
    We are told pi r^2 = 4000 in the question.
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    (Original post by Muttley79)
    We are told pi r^2 = 4000 in the question.
    ah ok, so we didn't need to solve dV/dh but just needed to know that dV/dh = A = 4000 ??

    thanks
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    (Original post by Maths&physics)
    ah ok, so we didn't need to solve dV/dh but just needed to know that dV/dh = A = 4000 ??

    thanks
    In this question we were given the area of cross section which is constant in a cylinder. Quite often we are given a specific radius [when it changes e.g. a circle] or time to get the rate of change - this questiom is just a bit different.
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    (Original post by Muttley79)
    In this question we were given the area of cross section which is constant in a cylinder. Quite often we are given a specific radius [when it changes e.g. a circle] or time to get the rate of change - this questiom is just a bit different.
    thank you
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    (Original post by DisneylandChina)
    Yes, a constant is usually used. If they are proportional, it is very different from being equal.
    (Original post by RDKGames)
    ....

    (Original post by Muttley79)
    ....
    in the next part of the question we're solving the the change in volume or height?

    can we use the dh/dt equation we've just solved? or use dV/dt

    I would think because this is the amount leaving, we do: 400 = - C\sqrt h ??
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    (Original post by Maths&physics)
    in the next part of the question we're solving the the change in volume or height?

    can we use the dh/dt equation we've just solved? or use dV/dt
    What do you think? Try it first
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    (Original post by Maths&physics)
    in the next part of the question we're solving the the change in volume or height?

    can we use the dh/dt equation we've just solved? or use dV/dt
    No because it tells us nothing about the rate at which the height changes.

    From part (i) we know it leaks out at a rate of C\sqrt{h} which at h=25 is 5C. We know now that this is 400. We also know that k = \frac{C}{4000} from previous work, so you can show the result.
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    (Original post by Muttley79)
    What do you think? Try it first
    because its the water leaving, I did:

    -400 = - C\sqrt h

    a negative 400 because its water leaving.

    -400 = - C\sqrt 25
    -400 = - C(5)
    80 = C
    from the first part:
     k = C/4000
    k = 80/4000
    k = 1/50
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    (Original post by RDKGames)
    No because it tells us nothing about the rate at which the height changes.

    From part (i) we know it leaks out at a range of C\sqrt{h} which at h=25 is 5C. We know now that this is 400. We also know that k = \frac{C}{4000} from previous work, so you can show the result.
    thanks, I figured that part out. see above
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    (Original post by Maths&physics)
    because its the water leaving, I did:

    -400 = - C\sqrt h

    a negative 400 because its water leaving.

    -400 = - C\sqrt 25
    -400 = - C(5)
    80 = C
    from the first part:
     k = C/4000
    k = 80/4000
    k = 1/50
    Well done - you completed it without help
 
 
 
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