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    Need help with this Q9 in exercise 4E, it just aint happening for me.
    Fluid flows out of a cylindrical tank with constant cross section. At time t minutes, t > 0, the volume of fluid remaining in the tank is V m^3. The rate at which the fluid flows in m^3/min is proportional to thr square root of V. Show that the depth h metres of fluid in the tank satisfies the differential equation dh/dt = -ksqroot(h) where k is a positive constant.

    Help would greatly be appreciated.
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    You'll need the chain rule, I think. How could you represent the flow in differential form?
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    All I can get is dh/dt = dh/dv * dV/dt and also I know that dV/dt = -ksqroot(V)
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    You know that \dfrac{dV}{dt} = k\sqrt{V}, but you also know that, because it's a cylinder, V = \pi r^2 h. From this, you can find an equation for \dfrac{dV}{dh} and then flip it to find \dfrac{dh}{dV}... combine these to get your answer.

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    Since the radius and \pi are constant, you can say that \dfrac{k}{r\sqrt{\pi}} = -k'
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    (Original post by Junker)
    Need help with this Q9 in exercise 4E, it just aint happening for me.
    Fluid flows out of a cylindrical tank with constant cross section. At time t minutes, t > 0, the volume of fluid remaining in the tank is V m^3. The rate at which the fluid flows in m^3/min is proportional to thr square root of V. Show that the depth h metres of fluid in the tank satisfies the differential equation dh/dt = -ksqroot(h) where k is a positive constant.

    Help would greatly be appreciated.
    Can you see where I get the following from:

    \frac{dV}{dt} =  A\sqrt{V} (A a positive constant), and V=\pi r^2 h?

    If so, use this and teh fact that:

    \frac{dh}{dt} = \frac{\frac{dV}{dt}}{\frac{dV}{d  h}}

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    When you plug into the RHS of this equation you'll get \frac{A\sqrt{V}}{\pi r^2} which you can write as B\sqrt{V} since everything but V is a constant. Now plug in V=pi r^2 h into this again...
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    AHh. thanks both of you, i think what I struggled with was differentiaing V = pi r^2 h
    I was doing with respect to r, when I should have been with respect to h. But still why is dV/dh of V = pi r^2 h
    just pi r^2? Sorry for my latex ignorance
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    Because pi r^2 is just a constant, since it is a cylinder.

    So it's dV/dh of, a constant times h.
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    (Original post by Junker)
    AHh. thanks both of you, i think what I struggled with was differentiaing V = pi r^2 h
    I was doing with respect to r, when I should have been with respect to h. But still why is dV/dh of V = pi r^2 h
    just pi r^2? Sorry for my latex ignorance
    Because this is a cylinder we're considering r is just a fixed number (say, I dunno, 2.5 metres), so it acts like a constant when you're differentiating. Therefore, you're differentiating V with respect to h you're effectively just differentiating:

    Ah

    with respect to h, where A is a constant.

    Spoiler:
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    If you're still struggling, try answering the question again but adding in: "Water is flowing from a cylincer with radius 2m ...". You'll obviously then use the formula V=pi x 2^2 x h, which will give dV/dh = pi x 2^2.
 
 
 
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