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    heya there, I was wondering if anyone could help me with this question, I have almost no idea on how to do it, so any help is greatly appreciated:

    A funnel has a diameter at the top of 12cm and height of 8cm. When the deptch of liquid in the funnel is 4cm, the liquid is dripping out of the funnel at the rate of 0.22cm^3/s. At what rate is the depth of the liquid in the funnel failing at this instant?

    Thanks in advance
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    The funnel is presumably cone shaped, so the volume of liquid at any instant is V=pi(r^2)h. So r and h vary with time t.

    You are told that dV/dt = 0.22 when h=4. You are asked to find dh/dt when h=4.

    Can you express dh/dt as the product of two rates of change?
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    thanks for your reply,

    hm i thought that pi(r^2)h was the volume for a cylinder, not a cone?

    and im not really sure how to do what you are asking
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    Sorry, I meant (1/3)pi(r^2)h !!
    It's not an easy topic!

    dh/dt = dh/dV x dV/dt (can you see why?)
    We are told that dV/dt=0.22 when h=4.
    You need to find dh/dV when h=4.
    You can find dh/dV by considering dV/dh and then...
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    Ah yes, I remember learning about this now hehe, is it beacuse the dV's cancel out?

    Ok so the equation for volume at a given h is V=(12pi)h ?

    And so dV/dh = 12pi
    so dh/dV = 1/(12pi)

    Therefore 1/(12pi) = dh/dV x 12

    so dh/dt = 1/(12pi) x 0.22 which equals around 0.0058?

    Am i right?

    Also what would the units be then? cm^3 s^-2?
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    Whoa that looks like a pretty hard questions for C3! This might be completely wrong but this is how i would do it.

    The way I would approach this is first of all assume that the funnel is a cone. The volume of a cone is:-

     V = \frac {1}{3} \pi r^{2} h .

    You're looking for  \frac {dh}{dt} . If we look at the cone from a sideways position a triangle can be seen which can be bisected into 2 right angled triangles which consist of the radius and height (and slant height). We know the radius of cone at the top (6) and the height (8) the ratio between the height of the cone and the radius is 4:3. Or  r = \frac {3h}{4}.

    Therefore:-  V = \frac {1}{3}\pi (\frac {3h}{4})^{2}h or
     V =  \frac {3h^{3}}{16}\pi

     \frac {dV}{dh} = \frac {9h^{2}\pi}{16}

     \frac {dh}{dV} = \frac {16}{9h^{2}\pi}

    Therefore, when h = 4,  \frac {dh}{dV} = \frac {16}{9*4^{2}\pi}= \frac {1}{9\pi}

    we also know that  \frac {dV}{dt} = 0.22

    Using connective rates of change we can say that the rate of change of depth,  \frac {dh}{dt} = \frac {dV}{dt} * \frac {dh}{dV}

    Therefore,  \frac {dh}{dt} = 0.22*\frac {1}{9\pi} = 7.78 * 10^{-3}

    This might be completely wrong but hey worth a try,
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    haha wow, yeah you've gone about it a totally different way,

    I'm not really sure how the ratio part comes into the question, but then I'm not really in a position to say, since I have no idea myself O_o

    and yeah, we've just finished c3 and I can't remember eveer doing a question like this xP
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    Well I'm not sure how else to calculate the rate of change of Volume wrt height without getting radius out of the question...
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    do you know what the actual answer is?
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    (Original post by reaped)
    Well I'm not sure how else to calculate the rate of change of Volume wrt height without getting radius out of the question...
    V=(1/3)pir^2h

    dV/dh = (1/3)*pi [2r(dr/dh)h + r^2] ...using product rule and implicit differentiation.

    Using similar triangles, r=(3/4)h so dr/dh = 3/4. And when h=4, r=3.
    Evaluating will also give dV/dh = 9*pi

    Just much longer!
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    wait so im guessing you're right then reaped?
 
 
 
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