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    A funnel has a circular top of diameter 20cm and a height of 30cm. When the depth of liquid in the funnel is 15cm, the liquid is dripping from the funnel at a rate of 0.2cm^3 s^-1. At which rate is the dept of the liquid in the funnel decreasing at this instant?

    I have done:
    dx/dt = 0.2 when x = 15cm
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    (Original post by elvss567)
    A funnel has a circular top of diameter 20cm and a height of 30cm. When the depth of liquid in the funnel is 15cm, the liquid is dripping from the funnel at a rate of 0.2cm^3 s^-1. At which rate is the dept of the liquid in the funnel decreasing at this instant?

    I have done:
    dx/dt = 0.2 when x = 15cm
    What is x?? Define your variables first before using them.

    Let h be the depth of the liquid, and t be the time in seconds. Then we want to find \frac{dh}{dt}.

    Let x denote the volume of liquid at a given time. Then we know at some point, we have \frac{dx}{dt}=0.2.

    However, note that by the chain rule, we have \frac{dh}{dx} \cdot \frac{dx}{dt}=\frac{dh}{dt}

    So you want to find \frac{dh}{dx} which is the rate of change of the depth w.r.t the volume at the given time.
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    (Original post by elvss567)
    A funnel has a circular top of diameter 20cm and a height of 30cm. When the depth of liquid in the funnel is 15cm, the liquid is dripping from the funnel at a rate of 0.2cm^3 s^-1. At which rate is the dept of the liquid in the funnel decreasing at this instant?

    I have done:
    dx/dt = 0.2 when x = 15cm
    Check the sign of \frac{dx}{dt} - as above, it's really \frac{dV}{dt}, if you're using x for the height of liquid.

    It looks like they want you to assume a zero width at the bottom of the funnel. Not nice.
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    (Original post by RDKGames)
    What is x?? Define your variables first before using them.

    Let h be the depth of the liquid, and t be the time in seconds. Then we want to find \frac{dh}{dt}.

    Let x denote the volume of liquid at a given time. Then we know at some point, we have \frac{dx}{dt}=0.2.

    However, note that by the chain rule, we have \frac{dh}{dx} \cdot \frac{dx}{dt}=\frac{dh}{dt}

    So you want to find \frac{dh}{dx} which is the rate of change of the depth w.r.t the volume at the given time.
    Hi, i followed what you were saying up until the bit where i have to find dh/dt. How do i find dh/dt? Do i have to differentiate the volume or the surface area of the sphere to find this?
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    (Original post by elvss567)
    Hi, i followed what you were saying up until the bit where i have to find dh/dt. How do i find dh/dt? Do i have to differentiate the volume or the surface area of the sphere to find this?
    Sorry, ignore those last two lines, it doesn't go in the direction I thought it would. The expected approach would be to note that x=\frac{\pi}{3} r^2 h which is the volume of a cone (and r is the radius)

    Then determine what r is in terms of h at a given time, thus substitute it in to get x(h). Then differentiate both sides w.r.t t, applying the chain rule for the RHS. This leaves you with an equation to rearrange for \frac{dh}{dt}
 
 
 
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