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    Q. A funnel (cone shaped) has a circular top of diameter 20 cm and a height of 30 cm. When the depth of liquid in the funnel is 15 cm, the liquid is dripping from the funnel at a rate of 0.2 cm^3 s^-1. At what rate is the depth of the liquid in the funnel decreasing at this instant?
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    (Original post by brut)
    Q. A funnel (cone shaped) has a circular top of diameter 20 cm and a height of 30 cm. When the depth of liquid in the funnel is 15 cm, the liquid is dripping from the funnel at a rate of 0.2 cm^3 s^-1. At what rate is the depth of the liquid in the funnel decreasing at this instant?
    Let:
    V = Volume of water
    h = Depth
    t = Time
    r = Radius of cone =

    dh/dt = dh/dv x dv/dt [Chain rule]

    But V = (1/3)πr²h
    Thus dV/dh = (1/3)πr²
    And so dh/DV = 3 ÷ πr²

    Work out the radius when h = 15 based on h=30 when 2r = 20:
    I think I'm right in saying that the diameter will halve, hence r = 5 at this point

    Therefore, using dh/dV = 3 ÷ 25π and dv/dt = 0.2 (from data given):

    dh/dt = (3 ÷ 25π) x 0.2

    dh/dt = 3 ÷ 125π


    I hope that's right, anyway
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    dV/dt = -0.2 (given in the question)
    V=h/3 * πr² (volume of liquid in the cone)

    Ratio r/h is constant (steepness of the cone). Initially r=10 and h = 30 therefore r = h/3.

    Substitute back into V.
    V=(h/3)³π
    Differentiating:
    dV/dh=(h/3)²π

    We want dh/dt, which is dh/dV * dV/dt (chain rule)
    So, at dV/dt = -0.2 and h=15:
    dh/dt = (-0.2)/((15/3)²π )
    =-1/125π

    Hope that helps, let me know if you need any of it clearing up.

    Edit: Beaten to it, this forum's quick. JohnSPals, it looks like you differentiated with respect to h treating r as a constant. I don't think you can do that because r is variable with h.
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    (Original post by Robert602)
    dV/dt = -0.2 (given in the question)
    V=h/3 * πr² (volume of liquid in the cone)

    Ratio r/h is constant (steepness of the cone). Initially r=10 and h = 30 therefore r = h/3.

    Substitute back into V.
    V=(h/3)³π
    Differentiating:
    dV/dh=(h/3)²π

    We want dh/dt, which is dh/dV * dV/dt (chain rule)
    So, at dV/dt = -0.2 and h=15:
    dh/dt = (-0.2)/((15/3)²π )
    =-1/125π

    Hope that helps, let me know if you need any of it clearing up.

    Edit: Beaten to it, this forum's quick. JohnSPals, it looks like you differentiated with respect to h treating r as a constant. I don't think you can do that because r is variable with h.
    Thanks, the answer was 0.0025 cm s^-1.
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    (Original post by Robert602)
    Edit: Beaten to it, this forum's quick. JohnSPals, it looks like you differentiated with respect to h treating r as a constant. I don't think you can do that because r is variable with h.
    Ahh. Fair enough, that seems like a good point

    I think I took it at that single point. The question said "at that specific point" so I treat the radius as being constant.
 
 
 
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