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    a vessel in the shape of a cone is standing on its apex. water flows in at a steady rate, of 1m^3 per minute. the vessel has a height of 2m and a diameter of 2m when the vessel is 1/8 full find the rate at which the free surface area of the water is increasing. im trying to use the chain rule but dont know what do
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    (Original post by markosheehan)
    a vessel in the shape of a cone is standing on its apex. water flows in at a steady rate, of 1m^3 per minute. the vessel has a height of 2m and a diameter of 2m when the vessel is 1/8 full find the rate at which the free surface area of the water is increasing. im trying to use the chain rule but dont know what do
    The free-surface area:

    A = \pi r^2.

    The volume of water as a function of time:

    V(t) = \dfrac{t}{60}.

    Equate the volume of the water at the time when the vessel is one-eighths full with the volume of the water as a function of height of the cone:

    V(h) = \int_0^h \pi r^2 \ dh, where h and H are the variable and fixed height of the cone respectively.

    Use the relation r = \dfrac{h}{H}R.

    Apply the chain rule:

    \dfrac{dA}{dt} = \dfrac{\partial A}{\partial h} \cdot \dfrac{dh}{dt}.
 
 
 
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