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Math Help - One conical tank question and one sphere rate of change question

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    One conical tank question and one sphere rate of change question

    Hey guys, I have a few questions from my homework that I'm having trouble with. I would really appreciate some help! Here goes...

    1) Water is being poured into an inverted conical tank at a rate of 5000 cm^3/min. However, you also have a leak in your cone that is leaking water at a constant rate. The tank has height of 5m and the diameter at the top is 2m. If the water level is falling at a rate of 2cm.min when the height of the water is 1.5m, find the rate at which water is leaking out of your conical tank.

    2) The radius of a sphere is increasing at a rate of 4mm/sec. How fast is the volume increasing when the diameter is 80mm? how fast is the surface area increasing for the same diameter?
    Last edited by mr fantastic; July 17th 2011 at 09:44 PM. Reason: Title.
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    Re: One conical tank question and one sphere question. I need help!

    Quote Originally Posted by bennett2136 View Post
    1) Water is being poured into an inverted conical tank at a rate of 5000 cm^3/min. However, you also have a leak in your cone that is leaking water at a constant rate. The tank has height of 5m and the diameter at the top is 2m. If the water level is falling at a rate of 2cm.min when the height of the water is 1.5m, find the rate at which water is leaking out of your conical tank.
    first, note that V = \frac{\pi}{3}r^2 h

    make a sketch of the conical tank.

    get a proportional geometric relationship between the radius of the cone and the height ... solve for r in terms of h

    derive the volume of water in the tank strictly in terms of the height

    take the time derivative of your tank volume equation

    sub in your given values and determine the value of dV/dt = volume in - volume leaking


    2) The radius of a sphere is increasing at a rate of 4mm/sec. How fast is the volume increasing when the diameter is 80mm? how fast is the surface area increasing for the same diameter?
    V = \frac{4\pi}{3} r^3 ... you're given dr/dt and r ... take the time derivative and find dV/dt

    A = 4\pi r^2 ... same drill as above except find dA/dt
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