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anonymous

  • 5 years ago

water is being poured into a cylindrical tank at 9 ft/sec. If the tank has a base radius of 5ft and a height of 10ft, how fast is water being poured in when the height is at 6ft?

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  1. anonymous
    • 5 years ago
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    V = pi * r^2 * h dV = pi * r^2 * dh dV/dt = pi * r^2 * dh/dt since dh/dt=9 , r=5, we subsititute these into that eqauation dV/dt = pi * 25 *9 = 225pi if you meant 9 ft^3 / sec then that would be dV/dt and you would be trying to find dh/dt so 9/25pi

  2. anonymous
    • 5 years ago
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    I'm so sorry!! I meant tank shaped like a cone.

  3. anonymous
    • 5 years ago
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    V = pi * r^2 * h/3 dV = pi *(2r * dr * h + r^2 *dh) also 5r = 10h so r = 2h and dr = 2dh dV = pi *(2(2h) * (2dh) * h + (2h)^2 *dh) divide both sides by dt and plug in 9for dh/dt, and 6 for h

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