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  • 3 years ago

A vessel in the form of an inverted cone of height 10 feet and semi vertical angle 30 is filled with a solution. this solution is drained through an orifice at its bottom into a cylindrical beaker of radius sqrt6 feet in such a way that the height of the solution in the conical vessel decreases at a uniform rate of 2 inch/min. Find the rate at which the height of the solution increases in the beaker when the height of the solution column in the conical vessel is 6 ft?

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  1. experimentX
    • 3 years ago
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    |dw:1359555212522:dw| is this the figure?

  2. experimentX
    • 3 years ago
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    \[ V = \frac 1 3 \pi R^2 h, \;\;R = h \tan 30 \\ \frac{dV}{dt} = \frac 1 3 \pi R^2 \frac {dh}{dt}\] Put h = 6, this will give you rate of increase of volume. do same for cylinder.

  3. experimentX
    • 3 years ago
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    and put dh/dt = 2

  4. shubhamsrg
    • 3 years ago
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    The semi vertical angle is 30 @experimentX

  5. shubhamsrg
    • 3 years ago
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    |dw:1359560975070:dw|

  6. experimentX
    • 3 years ago
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    oh .. my bad!!

  7. shubhamsrg
    • 3 years ago
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    So the working eqns will be : V = pi/3 R^2 h => dV/dt = pi/3 (2R) (dR/dt) (dh/dt) We know cot30 = h/R => dR/dt cot30 = dh/dt Make that substitution. For cylinder, V = pi r^2 h dV/dt = pi r^2 dh/dt equate both dV/dt 's so find dh/dt for cylinder.

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