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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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experimentXBest ResponseYou've already chosen the best response.0
dw:1359555212522:dw is this the figure?
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
\[ V = \frac 1 3 \pi R^2 h, \;\;R = h \tan 30 \\ \frac{dV}{dt} = \frac 1 3 \pi R^2 \frac {dh}{dt}\] Put h = 6, this will give you rate of increase of volume. do same for cylinder.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
The semi vertical angle is 30 @experimentX
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
dw:1359560975070:dw
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
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.
 one year ago
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