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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
Yahoo! Group Title
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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experimentX Group TitleBest ResponseYou've already chosen the best response.0
dw:1359555212522:dw is this the figure?
 one year ago

experimentX Group TitleBest 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

experimentX Group TitleBest ResponseYou've already chosen the best response.0
and put dh/dt = 2
 one year ago

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

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

experimentX Group TitleBest ResponseYou've already chosen the best response.0
oh .. my bad!!
 one year ago

shubhamsrg Group TitleBest 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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