



 
Hi Kimberly. The two "related rates" are the rate of change of volume and the rate of change of water level. To solve the problem, you are dealing with the volume of a cone, so you should know that V = (1/3) π r^{2} h. There are three variables in this, but we can reduce that to two, because the radius and the height are related (think of the similar triangles involved). Thus r = 2 m when h = 6 m. So r/h = 1/3, thus r = h/3. Now we substitute this into the formula for V. V = (1/3) π (h/3)^{2} h, which simplifies to just V = π h^{3} / 27. We are interested in rates, so now we take the derivative of this with respect to time t. dV / dt = ( π / 27 ) [ d(h^{3}) / dt ] = ( π / 27) (3h^{2}) ( dh / dt ) = ( π h^{2} ) / 9 ( dh / dt). The change in volume is dV / dt, which is the amount flowing in (call this I) minus the amount flowing out (12000 cm^{3}/min). Now we really need to pay attention to units. We are mixing meters and centimeters and can easily get confused. Let's stick with centimeters. So we have this (using centimeters): h is given as 2 m, which is 200 cm. dh / dt is given as 20 cm/min, so we substitute those in: I  12000 = ( π 200^{2} / 9 ) (20) And solve for I to finish the problem. Cheers,  


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