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Water is leaking out of an inverted conical tank at a rate of $ 10,000 cm^3/min $ at the same time that water is being pumped into the tank at a constant rate. The tank has height $ 6 m $ and the diameter at the top is $ 4 m. $ If the water level is rising at a rate of $ 20 cm/min $ when the height of the water is $ 2 m, $ find the rate at which water is being pumped into the tank.

$$289.253 \mathrm{cm}^{3} / \mathrm{min}$$

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Campbell University

University of Michigan - Ann Arbor

University of Nottingham

Idaho State University

in order to secure up the rate at which the water is being pumped. We have to recall the volume of a cone is pi r squared times h over three. You know, 1/3 h. So let's plug in what we know. Remember ours h over three. Therefore, we have pi times h over three square exists pi r squared times h over three which gives us high over 27 cause three scores nine men times three is 27 h cubed which gives us Devi over. DT is pi over nine page squared times D h over DT. So some are taking the derivative, as you can see which gives us P minus 10,000 equals high over nine times 200 squared times 20 which simplifies to 2.89 times 10 to the fifth centimeters cubed per minute is our unit for this rate here