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October 23rd, 2007, 03:35 AM  #1 
Newbie Joined: Oct 2007 Posts: 4 Thanks: 0  i need this one for school today (please)
The base of a coneshaped tank is a circle of radius 5 feet, and the vertex of the cone is 12 below the base. The tank is filled to a depth of 7 feet, and water is flowing out of the tank at a rate of 3 cubic feet per minute. Find the rate of change of the depth of the water in the tank. 
October 23rd, 2007, 04:11 AM  #2 
Newbie Joined: Oct 2007 Posts: 4 Thanks: 0 
is it 0.11225 ft/sec

October 23rd, 2007, 05:32 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 19,977 Thanks: 1851 
Let h feet be the depth of the water in the tank and t minutes be elapsed time. The water's "open surface" has radius (5/12)h feet, so the volume of water in the tank is pi(5h/12)²h/3 cubic feet. The derivative of this with respect to t has to be 3 cubic feet per minute, so pi(25/144)h²dh/dt = 3, so dh/dt = −3(144/25)/(pi h²). Hence when h = 7, dh/dt = −3(144/25)/(49 pi). Hence the required value is −432/(1225 pi) feet per minute, i.e., . −0.11225 feet per minute (to five decimal places). (You'd probably give fewer decimal places in an exam.) 

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