
Calculus Calculus Math Forum 
 LinkBack  Thread Tools  Display Modes 
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: 20,303 Thanks: 1974 
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.) 

Tags 
school, today 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Today's Computer Quiz ?  siva11  Computer Science  1  April 2nd, 2013 05:20 PM 
Today's Google Doodle  trapezodial  Physics  1  February 22nd, 2012 10:31 AM 
I need help please for today :)  Queen  Elementary Math  2  September 11th, 2009 11:42 AM 