My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Reply
 
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 cone-shaped 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.
croatiandan is offline  
 
October 23rd, 2007, 04:11 AM   #2
Newbie
 
Joined: Oct 2007

Posts: 4
Thanks: 0

is it -0.11225 ft/sec
croatiandan is offline  
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.)
skipjack is offline  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
school, today



Search tags for this page
Click on a term to search for related topics.
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





Copyright © 2018 My Math Forum. All rights reserved.