# rate of change

• Oct 22nd 2007, 04:33 PM
Mr_Green
rate of change
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.
• Oct 22nd 2007, 07:42 PM
DINOCALC09
Mr_green... do u go to scn high?

This one has got me stumped too =[
• Oct 22nd 2007, 08:15 PM
Jhevon
Quote:

Originally Posted by Mr_Green
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.

...at the instant the height is 7 feet?...
• Oct 22nd 2007, 08:21 PM
DINOCALC09
h= 7

dV/dt= -3

r= (5h)/12

V= (1/3)*pi*(r)^2 * h = (pi*25h^3)/(432)
• Oct 22nd 2007, 08:39 PM
Jhevon
Quote:

Originally Posted by DINOCALC09
h= 7

dV/dt= -3

r= (5h)/12

V= (1/3)*pi*(r)^2 * h = (pi*25h^3)/(432)

correct so far, now differentiate implicitly with respect to t
• Oct 23rd 2007, 02:11 AM
DINOCALC09
dont understand..
• Oct 23rd 2007, 02:52 AM
galactus
• Oct 23rd 2007, 03:02 AM
DINOCALC09
there are a few final steps that it fails to explain.
• Oct 23rd 2007, 03:08 AM
DINOCALC09