i am lost..rate of change- trough problem please help!

ihatecalc

New member
Joined
Sep 11, 2006
Messages
22
the problem reads:
the trough shown in the figure above is 5 feet long, and its vertical cross sections are inverted isosceles triangles with base 2 and height 3 feet. water is being siphoned out of the trough at the rate of 2 cubic feet per minute. at any time, t, let h be the depth and V be the volume of water in the trough.

a.) find the volume of water in the trough when it is full
---> i think i got this one right. check it out for me?
V = (1/2)(2)(3)(5)
V= 15 cubic ft.

b.) What is the rate of change h at the instant when the trough is 1/4 full by volume?

c.) What is the rate of change in the area of the surface of the water at the instant when the trough is 1/4 full by volume?

I don't know how to go about B and C. I know that dv/dt = 2, and that's about it. help me please!!
 
Top