innanetaholic08
New member
- Joined
- Jan 1, 2007
- Messages
- 5
The trough in the figure is 5 feet long and its vertical cross sections are inverted isosceles triangles with base 2 feet and height 3 feet. Water is being siphoned out of the trough at the rate of 2 cupic feet per minute. At any time T, let h be the depth and V be the volume of the water in the trough.
a) Find the volume of water in the trough when it is full.
I found this to be 15 cubic feet. I think that is right.
b) What is the rate of change in h at the instant when the trough is 1/4 full by volume?
I was thinking I find the volume of a fourth of the trough... but I'm not sure where to go from there.
c) What is the rate of change in the area of the surface of the water at the insant when the trough is 1/4 full by volume?
Well I know its something by five because the length won't change, I think I need to multiply by a fourth of three, maybe... so that would be five times .75 and get 3.75 as the surface area... or else I'm totally loss.
Thank you for your help, I really appreciate it.
a) Find the volume of water in the trough when it is full.
I found this to be 15 cubic feet. I think that is right.
b) What is the rate of change in h at the instant when the trough is 1/4 full by volume?
I was thinking I find the volume of a fourth of the trough... but I'm not sure where to go from there.
c) What is the rate of change in the area of the surface of the water at the insant when the trough is 1/4 full by volume?
Well I know its something by five because the length won't change, I think I need to multiply by a fourth of three, maybe... so that would be five times .75 and get 3.75 as the surface area... or else I'm totally loss.
Thank you for your help, I really appreciate it.