a trough has a triangular cross section. The trough is 6 ft across the top, 6 ft deep, and 16 ft long. Water is being pumped into the trough at the rate of 4ft^3 per minute. Find the rate at which the height of the water is increasing at the instant that the height is 4 ft.
A similar problem that might be of interest is discussed at this thread.
An acknowledgement - Thanks for that!, say - of the given solution, followed by asking
would not be inappropriate here.
Also: TheEmptySet did all the work! Seriously - if you've taken the effort to actually understand his/her solution (as opposed to just copying it out) surely you can do that one small thing and work out the unit yourself.
I ask: How can dx/dt possibly have a unit of volume?
May I show you a short cut to these types of related rates problems?.
By making the observation that the rate of change of volume is equal to the cross-sectional area at that instant times the rate of change of the height.
We know that dV/dt=4.
The area of the water surface, A(t), when it is 4 feet deep is just a rectangle with area 16*4=64.
So, we have
see - this all looks like gibberish to me. My mind can not conceptualize math terms whatsoever. I have a 4.0, and calculus is killin me right now... i've never had something i couldn't figure out, but calculus baffles me. And it's extremely frustrating. So thanks for all your help. I only hope I can recreate it on the final. *crosses fingers*