Question

A rectangular trough is 8ft long, 2ft across the top, and 4 feet deep. if water flows in at a rate of 2 ft 3 /min, how fast is the surface rising when the water is 1 ft deep?

Answer 1

 let V be the current volume of the trough and H the current height of the water 
then we have 
V = 16H 
taking derivative in terms of time we get 
dV/dt = 16 dH/dt 
we know that dV/dt = 2 
thus 
dH/dt = 2/16 = 1/8 
thus the water is rising at 
1/8 ft per minute 
or 
1.5 inches per minute