A trough is 12 ft long and 3 ft across the top Its ends are isosceles triangles with altitudes of 3 ft If water is being pumped in at 2 cubic feet per min how fast is water rising when h is 1 ft deep?

Answer 1

Since the altitude of the triangle on the sides of the trough are equal, the area covered when the water is h feet deep is 1/2*h².

The volume of the trough filled at a given height will be given by V=B*h=h²/2*12=6h².

If you differentiate both sides of the above equation with respect to time you get

dV/dt=12*h*dh/dt

The rate of change of the volume is given to be 2 cubic feet per minute, and we are asked to determine the change of height when h=1

2=12*2*dh/dt

dh/dt=1/12 feet/minute