Question

A trough is 12 ft long and 3 ft across the top. Its ends are isosceles triangles with an altitude of 3 ft. Water is being pumped into the trough at 2 cubic feet per minute. How fast is the water level rising when it is one foot 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

@speedygonzales

Answer 2

Thanks!!!