Calculus
posted by EA on .
Jesse has constructed a huge cylindrical can with a diameter of 60 ft. The can is being filled with water at a rate of 450 ft3/min. How fast is the depth of the water increasing? (Hint: The volume of water in the cylinder is determined by πr2h where r is the radius and h is the depth of the water )

the rate of change of volume is the surface area times the rate of change of height
450 ft^3/min = surface area * dh/dt
surface area = pi r^2
450 = pi (30^2) dh/dt
dh/dt = .159 ft/min
You could do this by saying
V = pi r^2 h
dV/dh = pi r^2
dV/dh*dh/dt = pi r^2 dh/dt
chain rule
dV/dt = pi r^2 dh/dt
but most of us just look at the lake rising an inch and seeing that the added volume is the area times one inch 
Thanks Damon, that really clears it up for me