At what rate is water pouring into a rectangular bathtub with base 22 ft2 22 ft 2 if the water level rises at a rate of 0.5 0.5 ft/min.

Answer 1

Since this is a rectangular container the only information needed to know the rate at which water is poured in is the rate at which the water is rising.

For this shape V=Bh and

dV/dt=Bdh/dt

Since B and dh/dt are given

Ww have:

dV/dt=22ft²(0.5ft/min)

dV/dt=11ft³/min

Answer 2

Your bathtub is a rectangular prism, which means the volume is the area of the base times the height.

V = B h

Since the base is a constant rectangle which doesn't change regardless of how high up the bathtub you go, and liquid takes the shape of its container, then your water will also take the shape of a rectangular prism whose volume depends on the height of the water, but the base is a constant 22.

V = 22 ft² h

Implicitly differentiating everything with respect to time, we get

dV/dt = 22 ft² dh/dt

Since I think you wrote that dh/dt = 0.5 ft/min, (water level is rising), then

dV/dt = 22 ft² (0.5) ft/min = 11 ft³ /min

It's a good idea to include units in everything because you will know you are on the right track if the units you end up with are the correct units.