An inverted pyramid is being filled with water at a constant rate of 25 cubic centimeters per second.

Question

An inverted pyramid is being filled with water at a constant rate of 25 cubic centimeters per second. The pyramid, at the top, has the shape of a square with sides of length 3 cm, and the height is 6 cm. Find the rate at which the water level is rising when the water level is 4 cm.

Please, I need help with this question.

Answer 1

You've got a related rates problem here, so the first thing we want to do ( after drawing a diagram of the inverted pyramid) is to identify all the known info and the unknown part for which we want to solve.

For the rates, dv/dt = 25 cm³/sec and dh/dt is the unknown rate we need to solve for. Our "freeze-frame" moment ( that particular point at which we want to find the unknown rate) is when h = 4 cm.

Now we set up our key equation that will tie in all the known and unknown rates together. Since we're dealing with a pyramid filling with water, the key formula is the volume formula for a pyramid: V = 1/3 B h, where B = s² with "s" being the side of the square base. Before we can take the derivative of our key formula, however, we need to deal with the fact that we have too many variables in the key formula, as the "B" value will need to be expressed in terms of "h". From the given information, we see that the ratio between the height of the pyramid and the length of the side of the pyramid is 6:3 or 2:1, so that, at any point in the pyramid, the ratio of h/s = 2/1. Solving this ratio for s, we have s = 1/2 h, so now the volume formula for the pyramid can be rewritten as V = 1/3 (1/2h)²h or, more simply, V = 1/12h³.

Now we are ready to take the derivative of both sides: dV/dt = 1/4h² dh/dt. We then plug in the values that we have and solve for the unknown value: 25 = 1/2(4)²dh/dt. So then dh/dt = 25/8 cm/sec.