Quandaries
and Queries 

The base of a pyramidshaped
tank is a square with sides of length 9 feet, and the vertex of the pyramid
is 12 feet above the base. The tank is filled to a depth of 4 feet, and
water is flowing into the tank at a rate of 3 cubic feet per second. Find
the rate of change of the depth of water in the tank. (Hint: the volume
of a pyramid is V = 1/3 B h , where B is the base area and h is the height
of the pyramid.) Thanks in advance. 

Hi Annette, I drew a diagram of your tank with the assumption that pyramidshaped means that the vertex of the pyramid is above the center of the square base.I labeled three of the points, the vertex V, the point on the base directly below the vertex P and the midpoint of one of the sides of the base Q. The right triangle VPQ has height VP = 12 ft, and base PQ = ^{9}/_{2} ft.If h(t) is the height of the water at time t seconds then I can sketch the triangle and label it. In the diagram AB is the waterline. Triangles VPQ and VAB are similar and hence
and hence
The top surface of the water has area (2 s(t))^{2} = 4 s(t)^{2} and hence the volume of water in the tank is
Hence
V'(t) = 3 cubic feet per second and h(t) = 4 feet when you want to find h'(t), thus you can solve for h'(t). I hope this helps, 

