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 6 cm, and the height is 9 cm.
Find the rate at which the water level is rising when the water level is 4 cm.
Using similar triangles, we see that the side of the cross-section is always 2/3 the height of the pyramid.
when the water is y cm deep, the area of the surface of the water is 4/9 y^2
so now we have
v = 1/3 (4/9 y^2)*y = 4/27 y^3
dv/dt = 4/9 y^2 dy/dt
when y=4,
that gives us
4/9 * 4^2 dy/dt = 25
dy/dt = 225/64 ≈ 3.51 cm/s