Related Rates: Inverted Pyramid filled with water at 55cc's/sec. Top is a square with sides 2cm long, and the height is 10cm. Find the rate of water level rise when the water level is 7 cm.

Question

An inverted pyramid is being filled with water at a constant rate of 55 cubic centimeters per second. The pyramid, at the top, has the shape of a square with sides of length 2 cm, and the height is 10 cm.

Find the rate at which the water level is rising when the water level is 7 cm.

Answer 1

Here is a Physicist way of solving this. You won't see this in the usual Calculus texts.

y = depth of water (the "water level")

x = side of of the square water surface at depth y: x = (2/10) y = y/5 (by similar figures)

For time dt, a volume of water dV is poured in and it is added as a thin slab of thickness dy at the water surface, which is a square of side x:

dV = (slab area).(slab thickness) = x² dy = (y/5)² dy

Dividing by dt, instantly gives the rates:

dV/dt = (y/5)² dy/dt

You are given dV/dt = 55 cm³/sec, y = 7 cm. Solve for dy/dt (the answer will be in cm/s).

With this method, you don't need to write the total volume of the water and to differentiate it.

Answer 2

Assuming that the tank is initially empty.

As water is poured into the tank, the surface of the water at time t has the shape of a square with side length x. Let y be the depth of the water at time t and let V be the volume of water at time t

Given: dV/dt = 55. Find dy/dt when y = 7.

By similar triangles, 1/10 = (x/2) / y. So, x = (1/5)y.

V = (1/3)(Area of base)(height) = (1/3)x²y = (1/75)y³

dV/dt = (1/25)y²(dy/dt)

55 = (1/25)(7²)(dy/dt). So, dy/dt = (55)(25)/49 ≈ 28.06 cc/sec