Final answer:
The student's question about finding the rate at which the water level is rising in an inverted pyramid is addressed by using the concept of similar triangles, determining the volume formula in terms of the water level, and then applying calculus to find that rate. When the water level is 6 cm, the water level is rising at approximately 0.03255 cm/s.
Step-by-step explanation:
Rate of Increase in Water Level in an Inverted Pyramid
The problem presented involves finding the rate at which the water level is rising in an inverted pyramid when the water level is 6 cm high. The pyramid has a square top with sides of length 4 cm and a height of 9 cm. This is a classic problem of related rates in calculus, specifically regarding the volume of a pyramid, which is changing over time as water is added at a constant rate.
Volume of a pyramid: V = (1/3) * base area * height
We need to express the volume of water (V) in terms of the water level (h), so we need to find the relationship between the height and the base of the water surface as the water level changes. We can use similar triangles for the pyramid and the water at height h to find the side length (s) of the water surface. The ratio of the side lengths of the original pyramid to the water surface is the same as the ratio of their corresponding heights:
s/4 = h/9
s = (4/9)h
Now, we can express the volume of water (V) in terms of h:
V = (1/3) * s^2 * h = (1/3) * ((4/9)h)^2 * h = (64/243) * h^3
To find the rate at which the height of water increases (dh/dt), we differentiate with respect to time (t):
dV/dt = (64/243) * 3 * h^2 * dh/dt
Since the water is being poured at a constant rate, we know dV/dt = 25 cm^3/s. We can solve for dh/dt when h = 6 cm:
25 = (64/243) * 3 * (6)^2 * dh/dt
dh/dt = 25 / (((64/243) * 3 * 36)
dh/dt = 25 / (64 * 12)
dh/dt = 25 / 768
dh/dt ≈ 0.03255 cm/s
So, the rate at which the water level is rising when the water level is 6 cm is approximately 0.03255 cm/s.