Final answer:
The rate at which the water level rises when the height of the water is 3m is found by differentiating the volume formula for the cone and solving for dh/dt given that the volume increases at 2 m³/min. The resulting rate is 2 / (9π) meters per minute.
Step-by-step explanation:
The question involves applying related rates, which is a concept in calculus that deals with determining how one variable changes about another variable over time. Given that the tank is shaped like an inverted cone with a radius (r) half the height (h), we have r = ½ h. The volume of a cone is given by V = ⅓ π r² h. Since water is being pumped at a rate of 2 m³/min, this is our dv/dt, and we are tasked with finding dh/dt when h = 3 m. To find this, we first express the volume in terms of one variable:
V = ⅓ π (½ h)² h = ⅓ π ⅔ h³
Now we differentiate both sides with respect to time t:
⅓ π ⅔ h³ → dv/dt = π h² dh/dt
We are given dv/dt = 2, and we are solving for dh/dt when h = 3 m:
2 = π (3)² dh/dt
2 = 9π dh/dt
dh/dt = 2 / (9π)
Therefore, the rate at which the water level rises when the height of the water is 3m is 2 / (9π) meters per minute.