Question

A water tank has the shape of an inverted circular cone with base radius 2m and height 4m. If water is pumped into the tank at a rate of 2m^3/min, find the rate at which the water level is rising when the water is 3m deep.

Answer 1

The rate at which the water level is rising when the water is 3m deep is approximately 0.119 m/min.

Step-by-step explanation:

To find the rate at which the water level is rising, we need to use the concept of related rates. Let's start by finding the volume of the water in the tank.

Since the tank has the shape of an inverted circular cone, we can use the formula for the volume of a cone:

V = (1/3)πr^2h, where r is the base radius and h is the height of the cone.

Plugging in the values, we get V = (1/3)π(2^2)(4)

= (16/3)π m^3.

Now, let's differentiate both sides of the equation with respect to time: dV/dt = (16π/3)*(d/dt)(h).

We know that the rate at which water is being added to the tank is 2 m^3/min, so dV/dt = 2 m^3/min.

Plugging in this value, we get 2 = (16π/3)*(d/dt)(h).

Now, we can solve for (d/dt)(h), which represents the rate at which the water level is rising. Rearranging the equation, we get (d/dt)(h) = (2*3)/(16π)

= 3/(8π) m/min.

Therefore, the water level is rising at a rate of approximately 0.119 m/min when the water is 3 m deep.