Questions: Water is leaking out of an inverted conical tank at a rate of 10,000 cm^3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank.

Solution Steps

To solve this problem, we need to apply the concept of related rates in calculus. The volume of water in the conical tank is changing due to both the leakage and the pumping of water. We will use the formula for the volume of a cone and differentiate it with respect to time to find the rate at which water is being pumped into the tank.

1. Express the volume of the cone as a function of the water height.
2. Differentiate the volume with respect to time to relate the rates of change of volume and height.
3. Use the given rates and the geometry of the cone to solve for the unknown rate at which water is being pumped into the tank.

The volume \( V \) of an inverted conical tank is given by the formula: \[ V = \frac{1}{3} \pi r^2 h \] where \( r \) is the radius at the water level and \( h \) is the height of the water. Given the dimensions of the tank, the relationship between the radius and height is: \[ r = \frac{2}{3} h \] Substituting this into the volume formula, we have: \[ V = \frac{1}{3} \pi \left(\frac{2}{3} h\right)^2 h = \frac{4}{27} \pi h^3 \]

To find the rate of change of volume with respect to time, we differentiate \( V \) with respect to \( t \): \[ \frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt} \] Calculating \( \frac{dV}{dh} \): \[ \frac{dV}{dh} = \frac{d}{dh} \left(\frac{4}{27} \pi h^3\right) = \frac{4}{9} \pi h^2 \] Thus, we have: \[ \frac{dV}{dt} = \frac{4}{9} \pi h^2 \cdot \frac{dh}{dt} \]

We know:

• \( \frac{dh}{dt} = 20 \, \text{cm/min} \)
• \( h = 200 \, \text{cm} \)
• Water is leaking out at a rate of \( 10,000 \, \text{cm}^3/\text{min} \), so \( \frac{dV_{\text{leak}}}{dt} = -10,000 \, \text{cm}^3/\text{min} \).

Substituting these values into the differentiated equation: \[ \frac{dV}{dt} = \frac{4}{9} \pi (200)^2 (20) = \frac{4}{9} \pi (40000) = \frac{160000}{9} \pi \, \text{cm}^3/\text{min} \]

The total rate of change of volume is the sum of the rate at which water is being pumped in and the rate at which it is leaking out: \[ \frac{dV_{\text{pump}}}{dt} = \frac{dV}{dt} - \frac{dV_{\text{leak}}}{dt} \] Substituting the values: \[ \frac{dV_{\text{pump}}}{dt} = \left(\frac{160000}{9} \pi + 10000\right) \] Calculating this gives: \[ \frac{dV_{\text{pump}}}{dt} = 10000 + \frac{160000}{9} \pi \]

Final Answer