Answer:
Rate at which water is being pumped into the tank, [tex]R_{in} = 429933.33 cm^3/min[/tex]
Explanation:
Rate at which water is leaking out, [tex]R_{out} = 11, 0000 cm^3/min[/tex]
Height of the tank, h = 12 m
The top diameter of the tank, d = 8 m = 800 cm
The top radius of the tank, r = d/2 = 800/2 = 400 cm
The rate of change of water height, dh/dt = 30 cm/min
Height of water = 2 m
By carefully observing the diagram contained in the file attached to this solution, using the property of similar triangle:
h/r = 12/4
r = h/3
Since the tank is conical, volume of the water at time, t will be:
[tex]V = \frac{1}{3} \pi r^{2} h\\\\V = \frac{1}{3} \pi (\frac{h}{3} )^{2} h\\\\V = \frac{1}{27} \pi h^{3} \\[/tex]
Finding the derivative of the above with respect to t to get the rate of change in the volume of water.
[tex]\frac{dV}{dt} = \frac{1}{9} \pi h^{2} \frac{dh}{dt} \\\\\frac{dV}{dt} = R_{in} - R_{out} \\\\\frac{1}{9} \pi h^{2} \frac{dh}{dt} = R_{in} - R_{out}\\\\\frac{1}{9} \pi * 200^{2}* 30 = R_{in} - 11000\\\\ R_{in} = 429933.33 cm^3/min[/tex]
