Water is leaking out of an inverted conical tank at a rate of 10,000 cm3ymin 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 cmymin when the height of the water is 2 m, find the rate at which water is being pumped into the tank.

the rate at which water is being pumped into the tank dv/dt = 2.89×10^5cm/min

Step-by-step explanation: Please find the attached file for the solution

The rate at which water is being pumped into the tank = 289252.68 cm³/min or 2.893 x 10^(5) cm³/min

Step-by-step explanation:

In the question, we are given the following;

diameter = 4 m = 400cm

radius = 200 cm

height (h) = 6 m = 600 cm

dh/dt = 20 cm/min

rate out = 10000 cm³/min

Now, by property of similar triangles, we have;

h/r = 6/2

r = h/3

Now, volume at time is given as volume of cone which is;

v = (1/3)πr²h

Substitute h/3 for r;

= (1/3)π(h/3)²h

= (1/27)(πh³)

dv/dt = (3/27)(π)h² dh/dt = (1/9)(π)h² dh/dt

(rate of water into tank) - 10000 = (1/9)(πh²)dh/dt

rate into tank = 10000 + (1/9)(πh²)dh/dt

Substituting the relevant values we get;

rate into tank = 10000 + (1/9)(π) (200²)(20)

= 10000 + 800000(π)/9 = 10000 + 279252.68 = 289252.68 cm³/min or 2.893 x 10^(5) cm³/min

tangent

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