Water is leaking out of the bottom of an inverted conical tank at the rate of 1/10 m^3/min, and at the same time is being pumped in the top at a constant rate of k m^3/min.

Question

Water is leaking out of the bottom of an inverted conical tank at the rate of 1/10 m^3/min, and at the same time is being pumped in the top at a constant rate of k m^3/min. The tank has a height of 6m and the radius at the top is 2m. Determine the constant k if the water level is rising at the rate of 15 m/min when the height of the water is 2m.

Answer 1

V = (1/3)pi(r^2)h

r=h/3

V= (1/3)pi(h/3)^2(h) = (pi/3)(h^2/9)h = (pi/27)h^3

V' = (pi/27)h^2(h')/2 = (pi/27)(2)^2(15)/2 = 20pi/9 m^3/minute

= k-.1

k = 20pi/9 + 1/10 = about 21.04 m^3/minute

Answer 2

dV/dt = k-.1 (Rate of accumulation of V is input - output rate

V = 1/3 πr²h

From the cone dimensions r = 2/6 h = h/3

V = (π/3) (h/3)²h = πh³/27 (subbing in for r in terms of hπ)

dV/dt = πh²/9 * dh/dt (Taking the derivative of V expression with time)

k-1 = πh²/9 * dh/dt (Using the first equation for dV/dt)

Now solve for k.as you are given h and dh/dt

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