** water runs out of an inverted conical tank at a rate of cubic centimeters per min at the same time? **

Water emerges from an inverted conical vessel with a rate of 8600.00 cm Â³ per minute at the same time that water is pumped into the vessel at a constant rate. The tank has height of 7,000 meters and the diameter at the top is 5.5000 meters. If the water level of 25,000 centimeters per minute increases as the height of the water column is 2.000 m, find the rate is pumped to the water in the tank in cubic centimeters per minute ** Best Answer (s):. **

* reply by Lawrence *

Given … R = D / 2 = 5.5 m / cm 2 = 550/2 = 275 cm H = 7 m = 700 cmh = 2m = 200 CMDh / dt = 25 cm / MinDV / dt (out) = 8600 cc / min —————— Apply similar triangles … r / R = h / Hr/275 = h/700dannr = 11/28 hThe tank volume is defined by … V = Ï€ r Â² h / 3To the dependent variable, r eliminate, substitute r = 11/28 hV = (Ï€ / 3) (11 / 28 h) Â² (h) V = (Ï€ / 3) (11/28) Â² (h) Â³ V = (121Ï€ / 2352) (h) Â³ Take derivative dV / dh findendV / dh = (363Ï€ / 2352) (h Â² ) Since dV / dt = (dV / dh) (dh / dt), danndV / dt (tank) = (363Ï€ / 2352) (h Â²) (dh / dt) dV / dt (tank) = (363Ï€ / 2352) (200 cm Â² Â²) (25 cm / min), dV / dt (tank) = (363Ï€ / 2352) (40000 cm Â²) (25 cm / min), dV / dt (tank) = (363Ï€ / 2352) (cm 1000000 / min Â³) dV / dt (tank) = 484,863 cm / min Â³, since dV / dt (tank) = dV / dt (in) – dV / dt (out), danndV / dt (in) = dV / dt (Tank) + dV / dt (out) dV / dt (in) = 484 863 cm Â³ / min + 8600 cc / MinDV / dt (in) = 493 463 cm Â³ / min

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