A water tank has the shape of an inverted right circular cone with its axis vertical and vertex lowermost. Its semi-vertical angle is tan inverse(0.5). Water is poured into it at a constant rate of 5 cubic metre per hour. Find the rate at which the level of water is rising at the instant when the depth of water in the tank is 4 m.

Dear Student,We know, tanα = rh .......1Here, r - radiush- hieghtα-semi vertical angle Also, α=tan-10.5Thus, Tan(tan-10.5)=rh ...........Using 10.5=rhThus, r=h2Also, Volume of Cone, V = 13πr2h= 13πh22(h)= 112πh3Now, dVdt=dVdh×dhdt..............Using Chain RuledVdt=d112πh3dr×dhdt=14πh2dhdtNow, It is given that Water is poured into it at a constant rate of 5 cubic metre per hourThus, dVdt=5 m3/hrThus, 14πh2dhdt=5dhdt=20πh2Thus, dhdt at h=4mdhdt=20π(4)2=54π=0.39 m/hThus, At the instant when h=4m , Water level is rising at a rate of 0.39m per hour.



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