A conical water tank with its vertex at bottom has radius 20m and depth 20m.Water is being pumped into the tank at rate 40m^3/min.how fast is the level of the water rising when water is 8m deep? ( v=(pi x r² x h)/3 )

Question

1604 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-12T16:34:12

The level is rising at the rate of #=0.199m/(mn)#

The volume of water in the tank is

#V=pi/3r^2h#

The volumetric flow is #q=(dV)/dt=40m^3mn^-1#

In the cone

#r/20=h/20#

#=>#, #r=h#

Therefore,

#V=pi/3r^2h=pi/3h^3#

Differentiating wrt #t#

#(dV)/dt=pi/3*3h^2*((dh)/dt)#

#h=8m#

Plugging in the values

#40=pi/3*3*8^2*((dh)/dt)#

#((dh)/dt)=40/(64pi)=5/(8pi)=0.199m/(mn)#