Water is being drained from a cone-shaped reservoir 10 ft. in diameter and 10 ft. deep at a constant rate of 3 ft3/min. How fast is the water level falling when the depth of the water is 6 ft?

1 Answer
Apr 14, 2015

The ratio of radius,#r#, of the upper surface of the water to the water depth,#w# is a constant dependent upon the overall dimensions of the cone
#r/w = 5/10#
#rarr r=w/2#
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The volume of the cone of water is given by the formula
#V(w,r) = pi/3 r^2w#
or, in terms of just #w# for the given situation
#V(w) = pi/(12)w^3#

#(dV)/(dw) = pi/4w^2#
#rarr (dw)/(dV) = 4/(piw^2)#

We are told that
#(dV)/(dt) = -3# (cu.ft./min.)

#(dw)/(dt) = (dw)/(dV)*(dV)/(dt)#

#= 4/(piw^2)*(-3)#

#=(-12)/(piw^2)#

When #w=6#
the water depth is changing at a rate of
#(dw)/(dt)(6) = = (-12)/(pi*36) = -1/(3pi)#

Expressed in terms of how fast the water level is falling, when the water depth is #6# feet, the water is falling at the rate of
#1/(3pi)# feet/min.