Question
Application of Integrals
Water in an open bowl evaporates at a rate proportaional to the area of the surface of the water. (This means that the rate of decrease of the volume ispropotional to the area of the surface.) Show that the depth of the water decreases at a constant rate, regardless of the shape of the bowl.
Answers
At any height "h" of water level, let the area be A(h) which depends on h for arbitrary shape
V= volume of water
so, dV/dt= change of volume per unit time
= -A(h)*c where c= constant ( because this rate is proportional to A(h) )
and dV/dt= -A(h) * (dh/dt) where dh/dt= rate of change of height
so,
-A(h) *dh/dt = -c*A(h) or, dh/dt = c = constant
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