Question

Water is being withdrawn from a conical reservoir 2 feet in radius and 10 feet in depth (the point of the cone is down) at a rate of 4 ft^3/min. how fast is the area of the surface of the water changing when the water is 6 feet deep

Answer 1

This is question of differential change or d/dt change by similar triangle we know r/h=2/10 so when h= 6 r=1.2 for cone v=.33*pi*r^2*h and area a=pi*r^2 given dv/dt=-4/3 so dv/dt=(4/300)*pi*3*h^2*dh/dt equating above two values for h=6 dh/dt=-100/(81*pi) ft/min so for area da/dt=-4/3 ft^2/min