Water is being withdrawn from a conical reservoir 2 feet in radius and 10 feet in depth (the point of the cone is down) at a rate of 4 ft^3/min.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
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
