Question

The radius of a conical tank is 2.7 meters and the height of the tank is 4.3 meters. Water is flowing into the tank at a constant rate of 59.5 m3/minute. At the instant the the depth of the water is 0.6 meters, answer the following:

(A) At what rate is the depth of the water changing?

(B) At what rate is the radius of the waters surface changing?

(C) At what rate is the surface area of the exposed water changing?

Answer 1

depth = h
dV/dh = area of surface (draw it :)

dV/dh = pi r^2 = surface area

r = (2.7/4.3)h

remember to use chain rule.
for example
dV/dt = 59.5 m^3/min
but
dV/dt = dV/dh dh/dt
so
dh/dt = 59.5/ (pi r^2)
etc