water is pouring into a conical cistern at the rate of 8 m^3/minute. If the height of the inverted cone is 12 meters and the radius of its circular opening is 6 meters, how fast is the water level rising when the water is 4 meters depth?

Answer 1

it is clear that h = 2r, so

v = 1/3 πr^2 h = π/12 h^3

dv/dt = π/12 h^2 dh/dt
so, now just crank it out:

π/12 * 8^2 dh/dt = 8
dh/dt = 3/(2π) m/min

Answer 2

we know dv/dt, we want dh/dt
V= 1/3 (pi)r^2 h
we need to get the value of r. how? let's get the ratio of
r/h = 6/12
so, r= h/2
then Substitute
V= 1/3 (pi) (h/2)^2 h
V= 1/12 (pi) h^3

then differentiate
dv/dt= pi/4 h^2 dh/dt
sub
8 = pi/4 (16)dh/dt
8 / 4pi =dh/dt
2/pi = dh/dt