Rate of change

a) The water is also in conical shape, being similar to the shape of the tank, i.e. the height is double the radius.

So we have:

V = πhr2/3

= πh(h/2)2/3

= πh3/12

dV/dt = (πh2/4) dh/dt

So when h = 6 (one-third full)

9 = (π x 62/4) dh/dt

dh/dt = 1/π m/min

b) When the cone is 1/8 full, by similar shape, the depth of water is 1/2 that of the tank, i.e. 9 m

Hence 9 = (π x 92/4) dh/dt

dh/dt = 4/(9π) m/min

(a)
Let V be the volume, r be the radius and h be the height of the water at the time t.

Considering the proportion, we obtain
r/9 = h/18
h = 2r

So, V = 1/3 (pie) r^2 h
= 1/3 (pie) (h/2)^2 h
= (pie)/3 (h^2 / 4) h
= (pie)/12 h^3

dV/dt = (pie)/12 * (3h^2) dh/dt
= (pie)/4 h^2 dh/dt

When the water level is one third up the cone,
h = 1/3 * 18m = 6 m
r = h/2 = 6/2 = 3 m

dV/dt = 9 = [(pie)/4 * (6)^2] dh/dt
9 = 9(pie) dh/dt
dh/dt = 1/(pie)

The water level is rising at a rate of 1/pie m per minute.


(b)
V = (pie)/12 h^3

When the cone is 1/8 full,
[(pie)/12 * h^3] / [(pie)/12 * 18^3 = 1/8
h^3 / 18^3 = 1/8
= (1/2)^3
h/18 = 1/2
h = 9 m

dV/dt = 9 = [(pie)/4 * (9)^2] dh/dt
dh/dt = 4 / 9(pie)

The water level is rising at a rate of 4 / 9(pie) m per minute