. A water tank is in the form of an inverted frustum as shown in Figure Q5: The radii of the top and the bottom circles are 80 cm and 36 cm respectively. Let h cm be the height of the water in the tank at time t second (s): 80 cm- 450 h m- 36 cm. Figure Q5 (a) Express the volume V cm of the water in the tank as a function of h. Hence, find the rate of change of V in terms of h and the rate of change of h. (4 marks) (b) There is a hole at the bottom of the tank so that the water flows out of the tank at a rate of 200Vh cm's'. In the mean time, a water tap fills the tank at a constant rate of 600 cm's'. Find the rate of change (4 marks) of h as a function of h. (c) Hence, find the minimum rate of change of h. (10 marks) (2 marks) (d) A student suggests that when the rate of change of h is minimum, the volume of water in the tank is also minimum. Do you agree with the student? Explain your answer.
1 answer
I'll take it to mean that the distance between the bases is 450. If so, then let H be the height of the entire cone, so using similar triangles,
H/80 = (H-450)/36
H = 9000/11 ≈ 818
That makes the height of the missing tip 818-450 = 368
That seems a bit odd, but maybe I misinterpreted the explanation. If so, maybe you can follow my logic with the real numbers.
(You know, your notation is weird. "200Vh m's'" ??? Why not just say 200 cm^3/s)
In fact, you seem to b e mixing units. A tank with radius 80cm and height 450m ? I doubt it. I'm assuming cm throughout.
So, the volume of the missing tip of the cone is
1/3 π * 36^2 * 368 = 238464 cm^3
(a) consider the frustrum as the base of a cone with base radius 80 and height H.
Using similar triangles again, the radius of the surface of the water can be found using
(h+368)/450 = r/80
r = 8/45 (h+368)
So the volume of water when it has height h is
V = 1/3 π * (8/45 (h+368))^2 * (h+368) - 238464 = 64/6075 (h+368)^3 - 238464
(b) now you can find dV/dh and answer (c)
(d) Since dV/dt is constant, the water level rises more slowly as the tank fills...