. 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.
(a) To express the volume V of the water in the tank as a function of h, we need to find the formula for the volume of a frustum.
The formula for the volume of a frustum of a cone is given by:
V = (1/3) * π * (r1^2 + r2^2 + r1 * r2) * h
where r1 and r2 are the radii of the top and bottom circles of the frustum, and h is the height of the frustum.
In this case, r1 = 80 cm = 0.8 m, r2 = 36 cm = 0.36 m, and the height of the frustum is given by h cm.
Plugging in the values, we get:
V = (1/3) * π * (0.8^2 + 0.36^2 + 0.8 * 0.36) * h
V = (1/3) * π * (0.64 + 0.1296 + 0.288) * h
V = (1/3) * π * 1.0576 * h
V = 0.352 * π * h
Therefore, the volume V cm of the water in the tank as a function of h is given by V = 0.352 * π * h.
To find the rate of change of V in terms of h, we take the derivative of V with respect to h:
dV/dh = 0.352 * π
Therefore, the rate of change of V in terms of h is constant and equal to 0.352 * π cm.
(b) The water flows out of the tank at a rate of 200Vh cm/s.
The rate of change of volume with respect to time is given by dV/dt = -200Vh cm/s, where the negative sign indicates that the volume is decreasing.
The water tap fills the tank at a constant rate of 600 cm^3/s, which can be written as 0.6 cm/s.
Therefore, the rate of change of volume with respect to time is given by:
dV/dt = 0.6 - 200Vh
(c) To find the minimum rate of change of h, we need to find the critical points of h. To do this, we set dV/dt = 0:
0.6 - 200Vh = 0
200Vh = 0.6
h = 0.003/V
To find the minimum rate of change of h, we differentiate the rate of change of volume with respect to time with respect to h and set it equal to 0:
(dV/dt) = -200Vh
d(dV/dt)/dh = -200V
Therefore, the minimum rate of change of h occurs when d(dV/dt)/dh = -200V equals 0, which means that V = 0.
(d) The student suggests that when the rate of change of h is minimum, the volume of water in the tank is also minimum. However, this is not correct.
From part (c), we found that the minimum rate of change of h occurs when V = 0. This means that there is no water in the tank.
So, when the rate of change of h is minimum, the volume of water in the tank is actually 0, not minimum. Therefore, the student's suggestion is incorrect.
(a) To express the volume V cm of the water in the tank as a function of h, we need to use the formula for the volume of a frustum. The formula is:
V = (1/3) * π * h * (r1^2 + r1*r2 + r2^2)
where h is the height of the water, r1 is the radius of the top circle (80 cm), and r2 is the radius of the bottom circle (36 cm).
Substituting the given values:
V = (1/3) * π * h * (80^2 + 80*36 + 36^2)
Simplifying the equation:
V = (1/3) * π * h * (6400 + 2880 + 1296)
V = (1/3) * π * h * 10576
V = π * h * 3525.33
So the volume V cm of the water in the tank as a function of h is V = π * h * 3525.33.
To find the rate of change of V with respect to h (dV/dh), we can take the derivative of the volume function with respect to h:
dV/dh = π * 3525.33
So the rate of change of V in terms of h is dV/dh = π * 3525.33.
(b) The water flows out of the tank at a rate of 200Vh cm's' and the water tap fills the tank at a constant rate of 600 cm's'. To find the rate of change of h as a function of h, we need to subtract the outflow rate from the inflow rate:
rate of change of h = inflow rate - outflow rate
rate of change of h = 600 - 200Vh
Substituting the value of V from part (a), we get:
rate of change of h = 600 - 200(π * h * 3525.33)
rate of change of h = 600 - 7050.66πh
So the rate of change of h as a function of h is rate of change of h = 600 - 7050.66πh.
(c) To find the minimum rate of change of h, we need to find the critical points of the rate of change function. We can do this by taking the derivative of the rate of change function with respect to h and setting it equal to zero:
d(rate of change of h)/dh = -7050.66π = 0
Solving for h:
h = 0
So the minimum rate of change of h occurs at h = 0.
(d) No, I disagree with the student. The minimum rate of change of h does not necessarily mean that the volume of water in the tank is also minimum. The minimum rate of change of h only indicates the point where the water level is changing at the slowest rate. The volume of water in the tank depends on the height of the water and the shape of the tank, so it is not directly related to the rate of change of h.
Not quite sure what "80 cm- 450 h m- 36 cm." means.
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...