# P3 - Connected Rates of ChangeWatch

This discussion is closed.
#1
Could someone explain how this is done? I looked at the mark scheme and it looks confusing.

A cylindrical tank with a horizontal circular base is leaking. At time t minutes the depth of oil in the tank is h metres. It is known that h=10 when t=0 and that h=5 when t=40.

Alan assumes that the rate of change of h with respect to t is constant.

a) Find an expression for h in terms of t.

Bhavana assumes that the rate of change of h with respect to t is proportional to h.

b) Form a differential equation and, using the conditions given, solve it to find Bhavana's expression for h in terms of t.

c) Find, in each case, the value of h when t=60.

d) Briefly explain which assumption you would use.
0
13 years ago
#2
A cylindrical tank with a horizontal circular base is leaking. At time t minutes the depth of oil in the tank is h metres. It is known that h=10 when t=0 and that h=5 when t=40.

Alan assumes that the rate of change of h with respect to t is constant.

a) Find an expression for h in terms of t.
dh/dt = k
-> dh = k dt
-> h = kt + c

h = 10, t = 0:
-> 10 = c

h = 5, t = 40:
-> 5 = 40k + 10
-> 40k = -5
-> k = -1/8

-> h = 10 - t/8

Bhavana assumes that the rate of change of h with respect to t is proportional to h.

b) Form a differential equation and, using the conditions given, solve it to find Bhavana's expression for h in terms of t.

dh/dt = kh
-> 1/h dh = k dt
-> lnh = kt + c
-> h = e^[kt + c] = M.e^(kt)

h = 10, t = 0:
-> 10 = M

h = 5, t = 40:
-> 5 = 10.e^(40k)
-> e^(40k) = 1/2
-> 40k = ln(1/2)
-> k = (1/40)ln(1/2)

-> h = 10.e^[(1/40)ln(1/2)t]

c) Find, in each case, the value of h when t=60.

Case 1 (Adam): When t = 60:
-> h = 10 - 60/8 = 5/2 m

Case 2 (Bhavana): When t = 60:
-> h = 10.e^[(1/40)ln(1/2).60] = 3.54 m (3.S.F)

d) Briefly explain which assumption you would use.

Adam's, as the width of the tank is constant throughout its entire length and thus the rate at which the oil leaks should be constant.
0
13 years ago
#3
(Original post by Nima)
...
d) Briefly explain which assumption you would use.

Adam's, as the width of the tank is constant throughout its entire length and thus the rate at which the oil leaks should be constant.
I would say Bhavana's since at time t = 90, Alan's assumtion gives

h = 10 - 90/8
h = -1.25 - a negative result!!

Bhavana's eqn
h = 10.e^[(1/40)ln(1/2)t]
at t = 90
h = 2.1

This is what usually happens in practice in such a situation.
The leak from the tank is probably through a simple hole and the rate of flow through that hole wil be proportional to the (oil) pressure at the hole, and the pressure will be proportional to the height.
So the rate of loss of oil (volume) is proportional to the height.
Since the actual volume of oil in the tank is directly proportioanl to the height, then the rate of change in height will be proportional to the height - Bhavana's assumption.
0
13 years ago
#4
Yeah, I'd probably go with bhavanas approximation too because of the pressure. It wouldn't surprise me if you could get the mark for either as long as you put a fairly valid reason because the pressure thing is more physics than maths so they can't expect you to know it.
0
13 years ago
#5
Bhavana - more height means greater pressure means faster leakage...
0
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