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帮忙解答数学试题2
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这是一道去年STPM的一道数学题。。。我想了很久但还是无法解答。请你们帮帮我。
问题如下:
A 50 litre tank is initially filled with 10 litres pf brine solution containing 20 kg of salt. Starting from time t=0, distilled water is poured into the tank at a constant rate of 4 litres per minute. At the same time, the mixture leaves the tank at a constant rate of k^(1/2) litre per minute, where k^(1/2) >0. The time taken for overflow to occur is 20 minutes.
(a) Let Q be the amount of salt in the tank at time t minutes. Show that the rate of change of Q is given by:
dQ/dt= (-Qk^(1/2))/(10+(4-k^(1/2))t)
* k^(1/2) means square root of k,
Hence, express Q in term of t,
(b) Show that k = 4, and calculate the amount of salt in the tank at the instant outflow occurs.
(c) Sketch the graph of Q against t for 0 < t < 20
谢谢。。。。 |
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发表于 30-5-2009 06:24 PM
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我不懂对不对,不过前面7分好像要explain怎样有那个eqn。
然后integrate就行了。
k=4要用 (10+(4-k^(1/2))t)来做。
overflow occur after 20min。
overflow = tank full = 50
(10+(4-k^(1/2))t) = 50
when t=20, (10+(4-k^(1/2)20)) = 50
(4-k^(1/2)20) = 40
4-k^(1/2) = 2
-k^(1/2) = -2
k = 4
接下来是画graph,只要solve了那个eqn就可以画了。
应该是个直线。
加油! |
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楼主 |
发表于 30-5-2009 09:46 PM
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谢谢。我觉得proof可以用formula:
Q/t = -Qk^(1/2)
dV/dt = 10liter(in the tank) + Rate of inflow - Rate of outflow
= 10 + 4t -(k^(1/2))t
但是我无法把两个算式联合。
无论如何,谢谢。
希望大家可以帮帮忙。 |
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发表于 31-5-2009 05:47 PM
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我去年也栽在这题,研究了是这样show:
下面那part是没问题的吧! |
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楼主 |
发表于 2-6-2009 11:29 AM
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回复 4# zfc 的帖子
谢谢。你们非常厉害。解决了我研究很久的问题。因为本学校是第一年开中六,所以老师没经验,全都要自己来。希望将来大家多多帮忙。 |
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发表于 2-6-2009 10:13 PM
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回复 5# MK705 的帖子
不用客气,加油! |
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楼主 |
发表于 4-6-2009 10:45 PM
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我已经得到答案了,答案如下:
>A 50 litre tank is initially filled with 10 litres pf brine solution
>containing 20 kg of salt. Starting from time t=0, distilled water is
>poured into the tank at a constant rate of 4 litres per minute. At
>the same time, the mixture leaves the tank at a constant rate of k^
>(1/2) litre per minute, where k^(1/2) >0. The time taken for
>overflow to occur is 20 minutes.
>
>(a) Let Q be the amount of salt in the tank at time t minutes.
I will ignore their unhelpful suggestions and follow the usual way of
doing these problems.
At time t mins after the start the volume of solution in the tank is
10 + (4 - sqrt(k)).t
when t=20 this volume has increased to 50 litres.
So 10 + 20(4-sqrt(k)) = 50
20(4-sqrt(k)) = 40
(4-sqrt(k)) = 2 and so sqrt(k) = 2 <----
Volume of solution in the tank at time t = 10+2t litres
And 2 litres per minute of mixture leaves the tank.
If Q = amount of salt at time t (and assuming perfect mixing) the
mass of salt leaving per minute = Q.2/(10+2t)
dQ/dt = -Q/(5+t)
dQ -dt
---- = ----- and integrating both sides
Q 5+t
ln(Q) = -ln(t+5) + ln(A) where A = constant
ln(Q) = ln[A/(t+5)]
Q = A/(t+5)
at t=0 Q=20 20 = A/5 so A = 100
Therefore Q = 100/(t+5) <--------
>Hence, express Q in term of t, (see above) <-------
>
>(b) Show that k = 4, and calculate the amount of salt in the tank at
>the instant outflow occurs.
We had sqrt(k) = 2 so k = 4 <--------
Also at t=20 Q = 4 <------ (overflow starts)
>
>(c) Sketch the graph of Q against t for 0 < t < 20
Starts at (0,20) and descends in a smooth curve to (20,4)
对不起没有翻译,希望大家看得明白。 破案!!!哈哈哈!!! |
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