帮忙解答数学试题2

MK705

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。。。。

四月一日的小皮

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！

MK705

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。

zfc

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！

MK705

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。

zfc

不用客气，加油！

MK705

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！！！