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April 25, 2024, 04:06:43 am

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2173819 times)  Share 

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Special At Specialist

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Re: Specialist 3/4 Question Thread!
« Reply #180 on: January 20, 2012, 06:30:13 pm »
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I hate these extremely long worded questions...

"A tank initially contains 400 litres of water in which are dissolved 10 kg of salt. A salt solution of concentration 0.2 kg/L is poured into the tank at a rate of 2 litres per minute; the mixtury, which is kept uniform by stirring, flows out at a rate of 2 litres per minute.
If the mass of salt in the tank after t minutes is x kg, set up and solve the differential equation for x in terms of t."

My thoughts are that every minute, 0.4kg of salt is being added and 1/200 times the current concentration is being removed from the tank. How would I express this?

Someone please help me!
« Last Edit: January 20, 2012, 06:38:01 pm by Special At Specialist »
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moekamo

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Re: Specialist 3/4 Question Thread!
« Reply #181 on: January 20, 2012, 06:44:11 pm »
+1
let Q=amount of salt at time t.

Then dQ(in)dt=0.2 kg/L * 2L/min = 0.4 kg/min . That should make sense, I'm putting in 2L per minute and if the solution has 0.2kg per litre, then there are 0.4 kg/min going in.

Now coming out, the concentration is not constant, it is Q/V where Q is the amount and V is the volume, which is actually just 400L since there is 2L/min going in and out.
So dQ(out)/dt=Q/400 kg/L * 2L/min = Q/200 kg/min.

so overall, dQ/dt=dQ(in)/dt - dQ(out)/dt = 2/5 - Q/200. now you should be able to solve this, put it on a common denominator and flip both sides etc.

then you use the initial amount of salt, 10kg to solve for the constant of integration

hope this helped :)
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Re: Specialist 3/4 Question Thread!
« Reply #182 on: January 20, 2012, 06:59:44 pm »
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Thanks, but it gets harder:

"If, instead, the mixture flows out at 1 litre per minute, set up, but do not solve, the differential equation for the mass of salt in the tank."

let Q=amount of salt at time t.
Now coming out, the concentration is not constant, it is Q/V where Q is the amount and V is the volume

I don't understand this part...
I get that if the volume is constantly changing, then the percent of input/output in relation to the volume will not be a constant value, but how do I work it out, and what does Q/V actually mean?
« Last Edit: January 20, 2012, 07:03:25 pm by Special At Specialist »
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Re: Specialist 3/4 Question Thread!
« Reply #183 on: January 20, 2012, 07:05:42 pm »
+1
Thanks, but it gets harder:

"If, instead, the mixture flows out at 1 litre per minute, set up, but do not solve, the differential equation for the mass of salt in the tank."
Do the same thing as above for finding the equation but for the flow out part, instead of using , the bottom part is your volume will change as time changes. So initally it was 400 and the volume us increasing at L/min.
So your volume will be 400+(1)t
i.e.

Then do the rest, i.e. dxin/dt-dxout/dt
« Last Edit: January 20, 2012, 07:08:58 pm by b^3 »
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Re: Specialist 3/4 Question Thread!
« Reply #184 on: January 20, 2012, 07:13:33 pm »
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Do the same thing as above for finding the equation but for the flow out part, instead of using , the bottom part is your volume will change as time changes. So initally it was 400 and the volume us increasing at L/min.
So your volume will be 400+(1)t
i.e.

Then do the rest, i.e. dxin/dt-dxout/dt

Thank you. I sort of get this, but not fully...
Why is it (400 + 1t) as the denominator?
What sort of situation would make it (400 + 2t) or (400 + 5t)?
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Re: Specialist 3/4 Question Thread!
« Reply #185 on: January 20, 2012, 07:36:09 pm »
+1
This is from a couple of pages back in this thread
For these questions where the flow in is not equal to the flow out, it is good to use this formula.











Hope that helps.

EDIT: Explaining the equation
The first part of the equation is giving how much of "x" is following in per second.
The second part of the equation is giving how much of "x" is following out per second.
You can see this by "rate"*"concentration"
In the second half of the equation, the concentration is  , think of it as c=mass/volume, mass is x and the volume is changing as t changes.
So basically if you do chem you know that c=n/v (in M) or c=m/v
So that is what we are doing, the flow in of x is  equal to m=cv.
now the conentration of x in we know as well as the volume in.
The flow out of x is m=c*v flow out, but we dont know the concentration or volume of the solution
The volume out will be give by inital volume +(flow rate in - flow rate out)*t, as we have the intial volume and every minute we add so much to it (flow rate in - flow rate out).
Now the concentration out will be c=m/v, so x represents the amouhnt of x (i..e. m) and v is the changing volume.
so from that we have the concentration of the solution as
Now we times that by the volume out to get the flow out of x.
Then

I think I may have over complicated a simple explanation but maybe someone else can come along and give a more concise explanation. Whenever you get to these types of questions , work out the flow rate of x in and the flow rate of x out, and minus the second from the first. Hope it helps though.
« Last Edit: January 20, 2012, 07:50:12 pm by b^3 »
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Re: Specialist 3/4 Question Thread!
« Reply #186 on: January 20, 2012, 09:31:07 pm »
+1
if anyone is interested, the general solution to these problems is as follows:

you have a tank with volume, initially with kg of salt present. You pump a solution of salt of concentration in at a rate of . The tank is continuously stirred and at any time there is kg of salt present. The mixture is pumped out at a rate of .

Now you differential equation is:



the second one is just an abbreviated form of the first one, so i don't have too many fractions everywhere...

Now, to get the general solution we use the Integrating Factor.

Basically, let

then using the formula on that wiki page,

then crank the handle to get:



where

pretty sure all of that is right, if anyone wants to check for any errors that'd be awesome.

This should only be used to check answers, the methods used are out of the VCE scope and they will never ask you to solve one of these without prompting the solution first and you showing that it is indeed a solution. also its pretty cool i reckon haha :)

and of course it reduces to the general solution when which is the ones you have to know how to solve, actually it might be a bit more complex given the in the exponent, but using some identities it should reduce haha
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Re: Specialist 3/4 Question Thread!
« Reply #187 on: January 20, 2012, 09:36:52 pm »
+1
^ more like this won't be used at all. They will at most ask you to plug the solution back into the DE to check that it's right. Good job anyway.
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Re: Specialist 3/4 Question Thread!
« Reply #188 on: January 20, 2012, 09:53:02 pm »
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haha yea, i just remember back in vce i wondered how you would solve the ones with changing volumes, or if you could at all, now that im in uni and know i may as well show other spec kids if they're wondering, also it can never hurt to learn a bit more maths :)
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Re: Specialist 3/4 Question Thread!
« Reply #189 on: January 21, 2012, 05:10:02 pm »
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I have another differential equation I can't solve :(

"A conical tank has a radius length at the top equal to its height. Water, initially with a depth of 25cm, leaks out through a hole in the bottom of the tank at a rate of 5*sqrt(h) cm^3 /min where the depth is h cm at time t minutes.
Construct a differential equation expressing dh/dt as a function of h, and solve it."

Please help me interpret this information.
I have to use weird rules like dh/dt = dh/dr * dr/dV * dV/dt for example (just an example, not necessarily this exact formula).
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Re: Specialist 3/4 Question Thread!
« Reply #190 on: January 21, 2012, 09:54:45 pm »
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Nvm I found the answer to the last question.
Now I'm doing Euler's Method and I don't fully understand it...

If dy/dx = cos(x), given that y-sub-0 = y(0) = 1, find y-sub-3 using h = 0.1
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Re: Specialist 3/4 Question Thread!
« Reply #191 on: January 22, 2012, 12:22:50 am »
+2
so you use the formula where

using then

now using then

now using then finally,

this seems correct as when you solve the DE analytically, you get and then , so we are pretty accurate using eulers method and haven't made any serious blunders :)

these problems you just have to avoid silly mistakes so do a few until you're confident with the formula and how it works.
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Re: Specialist 3/4 Question Thread!
« Reply #192 on: January 25, 2012, 03:38:50 pm »
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Thanks moekamo.
Here's another one:

dy/dx = sqrt(x), y-sub-0 = y(1) = 0, h = 0.01, find y-sub-4.
Does that mean I have to start at y(1) then go to y(1.01) then y(1.02)?
So would y'-sub-0 = y'(1) = sqrt(1) = 1? Is that right?
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Re: Specialist 3/4 Question Thread!
« Reply #193 on: January 25, 2012, 11:33:28 pm »
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yep, so then continue like that until you get to
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Re: Specialist 3/4 Question Thread!
« Reply #194 on: January 27, 2012, 10:47:17 pm »
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Just wondering if we need to know Derivates of inverse circular functions?
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