Edexcel C4 June 2009 Thread :) Watch

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kflynn
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#321
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#321
(Original post by MushyPeas)
Ohh those spawns of satan!!
You really just have to keep practicing them. Remember to write down EVERYTHING the question gives you.
For example:
Liquid is pouring into a container at a constant rate of 20 cm3 s–1 and is leaking out at a rate proportional to the volume of liquid already in the container. that is the question 8 on the june 2005 paper you sent link to.
Notice the words that i have underlined. So for part a) you know that it is a dV/dt because it is talking about the rate of change to do with the volume (V) of liquid in a certain time (t). Then you can see that there is a constant being put in of 20 but some is coming OUT so it will be dV/dt = 20 - ... And because it is proportional what is coming out to what the volume (V) already is you need k as a constant. So you come out with:
dV/dt = 20 - kV -- which is what the question asked you to prove :woo:


cheers that makes a bit more sense. i still hoping that one doesnt come up though. but its about 80% certain it will.
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KeineHeldenMehr
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(Original post by Waterstorm)
Draw it, OD = OA + AD, you know OA and BC, and cause it's a parallelogram you know BC = AD, you should then know AD.
thanks...
yeah, i drawed it but initiallly thought of using the pythagoras for 3 planes formula...that led to no where..thank you!
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Roboex
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#323
Ok, can someone give a summary for the Rate of Change and Vectors topic? They are the only questions that I get wrong in every paper, even though I've revised each topic separately!
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2710
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#324
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for part c of this question, in the answers it says that they use:

 \dfrac {dV}{dt} = 20 e^{-kt}

how do they get this?

Thanks
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MushyPeas
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(Original post by Roboex)
Ok, can someone give a summary for the Rate of Change and Vectors topic? They are the only questions that I get wrong in every paper, even though I've revised each topic separately!
I've just made a post on the previous page about rate of change =]
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Waterstorm
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#326
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#326
(Original post by 2710)
for part c of this question, in the answers it says that they use:

 \dfrac {dV}{dt} = 20 e^{-kt}

how do they get this?

Thanks
Separating the variables.
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Ballet_shoes
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#327
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#327
C4 can go die!
I hateee ittttt!!!!!!
Literally the only thing I can do is implicit differentiation.
I thought I could go partial fractions and binomial expansion as well- but apparently not!
Gap year/ gap life- here I come! =(
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MushyPeas
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#328
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#328
(Original post by 2710)
for part c of this question, in the answers it says that they use:

 \dfrac {dV}{dt} = 20 e^{-kt}

how do they get this?

Thanks
Part c)
They have differentiated the answer to part b)
V = 20/k - [20e^(-kt)]/k
We are differentiating with respect to t
Soo... dV/dt = 20e^(-kt)

Because when you differentiate e^..t with respect to t, you bring down any number or minus sign before it, so in this case you bring down the (-k) which cancels out the minus sign before this part of the equation and also the k it had been divided by
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My Alt
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#329
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(Original post by Ballet_shoes)
C4 can go die!
I hateee ittttt!!!!!!
Literally the only thing I can do is implicit differentiation.
I thought I could go partial fractions and binomial expansion as well- but apparently not!
Gap year/ gap life- here I come! =(
You can't give up now!






You are so close to sucess!
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MonkeyMajik
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#330
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#330
Panic stations now! I'm practising on those wordy differential equations
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2710
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#331
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(Original post by MushyPeas)
Part c)
They have differentiated the answer to part b)
V = 20/k - [20e^(-kt)]/k
We are differentiating with respect to t
Soo... dV/dt = 20e^(-kt)

Because when you differentiate e^..t with respect to t, you bring down any number or minus sign before it, so in this case you bring down the (-kt) which cancels out the minus sign before this part of the equation and also the k it had been divided by
But, why dont they just use the differential in part a) ? Its already differentiated there...

Thanks
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Waterstorm
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(Original post by 2710)
EDIT: Nevermind
My bad, I didn't see the question, I thought you were just trying to solve that.
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MushyPeas
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(Original post by 2710)
But, why dont they just use the differential in part a) ? Its already differentiated there...

Thanks
Because in the question it says when t = 5 so you need a differential equation with t in it.
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2710
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#334
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(Original post by MushyPeas)
Because in the question it says when t = 5 so you need a differential equation with t in it.
Ooh ok, I hate this ><
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MushyPeas
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#335
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(Original post by 2710)
Ooh ok, I hate this ><
Nearly over now =]
By this time tomorrow the exam will be over. YAY
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Waterstorm
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#336
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(Original post by 2710)
But, why dont they just use the differential in part a) ? Its already differentiated there...

Thanks
I've tried that before, and you can. You'll get the same answer.
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Roboex
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#337
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#337
In the real exam, will we get a paper as hard as the Solomon ones?
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Ballet_shoes
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#338
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(Original post by My Alt)
You can't give up now!






You are so close to sucess!
:curious: So close to success?! How?!
I'm probably going to get 15/75 max for this paper. I don't call that success. :p:
Why can't this paper be like C3? Thinking I can't do it, but then when I get down to it, it all comes naturally with common sense.

DIE DIE DIEEEE C4!!!!!!!!!!
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MR.PINK
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#339
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#339
Am so gona fail, i found c4 piss easy thoughout the year, but when it came to pastpapers, ill be proud if i got a u
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Sapphire Rose
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(Original post by fn_101)
You're handwriting s way better than mine!
Thank you!
Anytime

Let me know if you need any more help:yep:
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