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    does anyone remember what part (i) from q7 was?
    i think it read : dx/dt=x(1-x) with x=0.5 when t=0 and asked you to verify the solution : x= 1/(1-e^-t)
    but it could be x=1/(1+e^-t)? i cant remember. anyone ?
    feel free do post back with a worked solution. (it would be greatly appreciated XD)
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    (Original post by Domdude344)
    does anyone remember what part (i) from q7 was?
    i think it read : dx/dt=x(1-x) with x=0.5 when t=0 and asked you to verify the solution : x= 1/(1-e^-t)
    but it could be x=1/(1+e^-t)? i cant remember. anyone ?
    feel free do post back with a worked solution. (it would be greatly appreciated XD)
    \frac{dx}{dt}=x(1-x)

    \int \frac{1}{x(1-x)} dx = \int dt

    \int \{ \frac{1}{x} + \frac{1}{1-x} \} dx = \int dt

    t= \int \{ \frac{1}{x} - \frac{-1}{1-x} \} dx

    t = \ln(x) - \ln(x-1) + c

    t = \ln( \frac{x}{1-x} ) +c

    Ae^t = \frac{x}{1-x}

    x = \frac{Ae^t}{1+Ae^t}

    Now substitute in the initial condition that x is a half when t is zero to find the answer given

    It's really annoying that they said to "verify" the given answer; I spent a while just finding dx/dt and x(1-x), and then showing that they were the same - which must have been insufficient for some reason, given the requirements of part (ii).

    Still, I don't know - I hope this helps you in some way
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    (Original post by placenta medicae talpae)
    Haha, did you think it said "the mathematician spat" for ages too? :woo:

    I just got completely muddled up with the grammar, and in the end decided that I should check my answers instead
    I did and i wrote it as my final answer
    I did 20 letters instead of 16.
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    (Original post by placenta medicae talpae)
    \frac{dx}{dt}=x(1-x)

    \int \frac{1}{x(1-x)} dx = \int dt

    \int \{ \frac{1}{x} + \frac{1}{1-x} \} dx = \int dt

    t= \int \{ \frac{1}{x} - \frac{-1}{1-x} \} dx

    t = \ln(x) - \ln(x-1) + c

    t = \ln( \frac{x}{1-x} ) +c

    Ae^t = \frac{x}{1-x}

    x = \frac{Ae^t}{1+Ae^t}

    Now substitute in the initial condition that x is a half when t is zero to find the answer given
    Ah genius!
    See looking at that is something i know im capable of, but i just didnt think of seperating the 1/x(x-1)..i was sitting there for ages thinking how am i going to integrate this oh well only more marks down the drain..
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    (Original post by placenta medicae talpae)
    \frac{dx}{dt}=x(1-x)

    \int \frac{1}{x(1-x)} dx = \int dt

    \int \{ \frac{1}{x} + \frac{1}{1-x} \} dx = \int dt

    t= \int \{ \frac{1}{x} - \frac{-1}{1-x} \} dx

    t = \ln(x) - \ln(x-1) + c

    t = \ln( \frac{x}{1-x} ) +c

    Ae^t = \frac{x}{1-x}

    x = \frac{Ae^t}{1+Ae^t}

    Now substitute in the initial condition that x is a half when t is zero to find the answer given

    It's really annoying that they said to "verify" the given answer; I spent a while just finding dx/dt and x(1-x), and then showing that they were the same - which must have been insufficient for some reason, given the requirements of part (ii).

    Still, I don't know - I hope this helps you in some way
    Yes I did this, except the the 3rd from last line in your working I found c at that point then got to show the final answer (sorry can't do latex [really should learn ]. But I think that must be sufficient.
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    (Original post by Chriz M)
    Yes I did this, except the the 3rd from last line in your working I found c at that point then got to show the final answer (sorry can't do latex [really should learn ]. But I think that must be sufficient.
    Oh yeah, that'd deffo be sufficient!

    Marks aplenty for you, please!

    Does anyone reckon that this question might have been a "stretch and challenge" one? - I have no idea what these are in maths, tbh.
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    (Original post by sinesquared.)
    I did and i wrote it as my final answer
    I did 20 letters instead of 16.
    Haha, I think I did the first 17 for a while, so it read:

    THE MATHEMATICIANS
    And then realised

    Spoiler:
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    Or maybe it was the other way around - which would not be good!
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    (Original post by sinesquared.)
    Ah genius!
    See looking at that is something i know im capable of, but i just didnt think of seperating the 1/x(x-1)..i was sitting there for ages thinking how am i going to integrate this oh well only more marks down the drain..
    You know, I just realised something about this:

    \frac{1}{x(1-x)} = \frac{1}{x} \times \frac{1}{1-x}

    And:

    \frac{1}{x(1-x)} = \frac{1}{x} + \frac{1}{1-x}

    How cool is that!

    Anyway, yeah, don't worry about it - they should probably have split this question into parts since you could have done the rest of the question beyond the splitting of this.
    And yeah, I know exactly what you mean: sometimes, I just stare at things thinking how on earth, when it's really just exactly the same partial fractions stuff as usual that's required!
    Just lucky I pulled it off in the exam, I guess.
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    (Original post by placenta medicae talpae)
    \frac{dx}{dt}=x(1-x)

    \int \frac{1}{x(1-x)} dx = \int dt

    \int \{ \frac{1}{x} + \frac{1}{1-x} \} dx = \int dt

    t= \int \{ \frac{1}{x} - \frac{-1}{1-x} \} dx

    t = \ln(x) - \ln(x-1) + c

    t = \ln( \frac{x}{1-x} ) +c

    Ae^t = \frac{x}{1-x}

    x = \frac{Ae^t}{1+Ae^t}

    Now substitute in the initial condition that x is a half when t is zero to find the answer given

    It's really annoying that they said to "verify" the given answer; I spent a while just finding dx/dt and x(1-x), and then showing that they were the same - which must have been insufficient for some reason, given the requirements of part (ii).

    Still, I don't know - I hope this helps you in some way
    Hmm, that would have been a good idea but my mind went blank. Surley however, because it said verify it would have also been sufficient to differentiate the given value of x and then substitute that given value of x into the differential equation, to show that they are the same?
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    (Original post by The Question)
    Hmm, that would have been a good idea but my mind went blank. Surley however, because it said verify it would have also been sufficient to differentiate the given value of x and then substitute that given value of x into the differential equation, to show that they are the same?
    Yep, absolutely.
    I did this to begin with too.

    The only problems are that:
    (i) This question was worth 6 marks; you'd probably need a 3-mark question for that, I think.
    (ii) As far as I can see, you needed to have solved the differential equation in order to answer part (ii).

    But - if you have done as they asked - and have verified it, then they will have to give you all of the available marks for that part of the question.
    How you then go on to do part (ii), having merely verified it, I don't know.
    And that is why I think that OCR chose completely the wrong verb, and it should have been "show that", or something instead of "verify".
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    (Original post by placenta medicae talpae)
    Yep, absolutely.

    How you then go on to do part (ii), having merely verified it, I don't know.
    And that is why I think that OCR chose completely the wrong verb, and it should have been "show that", or something instead of "verify".
    wat was part 2 again i cant remember
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    (Original post by parthchill)
    wat was part 2 again i cant remember
    It was that thing about how long it would take for the population to get to 3/4 of its eventual size :cool:
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    (Original post by placenta medicae talpae)
    It was that thing about how long it would take for the population to get to 3/4 of its eventual size :cool:
    they already told you that

    x = 1/(1-(e^t))

    and now

    x= 3/4

    so just substitute it in and find t.
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    (Original post by The Question)
    they already told you that

    x = 1/(1-(e^t))

    and now

    x= 3/4

    so just substitute it in and find t.
    Oh yeah!
    Gee, I can be really thick sometimes.

    Anyway, what does everyone think the boundaries will be?
    It's just done on an out-of-90 basis, rather than setting separate boundaries for the two papers, like they do with C3, isn't it?
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    (Original post by placenta medicae talpae)
    \frac{dx}{dt}=x(1-x)

    \int \frac{1}{x(1-x)} dx = \int dt

    \int \{ \frac{1}{x} + \frac{1}{1-x} \} dx = \int dt

    t= \int \{ \frac{1}{x} - \frac{-1}{1-x} \} dx

    t = \ln(x) - \ln(x-1) + c

    t = \ln( \frac{x}{1-x} ) +c

    7. Ae^t = \frac{x}{1-x}

    8. x = \frac{Ae^t}{1+Ae^t}
    I would appreciate it if you could tell me how you get from line 7 to line 8, I did exactly that in the exam but thought i was just fudging it
    (btw it was e^-t but it is right none the less)

    I'm thinking it's gonna be high grade boundries for this exam as everyone at my school took it in their strides probably about 75-77ish for an A
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    (Original post by TmanG)
    I would appreciate it if you could tell me how you get from line 7 to line 8, I did exactly that in the exam but thought i was just fudging it
    (btw it was e^-t but it is right none the less)

    I'm thinking it's gonna be high grade boundries for this exam as everyone at my school took it in their strides probably about 75-77ish for an A
    No problemo mi amigo

    Ae^t=\frac{x}{1-x}

    Ae^t(1-x)=x

    Ae^t-xAe^t=x

    Ae^t=x+xAe^t

    Ae^t=x(1+Ae^t)

    \frac{Ae^t}{1+Ae^t}=x

    Let me know if the above doesn't seem to follow

    Spoiler:
    Show
    Someone else pointed out that the differential equation need not be solved anyway!
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    (Original post by placenta medicae talpae)
    No problemo mi amigo

    Ae^t=\frac{x}{1-x}

    Ae^t(1-x)=x

    Ae^t-xAe^t=x

    Ae^t=x+xAe^t

    Ae^t=x(1+Ae^t)

    \frac{Ae^t}{1+Ae^t}=x

    Let me know if the above doesn't seem to follow

    Spoiler:
    Show
    Someone else pointed out that the differential equation need not be solved anyway!

    Haha i completely missed the e^-t numerator i think i wrote a 1 numerator the exam which was the problem well i'll get most of the marks at least.

    But yeah the grade boundries are gonna be quite high i reckon. Good luck with the c3/s3/m3 if your doing them
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    Hey, the answer turned out as x = 1 / 1+ e^-t (which you get from the last line on the above post by dividing by e^t.

    All in all this paper worried me because i found it waaaay to easy; much easier than previous years lol

    as for the comprehension the answer was THE MATHEMATICIAN (16 letters lol)

    should be 90%+ and op fully the same for C3 tomorrow (20/01/10) which would get me an A*

    anyway good luck to everyone with the rest of your exams

    craig
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    (Original post by TmanG)
    Haha i completely missed the e^-t numerator i think i wrote a 1 numerator the exam which was the problem well i'll get most of the marks at least.

    But yeah the grade boundries are gonna be quite high i reckon. Good luck with the c3/s3/m3 if your doing them
    Cheers - yeah I'm doing them.
    Are you doing those ones as well? :woo:
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    i have the C4 mock tomorrow and its the january 2010 one! i am so gonna fail it and i need to know what questions there are so i can be prepared for it XD pleaseee help me! =) much appreciated.
 
 
 
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