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Advanced Higher Maths 2011-2012 : Discussion and Help

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    (Original post by ukdragon37)
    :lolwut:

    You know something is clearly wrong when you integrate to get 0
    True, what did I do wrong?
  2. Offline

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    (Original post by Mr Dangermouse)
    True, what did I do wrong?
    You've sort of "half-integrated" the 1/(1+x) part in x/(1+x). Instead:

    \displaystyle\int\dfrac{xdx}{1+x  } = \displaystyle\int\left(1 - \dfrac{1}{1+x}\right)dx = x - \ln\left(1+x\right) + C
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    Pretty sure that's right

    Edit: typical ukd: posting the right answer just before anyone else.
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    I don't understand that at all.
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    (Original post by Mr Dangermouse)
    I don't understand that at all.
    If you differentiate ln(1+x) it is 1/(1+x). So x/(1+x) can't integrate to ln(1+x). Instead you need to reduce the fraction into a quotient and remainder first, which is 1 - 1/(1+x), so you can integrate each part of the sum.
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    The actual question that Paul was referring to has limits between 0 and 1, just thought I'd throw that out there.

    Anyway, I got ~0.386 with those limits?
  7. Offline

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    Thanks for the help, I watched a video explaining how it (or a very similar one) was done. I have been taught how to do, just the notation was a bit different to what I was used to.
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    (Original post by Mr Dangermouse)
    I don't understand that at all.
    This is how we were told to think of it, although it can only really be used if the degree of the numerator is equal to the degree of the denominator. You could also use long division like ukd done, but no doubt he has his own magical technique.

    \displaystyle \int \dfrac {x}{x+1}\\





= \displaystyle \int \dfrac {x+1-1}{x+1} \\





= \displaystyle \int \dfrac {x+1}{x+1} - \dfrac {1}{x+1}\\





= x - \ln (x+1)

    Also, can anyone tell me why my latex is so close together vertically, and why my integral signs are tiny?
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    I'd appreciate if someone could help me with this question, it's from the 2002 paper. I think I have part (a) correct but part (b) looks tricky.

    2002 - A9.

    Functions x(t) and y(t) satisfy

    \frac{dx}{dt} = -x^{2}y , \frac{dy}{dt} = -xy^{2}

    When t = 0, x = 1, and y = 2.

    a) Express dy/dx in terms of x and y and hence obtain y as a function of x. [5 marks]

    b) Deduce that \frac{dy}{dx} = -2x^{3} and obtain x as a function of t for t \geq 0. [5 marks]
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    (Original post by Quintro)
    The actual question that Paul was referring to has limits between 0 and 1, just thought I'd throw that out there.

    Anyway, I got ~0.386 with those limits?
    I might be wrong, but I think the SQA want you to leave things like that as absolute values. So instead of whatever the result for ln1 is, just write ln1. I think I saw that in a marking scheme somewhere (but I could be wrong!).
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    (Original post by Quintro)
    This is how we were told to think of it, although it can only really be used if the degree of the numerator is equal to the degree of the denominator. You could also use long division like ukd done, but no doubt he has his own magical technique.

    \int \dfrac {x}{x+1}





= \int \dfrac {x+1-1}{x+1}





= \int \dfrac {x+1}{x+1} - \dfrac {1}{x+1}





= x - \ln (x+1)

    Also, can anyone tell me why my latex is so close together vertically, and why my integral signs are tiny?
    Add "\displaystyle" in front of integral signs to make them bigger. You can use "\\" for line breaks or use separate latex blocks for each line.

    Here's the latex fixed:

    \displaystyle\int \dfrac {x}{x+1}\\\\





= \displaystyle\int \dfrac {x+1-1}{x+1}\\\\





= \displaystyle\int \dfrac {x+1}{x+1} - \dfrac {1}{x+1}\\\\





= x - \ln (x+1)
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    (Original post by JordanR)
    I might be wrong, but I think the SQA want you to leave things like that as absolute values. So instead of whatever the result for ln1 is, just write ln1. I think I saw that in a marking scheme somewhere (but I could be wrong!).
    Yeah, I wrote the exact value above it: ln(4) - 1, so hopefully they wouldn't deduct any marks for it, but I think I'll just write the exact value from now on, thanks!

    Anyway I'm sure that for ln(1) you'd be ok with writing 0
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    (Original post by ukdragon37)
    ..
    Got it now, thanks
  14. Offline

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    (Original post by Quintro)
    I'd appreciate if someone could help me with this question, it's from the 2002 paper. I think I have part (a) correct but part (b) looks tricky.

    2002 - A9.

    Functions x(t) and y(t) satisfy

    \frac{dx}{dt} = -x^{2}y , \frac{dy}{dt} = -xy^{2}

    When t = 0, x = 1, and y = 2.

    a) Express dy/dx in terms of x and y and hence obtain y as a function of x. [5 marks]

    b) Deduce that \frac{dy}{dx} = -2x^{3} and obtain x as a function of t for t \geq 0. [5 marks]
    i'm pretty sure i have the answer to this somewhere if you're still needing it?
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    (Original post by katehhhh)
    i'm pretty sure i have the answer to this somewhere if you're still needing it?
    Yeah that would be great, thanks.
  16. Offline

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    (Original post by Quintro)
    I'd appreciate if someone could help me with this question, it's from the 2002 paper. I think I have part (a) correct but part (b) looks tricky.

    2002 - A9.

    Functions x(t) and y(t) satisfy

    \frac{dx}{dt} = -x^{2}y , \frac{dy}{dt} = -xy^{2}

    When t = 0, x = 1, and y = 2.

    a) Express dy/dx in terms of x and y and hence obtain y as a function of x. [5 marks]

    b) Deduce that \frac{dy}{dx} = -2x^{3} and obtain x as a function of t for t \geq 0. [5 marks]
    Well you have y = f(x) and you have dy/dx in terms of x and y. Substitute y in dy/dx for f(x). And then do the same for dx/dt (substitute y for f(x)) and then do differential equations for x.
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    (Original post by soup)
    Well you have y = f(x) and you have dy/dx in terms of x and y. Substitute y in dy/dx for f(x). And then do the same for dx/dt (substitute y for f(x)) and then do differential equations for x.
    Yeah, I was doing that but my function for y in part (a) was wrong. I'll go back and redo part (a) later and see if it works after that. Thanks for the help.

    EDIT: Got it now, it was my function from part A that was off.
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    Sorting out my last wee bit of revision


    unit 1.

    1. Algebra( )
    2. Differentiation ( )
    3. Integration ( )
    4. Functions ( ) (As long as local and global maxima and minima doesn't arise)
    5. Systems of linear equations ( ) (As long as ill-conditioning doesn't arise)


    Unit 2

    1. Further differentiation ( )
    2. Further integration ( )
    3. Complex numbers ( )
    4. Sequences and series ( )
    5. Elementary number theory and methods of proof ( )

    Unit 3

    1. Further ordinary differential equations ( ) (Need to revise the Yc for each of the 3 different forms)
    2. Vectors (They're easy but I need a bit of a revise)
    3. Matrices ( :mad: )
    4. Further sequences and series ( :mad: )
    5. Further proofs and number theory ( :mad: )


    I am very poor on unit 3.
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    Okay, time to start revising maths. Where to even begin?
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    2001-2008 done. Just 2009-2011 to do and three days left! (Or less actually). Gotta say 2006 and 2007 weren't very nice papers but 2008 was decent. Or maybe I was just getting into the swing of things.

    Still need to recap the little things like ill-conditioning and redundancy (always get them mixed up), proofs (always loose marks on them unless it's an alright proof by induction), maclaurins (even though it's pretty straight forward, it's those questions which I have been loosing marks on) and some complex number stuff. I think the rest I'm alright with.

    Feeling better about stats too now that I've done a few papers. Thank goodness that the pure maths part of the paper is worth like 1/3 of the marks.
Updated: August 12, 2012
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