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    (Original post by Mr Nonsense)
    I'll try and do it tomorrow, although I've been having a few problems with the scanner recently, but I'll do my best (I need to sort out the stupid machine anyway...).
    Any topics of particular interest?
    Thanks

    Hmmm Mechanics :teeth: electromagnetism, circuits. but anything would be good
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    (Original post by AsakuraMinamiFan)
    I have quite a few questions from Oxford + Cambridge scholarship papers dated at the very latest 1963. Only problem is that I have no idea how many of them are per paper or what the time limit is. Here's one for now that I have wrote down:

    1. (Oxford.) By means of the substitutions , or otherwise, find:



    I'll try and bring some more later today.
    After quite a while of effort, I put the first one into Wolfram and it came back as

    \displaystyle{\frac{x(x^2+2)}{6(  x^4+x^2+1)}+\frac{(7i+\sqrt3)\ar  ctan(\frac x2 (\sqrt3 -i))}{6\sqrt{6+6i\sqrt3}}+\frac{(-7i+\sqrt3)\arctan(\frac x2 (\sqrt3 +i))}{6\sqrt{6-6i\sqrt3}}}

    Did they honestly get that it '63? Maybe I put it into Wolfram wrong, but I don't think so!
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    Meatball- Programs like wolfram and mathcad don't always give simple answers. The substitution will make it all fall out nicely.
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    (Original post by tingtongdingdong)
    Meatball- Programs like wolfram and mathcad don't always give simple answers. The substitution will make it all fall out nicely.
    Well, I've had a look, and I don't see how it falls out "nicely". Care to post workings?
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    As I said, I did spend a while trying to do it before using wolfram, but to no avail.
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    (Original post by DFranklin)
    Q6 (i)
    Let





    Deduce

    Then and as required.


    (ii)




    So the integral becomes

    So we get
    I was just doing (ii) now, and took a different approach, but I left out a ^(3/2) along the way and ended up with ln2 at the end. At least I got (i) though...
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    I'm afraid this level of maths is beyond me DFranklin so i didn't get very far. I should have said: "I'd assume the substitution makes it all fall into place". I was just making the point that maths programs don't always give you what you want. Still very handy though.
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    (Original post by meatball893)
    After quite a while of effort, I put the first one into Wolfram and it came back as

    \displaystyle{\frac{x(x^2+2)}{6(  x^4+x^2+1)}+\frac{(7i+\sqrt3)\ar  ctan(\frac x2 (\sqrt3 -i))}{6\sqrt{6+6i\sqrt3}}+\frac{(-7i+\sqrt3)\arctan(\frac x2 (\sqrt3 +i))}{6\sqrt{6-6i\sqrt3}}}

    Did they honestly get that it '63? Maybe I put it into Wolfram wrong, but I don't think so!
    I'm gonna have another go at it and upload my workings, but I'm halfway through the first one and have found t=x+x^{-1} doesn't give any way to write in terms of t - I'm assuming the question means the substitution is either t=x+x^{-1} OR t=x-x^{-1}? I recongise this is to do with palindromic polynomials but keep getting random other terms which are impossible to get in terms of t when taking into account dt/dx.

    edit: unless I'm missing something massive (perhaps adding the integrals or something?) it's impossible unless there's some way to do it with t=x \pm x^{-1} which makes the \pm symbols cancel out.
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    It would work a lot more nicely without the square on the denominators - I am somewhat suspicious that they shouldn't be there.
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    (Original post by DFranklin)
    It would work a lot more nicely without the square on the denominators - I am somewhat suspicious that they shouldn't be there.
    I tried the first one without the square on the denominator (and having done hardly any integration before ) and got an answer of
    Spoiler:
    Show
    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \displaystyle\frac{1}{\sqrt{3}}[{\arctan\frac{x - x^{-1}}{\sqrt{3}}]
    which is probably most likely wrong but oh well :<
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    No, that's right I think.
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    Well if the original question did have a square in the denominator I suppose I would be less impressed with myself.
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    Sorry! I've most probably wrote it incorrectly, and, failing that, the book I found it in has an error. It makes much more sense to find the first two integrals (without the square) by each substitution and then consider those two to find the third.

    I haven't had chance to go to the library to get more (I will soon!), but if anybody is curious as to where they're from then you can find them in Tratner's Techniques of Mathematical Analysis.
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    Q9 i)


    For a line to be on the plane,

     \begin{pmatrix} A \\ B \\ C \end{pmatrix} . \begin{pmatrix} l \\ m \\ n \end{pmatrix} = 0

     Aa + Bb + Cc = D

    That is, the normal to the plane must be perpendicular to the line in the plane

    The point in the plane (a,b,c) must satisfy the vector plane equation.


    Q9 ii)

    The vector line equation that is perpendicular to the plane and goes through the origin is t(2i + j + 2k)

    Sub that back into the the vector plane equation

    9t = 27
    t = 3

    Subbing t =3 to get the foot of the perpendicular as 6i + 3j + 6k.



    Q9 iii)


    The next is where I'm sure about the answer.

    What I infer from the last part is that we need to look for equations in the plane that intersect N in the plane and must intersect the line x - y = 3, z = 6 at the 60 degrees also. I say intersect the line x - y = 3, z = 6, because I do not think you could make an angle of 60 with any line in the plane without intersecting it.

    So starting with x - y = 3 z = 6

     \begin{pmatrix} x \\ x- 3 \\ 6 \end{pmatrix} ->  \begin{pmatrix} 0 \\ -3 \\ 6 \end{pmatrix} + t\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} , where i've said x = t because it's the only thing that varies, and i like t better. I think i'm okay to do that.

    So this line must intersect the plane to make an angle

    Subbing  \begin{pmatrix} 0 \\ -3 \\ 6 \end{pmatrix} + t\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} into the vector plane equation

    2t + t - 3 + 12 = 27

    3t = 18

    t = 6

     \begin{pmatrix} 6 \\ 3 \\ 6 \end{pmatrix}

    So the line must intersects at N in the plane.

    For the line to make an angle of 60 with the vectors inside the plane

    it must also make 30 degrees with the perpendicular vector to the plane, since it intersects the plane at N.

    Finding |a|,|b| and a.b with theta = 30 we get

     3\frac{\sqrt{3}}{2} \sqrt{t^2 + (t - 3)^2 + 36}  = 3t + 9

     \sqrt{3t^2 + 3(t - 3)^2 + 108} = 2t + 6

     6t^2 - 18t + 135 = 4t^2 + 24t + 36

     2t^2 - 42t + 99 = 0

     t = \frac{42 \pm \sqrt{972}}{4}

     t = \frac{42 \pm 18\sqrt{3}}{4}

     t = \frac{21 \pm 9\sqrt{3}}{2}

    So the equations of the lines in the plane that intersect N and make an angle of 60 degrees with the line (x - y = 3, z = 6) are

     \begin{pmatrix} 6 \\ 3 \\ 6 \end{pmatrix} + \frac{21 + 9\sqrt{3}}{2}\left[ \begin{pmatrix} X - 6 \\ Y - 6 \\ Z - 6 \end{pmatrix} \right]

    and

     \begin{pmatrix} 6 \\ 3 \\ 6 \end{pmatrix} - \frac{21 + 9\sqrt{3}}{2}\left[ \begin{pmatrix} X - 6 \\ Y - 6 \\ Z - 6 \end{pmatrix} \right]

    Where X,Y,Z can be any point on the plane.



    Really really not sure about that last part but might as well put it up for you to pick at .

    It's been almost 2 years since I've done any FP3 vectors work. I feel for Dave.
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    (Original post by AsakuraMinamiFan)
    Sorry! I've most probably wrote it incorrectly, and, failing that, the book I found it in has an error. It makes much more sense to find the first two integrals (without the square) by each substitution and then consider those two to find the third.

    I haven't had chance to go to the library to get more (I will soon!), but if anybody is curious as to where they're from then you can find them in Tratner's Techniques of Mathematical Analysis.
    I tried the second one too (without the square in the denominator) and got
    Spoiler:
    Show
    \displaystyle-\frac{1}{2}\ln{(x + x^{-1} + 1)} + \frac{1}{2}\ln{(x + x^{-1} - 1)} + c
    but wolfram says different (only slightly different, with x^2 +- x + 1 in the logs). If I am wrong I cant see where I've gone wrong .

    *edit* Well I differentiated it again (
    Spoiler:
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    \displaystyle-\frac{1}{2}\ln{(x + x^{-1} + 1)} + \frac{1}{2}\ln{(x + x^{-1} - 1)} + c
    ) and got back to (x^2 - 1)/(x^4 + x^2 + 1) so I'm assuming I'm right unless I've made 2 catastrophic errors on both integrating/differentiating. Why would wolfram give a different answer to me if I was right?
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    Wolfram have simplified x + 1/x - 1 to (x^2 + 1 - x)/x. Then bottom x cancels out with the other logarithm. I think you assumed they had it in the same form as you. :p:
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    (Original post by AsakuraMinamiFan)
    Wolfram have simplified x + 1/x - 1 to (x^2 + 1 - x)/x. Then bottom x cancels out with the other logarithm. I think you assumed they had it in the same form as you. :p:
    What do you mean cancels out? As in log (x^2 + 1 - x)/x = log (x^2 + 1 - x) - log(x) ?

    I'll have a go at the third one now (assuming I'm getting these right).
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    Yes. There's two logarithm terms remember? You've definitely got it right! You just need to do a bit of simplifying to get Wolfram's answer.

    1/2 [ log( (x^2 + 1 - x)/x ) - log( (x^2 + 1 + x)/x) ]
    = 1/2 [ log (x^2 + 1 - x) - log(x) + log(x) - log(x^2 + 1 + x) ]
    = 1/2 [ log (x^2 + 1 - x) - log(x^2 + 1 + x) ]
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    (Original post by AsakuraMinamiFan)
    Yes. There's two logarithm terms remember? You've definitely got it right! You just need to do a bit of simplifying to get Wolfram's answer.

    1/2 [ log( (x^2 + 1 - x)/x ) - log( (x^2 + 1 + x)/x) ]
    = 1/2 [ log (x^2 + 1 - x) - log(x) + log(x) - log(x^2 + 1 + x) ]
    = 1/2 [ log (x^2 + 1 - x) - log(x^2 + 1 + x) ]
    What level is this question on anyway? It's the first integral question ive actually ever done so I don't know anything in terms of difficulty.
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    x+3=2
    solve for x.

    P.S. Well, in twenty years time I bet that would be on a C1 paper.
 
 
 
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