The Edexcel C4 (21/06/12 - PM) Revision Thread

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  1. raheem94's Avatar
    • TSR Demigod
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    Re: The Edexcel C4 (21/06/12 - PM) Revision Thread
    (Original post by gadzoinks)
    Help ? vectors are my weakest section by far


    The point A has coordinates (7,-1,3) and the point B has coordinates (10,-2,2). The line l has vector equation r= i + j + k + lambda( 3 i - j + k) where lambda is a real parameter.

    a) show that A lies on the line l.
    b)find the length of AB
    c)find the size of the acute angle between the line l and the line segment
    d)hence, calculate the perpendicular distance from B to the line


    I can get a and b.
    I'm struggling with c and d!
    Could someone help me through them?
    This question has been answered on this thread.
  2. Jack_Smith's Avatar
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    Re: The Edexcel C4 (21/06/12 - PM) Revision Thread
    is it me or does the elmwood papers for c4 have c3 questions as well ?
  3. Machoo's Avatar
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    Re: The Edexcel C4 (21/06/12 - PM) Revision Thread
    (Original post by geeogi)
    ah sorry bro but arcsin doesn't = 1/sin(x)

    it's an inverse function rather than reciprocal

    1/sin(x) = cosec(x)
    This may be a big reason why my grades have not been so good! Thanks man!
  4. kozo's Avatar
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    Re: The Edexcel C4 (21/06/12 - PM) Revision Thread
    (Original post by Jack_Smith)
    is it me or does the elmwood papers for c4 have c3 questions as well ?
    Different boards have different topics on C3 and C4 although generally overlap.
    It may either be (i) The papers are for another board or (b) They are expecting (which is fair) you can apply C3 knowledge in C4.

    Elmwood papers, IMO, are the last resort when it comes to papers to give to pupils. They are very same ol, same ol and not really challenging.
    The best, IMO, are the original practice papers set out by Edexcel from ~2004 then Zig Zag, then Solomon.
  5. exe's Avatar
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    Re: The Edexcel C4 (21/06/12 - PM) Revision Thread
    for c4 soloman are kinda bad. for c1 and c2 elmwood were great... just the regular papers seem to be the hardest for c4.
  6. <3 inbetweener =D's Avatar
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    Re: The Edexcel C4 (21/06/12 - PM) Revision Thread
    for the final integration q on june 08, I didn't use substitution, I thought it was easier to use a different method in the textbook (one where you spot if f(x) and [fx)]^n are next to each other blaa), but I didn't need to change the limits or anything... i got the right answer but do you get all the marks?



    also june 2008 q8a driving me insane...

    Curve C with parametric equations x = 8cost and y=4sin2t, 0 < t <pie/2

    Point P lies on C and has coordinates (4, 2 root 3)
    Find the value of t at the point P

    Why do I get two values of t but the markscheme only has 1 ?
  7. The Polymath's Avatar
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    Re: The Edexcel C4 (21/06/12 - PM) Revision Thread
    (Original post by <3 inbetweener =D)
    for the final integration q on june 08, I didn't use substitution, I thought it was easier to use a different method in the textbook (one where you spot if f(x) and [fx)]^n are next to each other blaa), but I didn't need to change the limits or anything... i got the right answer but do you get all the marks?



    also june 2008 q8a driving me insane...

    Curve C with parametric equations x = 8cost and y=4sin2t, 0 < t <pie/2

    Point P lies on C and has coordinates (4, 2 root 3)
    Find the value of t at the point P

    Why do I get two values of t but the markscheme only has 1 ?
    Oh, it's because you used y = 4sin2t to find it, didn't you? That gives you Pi/3 and Pi/6, and if you look at the curve those represent "near the start" and "near the end" of the first /\ part of the sin curve. However, only one of them works for x = 8cost too (can't remember which one, can't be bothered to apply myself and work it out ) I remember wondering which one to go for, and then realised only one worked for them both.
  8. Jack_Smith's Avatar
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    Re: The Edexcel C4 (21/06/12 - PM) Revision Thread
    i generally find substitution very difficult. Is there any general help for substitution questions.
  9. Azii's Avatar
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    Re: The Edexcel C4 (21/06/12 - PM) Revision Thread
    Anyone done question 3 on the examination style paper in the back of the c4 edexcel book? I havent got a clue how to solve it
  10. Jack_Smith's Avatar
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    Re: The Edexcel C4 (21/06/12 - PM) Revision Thread
    (Original post by Azii)
    Anyone done question 3 on the examination style paper in the back of the c4 edexcel book? I havent got a clue how to solve it
    im trying this question cause i struggle with substitution, but icant get the limits without it being a decimal.
    3= sec theta
    sec = 1/cos
    cos = 1/3
    theta= 1.23.... :/ im on radian mode aswell
  11. Azii's Avatar
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    Re: The Edexcel C4 (21/06/12 - PM) Revision Thread
    (Original post by Jack_Smith)
    im trying this question cause i struggle with substitution, but icant get the limits without it being a decimal.
    3= sec theta
    sec = 1/cos
    cos = 1/3
    theta= 1.23.... :/ im on radian mode aswell
    Maybe you gotta use decimals? I dont think substitutions that bad tbh, i find differential equations more trickier..
  12. QwertyG's Avatar
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    Re: The Edexcel C4 (21/06/12 - PM) Revision Thread
    (Original post by kozo)
    The best, IMO, are the original practice papers set out by Edexcel from ~2004 then Zig Zag, then Solomon.
    Why Zig Zag over Solomon papers?
  13. Arva's Avatar
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    Re: The Edexcel C4 (21/06/12 - PM) Revision Thread
    (Original post by Azii)
    Anyone done question 3 on the examination style paper in the back of the c4 edexcel book? I havent got a clue how to solve it
    If you just need a hint, remember to multiply through by the differential of sec\theta and just try to simplify.

    Solution in spoiler:
    Spoiler:
    Show

    u = sec \theta 



du/dx = sec\theta tan \theta
    Therefore:

    \displaystyle\int^3_2 1/(x^2-1)^{3/2}\ dx = \displaystyle\int^3_2 sec\theta tan \theta/(tan^2 \theta)^{3/2}\ d\theta

    Of course  (tan^2\theta)^{3/2} = tan^3 \theta

    So what we need to integrate is :sec \theta / tan^2 \theta
    Which = cot \theta cosec \theta
    This integrates to -cosec \theta
    I have to go now, so I can't change the limits for you. You can do that bit right?
    Last edited by Arva; 10-06-2012 at 20:15.
  14. <3 inbetweener =D's Avatar
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    Re: The Edexcel C4 (21/06/12 - PM) Revision Thread
    (Original post by Junaid96)
    Oh, it's because you used y = 4sin2t to find it, didn't you? That gives you Pi/3 and Pi/6, and if you look at the curve those represent "near the start" and "near the end" of the first /\ part of the sin curve. However, only one of them works for x = 8cost too (can't remember which one, can't be bothered to apply myself and work it out ) I remember wondering which one to go for, and then realised only one worked for them both.
    Oh so because pi/3 is the only one which works if you put it in the x and y equation, you know that t is that value?
  15. grazie's Avatar
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    Re: The Edexcel C4 (21/06/12 - PM) Revision Thread
    (Original post by Arva)
    If you just need a hint, remember to multiply through by the differential of sec\theta and just try to simplify.

    Solution in spoiler:
    Spoiler:
    Show

    u = sec \theta 



du/dx = sec\theta tan \theta
    Therefore:

    \displaystyle\int^3_2 1/(x^2-1)^{3/2}\ dx = \displaystyle\int^3_2 sec\theta tan \theta/(tan^2 \theta)^{3/2}\ d\theta

    Of course  (tan^2\theta)^{3/2} = tan^3 \theta

    So what we need to integrate is :sec \theta / tan^2 \theta
    Which = cot \theta cosec \theta
    This integrates to -cosec \theta
    I have to go now, so I can't change the limits for you. You can do that bit right?
    Your integral solution is correct, but the precision is a bit off in your workings and could lead to confusion. Also, working out how to change the limits is none standard in this question and doing so without using decimal fractions (as per the book answer) is an elegant approach that is rarely (if ever) explained in the book.

    Given the integral with limits

    I=\left[-cosec\theta\right]_{sec\theta=2}^{sec\theta=3}

    Refer to attached file for getting exact cosec values

    I=\left[(-\frac{3\sqrt{2}}{4})-(-\frac{2\sqrt{3}}{3})\right] = \frac{2\sqrt{3}}{3} - \frac{3\sqrt{2}}{4} = \frac{8\sqrt{3} - 9\sqrt{2}}{12}
    Attached Thumbnails
    Click image for larger version. 

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    Last edited by grazie; 11-06-2012 at 09:20.
  16. Arva's Avatar
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    Re: The Edexcel C4 (21/06/12 - PM) Revision Thread
    (Original post by grazie)
    Your integral solution is correct, but the precision is a bit off in your workings and could lead to confusion. Also, working out how to change the limits is none standard in this question and doing so without using decimal fractions (as per the book answer) is an elegant approach that is rarely (if ever) explained in the book.

    Given the integral with limits

    I=\left[-cosec x\right]_{sec\theta=2}^{sec\theta=3}

    Refer to attached file for getting correct cosec values

    I=\left[(-\frac{3\sqrt{2}}{4})-(-\frac{2\sqrt{3}}{3})\right] = \frac{2\sqrt{3}}{3} - \frac{3\sqrt{2}}{4} = \frac{8\sqrt{3} - 9\sqrt{2}}{12}
    Sorry, I was in a rush to go to the gym and I'm also terrible at laTexing, so I tried to leave out steps that I thought wouldn't cause confusion - in future I'll post fully fleshed out solutions.
    I like your changing of the limits by the way, I would have been much lazier and more clumsy, simply using long decimal approximations.
  17. sceptic_medic's Avatar
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    Re: The Edexcel C4 (21/06/12 - PM) Revision Thread
    (Original post by QwertyG)
    Why Zig Zag over Solomon papers?
    Lol, I second that.
  18. Mathematix's Avatar
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    Re: The Edexcel C4 (21/06/12 - PM) Revision Thread
    (Original post by grazie)
    Your integral solution is correct, but the precision is a bit off in your workings and could lead to confusion. Also, working out how to change the limits is none standard in this question and doing so without using decimal fractions (as per the book answer) is an elegant approach that is rarely (if ever) explained in the book.

    Given the integral with limits

    I=\left[-cosec x\right]_{sec\theta=2}^{sec\theta=3}

    Refer to attached file for getting correct cosec values

    I=\left[(-\frac{3\sqrt{2}}{4})-(-\frac{2\sqrt{3}}{3})\right] = \frac{2\sqrt{3}}{3} - \frac{3\sqrt{2}}{4} = \frac{8\sqrt{3} - 9\sqrt{2}}{12}
    That is simply beautiful! :eek: +rep
  19. StraightUpG's Avatar
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    Re: The Edexcel C4 (21/06/12 - PM) Revision Thread
    (Original post by marc_h94)
    Lol, I second that.
    Third!
  20. Pritesh.Mistry's Avatar
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    Re: The Edexcel C4 (21/06/12 - PM) Revision Thread
    Does anyone have a link to all C4 equations we need to know?

    Occasionally when doing a past paper I come across a Volume or Area question on a certain type of shape where I'm not 100% confident what the equation is. (I know, this is like year 6 stuff... but I hate shapes and haven't done them in detail for so long!)

    Thanks
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