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Edexcel C4 Revision Thread (28/06 14:00) watch

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    (Original post by nas7232)
    you should differentiate it via the product rule.
    Product & Chain Rule at the same time
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    (Original post by TheWolf)
    Yes I've got that, but the answer (q2 mock) says another thing? http://imaad.pwp.blueyonder.co.uk/mockanswer1.jpg

    naz i do use the product rule
    ah sorry i thought you said you used the chain rule
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    (Original post by samdavyson)
    Product & Chain Rule at the same time
    ah :cool:
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    (-5ln(2/3) - 2ln(5/3)) - (-5ln(5/6) - 2ln(4/3))

    => -5ln(2/3) + 5ln(5/6) - 2ln(5/3) + 2ln(4/3)

    => 5ln([5/6]/[2/3]) + 2ln([4/3]/[5/3])

    => 5ln(5/4) + 2ln(4/5)

    Argh! The question disappeared!
    • Thread Starter
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    Thanks. **** i totally missed that ='(
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    Good Luck All.

    Good night.
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    (Original post by samdavyson)
    Good Luck All.

    Good night.
    Good luck to you also!
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    Q2 - Mock paper.

    Ok, so when you get to:

    3x²-2x dy/dx + 2y - 4 + 3y² dy/dx = 0

    Which way do you rearrange that?
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    Hello all, i've been doing some questions from the Edexcel Papers of the Mathmo gmail account...got stuck on this question:

    int 1/(1-x²)^³/² dx

    =

    X/(1-x²)^½ + C

    Using X=sinθ, show....etc

    In the marksceme they have used:

    int cosθ/cos³θ = int sec²...

    And so on, but i am unsure of how they got there. Any hints?

    :tsr:
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    (Original post by bavman)
    Hello all, i've been doing some questions from the Edexcel Papers of the Mathmo gmail account...got stuck on this question:

    int 1/(1-x²)^³/² dx

    =

    X/(1-x²)^½ + C

    Using X=sinθ, show....etc

    In the marksceme they have used:

    int cosθ/cos³θ = int sec²...

    And so on, but i am unsure of how they got there. Any hints?

    :tsr:
    int 1/(1-x²)^³/² dx
    x = sinθ
    dx/dθ = cos θ
    dx = cosθ dθ

    (1-x²)^3/2 = (1-sin²θ )^3/2 = (cos²θ )^3/2 = cos³θ

    ∫ 1/cos³θ x cosθ dθ
    ∫ cosθ/cos³θ dθ
    ∫ sec²θ dθ

    and take it away from there!
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    isnt C4 at 1pm
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    Depends. Anything from 1 to 2, depending on the centre.
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    Specimen 7b =0 how to di it?
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    (Original post by TheWolf)
    Specimen 7b =0 how to di it?
    look at this thread dealing with the exact same question:
    http://www.thestudentroom.co.uk/t126670.html
    • Thread Starter
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    How to approach part a?
    Attached Images
     
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    I'd say find the equation of the line AB in the form r = a +λb, where a = (9i-8j+2k) and b = (4j+k) - (9i-8j+2k).
    Then equate the two line equations, and equate coefficients of i, j and k, giving 3 simultaneous equations. Finally, solve, and sub either λ or µ back into one of the line equations giving the position vector of C (i.e. the point where the two roads intersect).
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    Cheers!. :3
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    say if i found the gradient to be 2cos2t/2sint and I wanted to find what the gradient was at pi over 3, how would I type this into my calculator, starting with the top first? would be 2 times pi over 3, cos the answer and then times this answer by 2 for the top row?
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    (Original post by dnbhead)
    say if i found the gradient to be 2cos2t/2sint and I wanted to find what the gradient was at pi over 3, how would I type this into my calculator, starting with the top first? would be 2 times pi over 3, cos the answer and then times this answer by 2 for the top row?
    That sounds okay.
    You could alternately abuse brackets and do (2cos(2pi/3))/(2sin(2pi/3)) but a calculator isn't essential has cos(2pi/3)=-1/2 and sin(pi/3)=rt3/2 IIRC
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    (Original post by dnbhead)
    say if i found the gradient to be 2cos2t/2sint and I wanted to find what the gradient was at pi over 3, how would I type this into my calculator, starting with the top first? would be 2 times pi over 3, cos the answer and then times this answer by 2 for the top row?
    i would say, just dont use a calculator!
    cos (2pi/3) = -cos(pi/3) = -0.5
    sin (pi/3) = √3/2
    then just plug it in!
 
 
 
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