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    (Original post by endeavour)
    Do you just want help with the differentiation?

    M = 10(0.98)t
    lnM = ln10 + tln0.98

    differentiating implicitly wrt t:
    (1/M)(dM/dt) = ln0.98
    dM/dt = M ln0.98
    thanks. i didnt actually get what the question meant and what i was supposed to do.
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    (Original post by TheWolf)
    What new formulas (other than c3 ones) do we need to learn? I find the half angle ones v.useful.
    I have made a list of integrals for C4 (link) and other than those you need to know:

    -- Scalar Product Formula

    -- Angle between two Vectors (rearrangement of SP formula)

    -- Separation of Variable Formula (not a formula as such)

    -- Volume Formula (again not really a formula)

    -- Double Angle Formulae (V. Useful as you say)

    That is it.

    So basically: nothing.
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    (Original post by nas7232)
    can someone do this question for me

    evaluate
    ∫[2/(x-3)] - [1/(x+4)] ∂x

    Between 2 and 0

    plz do u have answers to mock c4 that u offer???? :rolleyes:

    if so plz send it to me on my mail : [email protected]
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    (Original post by ABUSHOMAL///M3)
    plz do u have answers to mock c4 that u offer???? :rolleyes:

    if so plz send it to me on my mail : [email protected]
    Just from this thread alone:

    http://www.thestudentroom.co.uk/show...9&postcount=25
    http://www.thestudentroom.co.uk/show...9&postcount=35
    http://www.thestudentroom.co.uk/show...9&postcount=44

    and now:

    http://www.thestudentroom.co.uk/show...9&postcount=63

    All people requesting the mock paper solutions.

    Is there an official mark scheme? I don't think so.

    I am yet to do the paper myself, but when I do (tomorrow) I think I will type up my answers (not the mark scheme just the numbers) to help all of you folk out.

    In the meantime why don't any of you do the same and then compare and constrast.
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    just started my c4 revision and this has got to b the best thread ever

    i :love: The Wolf

    thanks all
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    it says edexcel praccies file is invalid or currupted :confused:

    how close r the zigzag ones to the syllabus cos they work
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    Here is a very basic integration by substitution question but I'm not sure what I'm doing.

    (1 + sinx)/cosx)dx ; u = sinx
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    (Original post by ntrik)
    Here is a very basic integration by substitution question but I'm not sure what I'm doing.

    (1 + sinx)/cosx)dx ; u = sinx
    any limits?

    integrate with respect to u

    du/dx = cosx
    dx = du/cosx

    x= arcsin u

    substitute that in
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    (Original post by nas7232)
    any limits?

    integrate with respect to u

    du/dx = cosx
    dx = du/cosx

    x= arcsin u

    substitute that in
    No limits. Where would you substitute x = arcsin u? in both sinx and cosx?
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    (Original post by ntrik)
    Here is a very basic integration by substitution question but I'm not sure what I'm doing.

    ∫((1 + sinx)/(cosx))dx ; u = sinx
    ∫((1 + sinx)/(cosx))dx = I

    du/dx = cosx

    dx/du = 1/cosx

    dx = (1/cosx)du

    So when the dx is replaced you'll get cos²x on the bottom.

    Find cos²x in terms of u.

    cos²x = 1 - sin²x = 1 - u²

    So: I = ∫(1 + u)/(1 - u²)du

    I = ∫(1 + u)/(1 + u)(1 - u)du (diff of 2 squares)

    I = ∫1/(1 - u)du

    I = -ln(1 - u) + C

    I = -ln(1 - sinx) + C

    EDIT: making it correct.
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    (Original post by samdavyson)
    So: I = ∫(u)/(1 - u²)du
    Thanks, I got your pm but at this point, isn't it I = ∫(1 + u)/(1 - u²)du ?
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    hey guys~ kinda stuck in Jan 2002 Q.8 part (c). can't figure out the area of the parallelogram~0_0..help plz~
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    (Original post by ntrik)
    Thanks, I got your pm but at this point, isn't it I = ∫(1 + u)/(1 - u²)du ?
    Sorry at that point I was thinking 1 + sinx = u.

    I have corrected the post above.
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    (Original post by im37)
    hey guys~ kinda stuck in Jan 2002 Q.8 part (c). can't figure out the area of the parallelogram~0_0..help plz~
    What is that Jan 02 P3?
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    (Original post by samdavyson)
    What is that Jan 02 P3?
    yes, forget to specify
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    (Original post by im37)
    yes, forget to specify
    Umm.

    Parallelogram has area 1/2 x base x height.

    You find the base and height lengths from the equations of the normals to the ellipse at the given values of the parameter.

    (You will only need to do one normal due to the symmetry).

    [ paper link ]
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    i see wt u mean, but i dunno how to find the lengths of base & height in this case....
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    (Original post by im37)
    i see wt u mean, but i dunno how to find the lengths of base & height in this case....
    Find equations of normals.

    Sub in y = 0 for x axis intersection, and x = 0 for y axis intersection.

    Times both values by 2.

    These are the base and height.
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    thx alot~ i got it!
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    there is an official mark shecem for the mock , ill scan it in later today and post it, prob in the evenin say 11ish
 
 
 
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