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    how did everyone find the P3 exam today??

    i HATED the vector question. *urgh*
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    can any 1 remember what they got?

    and working for qu 6
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    (Original post by [email protected])
    can any 1 remember what they got?

    and working for qu 6
    q 6 was??
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    differentiation,
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    hm... think i was pretty unsuccessful with that! only managed to make it thru a couple of the questions, most of them i had to leave the last parts of it joyous!
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    Question 2 was also exteremly difficult esp. the first part
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    left out most of q 3
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    does anyone remember the answer for the last part of the very last ques of the paper.....it was to find the value of y
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    the only thing i know is that the x cord of q was 11.75. 6/75 for me!!!
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    I got y=1/[e^(pi/6)-1]
    Area of parallelogram = root(675)
    On the parametric question: x coordinate was 47/4. I realise that I cocked the area up but what I did was integrate between curve and y axis (int x dy) and then take that from the area of the quadrilateral bit. I knew what I had to do and I drew it on the diagram and then I went and wrote Integral y dx. Arrrggh. C'est la vie.
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    (Original post by SsEe)
    I got y=1/[e^(pi/6)-1]
    Area of parallelogram = root(675)
    On the parametric question: x coordinate was 47/4. I realise that I cocked the area up but what I did was integrate between curve and y axis (int x dy) and then take that from the area of the quadrilateral bit. I knew what I had to do and I drew it on the diagram and then I went and wrote Integral y dx. Arrrggh. C'est la vie.
    i got the same area for the parallelogram as well i think

    yup i got the same x-cor-ord too

    but the paper was messed up.
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    (Original post by Sinan)
    does anyone remember the answer for the last part of the very last ques of the paper.....it was to find the value of y
    thought the exam was well hard.... the vector question part c and d were well dodgy.

    So was the differential equation bit. i done half of the proof part but then left it. The second part was wierd as well. Got y/y+1 = e (something) but how the hek r u meant to have only y =??? i found c to be something along the lines 1/2ln2 -ln 2????

    i think i screwed up the other integration question as well even tho it was simple.... the last part of that one i got something like 9/8 for the area????

    ther was a nine mark question as well wasn't there? n e one remember what that was about?
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    (Original post by DaKe)
    thought the exam was well hard.... the vector question part c and d were well dodgy.

    So was the differential equation bit. i done half of the proof part but then left it. The second part was wierd as well. Got y/y+1 = e (something) but how the hek r u meant to have only y =??? i found c to be something along the lines 1/2ln2 -ln 2????

    i think i screwed up the other integration question as well even tho it was simple.... the last part of that one i got something like 9/8 for the area????

    ther was a nine mark question as well wasn't there? n e one remember what that was about?
    *sigh*
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    It was harder than any past paper I'd done. For the parallelogram I cheated my ass off. As I'm a big swot I read ahead and I've read the vectors chapter in P6. The area of the parallelogram was 2*(0.5|OB||OD|sin(t)) or something. I recognised that as the modulus of the vector product (like dot product only you get a vector as your answer) and there's a way of calculating that from the determinant of a matrix. I did the question in about 3 lines (3 lines of a 3x3 matrix that is)
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    (Original post by SsEe)
    I got y=1/[e^(pi/6)-1]
    Area of parallelogram = root(675)
    On the parametric question: x coordinate was 47/4. I realise that I cocked the area up but what I did was integrate between curve and y axis (int x dy) and then take that from the area of the quadrilateral bit. I knew what I had to do and I drew it on the diagram and then I went and wrote Integral y dx. Arrrggh. C'est la vie.
    for the integration one i integrated up to P(5,9) or something, then added the area of the triangle from 5 to 11.75 with height 9.... any one know if that sounds rite???
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    (Original post by DaKe)
    thought the exam was well hard.... the vector question part c and d were well dodgy.

    So was the differential equation bit. i done half of the proof part but then left it. The second part was wierd as well. Got y/y+1 = e (something) but how the hek r u meant to have only y =??? i found c to be something along the lines 1/2ln2 -ln 2????
    y/(y+1) = e

    y = ey + e

    y - ey = e

    y(1-e) = e

    y = e/(1-e)

    I don't know what the question was, or if it would matter, though.

    Ben
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    (Original post by Ben.S.)
    y/(y+1) = e

    y = ey + e

    y - ey = e

    y(1-e) = e

    y = e/(1-e)

    I don't know what the question was, or if it would matter, though.

    Ben
    at least we're not alone with messing up. hope the grade boundaries would go down.
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    I left all Q2 except half the first part blank. I needed more time!

    In the last question I got y=sqrt[2]/(e^pi/6 - sqrt[2]), but I probably made an arithmetic slip somewhere cause I did it in like 5mins.
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    (Original post by Ben.S.)
    y/(y+1) = e

    y = ey + e

    y - ey = e

    y(1-e) = e

    y = e/(1-e)

    I don't know what the question was, or if it would matter, though.

    Ben
    shite...... how did i not see that?????
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    The area can't have been 9/8. Far too small. let's see if I remember:
    x = t² + 1
    y = 3t + 3
    Area = 9[(5 + 47/4)/2] - Int (from y = 0 to 9) x dy
    = 603/8 - Int(from t = -1 to 2) x dy/dt dt
    = 603/8 - Int(from t = -1 to 2) (3t²+3) dt
    = 603/8 - [ ((2)³+3(2)) - ((-1)³+3(-1))
    = 603/8 - [ 14 - (-4)]
    = 459/8
 
 
 
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