Hey there! Sign in to join this conversationNew here? Join for free
    Offline

    0
    ReputationRep:
    (Original post by Wunderbarr)
    If you draw the lines, it can be used to explain why the w3 +w2 + w = 0 thing is true, by looking at the vector sum of the lines (if drawn as vectors going from the origin to Z).
    What is the w3 +w2 + w = 0 thing?
    Offline

    13
    ReputationRep:
    (Original post by aproei)
    What is the w3 +w2 + w = 0 thing?
    https://youtu.be/WlM1WjmAkA0?t=14m53s

    Maybe I should've written w2+ w +1 = 0 instead.
    Offline

    0
    ReputationRep:
    (Original post by Wunderbarr)
    https://youtu.be/WlM1WjmAkA0?t=14m53s

    Maybe I should've written w2+ w +1 = 0 instead.
    Gotcha
    Offline

    7
    ReputationRep:
    anyone help on jan 06 4Cii? dont understand what to do after I intergrated by parts

    Posted from TSR Mobile
    Offline

    13
    ReputationRep:
    (Original post by HFancy1997)
    anyone help on jan 06 4Cii? dont understand what to do after I intergrated by parts

    Posted from TSR Mobile
    You get to the necessary result, then realise you need to change the 2arsinh(4/3) -1 into a natural log somehow...

    So, look in the formula booklet. Page 2, bottom right corner.
    Offline

    7
    ReputationRep:
    (Original post by Wunderbarr)
    You get to the necessary result, then realise you need to change the 2arsinh(4/3) -1 into a natural log somehow...

    So, look in the formula booklet. Page 2, bottom right corner.
    Sorry I shouldve been more clear,

    I got uv but dunno what to do with the -intergral(Vdu/dx), I just dont see how went to the next step

    Posted from TSR Mobile
    Offline

    13
    ReputationRep:
    (Original post by HFancy1997)
    Sorry I shouldve been more clear,

    I got uv but dunno what to do with the -intergral(Vdu/dx), I just dont see how went to the next step

    Posted from TSR Mobile
    Rewrite the integral as brackets and powers (so no fractions or square roots).

    You should be able to tell what type of integration is done more clearly.

    The rewritten integral should appear something like:
    Spoiler:
    Show
    2x(9+4x2)½
    Offline

    7
    ReputationRep:
    (Original post by Wunderbarr)
    Rewrite the integral as brackets and powers (so no fractions or square roots).

    You should be able to tell what type of integration is done more clearly.

    The rewritten integral should appear something like:
    Spoiler:
    Show
    2x(9+4x2)½
    damn man, I still dont see it Lol? i wouldve done substitution again for that but isnt that too complicated?

    Posted from TSR Mobile
    Offline

    13
    ReputationRep:
    (Original post by HFancy1997)
    damn man, I still dont see it Lol? i wouldve done substitution again for that but isnt that too complicated?

    Posted from TSR Mobile
    It's a simple integration. Think about reversing the chain rule.
    Offline

    13
    ReputationRep:
    (Original post by HFancy1997)
    damn man, I still dont see it Lol? i wouldve done substitution again for that but isnt that too complicated?

    Posted from TSR Mobile
    I'm going off now, and you should too soon!

    Happy hunting tomorrow everyone!
    Offline

    15
    ReputationRep:
    (Original post by HFancy1997)
    damn man, I still dont see it Lol? i wouldve done substitution again for that but isnt that too complicated?

    Posted from TSR Mobile
    I just watched this video on the topic which helped but I think in the exam I will use substitution even though it takes longer.

    https://www.youtube.com/watch?v=lvkz1fKlxwQ
    Offline

    2
    ReputationRep:
    Could not finish off that last question. I always struggle with actual numbers and tedious expansions. Left my answer as a long string of (3+sqrt(5))^2 and such. Think I got everything else.
    Offline

    16
    ReputationRep:
    Damn that was tough, two new styles of questions never done before too. Annoying though, forgot something very basic during the exam
    Offline

    7
    ReputationRep:
    last question you had to change sinh to exponentials then it gives you a surd numbee, not sure if its right but I did that

    Posted from TSR Mobile
    Offline

    1
    ReputationRep:
    (Original post by Bunderwump)
    Could not finish off that last question. I always struggle with actual numbers and tedious expansions. Left my answer as a long string of (3+sqrt(5))^2 and such. Think I got everything else.
    Yes me too I just ran out of time
    Offline

    1
    ReputationRep:
    I think tha exam is the hardest ive ever done. Im just praying on low grade boundaries or im kissing uni goodbye. How did everone else find it.
    Offline

    1
    ReputationRep:
    Someone do an unofficial markscheme plzz

    Here are fragments from my memory:

    For Q1 I remember getting pi/6 as the integration

    Q2: in part (i) it simplified to 1 - z

    root arguments were pi/18, 13pi/18, -11pi/18 and modulus was root 2

    Q3: M-1^n tends to infinity

    Q4: show that..
    x = plus/ minus ln 3+root5 /2

    curve had y intercept at 2 and was symmetrical for plus and minus x

    Yh thats all I can remember atm and these are just my answers so may be wrong
    Offline

    7
    ReputationRep:
    (Original post by Sophieoo1)
    Someone do an unofficial markscheme plzz

    Here are fragments from my memory:

    For Q1 I remember getting pi/6 as the integration

    Q2: in part (i) it simplified to 1 - z

    root arguments were pi/18, 13pi/18, -11pi/18 and modulus was root 2

    Q3: M-1^n tends to infinity

    Q4: show that..
    x = plus/ minus ln 3+root5 /2

    curve had y intercept at 2 and was symmetrical for plus and minus x

    Yh thats all I can remember atm and these are just my answers so may be wrong
    got all these

    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    (Original post by Sophieoo1)
    Yes me too I just ran out of time
    My answers, by no means right, just want to compare with people :

    1)i) 1 - x^2 + x^4

    ii) x - 1/3 x^3 + 1/5 x^5

    iii) pi/6

    iv) Draw the graph, sort of like a loop but starting at pi/4 and ending on the horizontal axis

    As the angles tends to 0 r tends to infinity

    v) a^2 ln(2 root2) or 1/2 a^2 ln 8

    2)i) 1-z

    ii) Show that C + jS thing

    iii) Modulus of the cube roots was root2 the angles were pi/18 13pi/18 25pi/18


    3)i) Eigenvalues were -1/6, 1 eigenvectors were (1 1) and (3 -4) I think, cant quite remember the order

    ii) M^n tended towards (3 4) or something like that order might be different
    ................................ .(3 4)
    cant quite remember

    iii) Not too sure on this, I think it didnt tend to a limit (maybe infinity), as the elements of the matrix were greater than 1 but some were negative so it could have positive or negative infinity

    4)i) Show that arcosh thing

    ii) ln((3+root5)/2) and ln((3-root5)/2)

    iii) The graph sort of looked like x^2 but started at y=2

    The area bound by the curve was 5root5 / 2
    The area bound by the line y = 5 was 5ln((7+3root5)/2)

    Then take one away from the other.


    I think that was all the questions, not sure how many are right, just what I got
    Offline

    16
    ReputationRep:
    I ****ed that up. **** **** **** ****
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you rather give up salt or pepper?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Write a reply...
    Reply
    Hide
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.