Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    Offline

    3
    ReputationRep:
    (Original post by oinkk)
    I'd use the symmetry of the curves (i.e., deal with everything above the initial line, then double at the end).

    Work out the area bounded by the sector above the initial line (C2 formula), and then integrate the curve C1 between pi/3 and pi.

    ... and then double your answer to that.
    Thank you!
    Offline

    3
    ReputationRep:
    (Original post by economicss)
    Hi, does anyone know please whether the integral in question 6 here is on the spec http://madasmaths.com/archive/maths_...stitutions.pdf as I've not seen an integral that leads to arctan as the answer before? Thanks
    It's covered in FP3, probably won't show up in FP2. It's pretty simple to do just using C3/4 knowledge, though - just think of an appropriate trig substitute.
    Offline

    3
    ReputationRep:
    Please could anyone explain question 11, especially part c, really struggling with these complex types! Thank you ZackenName:  image.jpg
Views: 180
Size:  452.1 KB
    Offline

    3
    ReputationRep:
    (Original post by economicss)
    Please could anyone explain question 11, especially part c, really struggling with these complex types! Thank you ZackenName:  image.jpg
Views: 180
Size:  452.1 KB
    From part a(i) and a(ii) you get lambad =3 and mu=-sqrt(3)
    From the sketch you did in part b, you have 4 points that belong on the circle.
    So, you can subistute the cartesian equivlents of these points into the equation for a circle (x-a)^2 + (y-b)^2 = r^2 and then solve silmutaneously. Because the 4 points lie on either the x or y axis, this makes it a lot easier to solve.
    you should get centre (-0.5(1+sqrt(3)), 0.5(1+sqrt(3)))
    Offline

    3
    ReputationRep:
    (Original post by kennz)
    From part a(i) and a(ii) you get lambad =3 and mu=-sqrt(3)
    From the sketch you did in part b, you have 4 points that belong on the circle.
    So, you can subistute the cartesian equivlents of these points into the equation for a circle (x-a)^2 + (y-b)^2 = r^2 and then solve silmutaneously. Because the 4 points lie on either the x or y axis, this makes it a lot easier to solve.
    you should get centre (-0.5(1+sqrt(3)), 0.5(1+sqrt(3)))
    Thanks for your help how did you get the values for lambda and mu in the first part please? Thanks
    Offline

    2
    ReputationRep:
    Can anyone help me with part c please as it doesnt mention z so i dont know how to bring z-3i in
    Attached Images
     
    Offline

    11
    ReputationRep:
    (Original post by rm761)
    Can anyone help me with part c please as it doesnt mention z so i dont know how to bring z-3i in
    It must be a mistake, otherwise how can it transform from the z plane to the w plane if it doesn't contain z? Try w = 2i/z
    Offline

    3
    ReputationRep:
    (Original post by economicss)
    Thanks for your help how did you get the values for lambda and mu in the first part please? Thanks


    sorry for the late reply
    for part a(i), you can use argz= arctan(b/a) where b=complex part and a=real part.
    since the number lamba has no real part, subsitute z=bi
    split the arg(bi-i/bi+1) into arg(bi-i)-arg(bi+1) = pi/6
    arg(bi-i)=arg(i(b+1))=pi/2 so now
    pi/2-pi/6=pi/3=arg(bi+1)
    tan both sides as arg(bi+1) = arctan (b)
    therefore b=tanpi/3 = sqrt(3)

    for part a(ii) the logic is the same but substute z=a where a=real part as mu lies on the real axis so imaginary part =0 and then split the arg again.
    I can go through this if you want but I'll let you try part a(ii) first. Its quite an intuitive solution but Im pretty sure its correct.
    Offline

    9
    ReputationRep:
    Name:  fp2 june 2014 q6 b.jpg
Views: 135
Size:  354.4 KB June 2014 FP2 question 6 (b)

    For question 6 (b) in the second line of working where does the -3 come from in the first bracket?
    Offline

    3
    ReputationRep:
    (Original post by kennz)
    sorry for the late reply
    for part a(i), you can use argz= arctan(b/a) where b=complex part and a=real part.
    since the number lamba has no real part, subsitute z=bi
    split the arg(bi-i/bi+1) into arg(bi-i)-arg(bi+1) = pi/6
    arg(bi-i)=arg(i(b+1))=pi/2 so now
    pi/2-pi/6=pi/3=arg(bi+1)
    tan both sides as arg(bi+1) = arctan (b)
    therefore b=tanpi/3 = sqrt(3)

    for part a(ii) the logic is the same but substute z=a where a=real part as mu lies on the real axis so imaginary part =0 and then split the arg again.
    I can go through this if you want but I'll let you try part a(ii) first. Its quite an intuitive solution but Im pretty sure its correct.
    No problem, thank you I understand now up to splitting the args but how do you know the value of arg(bi-i) and arg(bi+1) I tried it for part ii aswell but got stuck at finding the values of the args again? Thanks for your help
    Offline

    3
    ReputationRep:
    (Original post by Music With Rocks)
    Name:  fp2 june 2014 q6 b.jpg
Views: 135
Size:  354.4 KB June 2014 FP2 question 6 (b)

    For question 6 (b) in the second line of working where does the -3 come from in the first bracket?
    Because we're told in the question that the equation of the circle is in the form (u-3) squared +v squared= k squared Hope that helps
    Offline

    9
    ReputationRep:
    (Original post by economicss)
    Because we're told in the question that the equation of the circle is in the form (u-3) squared +v squared= k squared Hope that helps
    ohhhhhhh yes that helps, I get it now. Thank you very much
    Offline

    3
    ReputationRep:
    (Original post by Music With Rocks)
    ohhhhhhh yes that helps, I get it now. Thank you very much
    No problem
    Offline

    3
    ReputationRep:
    (Original post by economicss)
    No problem, thank you I understand now up to splitting the args but how do you know the value of arg(bi-i) and arg(bi+1) I tried it for part ii aswell but got stuck at finding the values of the args again? Thanks for your help
    we know the value of arg(bi-i) as its purely imaginary and since b>1, the complex number (bi-i) lies on the positive imaginary axis so its argument is pi/2.

    we dont know the value of arg (bi+1) but we know the angle is arctan(b/1)=arctan(b).

    for part (ii)
    arg(a-i/a+1)=pi/6
    since a<-1, the complex number a+1 lies on the negative real axis so its agrument is pi, therefore arg(a-i)-pi=pi/6 so arg(a-i)=7pi/6
    now arg(a-i)=arctan(-1/a) so -1/a=tan(7pi/6)=-1/sqrt(3)
    therefore a=-sqrt(3)
    Offline

    0
    ReputationRep:
    Name:  jan 2003 q8.png
Views: 120
Size:  271.2 KB
    Can anyone explain part c to me please? The mark scheme says x=-1/4 a then WX = 2a +1/4 a but I dont understand how they have got this. Thanks
    Offline

    22
    ReputationRep:
    (Original post by lkara)
    Name:  jan 2003 q8.png
Views: 120
Size:  271.2 KB
    Can anyone explain part c to me please? The mark scheme says x=-1/4 a then WX = 2a +1/4 a but I dont understand how they have got this. Thanks
    The distance from O horiztontally to the left of the box is 1/4a

    The distance from O horizontally to the right of the box is r = a(1+1) = 2a by subbing in theta = 0 into the polar eq.

    Then asding these two distances gives you the total didtance from the left to the right of the box.
    Offline

    22
    ReputationRep:
    (Original post by Music With Rocks)
    Name:  fp2 june 2014 q6 b.jpg
Views: 135
Size:  354.4 KB June 2014 FP2 question 6 (b)

    For question 6 (b) in the second line of working where does the -3 come from in the first bracket?
    The question says (u -3)^2.
    Offline

    0
    ReputationRep:
    (Original post by Zacken)
    The distance from O horiztontally to the left of the box is 1/4a

    The distance from O horizontally to the right of the box is r = a(1+1) = 2a by subbing in theta = 0 into the polar eq.

    Then asding these two distances gives you the total didtance from the left to the right of the box.
    Ahhh that makes sense now, thank you!
    Offline

    6
    ReputationRep:
    Is it correct to assume that if you pick any two points on a circle and its perpendicular bisector of the line joining these points goes through the centre of circle?
    Offline

    4
    ReputationRep:
    Just a quick question;

    I was doing this first order differential equation...
    Name:  Screenshot_72.png
Views: 78
Size:  13.0 KBName:  Screenshot_71.png
Views: 97
Size:  23.6 KB
    and I don't know whether I'd get the last 2 marks or not. I got everything correct up until the last line, where, rather than putting the answer they give, I put the middle term as sin(2x)/4xsin(x), an unsimplified, but correct answer. But because I didn't simplify fully, would I lose those two marks? The question doesn't say simplify fully so I didn't, but I didn't want to assume just in case :P

    Thanks for your help in advance
    Attached Images
     
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: February 24, 2017
Poll
Do you agree with the proposed ban on plastic straws and cotton buds?

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.