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

Edexcel FP3 - 27th June, 2016 watch

  • View Poll Results: How did you find the Edexcel FP3 exam?
    Very hard
    41
    8.40%
    Hard
    69
    14.14%
    Normal
    156
    31.97%
    Easy
    165
    33.81%
    Very easy
    57
    11.68%

    Offline

    22
    ReputationRep:
    (Original post by samb1234)
    For diagonal matrices, does it matter where my numbers are in the matrix, e.g. the solution banks answer is 0 top left -1 mid 8 bottom right whereas mine is -1 top left 8 mid 0 bottom right. I'm assuming i've just defined my P matrix slightly differently to them hence the difference
    Nope, not a problem.
    Offline

    3
    ReputationRep:
    (Original post by target21859)

    Hello, I've got a few questions. For question 7 june 2014 FP3, for the surface area question why do you use 2pi and not pi for the surface area formula because the question says it is rotated through pi radians. Is it because the positive arc and negative arc of the circle both rotate through pi radians thus we they both add to give a total surface area of 2pi*(integral)? Also for question 7c why is pi/2 instead of just pi? Is it because y is positive so you only consider the positive quadrant so you're working out a quarter of the total circumference? Since if you squared both sides y could +1 or -1 when x=0 but the -1 is creating from squaring both sides therefore the domain of y is y=>0. That's what I think but I'm not sure. Lastly, do you think this year's paper will be as hard as the last two? I find it much better that way since the grade boundaries are so much lower.
    Thanks in advance Sorry if there are too many questions.
    b) Check your formula book for the surface area of revolution. It's always 2pi.
    c) C2 arc length formula (= rθ). You have a radius of 1 and an angle of pi/2 (from the x-axis anticlockwise to the y axis). Hence it's just pi/2.

    As for this year's exam, nobody can tell. Recently they've been really been discriminating the higher grades with difficult loci problems. FP3 will always be a challenging module, you've just got to prepare yourself for worst-case scenarios (exams are always worst case... low grade boundaries = hard paper, high grade boundaries = harder to get a really high grade).
    Offline

    12
    ReputationRep:
    (Original post by Zacken)
    Nope, not a problem.
    thanks i thought so just checking
    Offline

    2
    ReputationRep:
    (Original post by oinkk)
    b) Check your formula book for the surface area of revolution. It's always 2pi.
    c) C2 arc length formula (= rθ). You have a radius of 1 and an angle of pi/2 (from the x-axis anticlockwise to the y axis). Hence it's just pi/2.

    As for this year's exam, nobody can tell. Recently they've been really been discriminating the higher grades with difficult loci problems. FP3 will always be a challenging module, you've just got to prepare yourself for worst-case scenarios (exams are always worst case... low grade boundaries = hard paper, high grade boundaries = harder to get a really high grade).
    b) so it's 2pi if rotated through pi radians and 4pi of rotated through 2pi radians?
    c) why isn't the angle pi radians? If you go from 0 to 1 on the x axis you have y going from 1 to -1 so half the circumference of the circle
    Offline

    3
    ReputationRep:
    (Original post by target21859)
    b) so it's 2pi if rotated through pi radians and 4pi of rotated through 2pi radians?
    c) why isn't the angle pi radians? If you go from 0 to 1 on the x axis you have y going from 1 to -1 so half the circumference of the circle
    When we rotate an ordinary 2D curve, say the curve y = x^{2} about the x-axis, we are rotating it 2\pi radians. This forms a 3D shape.

    In this question, however, we have a complete 2D circle as opposed to a curve, so we only needed to rotate it \pi to produce the sphere.

    Try and visualise rotating both of these curves until you reach a 3D shape. Hopefully you can see the circle requires less of a rotation to achieve this. Why bother rotating it any more? You already have a 3D shape to find the area of.

    The actual reason why we use 2\pi in the formula is to do with how the formula is actually derived. If you have access to the FP3 textbook, page 82 explains this to an extent.

    Regardless of the shape and the angle of rotation, you still use 2\pi.

    Integration is all about summation of many little areas to find the overall area. That is to say, we are summing infinitesimally between two points on the curve, working along the entire curve (that is to say, between our bounds). And each of these two points produces a frustum (a cone with it's top bit cut off), with a specific area (given in the textbook).

    The 2\pi comes from the surface area of a frustum, not the angle of rotation. That is to say, the 2\pi in the formula is completely independent of the angle of rotation.

    This summation of all the little areas produced forms the equation for the surface area of revolution.

    The angle is not pi radians. Please see my attached image with an explanation. I've also attached the unit circle, which gives common angles on the axes.

    Name:  unit circle.png
Views: 473
Size:  303.2 KB
    Attachment 539033539035
    Attached Images
     
    Offline

    2
    ReputationRep:
    (Original post by oinkk)
    When we rotate an ordinary 2D curve, say the curve y = x^{2} about the x-axis, we are rotating it 2\pi radians. This forms a 3D shape.

    In this question, however, we have a complete 2D circle as opposed to a curve, so we only needed to rotate it \pi to produce the sphere.

    Try and visualise rotating both of these curves until you reach a 3D shape. Hopefully you can see the circle requires less of a rotation to achieve this. Why bother rotating it any more? You already have a 3D shape to find the area of.

    The actual reason why we use 2\pi in the formula is to do with how the formula is actually derived. If you have access to the FP3 textbook, page 82 explains this to an extent.

    Regardless of the shape and the angle of rotation, you still use 2\pi.

    Integration is all about summation of many little areas to find the overall area. That is to say, we are summing infinitesimally between two points on the curve, working along the entire curve (that is to say, between our bounds). And each of these two points produces a frustum (a cone with it's top bit cut off), with a specific area (given in the textbook).

    The 2\pi comes from the surface area of a frustum, not the angle of rotation. That is to say, the 2\pi in the formula is completely independent of the angle of rotation.

    This summation of all the little areas produced forms the equation for the surface area of revolution.

    The angle is not pi radians. Please see my attached image with an explanation. I've also attached the unit circle, which gives common angles on the axes.

    Name:  unit circle.png
Views: 473
Size:  303.2 KB
    Attachment 539033539035
    First off, thank you for the very detailed explanation. It is very much appreciated. I see now why the 2pi is there. In my lessons, my teacher usually went through the proofs for the various formulae we used but not this one. I find that going through the proof really helps as you can see where the formula came from. I found the explanation you said in my FP3 textbook and that really helped as well. I also understand now about the last part of the question. I was using the whole circle from part a and not the y=root(1-x^2) graph. The latter only has positive y values and that's why the angle is only pi/2 radians.
    I'm feeling fairly confident at the moment in FP2 and FP3, I'm doing well in the past papers but I keep getting the feeling that I'll panic in the exam and drop silly marks. The FP2 grade boundaries for an A* are so high (71 last year).The margin for error is so small which worries me. I've done quite a few past papers now and I usually finish in 45 mins which leaves ample time for checking so I should be fine.
    Offline

    3
    ReputationRep:
    (Original post by target21859)
    First off, thank you for the very detailed explanation. It is very much appreciated. I see now why the 2pi is there. In my lessons, my teacher usually went through the proofs for the various formulae we used but not this one. I find that going through the proof really helps as you can see where the formula came from. I found the explanation you said in my FP3 textbook and that really helped as well. I also understand now about the last part of the question. I was using the whole circle from part a and not the y=root(1-x^2) graph. The latter only has positive y values and that's why the angle is only pi/2 radians.
    I'm feeling fairly confident at the moment in FP2 and FP3, I'm doing well in the past papers but I keep getting the feeling that I'll panic in the exam and drop silly marks. The FP2 grade boundaries for an A* are so high (71 last year).The margin for error is so small which worries me. I've done quite a few past papers now and I usually finish in 45 mins which leaves ample time for checking so I should be fine.
    Glad I could help!

    And I know the feeling. I'm in the same boat as you... I've FP2 in less than two weeks and FP3 in one month tomorrow.

    I make little mistakes in every paper, so an A* in FP2 is going to be so tough. I'm hoping I can make that up in FP3, however, as boundaries are traditionally low to normal. Hopefully a high A in FP2 and a strong A* in FP3 will balance with whatever is my next best A2 module to get me that A*.

    Best of luck in your exams!
    Offline

    2
    ReputationRep:
    (Original post by oinkk)
    Glad I could help!

    And I know the feeling. I'm in the same boat as you... I've FP2 in less than two weeks and FP3 in one month tomorrow.

    I make little mistakes in every paper, so an A* in FP2 is going to be so tough. I'm hoping I can make that up in FP3, however, as boundaries are traditionally low to normal. Hopefully a high A in FP2 and a strong A* in FP3 will balance with whatever is my next best A2 module to get me that A*.

    Best of luck in your exams!
    Yeah good luck in yours as well. Are you doing maths at uni?
    Offline

    3
    ReputationRep:
    (Original post by target21859)
    Yeah good luck in yours as well. Are you doing maths at uni?
    A-levels at the moment, then Computer Science at Durham come October :-)

    Yourself?
    Offline

    2
    ReputationRep:
    (Original post by oinkk)
    A-levels at the moment, then Computer Science at Durham come October :-)

    Yourself?
    Maths at Warwick hopefully but I need 3A*s :/
    Offline

    3
    ReputationRep:
    (Original post by target21859)
    Maths at Warwick hopefully but I need 3A*s :/
    OMG ouch! I'm sure it's doable. Good luck!

    I have a reduced offer of AB (from A*AA), so I can fail one subject if I wanted

    But I'm still determined
    Offline

    19
    ReputationRep:
    (Original post by target21859)
    Maths at Warwick hopefully but I need 3A*s :/
    Not doing STEP?
    Offline

    2
    ReputationRep:
    (Original post by 13 1 20 8 42)
    Not doing STEP?
    No I thought it would be better to concentrate on my other exams. I am doing 4 A Levels so I've already got enough on my plate. I did go to STEP sessions for two years though. I enjoyed them because we did some interesting maths but I'd rather not do an exam if I had to. I've already got the A* in maths and the A* in French is pretty much guaranteed so I just need to do well in further maths ie not make silly mistakes like in the exams I did last year.
    Offline

    4
    ReputationRep:
    Bump.
    Offline

    20
    ReputationRep:
    (Original post by oinkk)
    When we rotate an ordinary 2D curve, say the curve y = x^{2} about the x-axis, we are rotating it 2\pi radians. This forms a 3D shape.

    In this question, however, we have a complete 2D circle as opposed to a curve, so we only needed to rotate it \pi to produce the sphere.

    Try and visualise rotating both of these curves until you reach a 3D shape. Hopefully you can see the circle requires less of a rotation to achieve this. Why bother rotating it any more? You already have a 3D shape to find the area of.

    The actual reason why we use 2\pi in the formula is to do with how the formula is actually derived. If you have access to the FP3 textbook, page 82 explains this to an extent.

    Regardless of the shape and the angle of rotation, you still use 2\pi.

    Integration is all about summation of many little areas to find the overall area. That is to say, we are summing infinitesimally between two points on the curve, working along the entire curve (that is to say, between our bounds). And each of these two points produces a frustum (a cone with it's top bit cut off), with a specific area (given in the textbook).

    The 2\pi comes from the surface area of a frustum, not the angle of rotation. That is to say, the 2\pi in the formula is completely independent of the angle of rotation.

    This summation of all the little areas produced forms the equation for the surface area of revolution.

    The angle is not pi radians. Please see my attached image with an explanation. I've also attached the unit circle, which gives common angles on the axes.

    Name:  unit circle.png
Views: 473
Size:  303.2 KB
    Attachment 539033539035
    That is incorrect.
    The reason we use 2 pi is because a circle isn't a function, you can't integrate non functions (not at A-Level). You're in actuality only integrating the positive bit of the circle, hence why you're turning it through 2 pi if the circle was turned through pi. In other words, if the circle was turned through pi/2, your integration should include (pi) and not (2pi). You can think about it this way, by turning a circle to form a volume, you're actually turning two half circles, thus the volume for each half circle is multiplied by 2. So the 2 pi is actually pi due to rotation multiplied by the 2 as you're turning two half circles.

    Take a look at this:
    http://www.math-prof.com/Calculus_1/Calc_Ch_32.asp
    Offline

    4
    ReputationRep:
    Anyone have any tricky reduction integrals? I'm happy that they're probably my strongest part of FP3 following the general hyperbolics/differentiation stuff.
    Offline

    11
    ReputationRep:
    (Original post by Craig1998)
    Anyone have any tricky reduction integrals? I'm happy that they're probably my strongest part of FP3 following the general hyperbolics/differentiation stuff.
    Given that \displaystyle I_n = \int_{0}^{2}x^n \sqrt{2x-x^2}\hspace{4pt}\mathrm{d}x, \hspace{15} n\geqslant0

    (a) find I_0

    (b) Show that \displaystyle I_n = \frac{2n+1}{n+2}I_{n-1}

    (c) Hence show that I_n = \displaystyle\frac{(2n+1)! \pi}{n!(n+2)!2^n}
    Offline

    4
    ReputationRep:
    (Original post by A Slice of Pi)
    Given that \displaystyle I_n = \int_{0}^{2}x^n \sqrt{2x-x^2}\hspace{4pt}\mathrm{d}x, \hspace{15} n\geqslant0

    (a) find I_0

    (b) Show that \displaystyle I_n = \frac{2n+1}{n+2}I_{n-1}

    (c) Hence show that I_n = \displaystyle\frac{(2n+1)! \pi}{n!(n+2)!2^n}
    I'll have a go at this tomorrow, but looking at it I'm guessing there's a subtitution for part a. Part b I'd change to I_n = \int_{0}^{2}x^{n-1} . x\sqrt{2x-x^2} dx, then use parts and a substitution u = 2x-x^2 to integrate the RHS bit.
    Offline

    11
    ReputationRep:
    (Original post by Craig1998)
    I'll have a go at this tomorrow, but looking at it I'm guessing there's a subtitution for part a. Part b I'd change to I_n = \int_{0}^{2}x^{n-1} . x\sqrt{2x-x^2} dx, then use parts and a substitution u = 2x-x^2 to integrate the RHS bit.
    How did this turn out for you? Did you get the result?
    Offline

    18
    ReputationRep:
    (Original post by A Slice of Pi)
    Given that \displaystyle I_n = \int_{0}^{2}x^n \sqrt{2x-x^2}\hspace{4pt}\mathrm{d}x, \hspace{15} n\geqslant0

    (a) find I_0

    (b) Show that \displaystyle I_n = \frac{2n+1}{n+2}I_{n-1}

    (c) Hence show that I_n = \displaystyle\frac{(2n+1)! \pi}{n!(n+2)!2^n}
    Very nice indeed.
    I cna see the method but can't be bothered to put pen to paper.


    Posted from TSR Mobile
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: August 20, 2016
Poll
Do you like carrot cake?

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.