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Edexcel FP2 Official 2016 Exam Thread - 8th June 2016 watch

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    (Original post by disabledalpaca)
    Could someone help me on how to solve part b, the markscheme doesnt make sense to me. Where does the -pi/9*2^2 part come from?

    Attachment 538063Attachment 538063538065
    It's the area "underneath" the equation r=2... When you find the area of r=1.5+sin3(theta), you get an excess area you don't want. In the marks scheme, they subtracted this area bounded by r=2 using the formula 1/2(r)^2(theta) You only want R, not the entire area underneath it.

    So here

    you want the area where only the blue lines are visible (region S), so you subtract region r=2 (which covers all the area in black) from the "upper" point at where they intersect to the "lower" point they intersect. These points are 5pi/18 and pi/18. Integrate r^2 = 2^2 = 4 between these two points and you should get 4pi/9. This is the area bounded by the line r=2 and the initial line, in other words, the area shaded by BOTH the blue and black lines. Sorry for my bad explanation. Hope it helps.
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    does anyone have resources (other than the FP2 past papers) for questions?
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    Thank you so much for that I understand now!!

    (Original post by tripleseven)
    It's the area "underneath" the equation r=2... When you find the area of r=1.5+sin3(theta), you get an excess area you don't want. In the marks scheme, they subtracted this area bounded by r=2 using the formula 1/2(r)^2(theta) You only want R, not the entire area underneath it.

    So here

    you want the area where only the blue lines are visible (region S), so you subtract region r=2 (which covers all the area in black) from the "upper" point at where they intersect to the "lower" point they intersect. These points are 5pi/18 and pi/18. Integrate r^2 = 2^2 = 4 between these two points and you should get 4pi/9. This is the area bounded by the line r=2 and the initial line, in other words, the area shaded by BOTH the blue and black lines. Sorry for my bad explanation. Hope it helps.
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    I meant S, not R by the way! Ahah, no worries
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    (Original post by tripleseven)
    ...
    I understand how areas work with polar coordinates for the most part, but how does integrating r=2 find between the coordinates find the excess area? Does it not form a sector with the origin as opposed to the second shape? I feel like I'm missing something.

    Edit: Is this because integrating the original curve also forms a sector?
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    (Original post by cjlh)
    I understand how areas work with polar coordinates for the most part, but how does integrating r=2 find between the coordinates find the excess area? Does it not form a sector with the origin as opposed to the second shape? I feel like I'm missing something.
    Not sure I fully understand, are you referring to the yellow lines in the diagram and how these don't cover "the whole region" in a sense?



    Edit: I think I know what you mean and although I'm not 100% sure so don't take my word for it (I'm an FP2 student myself :P), logically, you apply the same process when integrating the other curve, so yes, the original curve also forming a sector - "sector" of area found by integrating r=2 gives you area S.
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    (Original post by tripleseven)
    Not sure I fully understand, are you referring to the yellow lines in the diagram and how these don't cover "the whole region" in a sense?

    (image)

    Edit: I think I know what you mean and although I'm not 100% sure so don't take my word for it (I'm an FP2 student myself :P), logically, you apply the same process when integrating the other curve, so yes, the original curve also forming a sector - "sector" of area found by integrating r=2 gives you area S.
    Yeah I think you got it. It's gonna form straight lines from O to the boundaries for both curves, so you've got the same outline for the excess area. Cheers haha
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    Following.
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    Could a question come up where we have to set up the differential q? After last year I'm shook they could give anything.
    Also I remember in I think review ex 2 a q where you had to set up it was about embryos or something.
    Please any assistance.
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    Name:  ImageUploadedByStudent Room1464199173.043735.jpg
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    An interesting transformation question, a little confused about b)i) as I'm not sure where to plug in 'x +iy' or 'u +iv' to show the circle. Any help would be appreciated


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    (Original post by Rkai01)
    Could a question come up where we have to set up the differential q? After last year I'm shook they could give anything.
    Also I remember in I think review ex 2 a q where you had to set up it was about embryos or something.
    Please any assistance.
    Unfortunately yes, this is something that you could be asked to do.

    It's just what you've been doing in C4, however, with some cheeky FP2 twists.
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    (Original post by Louisb19)
    does anyone have resources (other than the FP2 past papers) for questions?
    Textbook questions? some of them are very challenging so can be a good test but some are just plain awkward
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    Would you ever be asked to prove  z=re^i ^\theta ?
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    (Original post by Music With Rocks)
    Would you ever be asked to prove  z=re^i ^\theta ?
    I do not think so, but it is very easy to do so with the series expansions of sin \theta, cos \theta and e^{x}, where  x = i\theta.

    You only really need to use it.
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    (Original post by oinkk)
    I do not think so, but it is very easy to do so with the series expansions of sin \theta, cos \theta and e^{x}, where  x = i\theta.

    You only really need to use it.
    ah okay, thank you

    Is that a Maclaurin and Taylor series expansion? I am working my way through FP2 and haven't reached that chapter yet
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    (Original post by Music With Rocks)
    ah okay, thank you

    Is that a Maclaurin and Taylor series expansion? I am working my way through FP2 and haven't reached that chapter yet
    It certainly is! It'll make so much more sense when you've done that chapter.
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    Name:  fp2 capture 3.JPG
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Size:  38.6 KB can anyone help with part b? (the area). Can't see how to get the region because of the slim bits that the area of the sector misses out
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    (Original post by Alby1234)
    Name:  fp2 capture 3.JPG
Views: 127
Size:  38.6 KB can anyone help with part b? (the area). Can't see how to get the region because of the slim bits that the area of the sector misses out
    The 'slim bits' are from the loop of the second curve. Work out the polar coordinates of where the curves meet first, and then use integration to find the area of one of these bits (e.g. take the smaller of the two angles from part (a) and 0 as the limits). By symmetry you only need to work out the area of one of the two slim pieces and then just double the answer, then add it on to the area of the sector.
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    (Original post by Alby1234)
    Name:  fp2 capture 3.JPG
Views: 127
Size:  38.6 KB can anyone help with part b? (the area). Can't see how to get the region because of the slim bits that the area of the sector misses out
    Draw a line to where they meet, see that integrating the polar cureve between 0 and the angle at where they meet gives that little patch outside the sector.


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    (Original post by physicsmaths)
    Draw a line to where they meet, see that integrating the polar cureve between 0 and the angle at where they meet gives that little patch outside the sector.


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    Got it, thank you!
 
 
 
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