The Student Room Group

Edexcel FP2 Official 2016 Exam Thread - 8th June 2016

Scroll to see replies

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?

Screenshot 2016-05-25 12.45.33.png
Attachment not found


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.
(edited 7 years ago)
does anyone have resources (other than the FP2 past papers) for questions?
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.
I meant S, not R by the way! Ahah, no worries :smile:
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?
(edited 7 years ago)
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.
(edited 7 years ago)
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
Reply 427
Following.
Reply 428
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.
ImageUploadedByStudent Room1464199173.043735.jpg

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


Posted from TSR Mobile
Reply 430
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.
(edited 7 years ago)
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
Would you ever be asked to prove
Unparseable latex formula:

z=re^i ^\theta

?
Reply 433
Original post by Music With Rocks
Would you ever be asked to prove
Unparseable latex formula:

z=re^i ^\theta

?


I do not think so, but it is very easy to do so with the series expansions of sinθsin \theta, cosθcos \theta and exe^{x}, where x=iθ x = i\theta.

You only really need to use it.
Original post by oinkk
I do not think so, but it is very easy to do so with the series expansions of sinθsin \theta, cosθcos \theta and exe^{x}, where x=iθ 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 :tongue:
Reply 435
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 :tongue:


It certainly is! It'll make so much more sense when you've done that chapter.
fp2 capture 3.JPG 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
Original post by Alby1234
fp2 capture 3.JPG 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.
(edited 7 years ago)
Original post by Alby1234
fp2 capture 3.JPG 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.


Posted from TSR Mobile
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.


Posted from TSR Mobile


Got it, thank you!

Quick Reply

Latest

Trending

Trending