Lucasium
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A couple harder polar curves questions. I didn't make any useful progress
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Can anyone do questions 7 and 8? 7 b ii onwards
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Zacken
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(Original post by Lucasium)
A couple harder polar curves questions. I didn't make any useful progress
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Can anyone do questions 7 and 8?
Any specific parts you're stuck on?
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the bear
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the first part of q7 is easier if you reflect the curve in the line y = x ... then it is just a simple integral wrt x. the limits are not hard to find.
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Lucasium
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(Original post by Zacken)
Any specific parts you're stuck on?
All apart from 7 a and 7bi
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Zacken
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(Original post by Lucasium)
All apart from 7 a and 7bi
7(b) (ii) is simple a matter of recognising that r^2 = x^2 + y^2 and x = r\cos \theta so your equation becomes r^2 = (2 - r\cos \theta)^2 after which, some re-arranging and simplifying and solving for r in terms of \theta gets you where you need to be.
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Lucasium
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(Original post by Zacken)
7(b) (ii) is simple a matter of recognising that r^2 = x^2 + y^2 and x = r\cos \theta so your equation becomes r^2 = (2 - r\cos \theta)^2 after which, some re-arranging and simplifying and solving for r in terms of \theta gets you where you need to be.
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Olipoo
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(Original post by Lucasium)
A couple harder polar curves questions. I didn't make any useful progress
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Can anyone do questions 7 and 8?
1=cos(0) from here you can use the (half?) angle formulas that are included in C3 but were most likely not taught in detail.
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Zacken
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(Original post by Lucasium)
I tried but only got this far
Bit unwieldy?

r^2 = (2-r\cos \theta)^2 \Rightarrow \pm r = 2-r\cos \theta \Rightarrow r (\cos \theta \pm 1) = 2 \Rightarrow r = \frac{2}{\cos \theta \pm 1}

From which it suffices to see that r must be positive so you take r = \frac{2}{\cos \theta + 1}
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Lucasium
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(Original post by Olipoo)
1=cos(0) from here you can use the (half?) angle formulas that are included in C3 but were most likely not taught in detail.
For which question?
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Zacken
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(Original post by Lucasium)
For which question?
Given that you've repped my post, I take it you've understood - in which case, can you then see how the integral is a multiple of \frac{1}{2}\int r^2 \, \mathrm{d}\theta which is the area of something you've already found so adjusting for the multiple gets you the required answer.
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Olipoo
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(Original post by Lucasium)
For which question?
7c
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Lucasium
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Good work guys, I appreciate it. Let's move on to question 8
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Zacken
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(Original post by Lucasium)
Good work guys, I appreciate it. Let's move on to question 8
Same thing, write x^2 + 3y^2 = x^2 + y^2 + 2y^2 = r^2 + 2(r\sin \theta)^2 re-arrange, blah blah.
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Lucasium
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(Original post by Zacken)
Same thing, write x^2 + 3y^2 = x^2 + y^2 + 2y^2 = r^2 + 2(r\sin \theta)^2 re-arrange, blah blah.
These questions are harder than usual aren't they? I haven't had such an issue before
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Zacken
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(Original post by Lucasium)
These questions are harder than usual aren't they? I haven't had such an issue before
I wouldn't know - I'm not on your exam board, so I'll defer to your judgement.
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Lucasium
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(Original post by Zacken)
I wouldn't know - I'm not on your exam board, so I'll defer to your judgement.
I've just received news that the booklet is a mix of aqa and MEI, so concern over. It's no excuse but it makes me feel a little better
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Argylesocksrox
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(Original post by Lucasium)
I've just received news that the booklet is a mix of aqa and MEI, so concern over. It's no excuse but it makes me feel a little better
Could you link this booklet?
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Lucasium
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(Original post by Argylesocksrox)
Could you link this booklet?
I suppose I'll take and upload pictures tomorrow, it's a big booklet
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Argylesocksrox
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(Original post by Lucasium)
I suppose I'll take and upload pictures tomorrow, it's a big booklet
Thanks mate
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