* Honours Maths Paper 2 " Let's fight back :)

hunii07 · 11-06-2011 01:39PM

PJelly wrote: »

Don't sob about yesterday!
This is the paper 2 thread, this is the future! Be positive.
I kept thinking on how badly I did on paper one, even the simple bits, but it's over and no amount of moping will change it.

Very True

jamesr1775 · 11-06-2011 01:53PM

pretty cool thing for circle. appeard in 2004 i think but you can use if an axis is a tangent to circle.
if y axis is tangent to circle then f^2 = c
if x axis is a tangent to circle then g^2 = c

epicwinning · 11-06-2011 01:59PM

jamesr1775 wrote: »

pretty cool thing for circle. appeard in 2004 i think but you can use if an axis is a tangent to circle.
if y axis is tangent to circle then f^2 = c
if x axis is a tangent to circle then g^2 = c

If an axis is a tangent, then the centre of the circle is going to have the point of contact as part of the point (eg. if x=0 is a tangent at (3,0), the centre point is (3,f)). That solves one variable right off the bat.

jamesr1775 · 11-06-2011 02:05PM

yep and then u can get the C variable by subbing (0,f) in to equation of circle with the 2gx and 2fy

jamesr1775 · 11-06-2011 02:06PM

or sorry watever the point is thts in contact with the axis

GV_NRG · 11-06-2011 02:20PM

Shayanzadeh wrote: »

It's -A/B not -B/A. So in your 2x-y=0 you would get 2 as slope. Easy to mix up with -B/A which is from alpha + betha.

can someone please explain this -A/B methoo please?

jamesr1775 wrote: »

yep and then u can get the C variable by subbing (0,f) in to equation of circle with the 2gx and 2fy

so it would look something like this...

x^2 + y^2 + 2gx + 2fy + c = 0 @ (0 , f)

0 + f^2 + 0 + 2f^2 + c = 0

3f^2 = -c

f^2 = -c/3

^^ is this correct?

PJelly · 11-06-2011 02:35PM

GV_NRG wrote: »

can someone please explain this -A/B methoo please?

Take the three parts of a line like A B and C
Ax+By+C you could say. Or Ay+Bx+c.
To get the slope, put -A/B
Sooo. 4x-6y+3 would have a slope of 4/6 = 2/3 (Minus's cancel)
And 7x+13y-9 would have a slope of -7/13
Geddit?

jamesr1775 · 11-06-2011 02:39PM

sry yea the point would be (o,-f)
so you woud get 0+f^2+0-2f^2+c=0
then its -2f^2+f^2 = -c simple algebra

then -f^2=-c => f^2=c
Check 2004 circle question if u want.

PJelly · 11-06-2011 02:50PM

There's one part of McLaurin series that I struggle to grasp.
When you need to find the value of X that you sub in for say, the 1997 example. You need root ten and you have root 1 + x. But you cant let X=9. So you need to rejig the equation around. There was a similar question like it somewhere where you need to get root seventeen.

I always struggle with the actual re-jigging of the equation.

GV_NRG · 11-06-2011 03:01PM

PJelly wrote: »

Take the three parts of a line like A B and C
Ax+By+C you could say. Or Ay+Bx+c.
To get the slope, put -A/B
Sooo. 4x-6y+3 would have a slope of 4/6 = 2/3 (Minus's cancel)
And 7x+13y-9 would have a slope of -7/13
Geddit?

ya i get it now cheers for that!

Daniel S · 11-06-2011 04:11PM

canister94 wrote: »

eq of tangent of circle x1x+y1y=r squared
difference equations un=la power of n +mb power n

Can you show me what you mean by these?
I've never heard of the second one (I think) and I don't do the circle, but if it came up, it would be handy to have (Grinds teacher also did some of the circle with me).

Daniel S · 11-06-2011 04:18PM

kpac wrote: »

I was going to say the difference equation wouldn't come up because it appeared last year, but after yesterday I wouldn't rule out anything.

Yea, the examiners might ask us where they're going on holidays, I know it's not on the curriculum, but sure that didn't stop them yesterday did it? :pac:

"Q6a

(i) There are loads of people complaining about paper one. What's the probability you passed?

"

cocopopsxx · 11-06-2011 04:21PM

mtb_kng wrote: »

Yea, the examiners might ask us where they're going on holidays, I know it's not on the curriculum, but sure that didn't stop them yesterday did it? :pac:

"Q6a

(i) There are loads of people complaining about paper one. What's the probability you passed?

"

lol..that question is so funny!!

partyatmygaff · 11-06-2011 04:30PM

PJelly wrote: »

There's one part of McLaurin series that I struggle to grasp.
When you need to find the value of X that you sub in for say, the 1997 example. You need root ten and you have root 1 + x. But you cant let X=9. So you need to rejig the equation around. There was a similar question like it somewhere where you need to get root seventeen.

I always struggle with the actual re-jigging of the equation.

First of all, you have to realise the use of a Maclaurin series. All it does is convert a function like sqrt(1+x) in to a simple series that can be summed arithmetically. The reason this is important is because calculators can only add or subtract. The more terms you get in a Maclaurin series the more accurate it becomes. This is why some calculators advertise "8 decimal point precision" e.t.c.

Now, look at this 1997 question. You're being asked to find the value of sqrt(10) using the Maclaurin series of sqrt(1+x). Essentially you're being asked to do the job your calculator would normally do. Now let's make this extremely easy.

sqrt(10) can be written as sqrt(9+1). You want to get one of those numbers down to 1 and the other between -1 and 1. So what you do is factorise. You then get sqrt(9(1+1/9)) which can also be rewritten as sqrt(9) x sqrt(1+1/9). Simplifying that you get 3sqrt(1+1/9).

The question is asking you to evaluate sqrt(10) using the series expansion. You now have 3.sqrt(1+1/9) and 1.sqrt(1+x). What you now do is sub x = 1/9 in to your expansion and then multiply the answer of that by three.

PJelly · 11-06-2011 04:36PM

partyatmygaff wrote: »

First of all, you have to realise the use of a Maclaurin series. All it does is convert a function like sqrt(1+x) in to a simple series that can be summed arithmetically. The reason this is important is because calculators can only add or subtract. The more terms you get in a Maclaurin series the more accurate it becomes. This is why some calculators advertise "8 decimal point precision" e.t.c.

Now, look at this 1997 question. You're being asked to find the value of sqrt(10) using the Maclaurin series of sqrt(1+x). Essentially you're being asked to do the job your calculator would normally do. Now let's make this extremely easy.

sqrt(10) can be written as sqrt(9+1). You want to get one of those numbers down to 1 and the other between -1 and 1. So what you do is factorise. You then get sqrt(9(1+1/9)) which can also be rewritten as sqrt(9) x sqrt(1+1/9). Simplifying that you get 3sqrt(1+1/9).

The question is asking you to evaluate sqrt(10) using the series expansion. You now have 3.sqrt(1+1/9) and 1.sqrt(1+x). What you now do is sub x = 1/9 in to your expansion and then multiply the answer of that by three.

So you can change the 1 then? Not just the X?
Everything suddenly makes more sense. Thank you

So is the general rule then kind of, if you would assume you sub it in for X (in this case 9), swap it for 1 and rejig?

krisityfer · 11-06-2011 04:41PM

This is helpful: http://www.mathopenref.com/trianglecircumcircle.html

Take note of the "For right triangles" bit. It's obvious once you think about it, but during an exam question, not that easy to think of.

See 2002, P2 Q1 b (ii), for an example of it in use. Could be really helpful if a similar question comes up, which is likely given the "Think about things" theme of the last paper.

partyatmygaff · 11-06-2011 04:47PM

The thing that's worrying me most about Paper II is question 8 (And 6 and 7). I'm almost certain they're going to toss in a a terrible Max/Min question. Does anyone have any tips for modelling equations?

LilMissCiara · 11-06-2011 05:36PM

partyatmygaff wrote: »

Now, look at this 1997 question. You're being asked to find the value of sqrt(10) using the Maclaurin series of sqrt(1+x). Essentially you're being asked to do the job your calculator would normally do. Now let's make this extremely easy.

sqrt(10) can be written as sqrt(9+1). You want to get one of those numbers down to 1 and the other between -1 and 1. So what you do is factorise. You then get sqrt(9(1+1/9)) which can also be rewritten as sqrt(9) x sqrt(1+1/9). Simplifying that you get 3sqrt(1+1/9).

The question is asking you to evaluate sqrt(10) using the series expansion. You now have 3.sqrt(1+1/9) and 1.sqrt(1+x). What you now do is sub x = 1/9 in to your expansion and then multiply the answer of that by three.

Can you not then equate 9+1 to X+1 and have X = 9 and put that in to what you worked out for sqrt(1+x). Why has it to be between -1 and 1.. I probably sound really stupid right now!

partyatmygaff · 11-06-2011 05:48PM

LilMissCiara wrote: »

Can you not then equate 9+1 to X+1 and have X = 9 and put that in to what you worked out for sqrt(1+x). Why has it to be between -1 and 1.. I probably sound really stupid right now!

That's because the series only converges when x is between -1 and 1. The Maclaurin series only works when the series is convergent.

A convergent series is one which has a limit. It always converges on to a definite number. A divergent series does not converge on to a definite number, it goes on to infinity which isn't of much use when you're trying to evaluate a function.

This is the reason why we do the ratio test. We want to find the range of values of x where the series is convergent so that we can make use of the series to evaluate functions.

This mightn't be the best of explanations as it's the result of me trawling through the theory behind it on the odd website and Wikipedia article but for the purposes of Q8, I think it's good enough.

LilMissCiara · 11-06-2011 05:58PM

partyatmygaff wrote: »

That's because the series only converges when x is between -1 and 1. The Maclaurin series only works when the series is convergent.

A convergent series is one which has a limit. It always converges on to a definite number. A divergent series does not converge on to a definite number, it goes on to infinity which isn't of much use when you're trying to evaluate a function.

Ok thanks, it makes sense...

But.. Just after looking at the 2009 question and it had something very similar. I understand how to get it to -1>X<1 but I'm confused by something..

You get the maclaurin series for Root(1+X) which is grand.

Then you want to approximate Root(17) so you break that down and it becomes 4Root(1/16 + 1) and you use 1/16 for X. But that's 4Root(1+X) and you are then putting X in to just Root(1+X).

I would multiply the answer by 4 but the solution says it's wrong.. Can anybody spread some light on that!

Solution = http://www.studentxpress.ie/papers/optionsoln2009.pdf

john1741 · 11-06-2011 06:04PM

partyatmygaff wrote: »

That's because the series only converges when x is between -1 and 1. The Maclaurin series only works when the series is convergent.

A convergent series is one which has a limit. It always converges on to a definite number. A divergent series does not converge on to a definite number, it goes on to infinity which isn't of much use when you're trying to evaluate a function.

This is the reason why we do the ratio test. We want to find the range of values of x where the series is convergent so that we can make use of the series to evaluate functions.

This mightn't be the best of explanations as it's the result of me trawling through the theory behind it on the odd website and Wikipedia article but for the purposes of Q8, I think it's good enough.

So say if it looks like x=6 would you use 1/6 instead? Putting a one over whichever number making it a fraction

partyatmygaff · 11-06-2011 06:06PM

I honestly wouldn't know why they didn't...

If worst comes to worst. Just put in sqrt(17) in to your calculator and compare it to what you evaluated and then adjust.

partyatmygaff · 11-06-2011 06:08PM

john1741 wrote: »

So say if it looks like x=6 would you use 1/6 instead? Putting a one over whichever number making it a fraction

Generally speaking yes. But technically speaking what you're doing is separating the root in to two parts.

PaddyP056 · 11-06-2011 06:14PM

luciemc wrote: »

anyone doing question 9?

YES! Ive been teaching it to myself over the past month, its easily the most underated question on the paper. Im so suprised more teachers dont teach it. How u set for it?

also heres some handy notes on it
http://www.projectmaths.com/cms/wp-content/uploads/2011/05/projectnotes2010.pdf

hairybacon · 11-06-2011 06:46PM

Just wondering what way do ye find best to study? I tried the papers but I think I may go back to basics- I am absolutely s**t at nearly every question

PJelly · 11-06-2011 06:53PM

LilMissCiara wrote: »

Ok thanks, it makes sense...

But.. Just after looking at the 2009 question and it had something very similar. I understand how to get it to -1>X<1 but I'm confused by something..

You get the maclaurin series for Root(1+X) which is grand.

Then you want to approximate Root(17) so you break that down and it becomes 4Root(1/16 + 1) and you use 1/16 for X. But that's 4Root(1+X) and you are then putting X in to just Root(1+X).

I would multiply the answer by 4 but the solution says it's wrong.. Can anybody spread some light on that!

Solution = http://www.studentxpress.ie/papers/optionsoln2009.pdf

I did that one about an hour ago, I had 4 on root 1 + 1/16
So X = 1/16
You get some mad fractions then, and as you have a 4 on the outisde, you multiply your answer by 4 at the end.
It says to give it in fraction form, but it's such a crazy fraction that my calculator refuses to. I'm getting the right decimal though.
May prove problematic if my calculator won't spit out the fractions on the actual day when needed.

River Song · 11-06-2011 06:54PM

I know all the proofs now, I think. The Perpendicular distance one is so easy to make mistakes in, though.

Thanks to whomever linked to the concurrent lines site, it was REALLY helpful

I'm going to spend tonight doing the tan inverse series and let that be that for today.

hunii07 · 11-06-2011 06:57PM

Could someone repost the link for the concurrent lines site please?

PJelly · 11-06-2011 06:58PM

Michael_E wrote: »

I know all the proofs now, I think. The Perpendicular distance one is so easy to make mistakes in, though.

Id LOVE for that to come up as it's a full part C.

jamesr1775 · 11-06-2011 07:04PM

guys checkout example 2 and example 4 on pg364/365 of text and tests 4(green one) .
Could be possible questions for circle since they're throwing up obscure questions.
Definately worth a look anyway peeps

* Honours Maths Paper 2 " Let's fight back :)

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