Advanced Higher Applied Maths 2009

For 1b I forget the actual number, but I got something like

1 / 0.3085 = 3.24
3.24 x 15 = 48.6 mins

Round up to 60 mins.

Didn't et 0.6 for 1b instead I got something ridiculously small. At least I'll get a mark for carry through for the number of minutes

Dataset 1 didn't look suspicious so I chose that. (Trend, constant variance, independece etc.)

Nah I couldn't write up solutions just now. Physics tomorrow and I can't afford to make an arse of that especially if I made an arse of the paper today. In fact if I don't get an A in Stats this year both of my choices gets declined

is it just a simple Normal Cumulative Frequency calculation?! I feel so dumb!

For 1b I forget the actual number, but I got something like

1 / 0.3085 = 3.24
3.24 x 15 = 48.6 mins

Round up to 60 mins.

*shrug* That's how I and at least one other person did it, and I don't see any fundamental reason it shouldn't work. You can't just double the calculation for one side, since it's no longer symmetrical, but it turns out the probability of a point lying below the limits is approximately 0.

I got stuck on 11 so I left 4 sheets blank (I use a lot of paper, used two full booklets) and came back to it later. I got it in one page though, so there's three blank sheets before I started section B I wrote

"Section B 3 pages ahead, I only left this space because Q11 was really hard"

Now that I think about it, the examiner is probably going to question my sanity

Also would anyone mind giving me a brief overview on how to do 10b for the mechanics? I tried using 4.5 = mv^2/2 + 2mg(1-costheta) But I didn't realised that that would be wrong as there is a dip from B to C

I reckon 10b was probably the hardest on the paper, this is a brief run through of how I did it:

I worked out the kinetic energy at the very lowest point which I called E (4.5+mg(2-2cos30))

I then said at some point P an angle θ anti-clockwise around from OE the height of P above E was 2-2cosθ hence the kinetic energy possesed by the sledge would be the kinetic energy at E minus mg(2-2cos&#952

So 0.5mv^2 = 4.5+mg(2-2cos30)-mg(2-2cos&#952 = 4.5+2mg(cosθ-cos30)

We know that when it looses contact the reaction force acting on the sledge by the track = 0

So the centripetal force is supplied only by the component of gravity acting towards the center of the circle.

-mgcosθ=(mv^2)/r
v^2=-grcosθ
v^2=-2gcosθ

You then equate this equation for v^2 with the equation for v^2 from the energy considerations and find θ.

The angle between OD and the horizontal is then θ-90. Which I got to be approx 10.8°

Sorry, I've just realised that definately wasn't brief and I may well have made some mistakes when typing it up. I'll also have to learn how to type up maths properly as this wordy answer is probably very hard to understand.

I get you don't worry. I didn't realised there was the flipping bend I know you're not an SQA marker but how many marks do you reckon i'd get for doing all of that, just not remembering about the extra energy?

I wished i'd messed up later in the question, because if I don't get any marks after doing essentially the same thing with the expression i'll be miffed!

It depends, I would suggest if you used the energy at c as if it was the energy at the bottom you would loose at most 2. You don't really need to work out the energy at the very bottom though (I just felt it would be most efficient). If you only ever used the 4.5J then you would just get a different expression for the height above this point at an angle theta than you would do if working out the height above the very bottom. It sounds to me like you shouldn't really be worrying about marks, I spoke to some of the more able mathematicians at school after the paper and nobody else had got 10b remotely correct (assuming what I did was correct), it sounds like you had a fair stab so I'm sure you'll be fine.

Thanks, that would be really good if only lost 2 for that. I just realised I got 11b wrong because i got the area of the triangle wrong after getting the integration on the curve right. My other mistake was factorising wrong

That was the story of the paper for me, struggling on easy bits and finding the hard bits fine!

Well done on full/near full marks though, and good luck with physics tomorrow! Good to see someone else with the double maths physics and chemistry combo

From mechanics perspective, personally It may have been slightly easier than normal, but not too much. My friend who sat stats and mechanics (weirdo that he is ) said that the stats was a bit harder.

Would you say the pass mark will be higher/lower than last year's (72)? I know you can only give the Mechanics persepective. It's just I'm very worried that I lost at least 10 on questions I did using totally wrong methods and perhaps more on the other questions

And what a combo it is. Are you not doing music as well though or have I just made that up. I did the stats course but then couldn't do that as well as mechanics and reckoned that mechanics was more working things out where as stats is just remembering how to do all of these different hypothesis tests and things like that. Good luck for physics! How was your investigation?

Are both stats and mechanics treated as one subject with the same grade boundaries as each other and stuff like that. I'd only done the 2006 and 2007 papers and I would have said this years was easier in general but with the odd question that was harder than anything in the other years.

Would you say the pass mark will be higher/lower than last year's (72)?

From mechanics perspective, personally It may have been slightly easier than normal, but not too much. My friend who sat stats and mechanics (weirdo that he is ) said that the stats was a bit harder.

How'd he manage that? We have a maths genius in the year below who's doing this next year - but at least one of the two is going to be an A level, from what I understand.