HL Math Paper 2- 24 Hours Later...
Discussion forum for International Baccalaureate students.
-
- Reputation:
- Junior Member
- Posts: 69
Re: HL Math Paper 2- 24 Hours Later...No no it was only 3 marks!!! I'm pretty sure you were supposed to use the equation solver on your GDC or use trial and error. The answer was 1.40(Original post by johndoe04)
This was another annoying thing. Substitution wouldn't work. I tried by parts and let dv/dx = 1 and u = 1/1+x^4. It was a long process.... -
- Reputation:
- Adored and Respected Member
- Posts: 401
Re: HL Math Paper 2- 24 Hours Later...I put 28 for part b as well.(Original post by Hyaline)
Yeah I put 28 I believe. Wait for part b) did you guys use n-1 or not?
The formula I had was Hn = 4*(0.95)^n.
The first bounce is 4, the second bounce is 5% less, so on so forth.
Make this a geometric sequence. R = 0.05. U1 = 4
Hence Sinf = 4/(1-0.05) = 4*20 = 80. -
- Reputation:
- Junior Member
- Posts: 69
Re: HL Math Paper 2- 24 Hours Later...Don't you mean v=tan(pi/4-t)... since you tan both sides to get rid of the arctanv(Original post by Saymyname!)
I got v=arctan(pi/4-t) -
- Reputation:
- Thread Starter
- Junior Member
- Posts: 42
Re: HL Math Paper 2- 24 Hours Later...Yeah you're absolutely right. I corrected myself at some point in this thread :P(Original post by ElationAndPathways)
Don't you mean v=tan(pi/4-t)... since you tan both sides to get rid of the arctanv -
- Reputation:
- Full Member
- Posts: 110
Re: HL Math Paper 2- 24 Hours Later...Wait what?!(Original post by arra)
Hence Sinf = 4/(1-0.05) = 4*20 = 80. -
- Reputation:
- Thread Starter
- Junior Member
- Posts: 42
Re: HL Math Paper 2- 24 Hours Later...Sorry bro, you might be slightly off on that one. R=0.95, and for that, U1=3.8, at least the was I did it. Sinf then =76m, so Sinf*2+4=156m.(Original post by arra)
I put 28 for part b as well.
The formula I had was Hn = 4*(0.95)^n.
The first bounce is 4, the second bounce is 5% less, so on so forth.
Make this a geometric sequence. R = 0.05. U1 = 4
Hence Sinf = 4/(1-0.05) = 4*20 = 80. -
- Reputation:
- Adored and Respected Member
- Posts: 401
Re: HL Math Paper 2- 24 Hours Later...We want the sum to infinity. If it is a geometric series described by H = 4(.95)^n, from 0 to infinity, the first term is 4 (obviously, as he drops it from this height) and the common ratio is 0.95.(Original post by Hyaline)
Wait what?!
If abs(R) < 1, we can say that the total sum = u1/(1-r) = 4/(1-0.95) = 4/(0.05) = 4/(1/20) = 80. -
- Reputation:
- Adored and Respected Member
- Posts: 401
Re: HL Math Paper 2- 24 Hours Later...I see what you're saying. I think I totally forgot to account for the up-distance. ****.(Original post by Saymyname!)
Sorry bro, you might be slightly off on that one. R=0.95, and for that, U1=3.8, at least the was I did it. Sinf then =76m, so Sinf*2+4=156m. -
- Reputation:
- Junior Member
- Posts: 69
Re: HL Math Paper 2- 24 Hours Later...Ooooh ok sorry!! Pretty sure I went through the whole thread but maybe I missed some out :P(Original post by Saymyname!)
Yeah you're absolutely right. I corrected myself at some point in this thread :P -
- Reputation:
- New Member
- Posts: 23
Re: HL Math Paper 2- 24 Hours Later...relax...that last part was worth only 2 or 3 marks!(Original post by arra)
I see what you're saying. I think I totally forgot to account for the up-distance. ****. -
- Reputation:
- New Member
- Posts: 6
Re: HL Math Paper 2- 24 Hours Later...RE: The bouncing ball question.
It doesn't matter whether you use n or n-1 so long as you keep that throughout. I used ar^n but specified that n was the number of bounces, which made things clearer, and a was 4. So after 0 bounces, a*r^0 = a = 4, u2 = 3.8 etc. It meant I didn't get confused with my ns, but you could just have easily have used a = 3.8 and n-1. Then when you do the sum to infinity, after you multiply by 2 for the up and down, the only difference is whether you add 4 for the missing drop if you used a = 3.8 or take away four for the additional rise if you used a = 4. Either answer gives Total distance as 156m. (a = 3.8, r = 0.95 => Sinf = 76, Sinf*2 + 4 = 156; a = 4, r = 0.95 => Sinf = 80, Sinf*2 - 4 = 156)
Banban: What was that one?Last edited by dawesbr; 05-05-2012 at 23:08. -
- Reputation:
- New Member
- Posts: 23
Re: HL Math Paper 2- 24 Hours Later...The first question from Part B.(Original post by dawesbr)
RE: The bouncing ball question.
It doesn't matter whether you use n or n-1 so long as you keep that throughout. I used ar^n but specified that n was the number of bounces, which made things clearer, and a was 4. So after 0 bounces, a*r^0 = a = 4, u2 = 3.8 etc. It meant I didn't get confused with my ns, but you could just have easily have used a = 3.8 and n-1. Then when you do the sum to infinity, after you multiply by 2 for the up and down, the only difference is whether you add 4 for the missing drop if you used a = 3.8 or take away four for the additional rise if you used a = 4. Either answer gives Total distance as 156m. (a = 3.8, r = 0.95 => Sinf = 76, Sinf*2 + 4 = 156; a = 4, r = 0.95 => Sinf = 80, Sinf*2 - 4 = 156)
Banban: What was that one?
the 14 mark normal distribution qs -
- Reputation:
- Full Member
- Posts: 110
Re: HL Math Paper 2- 24 Hours Later...I have a bad feeling part b) for the bouncing ball question was 27............ because I got 28 when I used n-1 -.- dammit. I find all distribution questions easy, its one thing I'm quite good at ^^. For the final part of the Q we had to mix binomial distribution with normal distribution right.
-
- Reputation:
- Junior Member
- Posts: 60
Re: HL Math Paper 2- 24 Hours Later...My answer to part (a) of the bouncing ball question was 27. And for part b I got 156m(Original post by Hyaline)
I have a bad feeling part b) for the bouncing ball question was 27............ because I got 28 when I used n-1 -.- dammit. I find all distribution questions easy, its one thing I'm quite good at ^^. For the final part of the Q we had to mix binomial distribution with normal distribution right.
