HL Math Paper 2- 24 Hours Later...
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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....
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 Hyaline
Yeah I put 28 I believe. Wait for part b) did you guys use n-1 or not?
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.
Original post by Saymyname!
I got v=arctan(pi/4-t)
Don't you mean v=tan(pi/4-t)... since you tan both sides to get rid of the arctanv
Original post by ElationAndPathways
Don't you mean v=tan(pi/4-t)... since you tan both sides to get rid of the arctanv
Yeah you're absolutely right. I corrected myself at some point in this thread :P
Original post by arra
Hence Sinf = 4/(1-0.05) = 4*20 = 80.
Wait what?!
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.
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.
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 Hyaline
Wait what?!
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.
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.
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.
I see what you're saying. I think I totally forgot to account for the up-distance. ****.
Original post by Saymyname!
Yeah you're absolutely right. I corrected myself at some point in this thread :P
Ooooh ok sorry!! Pretty sure I went through the whole thread but maybe I missed some out :P
Original post by arra
I see what you're saying. I think I totally forgot to account for the up-distance. ****.
relax...that last part was worth only 2 or 3 marks!
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?
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?
(edited 11 years ago)
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?
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 first question from Part B.
the 14 mark normal distribution qs
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.
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.
My answer to part (a) of the bouncing ball question was 27. And for part b I got 156m
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