P3 Mathematics Watch

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Darkness
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#21
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#21
(Original post by mensandan)
This is my first P3 exam, I really enjoyed it. I had a go at the June 2003 paper, I thought was fairly straight forward. I heard a big fuss about it last June, but I must say it was no harder than all the other papers, just different. Mensandan
Yeh, P3 is fun.

The problem with the June paper was not so much difficulty but rather a poor allocation of time (i.e. too much cramped into 90 minutes)
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mensandan
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#22
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#22
Yeah I remember doing that P3 as a practice and it did take a lot longer than any of the others. How did you score on it Darkness?
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Darkness
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#23
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#23
93 so I guess I really shouldn't complain lol. But (not to sound cocky) could have probably gotten 100 if it was any of the other papers.
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hattori
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#24
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#24
(Original post by mensandan)
Since it was a right angled triangle I used
(1/2)bh

b=root6
h=root21

So 0.5root126
= 0.5root9root 14
= 1.5root14

This is if the points were (3, 7, 3) and (4, 6, 8)

When given the 2 x values. Once you got these you got new values of mew and lamda. I think this gives what I got, not sure.
yeah i did it the same way used the vales of X to find lambda and mew then found the other coordinates. 3,7,3 is familiar so is 4,6,8 but the distance between those two points is 3root3, isn't it
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mensandan
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#25
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#25
Yeah it is ,but the side you are talking about is the hypotenuse. The 2 shorter sides are from these 2 points to the point of intersection of the lines L1 and L2 (this is the right angle since they are perpendicular) which was (2, 5, 4) if you work out the distance from these 2 points to this point then use 0.5 basexheight for a right angle triangle you get 1.5(root 14).
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Rich
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#26
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#26
(Original post by Darkness)
Because I'm bored and would like to have a go. Uni doesn't start till Monday.
You could always do some Haskell revision for Tuesday.
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hattori
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#27
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#27
(Original post by mensandan)
Yeah it is ,but the side you are talking about is the hypotenuse. The 2 shorter sides are from these 2 points to the point of intersection of the lines L1 and L2 (this is the right angle since they are perpendicular) which was (2, 5, 4) if you work out the distance from these 2 points to this point then use 0.5 basexheight for a right angle triangle you get 1.5(root 14).
sorry mate, got confused there but I didn't do that in the exam. I found length of QR and QP and then half base*height. but i can't remember what i put as my answer. so probably got the same unless i did one of those brain dead fck ups.
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mensandan
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#28
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#28
Here's hoping we both got it right, all the best mate. Mensandan
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visitor
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#29
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#29
(Original post by mensandan)
Yeah it is ,but the side you are talking about is the hypotenuse. The 2 shorter sides are from these 2 points to the point of intersection of the lines L1 and L2 (this is the right angle since they are perpendicular) which was (2, 5, 4) if you work out the distance from these 2 points to this point then use 0.5 basexheight for a right angle triangle you get 1.5(root 14).
I did P3 2day, quite easy. I got most of the same answers as u.
how did u do 5c and 6b?
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mensandan
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#30
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To do 5c the percentage error. I did,

(correct answer-estimate)/Correct multiplied by 100

This gives the factor you are out as a percentage of the total.

For 6b I found both sides to be logs. As such I could combine them

I had

ln y = 2ln(2x-3) - 3ln(x+1)

This becomes

lny = ln((2x-3)^2)/(x+1)^3))

So y = (2x-3)^2)/(x+1)^3)

Hope that makes sense dude.
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visitor
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#31
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#31
(Original post by mensandan)
To do 5c the percentage error. I did,

(correct answer-estimate)/Correct multiplied by 100

This gives the factor you are out as a percentage of the total.

For 6b I found both sides to be logs. As such I could combine them

I had

ln y = 2ln(2x-3) - 3ln(x+1)

This becomes

lny = ln((2x-3)^2)/(x+1)^3))

So y = (2x-3)^2)/(x+1)^3)

Hope that makes sense dude.
Thanx
for 5c i did (240(pi -1) - 45pi^2)/45pi^2 = 15.8% i dunno how many marks they will give me for that.

for 6b i got the same but i got + 108 at the end for the C which comes with infdefinite integrals. Did I do something wrong?
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mensandan
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#32
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#32
(Original post by visitor)
Thanx
for 5c i did (240(pi -1) - 45pi^2)/45pi^2 = 15.8% i dunno how many marks they will give me for that.

for 6b i got the same but i got + 108 at the end for the C which comes with infdefinite integrals. Did I do something wrong?

Yeah that's right you do have the extra term from integrating. I think C was ln 108 but I'm not sure if Iremebered to tidy it up. Anyway, sounds like you got it right, so well done dude. Mensandan.
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