# OCR MEI DE Friday 15th June 2018Watch

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#1
Well. F*ck.
0
1 year ago
#2
(Original post by Kaytra)
Well. F*ck.
What q's did you attempt?
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#3
(Original post by tedtheshred)
What q's did you attempt?
1, 2 were ok, tried 4 and realised it was going to be disgusting so went for 3 and didn’t read that k was 3 and didn’t know how I was going to do the integral. Ended up only answering 1 and 2 fully 😰
0
1 year ago
#4
I did questions 1 2 3 and 4 lol
finished all with about 3 mins to spare
question 4 was pretty horrible - solution for x had awkward coefficients and probably got the value for time wrong.
But betting on questions 1 2 and 3
1
1 year ago
#5
What did you get for the time, I think I got 1.39 something but not too sure
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#6
(Original post by Tsr74)
What did you get for the time, I think I got 1.39 something but not too sure
I think that's what I got too
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1 year ago
#7
Lol I definitely got that wrong. I kind of gave up a bit after getting the general solution for x.
Oh well I hope 1 2 and 3 went well for me
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1 year ago
#8
What was y when x is 1.8 for eulers in questions 3??
0
1 year ago
#9
For the two solution curves in question 3 did any one get a parabolic shaped curve through (0,1) and a cubic shaped curve through (1,0)
And also the Euler method question 0.099 something?
And for some other question the exact value of y to be -1/16 pi^3 + 1/16 pi^2 or something similar (can’t remember my signs exactly)
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1 year ago
#10
(Original post by W00113)
What was y when x is 1.8 for eulers in questions 3??
I thought it was x=0.8

Well I got like 0.099... something
0
1 year ago
#11
(Original post by xiaochen)
For the two solution curves in question 3 did any one get a parabolic shaped curve through (0,1) and a cubic shaped curve through (1,0)
And also the Euler method question 0.099 something?
And for some other question the exact value of y to be -1/16 pi^3 + 1/16 pi^2 or something similar (can’t remember my signs exactly)
yes i got 0.099 and the 1/16pi^3 + 1/16pi^2
0
1 year ago
#12
(Original post by W00113)
yes i got 0.099 and the 1/16pi^3 + 1/16pi^2
Yayyy
0
1 year ago
#13
(Original post by Tsr74)
What did you get for the time, I think I got 1.39 something but not too sure
The time for y to be 0 in Q4? Im pretty certain that was 0.0429? It was very easy to use the wrong solution to cos(...) = 1/2
0
1 year ago
#14
(Original post by xiaochen)
For the two solution curves in question 3 did any one get a parabolic shaped curve through (0,1) and a cubic shaped curve through (1,0)
And also the Euler method question 0.099 something?
And for some other question the exact value of y to be -1/16 pi^3 + 1/16 pi^2 or something similar (can’t remember my signs exactly)
I thought one of coefficients was 1/64 or something
0
1 year ago
#15
(Original post by TauBilly)
The time for y to be 0 in Q4? Im pretty certain that was 0.0429? It was very easy to use the wrong solution to cos(...) = 1/2
I got that I think??

I had to use rcos(theta+alpha) form to convert though...

Because I got y=e^4x (10cos3x +24sin3x -13)
0
1 year ago
#16
(Original post by xiaochen)
I got that I think??

I had to use rcos(theta+alpha) form to convert though...

Because I got y=e^4x (10cos3x +24sin3x -13)
Yes you're right
0
1 year ago
#17
(Original post by xiaochen)
I got that I think??

I had to use rcos(theta+alpha) form to convert though...

Because I got y=e^4x (10cos3x +24sin3x -13)
Lol wait I used the wrong value lmao and I noticed in the exam it was wrong but I did 4 questions so I had no time to correct - I remember now haha.

Oh well!
0
1 year ago
#18

TauBilly is right!! If PS for y is right.

Oh well if people used wrong cos solution we’ll probs only lose like 2 marks as it was 4 marks total
0
1 year ago
#19
(Original post by Kaytra)
1, 2 were ok, tried 4 and realised it was going to be disgusting so went for 3 and didn’t read that k was 3 and didn’t know how I was going to do the integral. Ended up only answering 1 and 2 fully 😰
Did 1,2,4. no problems there. 100 ums boundary?
0
#20
(Original post by tedtheshred)
Did 1,2,4. no problems there. 100 ums boundary?
Low I'm hoping but by the sounds of people on here I'm not to sure that'll be the case
0
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