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DE maths, OCR, 24/05/10

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Reply 20
Avatar for Apex9
Apex9
OP
8
jamesBT
I thought it was pretty good but I probably made loads of mistakes. I did question 1,2 and 4 (avoided the Euler's method :smile: ).
Did anyone who did question 4 get a constant that becomes 0 when y=2 and then when y became just less than 2 it wasn't 0 and changed the shape of the graph. If not then I may have screwed up a bit.:mad:
Overall not a bad exam!


Yes.
Look up. Right below the thread title. See the big red letters? Smart thread.
Reply 22
Avatar for Apex9
Apex9
OP
8
You'll have to be more specific.
my bad, I thought it was all exams where you weren't allowed to talk about them.
Bakes0011
Look up. Right below the thread title. See the big red letters? Smart thread.

That's only for international exams, OCR only does UK exams so it's fine
Reply 25
Avatar for Apex9
Apex9
OP
8
Smart comment.
I think it'll be like 58/72 or something for an A on the written paper, i.e. slightly lower than usual. I am expecting most people (including me) will probably have done the "weird bits" I said about badly or not at all
Reply 27
Avatar for chemcat
chemcat
For qu2, i got z=(e^-2t)(1+2t+t^2) as one of the answers.

Question 4, yes the constant is 0 so as t tends to infinity both tend to zero. If it is initially slightly less than 2 the coefficient of the e^t term becomes positive so it becomes very large as t tends to infinty.
Reply 28
Avatar for chemcat
chemcat
What did people get as the solution to the final part of 2a. Finding z in terms of t?
Reply 29
Avatar for Apex9
Apex9
OP
8
I think I had a quadratic of 't' multiplied by e^-2t.

I can't remember exactly, something similar to that.
Reply 30
Avatar for Apex9
Apex9
OP
8
chemcat
For qu2, i got z=(e^-2t)(1+2t+t^2) as one of the answers.

Question 4, yes the constant is 0 so as t tends to infinity both tend to zero. If it is initially slightly less than 2 the coefficient of the e^t term becomes positive so it becomes very large as t tends to infinty.


We happen to be communicating in the form of two threads but yes I agree with you.
Reply 31
Avatar for chemcat
chemcat
Apex9
We happen to be communicating in the form of two threads but yes I agree with you.


I feel i should now reply in the other thread! That's good that you agree, as i did that question in a weird way, and that was the only part of the paper i was unsure of.
Reply 32
Avatar for doivid
doivid
17
I did 2,3 and 4. I was the only person to do question 3 in my further maths class, everyone else was scared by the page of words. :biggrin:
Reply 33
Avatar for Apex9
Apex9
OP
8
For me, question 3 was probably more manageable than question 1 though.
I remember it was a quadratic in t multiplied by e^-2t and the coefficient of the t^2 term was 1/2
Reply 35
Avatar for Apex9
Apex9
OP
8
I agree.

What were your particular solutions for x and y in question 4?
Can't remember them at all
Reply 37
Avatar for chemcat
chemcat
Apex9
I agree.

What were your particular solutions for x and y in question 4?



something like:

x=e^-3t - e^-t

and

y=e^-3t + e^-t
chemcat
something like:

x=e^-3t - e^-t

and

y=e^-3t + e^-t

I remember it being a combination of e^-2t, e^-3t and e^t terms.
Reply 39
Avatar for chemcat
chemcat
matt2k8
I remember it being a combination of e^-2t, e^-3t and e^t terms.


Yeah probably, but with the e^t term being zero

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