This discussion is closed.
8 years ago
#1321
(Original post by v1oXx-)
I got to dy/dx = 1/sec^2y

Then I subbed in y=arctanx

Haha!! This would have been all correct if the question didn't ask for something specific!

What you needed to do was to think that you have a right angled triangle with an angle y whose tangent is equal to x / 1. So you have an opposite side of size x and an adjacent one of size 1. Hence the Hypotenuse is sqrt(1+x^2), and cos(y) = 1 / sqrt(1+x^2), so cos^2(y) = 1 / (1+x^2) as requested.

Note that 1 / sec^2(y) = cos^2(y).
0
8 years ago
#1322
(Original post by Doughboy)
I forgot the identity linking tan and sec^2 so I drew a triangle and found cosy and then sec^2[y] =)
This identity is so damn easy to remember. Always know sin^2(x)+cos^2(x)=1, and divide by parts by cos^2(x) to get this one, or by sin^2(x) to get the one linking cot to cosec.
0
8 years ago
#1323
(Original post by Zacula)
I agree that the question is very clear:

Part (i) Given that y = .... , find dy/dx.

Part (ii) Given that x = tan(y), find dy/dx.

It couldn't be any clearer!!
Well it wasn't really worded like that, quoting from that other guy

There's not much difference between what you said and this, but the 'show that' was pretty misleading for some of us, it could easily be suggesting that it's for the previous question. It's got to be remembered that not everyone thinks as clearly during an exam as they do out of it :s
Last edited by Sasukekun; 8 years ago
0
#1324
(Original post by Zacula)
This identity is so damn easy to remember. Always know sin^2(x)+cos^2(x)=1, and divide by parts by cos^2(x) to get this one, or by sin^2(x) to get the one linking cot to cosec.
Yeah, I know. But I got so caught up in the question that I forgot that -_-

My method isn't wrong though
0
8 years ago
#1325
i just worked out that i got 77.8% for this paper so what grade is it going to be?
0
8 years ago
#1326
(Original post by Zacula)
This identity is so damn easy to remember. Always know sin^2(x)+cos^2(x)=1, and divide by parts by cos^2(x) to get this one, or by sin^2(x) to get the one linking cot to cosec.
i lost a 5 mark on that question, because i forget to say let u=2x and work my way through using the trig identity which i did but never got to the point of answering the question fully.
0
8 years ago
#1327
(Original post by Remarqable M)
i lost a 5 mark on that question, because i forget to say let u=2x and work my way through using the trig identity which i did but never got to the point of answering the question fully.
Kinda tricky to work out but...
for most modules 60 is the A boundary which is exactly 80%, for c3 though it seems to be 63-64 for average papers.
But i think the boundary will be 58~55 because many found this hard.
So hopefully you could be in for ~80 UMS
i think i got around 59-65 depending on how many working marks i get so i'm sorta in the same place as you.
I can only hope for low low boundaries
0
#1328
One time 73% raw (55/75) was needed for an A.
0
8 years ago
#1329
(Original post by Doughboy)
One time 73% raw (55/75) was needed for an A.
really? thats great then maybe for this paper 70% raw percentage is only needed to get an A
0
8 years ago
#1330
(Original post by v1oXx-)
I got to dy/dx = 1/sec^2y

Then I subbed in y=arctanx

i did exactly that, hopefully they give a mark for changing x=tany in terms of x
0
8 years ago
#1331
Any of Ya'll gonna be kind enough to supply a link to the paper?
0
8 years ago
#1332
1337 posts

...oh wait, i just ruined that!
0
8 years ago
#1333
(Original post by Mayurgami1992)
Any of Ya'll gonna be kind enough to supply a link to the paper?

http://www.studentforums.biz/index.p...ic,5846.0.html
0
#1334
(Original post by taherq)

http://www.studentforums.biz/index.p...ic,5846.0.html
Thank you.
0
#1335
6665 Core Mathematics 3: 59 52 45 39 33 27 0

Those are the boundaries.
A 59
B 52
C 45
D 39
E 33
N 27
U 00

100 UMS is 73/75. I got a B most likely.
0
8 years ago
#1336
(Original post by Doughboy)
6665 Core Mathematics 3: 59 52 45 39 33 27 0

Those are the boundaries.
A 59
B 52
C 45
D 39
E 33
N 27
U 00

100 UMS is 73/75. I got a B most likely.
The boundaries are much higher than expected for one very good reason. They can only have x% of people getting an A*, an apparently way too many people would've achieved an A* if they bought the boundaries down any further than that.
0
8 years ago
#1337
no A in c3 now for me
0
8 years ago
#1338
Hmmm I'm going to have to smash C4 and get around 98-100%, or no A* for mee
0
8 years ago
#1339
(Original post by Doughboy)
6665 Core Mathematics 3: 59 52 45 39 33 27 0

Those are the boundaries.
A 59
B 52
C 45
D 39
E 33
N 27
U 00

100 UMS is 73/75. I got a B most likely.
8 years ago
#1340
I have to hold my hands up and say I was way off with my UMS prediction. All of them seem a lot higher than I expected. The good news is that as a cohort of national results the raw scores must have pretty good.

The A* issue has absolutely nothing to do with these grade boundaries. There will be an almost identical percentage of A*-A in 2010 as there were percentage of grade As in 2009.
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