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Edexcel C3,C4 June 2013 Thread

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Reply 3560
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Giant
11
Original post by Theafricanlegend
if you look at the question it didnt specify whether they want it obtuse or acute but mark scheme says 109.47??!


That means if you had done the a.b=|a||b|costheta, the costheta should have been negative. This would mean that the angle would be obtuse.
Reply 3561
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Giant
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Original post by Supes180
Why? How do you know?


It's just standard practice. Degrees are a completely arbitrary system whilst radians are "natural". I'm not very good at explaining so I'll quote Yahoo Answers.

The reason people use radians is convenience. Degrees is a totally artifitial unit, which is hard to relate to anything geometrical.
If you draw a unit circle (radius=1), and consider different angles from it's center, the the length of the chord will be equal to the value of the angle in radians.
Also, if angle is very small, then you can use it's value in radians instead of its sin... or 1-x^2 instead of cos.
Convinient.

You are right, you COULD use degrees the same way - just have to remember to drag that pi/180 everywhere. It's just easier when you don't have to.
I am trying to prepare for this exam i think overall everything is going to be okay, except those 5 marks question where they ask you to form a differential equation from the information. Does anyone have any how tackle these questions, or how to revise for them, or finally if anyone has a bunch of questions on the topic.
Original post by nukethemaly
I think its synoptic, so you're better off knowing all your identities from C3. Especially because Edexcel are bastards.


Original post by PoorLoser
By C4 identities, do you mean trig identities?

If so, just the same as in C3


I meant integration identities :smile:
Just tell your schools to start doing OCR MEI I do it and there's been no problem :smile:
Original post by AAA_
I am trying to prepare for this exam i think overall everything is going to be okay, except those 5 marks question where they ask you to form a differential equation from the information. Does anyone have any how tackle these questions, or how to revise for them, or finally if anyone has a bunch of questions on the topic.


You have to understand the information that thy are giving you.

Example dy/dx= (the rate of the amount that is going in) minus (the rate of the amount which is going out)

If they've used different letters, you separate the variables and integrate accordingly. If you still struggle, go on the exam solutions website. They have plenty of tutorials.
Reply 3566
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task
when region R is rotated around the x axis is it always rotated 2 pi radians? hence I can ignore it or will the change in value make a difference?


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Reply 3567
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MAD Phil
Original post by Zytex
Is this the original paper? Not the replacement one?


No, it's the "international paper", taken (literally) on the other side of the world. (International candidates on this side of the world take the UK paper.) It's a new system introduced in this session - up until now, the same paper was used all the way around the world, which made keeping its contents secret until everyone had taken it very difficult.

Edexcel originally said that the lost paper would be kept embargoed for weeks, but now it has been used at four centres, I don't know whether they are still insisting on that. I certainly haven't seen it.
Original post by BooAlphie
I got a high A, so hopefully it'll be okay :frown:


Yeah, see, it's still possible! As long as it's possible you have a reason to be motivated. There's no reason why you cannot smash the C4 exam :smile:
does anyone have a list of all the formulas for the various volumes and surface areas for the shapes we need to know?
Original post by Myocardium
I meant integration identities :smile:


Do you mean cosx = sinx etc?
Original post by physicso
does anyone have a list of all the formulas for the various volumes and surface areas for the shapes we need to know?


I might make a list later and post it :smile:
Original post by nukethemaly
I might make a list later and post it :smile:


Please do!
Original post by Chris-69
Yeah, see, it's still possible! As long as it's possible you have a reason to be motivated. There's no reason why you cannot smash the C4 exam :smile:


:smile:
Thanks! Made me feel better. No point in moping
Reply 3574
Avatar for MAD Phil
MAD Phil
Original post by qwertyuiopg
Thank you!! Do you have any guesses about grade boundaries??



Also in question 5c - do you think you'd lose a mark for not putting the plus or minus?


Probably 1, but they might be kind.

It really should be there, though. You can see dx/dy takes both positive and negative values, so so does dy/dy, but the expression involving the root only takes positive values.

(We're talking about the international C3 paper, in case anyone is wondering.)
what are some of the hardest real C4 past papers?
are there any particular solomons that are quite challenging and worth a look?

also: do we need to know shape areas/volumes? any particulars I should memorise? always hate when those show up in papers
(edited 10 years ago)
On a question in a book, it's says that:

A clylindrical block of radius rcm and Heigh Hcm is worn down at one circular end at a rate of 0.5mmh-1.

Find dV/dT



Now, i know that the volume of a cylinder is Pi(r^2)h, So surely dV/dH is 2(Pi)r right?

The book says it is (Pi)r^2 though :s


Can anyone help? :smile:
Original post by nukethemaly
Do you mean cosx = sinx etc?


Yeah and all the other identities such as inegral of 1/ax+b = 1/a ln(ax+b)
Reply 3578
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task
ImageUploadedByStudent Room1371300323.071705.jpg can someone please help me with 3)a) I know how to do it but I get the a wrong answer?
I get the coefficient of x^2 as 8/27 instead of 4/27 and x^3 as 40/243 instead of 40/729!
thank you


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Reply 3579
Avatar for MasterYi
MasterYi
Original post by physicso
On a question in a book, it's says that:

A clylindrical block of radius rcm and Heigh Hcm is worn down at one circular end at a rate of 0.5mmh-1.

Find dV/dT



Now, i know that the volume of a cylinder is Pi(r^2)h, So surely dV/dH is 2(Pi)r(you forgot a 'h') right?

The book says it is (Pi)r^2 though :s


Can anyone help? :smile:



dV/dh means it is with respect to h(height). So so it is the 'x' in the equation.

If it was dV/dr then 2πrh would be correct.
(edited 10 years ago)

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