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AQA Core 4 Exam Discussion 14/06/2012 !Poll, paper and unofficial MS (first post)!

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Reply 100
Avatar for mojopin1
mojopin1
Can anyone help me with this vectors question (part c)?

vectors.JPG

I need to find the coordinates of D. Since AD and BC are equal, I found D simply by adding the difference from B to C if you know what I mean. But I want to know how to do the method they used in the mark scheme (OD = OC + BA). How do I do this? How do I know where the origin is?

http://store.aqa.org.uk/qual/gceasa/qp-ms/AQA-MPC4-W-MS-JAN07.PDF
Original post by mojopin1
Can anyone help me with this vectors question (part c)?

vectors.JPG

I need to find the coordinates of D. Since AD and BC are equal, I found D simply by adding the difference from B to C if you know what I mean. But I want to know how to do the method they used in the mark scheme (OD = OC + BA). How do I do this? How do I know where the origin is?

http://store.aqa.org.uk/qual/gceasa/qp-ms/AQA-MPC4-W-MS-JAN07.PDF


If you look at the attachment, using the parallel sides, Vector OC (which is given) + Vector BA (which you have already worked out) will give the co-ordinates of "D"
(edited 11 years ago)
maths.jpg28.3KB
Reply 102
Avatar for Quexx
Quexx
Original post by JulietheCat
And I don't remember the triple angle formulae. I still need to learn them.


Two things...
In this C4 exam you don't need to learn the triple angle formula because they are not used.
And if you have to use the triple angle formula, you can just use the double angle formula twice.
Eg: Cos(3X) = cos(2X+X)
= cos(2X)cos(X) - sin(2X)sin(X)

Cos(2X) = 2cos^2(X) - 1
Sin(2X) = 2sin(X)cos(X)

Cos(3X) = [2cos^2(X) - 1]cos(X) - [2sin(X)cos(X)]sin(X)
= 2cos^3(X) - cos(X) - 2sin^2(X)cos(X)
= 2cos^3(X) - cos(X) - 2[1 - cos^2(X)]cos(X)
= 2cos^3(X) - cos(X) - 2cos(X) + 2cos^3(X)
= 4cos^3(X) - 3cos(X)

It works out as well if you do the same with sine and tangent.
They are quite easy to remember if you want to remember them though xD
Reply 103
Avatar for f1mad
f1mad
13
Original post by Quexx
Two things...
In this C4 exam you don't need to learn the triple angle formula because they are not used.
And if you have to use the triple angle formula, you can just use the double angle formula twice.
Eg: Cos(3X) = cos(2X+X)
= cos(2X)cos(X) - sin(2X)sin(X)

Cos(2X) = 2cos^2(X) - 1
Sin(2X) = 2sin(X)cos(X)

Cos(3X) = [2cos^2(X) - 1]cos(X) - [2sin(X)cos(X)]sin(X)
= 2cos^3(X) - cos(X) - 2sin^2(X)cos(X)
= 2cos^3(X) - cos(X) - 2[1 - cos^2(X)]cos(X)
= 2cos^3(X) - cos(X) - 2cos(X) + 2cos^3(X)
= 4cos^3(X) - 3cos(X)

It works out as well if you do the same with sine and tangent.
They are quite easy to remember if you want to remember them though xD


Re(cosx+isinx)^3= Re(cos^3x + 3(cos^2x(isinx)) + 3(cosx(isinx)^2)+ (isinx)^3)

cos(3x) = cos^3(x)+ 3cosx(-sin^2(x))
= cos^3(x) -3cosx(1-cos^2x)

= 4cos^3(x)-3cosx

:tongue:
Original post by Quexx
Two things...
In this C4 exam you don't need to learn the triple angle formula because they are not used.
And if you have to use the triple angle formula, you can just use the double angle formula twice.
Eg: Cos(3X) = cos(2X+X)
= cos(2X)cos(X) - sin(2X)sin(X)

Cos(2X) = 2cos^2(X) - 1
Sin(2X) = 2sin(X)cos(X)

Cos(3X) = [2cos^2(X) - 1]cos(X) - [2sin(X)cos(X)]sin(X)
= 2cos^3(X) - cos(X) - 2sin^2(X)cos(X)
= 2cos^3(X) - cos(X) - 2[1 - cos^2(X)]cos(X)
= 2cos^3(X) - cos(X) - 2cos(X) + 2cos^3(X)
= 4cos^3(X) - 3cos(X)

It works out as well if you do the same with sine and tangent.
They are quite easy to remember if you want to remember them though xD


What unit are the triple angle formulae used in?
Reply 105
Avatar for Quexx
Quexx
Original post by 1platinum
What unit are the triple angle formulae used in?


Might come up in the FP3 exam.
It could have come up during the FP2 exam, but luckily it didn't... =P
Original post by Quexx
Two things...
In this C4 exam you don't need to learn the triple angle formula because they are not used.
And if you have to use the triple angle formula, you can just use the double angle formula twice.
Eg: Cos(3X) = cos(2X+X)
= cos(2X)cos(X) - sin(2X)sin(X)

Cos(2X) = 2cos^2(X) - 1
Sin(2X) = 2sin(X)cos(X)

Cos(3X) = [2cos^2(X) - 1]cos(X) - [2sin(X)cos(X)]sin(X)
= 2cos^3(X) - cos(X) - 2sin^2(X)cos(X)
= 2cos^3(X) - cos(X) - 2[1 - cos^2(X)]cos(X)
= 2cos^3(X) - cos(X) - 2cos(X) + 2cos^3(X)
= 4cos^3(X) - 3cos(X)

It works out as well if you do the same with sine and tangent.
They are quite easy to remember if you want to remember them though xD


Thank you so much! I guess it means that just like linear interpolation in S1 and standard integrals in C4, just because it's in the textbook doesn't mean it would come up.
Reply 107
Avatar for Z REFAN Z
Z REFAN Z
Useful in understanding the binomial expansion. How to rearrange equations in the form of (1+ax)^n
and also describes how to work out the range of values for which the binomial expansion is valid.

http://www.youtube.com/watch?v=_5CGSTLAjeQ
Reply 108
Avatar for ch0wm4n
ch0wm4n
10
Not feeling great about this exam :s-smilie:.
these vectors are killing me!
Original post by snoopydoops
these vectors are killing me!


tell me about it....
Reply 111
Avatar for mojopin1
mojopin1
Is anyone else struggling with cartesian equations? I know it's simply rearranging but I find it difficult getting it into the form they want.
Reply 112
Avatar for Cath-ay
Cath-ay
OP
Original post by mojopin1
Is anyone else struggling with cartesian equations? I know it's simply rearranging but I find it difficult getting it into the form they want.


For example...? :/
And do you mean the form they state in the question or in the mark scheme?
Reply 113
Avatar for mojopin1
mojopin1
Yes, when they ask you to 'show' that the cartesian equation can be written in the form. What is the best stratergy to this?
Reply 114
Avatar for Cath-ay
Cath-ay
OP
Original post by mojopin1
Yes, when they ask you to 'show' that the cartesian equation can be written in the form. What is the best stratergy to this?


It depends on the question... I'd say to attempt to pick out parts of the form which you can manipulate the x or y equation for.. Like if the form they ask for has x^2 somewhere, try squaring the x equation, maybe rearranging it a bit and substituting that in. Or if it's trig, possibly mess around with trig identities until it looks like it might fit the form. There can be quite a bit of trial and error involved. I find that the mark scheme provides some insight on how to do it too!
Reply 115
Avatar for ADA93
ADA93
Can anyone help me with jan 2012 question 1c? I don't understand how ln has been rearranged to get the answer! Thanks.
Reply 116
Avatar for Cath-ay
Cath-ay
OP
Original post by ADA93
Can anyone help me with jan 2012 question 1c? I don't understand how ln has been rearranged to get the answer! Thanks.


after you've intergrated and inserted the limits, you need to use the log rules which you learnt from C2 to help you rearrange (ln is the natural logarithm)
-3ln3=-ln3^3=-ln27
ln5 - ln27 = ln(5/27)
Reply 117
Avatar for ADA93
ADA93
Original post by Cath-ay
after you've intergrated and inserted the limits, you need to use the log rules which you learnt from C2 to help you rearrange (ln is the natural logarithm)
-3ln3=-ln3^3=-ln27
ln5 - ln27 = ln(5/27)


Got it, thanks very much! :smile:
Reply 118
Avatar for kenau
kenau
Hi Cathay:biggrin:
I only left one type of question don't know how to do it...
Can anyone help me the JUN 10 question 7c and 8?
Original post by kenau
Hi Cathay:biggrin:
I only left one type of question don't know how to do it...
Can anyone help me the JUN 10 question 7c and 8?


I'll start you off on question 8. To start with, you separate the variables:

dxdt=−15(x+1)12⇒∫1(x+1)12.dx=∫−15.dt\dfrac {dx} {dt}=-\dfrac {1} {5}\left( x+1\right) ^{\dfrac {1} {2}}\Rightarrow \int \dfrac {1} {\left( x+1\right)^{\dfrac {1} {2}} }.dx=\int \dfrac {-1} {5}.dt

Continued inside spoiler, but give the rest of part A a go before looking.

Spoiler


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