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pal_sch
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#41
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#41
A lot of trig and calc on the P1 exam today. I am OK with it, I prefer to visualise the shape or graph. This helped today as there was little space on the paper of diagrams and I wasn't going to ask for paper just unessicary drawings. Edexcel have a tendency to lose or not mark things like that.
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greenie787
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#42
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#42
i did the WJEC paper
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WerelWolf
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#43
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#43
(Original post by Jonatan)
They clean up lots of equations. If you want to use degrees you very often end up with a 360/2pi factor.

Also, when you need to raise a number to a complex exponent you probably want to use radians when expressing the complex numbers in polar form. This gives rise the following beautiful equation:

e^(i * pi) + 1 = 0

Also, do you take this task (Its a physics one, but still)

At what height do you find geostationary satelites?
Hey, I found the other day that and negative ln will give a complex number with the imaginary part being ( i * pi ) and asked school around if anyone knew anything.. Could anyone here care to explain or point me in the direction?

I thought the there would be a square root of negative pi squared, but this doesn't seem so, unless it comes from (root(-pi^2) somehow..

There's no HL Math here, not enough participation, and the years are semestered, so there's not much time to get all the subjects required; it would be great to know this.
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mikesgt2
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#44
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(Original post by sazzles)
I'm probably being really stupid here, but with your arguement isn't there the possibility of having a sector with a different angle, that has a larger area than the square. I thought I had phrased the question so it was prove that with a set perimeter, the maximum area of a sector is equal to the area of a square. I still can't see how your logic proves this (try using one syllable words - I'm kind of dumb).
Oh, ok. I think I understood the question wrong.
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LimpNoodle
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#45
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#45
ne one takin pure 1 and 2 on wednesday?

i am and i'm cacking myself
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sazzles
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#46
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(Original post by mikesgt2)
Oh, ok. I think I understood the question wrong.
It's probably my fault with the phrasing of the question, sentences that make sense aren't my strong point.
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LimpNoodle
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#47
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#47
hmmm yeah cos that makes sense?!!?
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sazzles
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#48
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#48
(Original post by WerelWolf)
Hey, I found the other day that and negative ln will give a complex number with the imaginary part being ( i * pi ) and asked school around if anyone knew anything.. Could anyone here care to explain or point me in the direction?

I thought the there would be a square root of negative pi squared, but this doesn't seem so, unless it comes from (root(-pi^2) somehow..

There's no HL Math here, not enough participation, and the years are semestered, so there's not much time to get all the subjects required; it would be great to know this.
I'm not sure, but it seems to be a logical consequence of
e^(i*pi)+1=0
rearrange
e^(i*pi)=-1
ln both sides
i*pi=ln(-1)
Any ln(-x) can be split into
ln(-1*x)=ln(-1)+ln(x)
ln(-1*x)=i*pi+ln(x)

Oops, I've just realised that you're right.
square root (-pi^2)=square root (-1) x square root (pi^2) =i*pi
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LimpNoodle
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#49
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not many people are posting here, and i'm seeing too many people making new topics asking about maths coursework and i need help in maths! well save time and ask quickly here!

Here is a problem: check it out
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LimpNoodle
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#50
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ooops sorry double post
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greenie787
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#51
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hi does any1 know any websites for AS physics revision? i'm particularly looking for stuff on forces, moments and speed. thanks
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LimpNoodle
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#52
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like what i know most stuff on physics i'm finishing off a-level this year
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greenie787
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#53
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#53
i've got a modular test on thurs on forces and stuff, and i dont get how you resolve the force in different directions. the teacher was explaining stuff about 'perpendicular' forces. i understood most of the stuff in class, but i dont know how to do it now. theres also stuff i dont know about how to find tension and resultant force
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LimpNoodle
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#54
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#54
yeah i see its forces on a line. Did your teacher mention torque? if so its

Torque = Forces x perpendicular seperation

Also it could have been all force up must equal the forces down, say like a see-saw and the pivots reaction force must equal the forces downwards (the weight of the people)

tell me if that doesnt help you
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greenie787
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#55
no we havent done anything to do with torque. i'm mostly confused over how to find tension and resultant force type questions. the exam is with edexcel if thats any help?
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crana
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#56
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hi someone asked me this the other day, any ideas? I cheated and used excel to work it all out but Id like to know how to do it properly!

tan1 x tan2 x tan3 x ... x tan89

In degrees......thanks!

Rosie xx
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It'sPhil...
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#57
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(Original post by crana)
hi someone asked me this the other day, any ideas? I cheated and used excel to work it all out but Id like to know how to do it properly!

tan1 x tan2 x tan3 x ... x tan89

In degrees......thanks!

Rosie xx
Group the product like so

(tan1.tan89) x (tan2.tan88) x ... x (tan45)

Here we are evaluating a product of tanx.tan(90-x) for 1<x<45

But tan(90-x) = cotx hence tanx.tan(90-x) = 1

Therefore the product is 1x1x1x1x1x1x...x1 = 1
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theone
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#58
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#58
(Original post by crana)
hi someone asked me this the other day, any ideas? I cheated and used excel to work it all out but Id like to know how to do it properly!

tan1 x tan2 x tan3 x ... x tan89

In degrees......thanks!

Rosie xx
There is a similar, albeit far more involved problem here:

http://www.nrich.maths.org/discus/me...tml?1072306437
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M47h5
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#59
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#59
(Original post by greenie787)
no we havent done anything to do with torque. i'm mostly confused over how to find tension and resultant force type questions. the exam is with edexcel if thats any help?
Hi I'm new here but think I can give this one a go!

As far as resolving forces goes:

You will most commonly be asked to resolve them either vertically or horizontally which you can denote by:

R(->) =
R(^) =
you do this to equate all forces in these directions so that you can add and subtract them from each other to give the overall direction of the force acting on a body. Taking up (for vert) and right (for horiz) as +ve you do this by taking each force acting on the body individually and multiplying them by either sinA (when moving away from the angle) or CosA (when moving through the angle) Where A= the angle the force is acting from the direction of resolution. once you have built up all these forces into a sum equate it!

eg if something is projected at 20 deg to the horiz and velocity = 5 ms-1

to resolve vert R(^) = vsinA = 5Sin20 = about 1.7 ms-1 (Sin as moving away from angle)
to resolve horiz R(>) = vCosA = 5Cos20 = about 4.7 ms-1 (COs as moving through angle)

Although you may be asked to resolve a force in a direction parallel or perpendicular to a plane it is exactly the same thing using the angle the force is between 2 perpendicular lines (those in which you are resolving)

This should help with resultant force as the two forces you equate from these resolutions can be made into a force triangle to obtain the direction and magnitude of the resultant.

tension is pretty similar and I could help with specific questions or queries if you like but i know have to revise for a math exam which is in the morn and I havent started yet

Hope this helps if not let me know and I'll try to clear up specifics.

Luda
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M47h5
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#60
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(Original post by LimpNoodle)
ne one takin pure 1 and 2 on wednesday?

i am and i'm cacking myself


I am taking P2 and I'm still 'jus bout to revise' ! LOL
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