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# Edexcel FP3 June 2015 - Official Thread watch

1. (Original post by A*desperate!)
Hi, glad to see a thread fp3!
Can anyone please suggest me some past papers to do? The tricky ones I mean, cz time is short! Thanks!
Standard paper from last year is the hardest that's been done
If I remember correctly the R paper from last year is slightly tricky as well
2. June 2013 question 3, how do they simplify the integral? the second line with (2sinh(2theta))^2 + (4cosh(theta))^2 ???

https://3a14597dd5c7aa2363f067571766...%20Edexcel.pdf
3. (Original post by mmms95)
June 2013 question 3, how do they simplify the integral? the second line with (2sinh(2theta))^2 + (4cosh(theta))^2 ???

https://3a14597dd5c7aa2363f067571766...%20Edexcel.pdf
Their way makes little sense to me. What they've done in the M1 box is multiply numerator and denominator by 2sinh(2theta) though. Could you not do the question or did you just do it another way and don't get their method?

edit: misread that didn't see the or

The integral is simplified using the identity for sinh(2theta) and cosh^2 - sinh^2
4. (Original post by mmms95)
June 2013 question 3, how do they simplify the integral? the second line with (2sinh(2theta))^2 + (4cosh(theta))^2 ???

https://3a14597dd5c7aa2363f067571766...%20Edexcel.pdf
You know from C4 sin(2theta) is 2sin(theta)cos(theta). It's the same with sinh(2theta) where sinh(2theta) = 2sinh(theta)cosh(theta). It's straight forward from here onwards.
5. So here we go... the last one for everyone else as well? Anyone find this easier than FP2 or is that idea preposterous?
6. (Original post by Skitee)
So here we go... the last one for everyone else as well? Anyone find this easier than FP2 or is that idea preposterous?
Yeah, straight after S2. I think it tends to be easier than FP2 but it also has more scope for a hard paper, e.g. last year's. The reduction formulae expected can be evil and vectors can be pretty unpleasant.
7. (Original post by Skitee)
So here we go... the last one for everyone else as well? Anyone find this easier than FP2 or is that idea preposterous?
I still got D2 on the 24th T_T. That's what all my teachers said but vectors is def a killer for me.
8. cant possibly do any more maths after step 2 today, as expected it was terrible
9. (Original post by 13 1 20 8 42)
Standard paper from last year is the hardest that's been done
If I remember correctly the R paper from last year is slightly tricky as well
Okay will do! Thankyou
good luck to all you peeps for the exam!
10. This might be a really stupid question, but how come for hyperbolic functions it doesn't matter whether you're working in degrees or radians, whereas it does for trig?
11. (Original post by smem_96)
This might be a really stupid question, but how come for hyperbolic functions it doesn't matter whether you're working in degrees or radians, whereas it does for trig?
Because hyperbolic functions are made from exponentials, so e^x is the same whether you are in degrees or radians...they're not actually trig funtions they just behave similarly so are sometimes treated as the same. but they are exponentials by definition
12. (Original post by luckystar1900)
Because hyperbolic functions are made from exponentials, so e^x is the same whether you are in degrees or radians...they're not actually trig funtions they just behave similarly so are sometimes treated as the same. but they are exponentials by definition
Thanks
13. (Original post by smem_96)
This might be a really stupid question, but how come for hyperbolic functions it doesn't matter whether you're working in degrees or radians, whereas it does for trig?
People work in degrees for trig often because degrees are such a common measure of angle and trig is fundamentally about the angles in a triangle.
With hyperbolics you're more considering just a number rather than a measure of an angle.
Also hyperbolics are closely linked to the complex number expressions for trig functions and in those cases we always have to work with radians
14. Can someone please help me with June 2013 question 8c?

https://3a14597dd5c7aa2363f067571766...%20Edexcel.pdf

I have a= -2i+5j+13k

and I have found an equation for where the lines meet: point: (15/13, -15/13,0) and direction lambda (-2/13i+5/13j+k)

How do I get B?
15. (Original post by smem_96)
This might be a really stupid question, but how come for hyperbolic functions it doesn't matter whether you're working in degrees or radians, whereas it does for trig?
Because hyperbolics are calculated by exponentials (remember the exponential identities for sinh/cosh) which have nothing to do with radians and degrees

Edit: seems like lots of people beat me to it haha
16. (Original post by Skitee)
Because hyperbolics are calculated by exponentials (remember the exponential identities for sinh/cosh) which have nothing to do with radians and degrees

Edit: seems like lots of people beat me to it haha
17. (Original post by Teddysmith123)
Can someone please help me with June 2013 question 8c?

https://3a14597dd5c7aa2363f067571766...%20Edexcel.pdf

I have a= -2i+5j+13k

and I have found an equation for where the lines meet: point: (15/13, -15/13,0) and direction lambda (-2/13i+5/13j+k)

How do I get B?
The general formula is (r-a)xb=0 they changed it to rxb=axb where a is a point on the line, crossed with it's parallel vector. So in this case, it would be the point of intersection (which lies on the line) crossed with wht your a (-2i + 5j + 13k)
18. (Original post by chemlover12)
The general formula is (r-a)xb=0 they changed it to rxb=axb where a is a point on the line, crossed with it's parallel vector. So in this case, it would be the point of intersection (which lies on the line) crossed with wht your a (-2i + 5j + 13k)
see I understand the formula and know that A is -2i+5j+13k but what would b actually be? Would the point of intersection be (15/13,-15/13,0) ?
19. (Original post by Teddysmith123)
see I understand the formula and know that A is -2i+5j+13k but what would b actually be? Would the point of intersection be (15/13,-15/13,0) ?
In the question, b is axb (it's the point of intersection crossed with your a)

I'm awful at explaining. RxB=AxB the a in the question is B in this equation. the b in the question is the AxB in this equation.
20. (Original post by chemlover12)
In the question, b is axb (it's the point of intersection crossed with your a)

I'm awful at explaining. RxB=AxB the a in the question is B in this equation. the b in the question is the AxB in this equation.
Ok I've attached my working because im still stuck on getting the final answer

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