Edexcel FP3 June 2015 - Official Thread

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

Reply 181

username1259045

17

June 2013 question 3, how do they simplify the integral? the second line with (2sinh(2theta))^2 + (4cosh(theta))^2 ???

https://3a14597dd5c7aa2363f0675717665774b02557b0.googledrive.com/host/0B1ZiqBksUHNYQWE5bVRTVE9BLW8/June%202013%20MS%20-%20FP3%20Edexcel.pdf

Reply 182

math42

20

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://3a14597dd5c7aa2363f0675717665774b02557b0.googledrive.com/host/0B1ZiqBksUHNYQWE5bVRTVE9BLW8/June%202013%20MS%20-%20FP3%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

(edited 8 years ago)

Reply 183

chemlover12

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://3a14597dd5c7aa2363f0675717665774b02557b0.googledrive.com/host/0B1ZiqBksUHNYQWE5bVRTVE9BLW8/June%202013%20MS%20-%20FP3%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.

Reply 184

Skitee

10

So here we go... the last one for everyone else as well? Anyone find this easier than FP2 or is that idea preposterous?

Reply 185

math42

20

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.

Reply 186

chemlover12

3

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.

Reply 187

Ilovemaths96

13

cant possibly do any more maths after step 2 today, as expected it was terrible

(edited 8 years ago)

Reply 188

A*desperate!

Original post by 1 8 13 20 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!

Reply 189

smem_96

1

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?

Reply 190

luckystar1900

2

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

Reply 191

smem_96

1

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

Reply 192

math42

20

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

Reply 193

Teddysmith123

19

Can someone please help me with June 2013 question 8c?

https://3a14597dd5c7aa2363f0675717665774b02557b0.googledrive.com/host/0B1ZiqBksUHNYQWE5bVRTVE9BLW8/June%202013%20QP%20-%20FP3%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?

Reply 194

Skitee

10

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

Reply 195

smem_96

1

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

Still helpful, thanks :smile:

Reply 196

chemlover12

3

Original post by Teddysmith123

Can someone please help me with June 2013 question 8c?

https://3a14597dd5c7aa2363f0675717665774b02557b0.googledrive.com/host/0B1ZiqBksUHNYQWE5bVRTVE9BLW8/June%202013%20QP%20-%20FP3%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)

Reply 197

Teddysmith123

19

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) ?

Reply 198

chemlover12

3

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.

(edited 8 years ago)

Reply 199

Teddysmith123

19

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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1434560787132.jpg42.2KB

Edexcel FP3 June 2015 - Official Thread

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