This URL takes you to the formula book which you get in the exams. Might be useful when you’re revising so you can see exactly what is there – and, more importantly, what is NOT there! Have fun 
http://www.edexcel.com/migrationdocuments/GCE%20Curriculum%202000/N38210A-GCE-Mathematical-Formulae-Statistical-Tables.pdf
Hello!
I haven’t been on here in ages!
Basicaly, I wondered if I could get some M1 help
Review Exercise 2
Question 7
A rough plane is inclined at an angle α to the horizontal where tanα = 3/4. A particle slides down with acceleration 3.3m/s(squared) down a line of greatest slope of this inclined plane. Calculate the coefficient of friction between the particle and the plane.
Right, so the things I know are:
- that I’ll have to solve by resolving
- that Friction = Fmax or μR and is acting up the slope
- that cosα = 4/5 and sinα = 3/5
- that I’ll have to substitute in Force = ma somewhere
But still it doesn’t seem as if I have enough information to work this out – for example the mass of the particle is never mentioned but I will need to work out R.
I think that the thing that stresses me out the most here is the fact that I should find this question relatively easy!
Thank you!
Grace
(I hope this is not too inconvenient a place to ask my question.. I chose the most recent yr13 (non-password protected) post)
Hi Grace,
Sorry for the delay – only just seen this.
You don’t need the value of m because it cancels out.
mgsinα is the force down the slope
R=mgcosα so F=μ x mgcosα
It accelerates down the slope so
down – up = 3.3m
The trig terms and g you know so the only unknown is mu.
JT
Grace – i’ll be around for M1 help tomorrow morning, so do come in if you want to.
Thanks for all your work on Friday – was great!
KB
Thanks, Mrs Tibble – got it. (I think I was being a bit lazy yesterday, and not properly exploring the question..)
Also, thanks Mr Brewin, I’ll see you soon!
Grace
Hello!!
I just wanted to know how you know when to integrate by parts twice cos I am kind of confused about it in exercise 6G
Thank you
from Bernice
Hi Bernice,
When you integrate by parts, the 2nd term on the right is another integral. If you can’t do it, then you’ll need to use parts again. The commonest type is when you have x-squared in the original integral. This is your u term and your du is an x term. When you do it again, the x term becomes a number and then you can do the integral.
Does that help?
JT
Hello
Im a bit stuck on a couple of vector questions so please may I have some help?
Im doing the mixed exercise 5K and am currently stuck on question 2 and 6. Im not sure how to work them out, especially question 6
Thank you
Jacynth
Hi Jac, just been looking at the book online, so can’t read the text very well. Think it says pv of A and E are 5i + 4j + k and -i + j -2k
and you need to find P such that AP = 2EP?
If so, AE = -6i – 3j – 3k
And if AP = 2EP then P is one third of the way along AE, so P = A + 1/3 of AE
Does that work?
For Q6, two vectors are perpendicular if their dot product is zero. So if you set up some vector ai + bj + ck then do dot product with the two vectors you have you should get 2 sim eqns and then can work out those values?
Let me know if you need further guidance!
Hi Jacynth,
I’ve just had a look at those questions in the textbook and in Q2 P lies on AB produced, which means you have to extend AB past B such that BP is half of AP ie B is the midpoint. OP will therefore be OA + 2AB or OB + AB (either will get you to P).
With Q3 do what Mr B said, but the problem is you only have 2 equations for 3 unknowns so you can’t solve to get a unique solution. There won’t be a unique solution for a general vector anyway as any multiple in the same direction will also work. So what you do is get a relationship between 2 of the letters, then let one letter be anything you like (1 is easiest – just don’t choose 0! ) then work out the other letter values and whatever values you get will be a vector that works (dot it to check). The answer they got in the book was from setting c equal to 1, but this is arbitrary and just chosen to make the numbers ‘easy’.
Hope you’ve managed to make sense of that; if not, come back for more help.
JT
thanks ms tibble!!!!!
Oh also how do you integrate ln x does it just stay the same???
Hi Bernice,
If you have to do it from first principles, you need to integrate by parts because you can only differentiate ln x. Use ln x as your u and put a 1 in for your dv/dx. It’s done in example 19 on page 101.
If you don’t have to do it from scratch, learn the result: x ln x – x + C
JT
Thank you ms. and also have you checked your school email????
From Bernice
Can someone please tell me the password to get onto the C4 papers?????
Thank you. L
Hi Laura,
Sorry I don’t know it, but you can get some past C4 papers here
https://eiewebvip.edexcel.org.uk/pastpapers/
You’ll need the school centre number which is on your statement of entry if you can’t remember it.
JT
Thank you.
L
Hello!
I was just wondering if you could give me a hand with some S2?
Review Exercise 1 Question 22. (If you don’t have the textbook, then I can type the question out for you, but for now I will presume that you do.)
I am stuck on the last part of the Question: “Given that each of 2 letters chosen at random from the day’s typing contains one or more errors, find, to 4 decimal places, the probability that one was typed by Pat and the other by Lyn”
The rest of the question has been fine – I’m just not quite sure how to go about this part.
Thanks,
Grace
Hi Grace,
Will let you know when I’ve pondered it lol! It sounds like a conditional probability from S1 but I haven’t been able to get their answer yet!
Back to the drawing board
JT
Hi Grace,
It is conditional probability so you can either draw a tree diagram to help you understand it or use the cond. prob equation in the formula book.
Either way, you should be able to show that P(Pat and errors)= 0.018
and P(Lyn and errors) = 0.047 (This is higher because she types loads more letters than the trainee)
Using the formula, the P that it’s Pat (given there are errors) = 18/65
and the P that it’s Lyn (given there are errors) = 47/65
Therefore the probability that the sample of 2 contains one from each of them =
18/65 x 47/65 + 47/65 x 18/65 = 0.4005 to 4 dp.
Hope you can follow that! It’s a bit mean to make you use the S1 stuff in S2!
JT
Yes, I was just doing another question that brought in conditional probability too – Very mean!
The general formula as i remember it is: P(A|B) = P(A and B)/P(A) Is that correct?
Right, I follow that apart from when you work out P(Pat and errors). Do you do just do P(pat)x her p(errors) or is there something else from S1 I need to remember? Doing P(Pat) x P(Errors) is actually wrong I know – becasue that would make P(A|B) = P(AandB) which is a special kind of distribution that I can’t remember the name of…
My overriding question is really do you think that they will bring in many things – like conditional probablity – from S1? I know they can expect your prior knowledge, but are they actually allowed to put such an S1 based question in?
Thanks for your help,
Grace
Hi Grace,
Well the formula is in the formula book so in theory they could. This is in the spec for S2:
Prerequisites
A knowledge of the specification for S1 and its prerequisites and associated formulae, together with a knowledge of differentiation and integration of polynomials, binomial coefficients in connection with the binomial distribution and the evaluation of the exponential function is assumed and may be tested.
However, we haven’t seen it tested on the past papers we’ve looked at – they’ve asked about skewness from S1 and calculation of mean and variance from 1st principles, but not this.
The formula you need to use is Bayes formula, although I don’t know if they give it that name in S1; it’s the backwards one!
P(A|B) = P(B|A) P(A) / {P(B|A)P(A) + P(B|A’)P(A’)}
JT
great thank you for all the help! i think i can make sense of the questions now! sorry for not managing to reply sooner!
Im also having trouble integrating 2/ (1-x)^2. i expect its quite simple but please may i have some help?
Hope you’re all enjoying your holidays!
Jacynth
Hi Jacynth,
Take the bracket to the top with a -2 power and then you’re there with the normal rules for integration.
Have fun
JT
Hi Mr Brewin or Mrs tibble
When you differentiate a constant in the example it goes from
lny = xlna
to
1/y dy/dx = ln a
i understand that you differntiate the lny y to get 1/y dy/dx but on the right hand side why has it gone from xlna to lna? where did the x go?
also when you differentiate 2^-x does that equal -2^-xln2? and if it was 2^2x would that be 4^2xln2?
Megan
Hi Megan,
For your first question, a is a constant not a variable so ln a is also a constant ie just a number. Just as when you differentiate 4x say and get 4, if you differentiate xlna you get ln a.
2nd question. Yes, except the way you’ve written it here looks confusing: the ln2 isn’t part of the power.
3rd question. The 4 is wrong – it should be a 2 because when you differentiate 2x you get 2. Rest is OK with same comment as before re the ln2.
Are you having fun?
JT
ooo okay i understand the first part.
But for the second part i’m still kinda confused.
so theres a rule that
y=a^x
dy/dx = a^x lna
so for y=4^x all you do is substitute in the numbers? so the answer would be:
dy/dx = 4^x ln4
but then when you have y= 2^-x
dy/dx = -2^-x ln2 (i this case you don’t just substitute numbers in. Do you differentiate the power and bring it forward? i understand why there is a negative when i do it long hand but i don’t understand it short hand)
when you do y=2^2x long hand does ln2^2x go to 2xln2 or xln4(xln2^2)?
I’m not sure whether that made much sense lol
Megan
Hi Megan,
Yes to the rule. When you differentiate e^x you get e^x ln(e) but ln(e)=1 so it’s just e^x. That’s why differentiating e^x doesn’t change it.
When the power isn’t just x you have to multiply by the derivative of the power as well.
If y=2^2x
then dy/dx = 2^2x times the derivative of the power x ln2 (because it’s 2 not e)
so it becomes 2ln2 times 2^2x
Hope that helps!!
JT
PS See section 4.3 Example 6
I think i get it. I will have a look at the example aswell
Also i have a question paper for June 2008 is there a mark scheme for this paper, i couldn’t find one on the edexcel website
Thankyou
Hi Megan,
It’s on the website at
http://www.edexcel.com/quals/gce/gce-leg/maths/Pages/default.aspx
under mark schemes paper 6666
JT
Thank you for your help!
Also how do I integrate (xe^x)^2 please? I have a feeling I need to do integration by parts (although I may be wrong) but I’m a bit stuck on actually how to solve it
Thanks
Jacynth
Hi Jacynth,
Start by squaring everything to get rid of the bracket, then integrate by parts twice.
JT
Hi, I was just doing a Normal approximation to Binomial distribution question and got a little bit stuck.
In the production of compact disks at a certain factory, the proportion of faulty disks is known to be 0.2. Each week, the factory produces 2000 disks. Estimate, to 2 significant figures, the probability that there will be at most 349 faulty disks produced in one week.
So when X is the r.v. “number of faulty disks produced in a week”
X~B(2000,0.2) so Y~N(400,4√5)
P(X<349) is P(Y<349.5)
I work that through to get P(Z<-5.64607…) But obviously table figures aren't available for 5.65 – do I have to work this out by symmetry of the curve or something or have I made an error in my working?
Secondly, I just wanted to quickly clarify something: in a question, when they say "more than 7" they mean 8 and above, don't they?
Thanks!
Grace
I’ve just done paper c of the solomon papers and i got a bit stuck on a few of the questions
Question 4 – vectors
we know: A(3,9,-7) B(13,-6,-2)
- in part a i worked out the line l that passes through A and B (which i could do)
- part b it says show c has the cordinates(9,0,-4)(which i could do)
c) the point D is the point on l closest to the origin 0.
- in the answer they did
[L = lambda]
3+2L 2
( 9-3L ) . ( -3 ) = 0
-7+L 1
but they havn’t told us D is perpendicular so how do we know the dot product equals 0? I would understand this method if 0D was perpendicular to line A
question 6 – parametric equation
x=3sint y=2sin2t
the curve meets the x axis at the origin and at the point A
a) Find the value of t at 0 and the value of t at A
- wouldnt at both places t=0? in the answer it says at a t=0 or pie/2 i dont understand how you get pie/2?
Thanks Megan
P.S. thanks for the mark scheme
wops i forgot to ask my last question
for part b of question 6 it says
the region enclosed by the curves is rotated through pie radians about the x axis.
-Do you do it exactly the same as if it was through 2pie radians, and if so why?
Megan
Hi Megan,
1)The shortest distance to a line is always the perpendicular distance.
2)On the x axis y=0 so put 2sin2t=0
thus sin2t=0 so 2t=0 or pi
so t=0 or pi/2
JT
PS Last bit of your question. 2pi radians would be a complete revolution around the x axis. Pi radians is half a revolution. Do it the same way and halve your answer.
JT
Hi Grace,
Sorry for the delay – your post went into spam for some reason!
Can you tell me which year that question is from as I don’t have my file here.
JT
Grace,
Your normal distribution parameters should be mean and variance: you’ve done mean and standard deviation. Does that help?
JT
PS Grace,
Yes, more than 7 means 8 or more for a discrete distribution.
JT
o right. Why is that, that the shortest distance to a line always is the perpendicular distance, or is it something you just learn?
also for the last part i don’t think they divided it by two.
In the answers they say
= volume when region above x-axis is rotated through 2pi?
what do they mean by that?
Yes Megan, the shortest distance is always perpendicular.
When a volume is rotated by 2 pi you get a solid…. think about spinning a pot on a potter’s wheel!
Megan, with the perpendicular thing, think about putting the point of your compasses on the point, and trying to find the smallest radius possible such that your arc meets the line. This will happen when the line is a tangent to your curve because the arc will only touch the curve, not cross it (twice). If the line is a tangent, then you know that tangent and radius always meet at 90 degrees.
Does that help?
JT
Yer thankyou Mrs Tibble that i kinda get i know.
For the Volume question i still kinda confused because it says it is rotated through pi radians not 2pi? but in the answer it still uses the normal method of multiplying it by pi? Why have they not divided it by two?
I’m not sure if its because the curve is identical on either side of the x axis, would that make a differnce?
Megan
Maybe it’s just a typo; I don’t remember seeing a question where the rotation was only through pi.
This was the question:
Figure 1 shows the curve with parametric equations
x = 3 sin t, y = 2 sin 2t, 0 ≤ t < π.
The curve meets the x-axis at the origin, O, and at the point A.
(a) Find the value of t at O and the value of t at A. (2)
The region enclosed by the curve is rotated through π radians about the x-axis.
(b) Show that the volume of the solid formed is given by
π2
0 ∫ 12π sin2 2t cos t dt. (3)
Shall i just assume it was a typo?
Megan
Yes!
Hello
I am wondering if I can get some help on C4 integration
How do integrate In (x over 2)? I was going to try integration by parts, but I could not work out u and what would be dV/dU.
Raslyn
Hi Raslyn,
You can’t integrate logs so you need to write it as 1xln(x/2) and integrate by parts with ln(x/2) as your u and 1 as your dv/dx.
JT
Thanks
By the way, when you differentiate In(x/2), would you use the quotient rule and then the chain rule
Raslyn