The Sydenham Maths Department Blog

ok, 2nd post with same info, anyone feel like helping me with the last parts of the questions from Ex5E??

Thank you

i’ll do that now, i think i’ve got the whole transformation things (finally) it’s the slightly more complex things that are :S but it’s going good

write tan as sin/cos

Read my last message above as a clue.

No. Move the cos term to the other side so it’s on the bottom.

OK start again!!

Write down the question.

Divide both sides by cos x

Now divide both sides by 5

Now tell me what you’ve got!

No: sin/cos = tan (C1!!)

cot = 1/tan = cos/sin (C3 current topic)

I was actually going AAAAAAAGGHHHHHHHH!

Why has that gone in above you?

Think the gremlins are playing with the clock!

Must be time for a cuppa and bed then.

“you looked dreadful” nice Mrs tibble. real nice.

And it’s true I did look awful… I’m very glad people noticed and they don’t think I always look like that! But given I certainly don’t have anything mathsy to say, I’m off to bed…

Ok any help? Generally I think these questions are ok, but can someone start me on the right track for these: (Ex 6D, Question 6)

show that:
a. sec^4x-tan^4x=sec^2x+tan^2x
e. (1-tan^2x)/(1+tan^2x)= 1-2sin^2x

Thank you!

A MATHS TRIP!?!?!

Actually Em, we do have a lot of other things to do, but still, I’ve NEVER been on a maths trip. You can’t help with the maths can you?

sin q x sin q = sin squared q

Sin squared q x sin squared q = sin to the power 4 q

Sorry – I tried to copy and paste from Word but it wouldn’t work!

Obtuse tells you which quadrant you’re in so you know the signs of the functions – your All Stations To Catford thingy!

Ok, unmaths related (except I’ll put in here that we have a test on friday emma, in case you read this before your emails!) but what is this new message board thing? With the a level section entitled (or is it intitled, I know I can’t spell ok!) tea talk or something? Is it a hint to take our ramblings off here on to there?

Oh and something else mathsy: did anyone else find the core stuff really difficult, without quite knowing why, as the examples in class are never hard?? However it is now given in, albeit with an apology rather than all the work, and what’s done is done

What is the sine of pi/3?

Sub that in and tidy up the sin terms and it matches the right hand side expansion. QED

From Wikipedia:
Q.E.D. is an abbreviation of the Latin phrase “quod erat demonstrandum” (literally, “which was to be demonstrated”, and figuratively, “I rest my case”). The phrase is written in its abbreviated form at the end of a mathematical proof or philosophical argument, to signify that the last statement deduced was the one to be demonstrated, so the proof is complete.

You’ve slipped a sign – be careful!

√3 – half√3 = half√3

>>>
1/ tanAtanB which is cot (AB) right?

NNNOOOOOOOOOOOOOOOOOOOOOOOO!

AAAAAAGGGGHHHHHHHHHHHHHHHHHHHH!

Well Mrs Tibble shouldn’t be here lol!!

Mr W said he was going to be on here this afternoon. Why is there never a man around when you want one?!!

Just realised it’s one of the ones KB posted.

Combine your answers to a and b as x has to satisfy both of these conditions. If you shade the two regions on a graph, it’s the overlap region.

The quadratic in (b) is >0 so you only want the + bits, not the middle bit below the axis (sketch it!)

So, that’s any x value above 3 or below 1/2. However, the restriction from part a limits x to values >1/4 so on the left hand side of the curve you only get a tiny region which satisfies both inequalities.
HINT: It really is a good idea to sketch the two inequalities to understand this question.

Ok, partial fractions… Ex1F 2. c/d but really they are questions about the theory of it, rather than problems with specific questions…
In c. If you have a repeated root 3 times do you end up with 3 fractions, one with the denominator to the power of 4 and the others as single x terms?
In d. The problem is more with algebraic long division… basically I can divide fine, but the division leaves me with x^2+2x etc. Ok, so when you divided and got a whole number then you wrote original fraction= number from long division + partial fraction1 + partial fraction 2 etc. Is it the same with x? I feel like I shouldn’t end up with some x which are no longer part of a fraction… but I don’t really know… Anyone? (Ashely?!)

…

I’ll take your word for it! Ashley promised she’d come online to help me but she didn’t…

At least the decision was ok! I’ve got the Feynman book for you at school- I’ll try and remember it, with your mugs!

Sorry I wasn’t on earlier – been marking Year 11 papers

AAAAAAAAAAGGGGHHHHHHHHHHHHHHH!!!!!!!!!

I’ve lost count of the number of times I’ve crossed out (x+y)squared = x^2 + y^2

what is 4t ^ -1 intergrated?

Could I please just ask how sin 2t = 2 sint cost
It’s not from a specific question, just an example at the bottom of P.14 I was just looking at.

And another…
If you’re differentiating something like y=-4cost why don’t you have to use the product rule and treat it as -4 x cost (getting 4sint +cost) but instead simply leave the -4 and only differentiate the cos bit?

Well I assume that’s what needs to be done… certainly to get the answer in the back anyway…

Only if you’ll give me answers to my questions!

Ex4A 1a,b,c,f,h,j 3
Ex 4B 1a,b,d,g,h 3

Don’t spend more than two hours!

Hi Mel,
My son could never remember the quotient rule so he always did what you suggested!! Yes it works

If you use the product rule for 4cost the 2nd term disappears because when you differentiate 4 you get zero!

Oooooh. I thought I must be missing something because that’s the nice thing about maths- all the little bits fit together.

Right I’ll go back to the question and see if knowing I’m going in the right direction makes it any easier… We’ll see!

Ok… we’re doing exercise 4F or something but basically I’m confused about differentiating y…
If you need to differentiate 3y^2 or something similar do you end up with 6y x dy/dx? what about 3y^3. Once again the notes in the book are helpful like “use the chain rule”. er. Ok… Basically you treat it as thought it were an x and then multiply the answer by dy/dx, regardless of what the other bits are?

And more specifically I was doing question 8 and I got to part d and was very confused. because I got part c and then I thought you could insert terms for x and y from the bit at the beginning, multiplying out the cos2t to make it all in terms of sin… And then I thought by using the -b 4ac thing right I would get one answer of 2, given in the question and a different answer that I was looking for… But I didn’t get 2 at all. And then I thought I’d try putting my values for x and y (2, 1.5) into the equation 6y-16x+23=0 to see what I got, and it wasn’t 0. So am I going about it in the wrong way?

Using the chain rule means:

d/dx = d/dy times dy/dx

so basically what you said – but don’t say it like that or Mr W will get very very upset

mrs tibble i’m so cufflufled again i thought i could do it but qu 2 has gone so wrong

i’m on the 3rd table but i’ve got y = 4 1/5 and x = 12

and i dont’ have a clue of how to get to z

help?

Sorry, that should be s,t,x in the 2nd and s,y,x in the 3rd.

i don’t have those as the rows in the 2nd one i have y,t,z as my rows and in the 3rd i have y,x,z

Hi Mrs Tibble,

I tried to find you earlier today but you were in a lesson. I am stuck! Certainly am not getting the right number of elephants!

I am on Q3 (making quick progress I know! lol) It is difficult to try and explain my problem without you seeing it… if you are free tomorrow morning (P1/2), could you please go through it AGAIN with me?… I am just getting really weird results in my second table…

Fiona – use Z not z for the objective function.

First tableau you should have s, t, u and Z rows

2nd tableau: s, t, x and Z rows

I’m free in period 3

points in space? and i have a lesson period 3 :(

YAY!! Well done

They’ve used P for the objective function in Q3 so the bottom row will be a P row rather than Z.

P=3x+6y+2z

so P-3x-6y-2z+0s+0t=0

In the 1st tableau you only have s and t as your basic variables because you start at the origin where x=y=z=0

Yes

You don’t have x or y either!!!

You start at the origin for the first tableau where x y and z are zero so only the slack variables are basic.

The answers come from the last tableau. Each time you generate a new tableau, one basic variable goes out and one non-basic comes in which is why they keep changing.
Remember you are moving round the vertices so the combinations are going to change.

One of the 3 is zero because it’s non-basic in the final tableau.

does anyone know miss spicers first initial??
and i know im not doing a level yet. *sigh*

Simplex must have finally clicked! YAY

Mrs Tibble! (Or Mr Williams, but I guess you are on the ski trip now…)

I really wanted some core paper to work on over the holidays and Mr W the photocopier was broken so he’d put them up on the blog for us to do instead, but he’s gone on the ski trip now hasn’t he, and won’t be back for a while… But the end of the holidays is when I have a lot of family over so I need to get my work done over the next week… I know he said he wanted to keep some papers to do as mocks, but is there anything online that I can do?! I swear for C1 and C2 we had millions of papers (lettered ones?) Please? Thank you!

Thank you! Shall I mark them as well do you think, or do you think Mr Williams would prefer them in?

i totally can’t find the decision thread anymore :S so i know this is in the wrong place but….

mrs tibble is there other places we can get more D1 papers? doing this one has made me realise i can’t remember any algorithms so i’m doing it open book but would like to do a close one

hope u had a good easter

mrs tibble just curious now, why hvae you become tea lady??? and thank you i’m finally getting prims algorithm…i hope lol

i no but i didn’t think you use to end blog thingys with it :S or maybe i am acutally that visually unaware

okay, i’m trying to do qu 5 on the C4 specimin paper, and i’m really confused, i’ve got the right values for mu and lamda but i don’t know how to then get the point of interception :S

Same as Fiona He also has my current book… and the text book doesn’t really have enough examples to help so I’m sort of already stuck before I’ve even begun…
I don’t mind picking them up from school if they are there
Hope you had a good Easter

How weird: I’ve just replied to you Fiona and it seems to have vanished into cyberspace

Sub your lambda and mu values into the 1st vector equation.

emma how and y are u getting messages from mrs tibble??

Telepathy lol!

….. I can hear voices…..

Yay! I’m back. Mr B to the rescue – I was being spammed

Fiona, are you sorted now?

Tea Lady

Mr B went into school today and found they’ve put those fuschia pink doors on the maths floor

Nooooooooooooooooooooooo!!

and mrs tibble i haven’t ACTUALLY tried doing it yet….knitting/myspace kinda distracted me

Let’s choose a maths colour and paint our own department!

PINK?!

Well at least we only have to look at them for another 2 months!
Have done the D1 paper you gave us (I think I’ve finally got my head around the critical path stuff- but when you schedule activities can you do it by sight or is there an algorithm you’re supposed to use?) But I found the flow one hard: I couldn’t find a maximal flow at all: I guess there is a backflow, but I put one in and still didn’t have a maximal… Oh well… apart from that it was ok

And I’ve been reading your book Mrs Tibble… I haven’t got that far yet, and it’s quite weird, but good…

Which paper was it Mel – Jan or June?

(i’m not in your class but yes the doors are HORRIBLE. not a good move with BROWN CARPETS. oh dear. well. practically brown. i can’t even remember all i know is IT DID NOT LOOK GOOD.

Hi Mel,

The idea of a cut is to completely separate S from T which you do by “removing” arcs. If those arcs are saturated then you can find a min cut. Have a look at your notes right at the beginning of this topic – it should say somewhere that it isn’t always possible to draw the cut.

Yes, give worker 1 the critical path, then allocate the rest. You aren’t allowed to have idle time if there is a job which could be started. Make sure you check the precedences ie don’t start a job unless all the things it depends on are complete.

If sin + cos = 0 then sin = -cos
therefore tan = -1

In function notation, f’(x) is the 1st derivative, f”(x) is the 2nd.

In Q1 the function is a smiley parabola so it will have a minimum y value, below which the function doesn’t exist. If you differentiate and put g’(x) =0 for the min, you get the x value which you can then plug in to get the y value. This is the 2nd method in the mark scheme. The other method finds the min by completing the square.

Hope this helps,

Mrs T

I really don’t actually… but I am praying that it is not Core.

You can’t do past papers, only the specimen.

You’ll need to get past papers from school.

em, there is another paper at the back of the text books, if you wanted to do that as well?

Hi Emma,

Not sure where Mr W is; have sent him a text.

hey, mel read ur e-mail but are we getting the papers online or at home? because if their not online i still can’t do them

I’m going in to school early afternoon if there’s anything you want.

actually you’ve probably already been… don’t worry… I’ve got enough other work to keep me more than entertained over the next few days… but do you know which maths paper we’ll doing on the 9th? (fingers crossed decision!)

Mel,

You’re going to be doing the D1 as a first paper and then the C3. Both of these will be the in the two maths EPS after school sessions. The C4 will be in lesson time.

MW

Well… people have exams but maybe it’s none of you guys… So no, I don’t know…

hiya

I was just wondering, when you are differentiating 0.5e^x, is it still just the same thing? Because differentiating e^x is still e^x?

Thanks Mrs Tibble

Ok cool, I’ll make sure to pick them up tomorrow… I keep looking at my poster

My art went well thank you! I managed to finish my piece in time which I was so relieved about because at one point I didn’t think I would… but now at least I can concentrate on Maths and Physics!

Hope you’ve had a good weekend

Oh I really wanted to help with D of E this weekend! But thought I’d better stay home and work… If I’d known you were going then I could have done Decision revision in the middle of a field and asked you for help!
Glad you had fun, especially with such lovely weather!
See you tomorrow

Mel,

Try taking logs.

Mr Riedel

The question was from the C3 practice paper 4. Question7.
The first part of the question asked you to find dy/dx which was fine. And then part b asked you to find the tangent where the curve crosses the………

Oh wait.

x co-ordinate at the y axis is zero

wooops.

And I spent a long time wondering about that!

Oh that’s fine now, I can just put zero into the dy/dx equation and I’ll have the gradient. And then I can just put two values in and get c.

I thought what I was trying to was almost impossible…

Well thanks for your attempts

Any chance I could come into school monday or tuesday and get some more papers? Probably about 2 more C3 and as many C4 as I can…? I’m all out of question now…

I’ve got a question!

Find the angle that the vector 9i-5j+3k makes with the axis. I could do the questions that invovled the angle between 2 known vectors, so I think you’re supposed to work out the vector that describes the x axis and I feel like it should be really easy… but… I tried using =ax where x is a unknown constant and then multiplying out, but I got too many unknowns. And then I tried thinking that at the x axis the y and z values are zero so =9i but then I got a cos value which was greater than 1, so I think that’s wrong too.
Why is this so difficult?!

But how do we know it’s one? why isn’t it just any constant?

eg if you did 2i

a.b would be 18
l b l would be 2
18/2 =9

What exactly do you mean by ‘area of vector stuff’? Do your vectors form a triangle?

Can you give me the question?

OK. I’ll try and describe this!!!!

Draw a circle centre (0,0) and a random diameter AB.
Let vector OB = b
therefore vector AO (watch direction) is also b

Now put some point P on the circumference. Join PA, PO, PB.

Let PA = v say, PB = w say, and PO = p

Therefore v = p – b
and w = p + b

Now do the scalar product v.w = (p-b)(p+b)
=p^2 – r^2

But p^2 = r^2 as p is a radius
therefore p^2 – r^2 = 0
therefore v.w = 0 so vectors are perpendicular!!

Hiya, I too have a question

In the C3 Practice paper 4, Q8 b) it asks to find the maximum and minimum values of 2 cos Q + 5 sin Q. I found R to be the square root of 29 and this answer is given as the maxium too, but I don’t understand why… does that make any sense??

Oh ok, yep thanks brilliant thank you

Just one more thing, sorry:

How does:

1 – cos 2Q / tan Q = sin 2Q

1 – cos 2Q = 2 sin^2 Q doesn’t it? So should I substitute that and turn tanQ into sinQ/cosQ?

Gosh – I’ve not seen that before.

Now to your question:

Do you mean
1 – cos 2Q / tan Q = sin 2Q

or (1 – cos 2Q) / tan Q = sin 2Q

I am guessing the second (they are different as I’m sure you’ll agree!)

Ok. I’ve got a question!

If f(x)= 2x^3 find f’(x). The mark scheme gave me ((1-x)/2)^1/3. I don’t see where the 1 has come from.
And if y= (8x^3-1)/(1-2^3) why doesn’t dy/dx= 30x^2 /(1-4x^3+4x^6) and therefore the stationary point be 30x^2=0 and therefore x=0? because I just used to formula from the formula booklet and now I’m confused

Re the first part – are you sure the question you’ve typed is correct? It doesn’t make sense to me!

Yep. It said f(x)= 2x^3 and I got f’(x)= (x/2)^1/3

Maybe the 1 got lost on the question paper?

Wait… is it asking for the derivative? I thought it was asking for the inverse…?!

i don’t know if this will be looked at before our exam and i’ll come up to school later i was just wondering how sin2AcosA + cos2AsinA becomes 2sinAcos^2A + (1-2sin^2A)sinA :S

ooooh that makes perfect sense thank you one more question? on january 07, qu 5 part b how comes sin(x + pie/3) = 1/2 i don’t understand where the half came from

If so then

x + pi/3 = arcsin (1/2)

You’ll need to use ALL SILVER TEA CUPS to get the full set of answers to arcsin (1/2).

Treat each one as a separate equation

MW

actually don’t worry i get it now

lol LF u are funny

Alright, C4 questions (this is so frustrating- full marks on 4 out of 8 questions and I still can’t get more than 60%… and I genuinely don’t understand why, I’ve been through every vector question and apart from that it’s always the crazy twists that through me…)

Ok.
1.What’s it asking when it says ‘Hence find the coordinates of the foot of the perpendicular from O to AB’ (question before was find the value of lambda for which OP is perpendicular to AB…) I didn’t even know what foot it was talking about…

2. [10(2-3x)]/(1-2x)(2+x)= A/(1-2x)+B/(2+x).
So 10(2-3x)= A(2+x)+ B(1-2x) so A= 2 but I don’t understand how B=-16… And is there any way of checking these except checking that the sides are equal when you put A and B in?

3. If you have a crossing vectors and you are wanting to find the angle between them, but they aren’t coming from the same point (see I did remember you had to check!) can you just multiply one of your vectors by -1 to give you the vector in the opposite direction to the way it was?

4. If you’re given x= 5cosQ y=4sinQ and they ask you to find by integration the area enclosed by the ellipse then how would you know how to go about it? I started by trying to find y=f(x) but it wasn’t very easy and the mark scheme said intergrate ydx/dQ dQ and I can see how that works, but I wouldn’t have thought of it…?

Thank you in advance

Hi Mel.

Q1 If OP is perpendicular to AB then you want the coordinates of where they cross.

Q2 B=16

Q3 Yes

Q4 See C4 top of page 16.

To eliminate the B you need to make 2x-1 = 0

So x =1/2

To eliminate the C make 2x+1=0

So x=-1/2

Hope that this helps

You know.

I just realised.

We’re no longer maths A level students!
No more blog

But on the other hand, no more maths that we can’t do therefore providing the reason for being on the blog

I’m sure you’ll all be fine!

MW