OCR C4 (not mei) 18th June 2013 revision

Wait, how do we know what x,y and z are equal to? Other than:
x=3-2t
y=2+t
z=3+t

you know from the scalar prod that:

-2x + y + z = 0

so you know x= 3-2t etc. so sub that into the above equation and you'll have an equation with just t and solve.

Well I understand that, and I also appreciate what the mark scheme says in terms of the answer, but as we've been given the parametric equations, do we literally just have to assume the graph has the same "limits" in terms of range and domain as the parametric equations?

The way I found the values was assuming the graph followed the same pattern as the sine graph in terms of highest value and value at y=0, but only because that's the trig function that was in the parametric equations..so lets say those were cos instead of sine but everything else the same (except for the graph shifting to the right appropriately) we'd assume theta = 0 and 3pi over 2?

I saw your post on the FP2 thread, congrats on slaying the beast

hows ur c4 revision going?

C4 is going well. The errors I am making now are stupid mistakes, not actual misunderstanding of the theory which is a great sign. What about you?

Haha nahh it will be fine dw. Same as you, only making silly errors, I hope all goes well and we get a nice paper.

haven't looked at the 12 or 13 papers yet so I don't know what topics are likely to be the big ones in our paper. Hoping that whatever it is will be fairly standard, nothing totally out of the blue or random

Hi, can anyone help with question 2ii from Jan 10

Points A, B, and C have position vectors -5i-10j+12k, i+2j-3k and 3i+6j+pk respectively.
Given that ABC is a straight line, find the value of p
The answer is -8

I understand how to do it using dot product, cos(theta)=-1, but it was only two marks, and the mark scheme mentioned using ratios. Does anyone know how to do it using ratios?

Thanks!

If you find the direction vectors AB and AC, you will see that AB * by a constant = AC, this means that they are parallel (as there directions are only different in terms of scale factor, not actual vector direction) and since both directions share a common point A, you can say that A, B and C are all on the same line

Hi everyone! Do we get penalised if we put our vectors in column format even though it involves letters in the question?

Well yeah it is a graph of the parametric equations so they are going to share the same properties. The highest y-value is 4 (if sin = 1) and since that is at Pi/2 .... and it is the 1st occurance of a maximum point, it has to be the smallest positive theta possible. While for B is is the 3rd intersection with the x-axis (3rd non negative anyway) so you pick the 3rd non-negative solution of when y = 0.

And yeah that would be right!

What topics do people find hard? Is it just me or is c4 more straightforward than c3 which was a bastard of a paper..

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I was really happy with the C3 paper but I agree that C4 is more straightforward.

The exact same stuff always comes up in the C4 paper... Its like every paper is identical in format just with different questions. I'd found that in the C3 papers more random stuff was thrown in and there was more places to slip up..

I'm sure I've got over 90UMS in C3 so hopefully I don't mess up on C4, that would annoy me so much

If you find the direction vectors AB and AC, you will see that AB * by a constant = AC, this means that they are parallel (as there directions are only different in terms of scale factor, not actual vector direction) and since both directions share a common point A, you can say that A, B and C are all on the same line

In the June 2009 Paper, on question 6ii. why does it not allow Aln(x-5), Bln(x-3) etc. I put the integral into wolfram alpha, and it got those answers which I did as well. Can someone explain please?!

The limits are 2 and 1 so if you integrate as Aln(x-5) and Bln(x-3) when you put your limits in you will get Aln(-3), Bln(-1), Aln(-4) and Bln(-2) respectively. I am sure you know that this is impossible (you can't have ln of a number that is less than or equal to 0)

What you do here is that you have to manipulate the + and - signs so you can flip it to 5-x and 3-x, or you could just note the results that the integral of 1/x is in fact ln|x| not just lnx. It is surprising how my teacher neglected to mention the modulus sign also, but in these cases it is better safe than sorry to write Aln|x-5| so if you get limits like in this question you will have Aln|-3| = Aln3 etc. which does indeed work