[BenUny]

I've just done a past paper and I'm struggling on a couple of questions. Here's the question paper: http://pdf.ocr.org.uk/download/pp_11..._gce_4724.pdf?
and the mark scheme: http://pdf.ocr.org.uk/download/ms_11...u4724_jun.pdf?

2) I understand that the modulus of the vector is 5, but I don't understand what the next bit (they multiplied the vector by 1/5), can someone please explain this?

3)ii) The only part of this that I don't understand is how x/((x^2)+3) integrates to .5ln((x^2)+3). Can someone please explain this.

Thanks in advance.

[etothepii]

2) You multiply it by 1/5 to get a modulus of 1 as the modulus of the unit vector is 1.

3. ii) You use the f'(x)/f(x) rule. But, you need to tweak the numerator so you get 2x, which is then the differential of the denominator, but to account for the change, you multiple it by .5.

[Mathlete 4 the win]

2) The unit vector is the vector given divided by its magnitude. That is how it is defined. So to find the unit vector you find the magnitude of the given vector, in this case 5 and then divide each component in the vector by it.

3)ii)
For this you use the derivative over a function rule. The top of the fraction is x, the derivative of the bottom is 2x.

So pull out of a half (or you could say multiply the top by 2 and bring out a half in front) to get (1/2) X 2x/(x^2+3).

This integrates to 1/2 ln(x^2+3)