PAT Solutions.

Q9 - 2009

I'm not sure how you can arrive at that answer unless you had $\frac{2}{16}L^2 + \frac{1}{16}L^2$ at the end, which, in that case, suggests you've made a sign error somewhere

Duuuude, not cool, your website charges for answers

2009-Q.01

technically not true becuase the graph doesn't exist when x =0. You should do the integral with limits tending to 0 but this still works

technically not true becuase the graph doesn't exist when x =0. You should do the integral with limits tending to 0 but this still works

That's a tad too complicated. Inverse trig functions aren't covered my most applicants. I think they expect you to use the simplest identities (tan as ratio of sin and cos & the sum of squares equals 1)

x^2=sin^2(t)=1-cos^2(t)
sin^2(t)=x^2
cos^2(t)=1-x^2

y^2=(sin^2(t))/(cos^2(t))=(x^2)/(1-x^2)
y=x/(1-x^2)^(1/2)

You must have an amazing teacher! Which modules are you doing in year 12?

She is a really good teacher C1 FP1 C2 D1 M1 S1.

No wonder you're dissatisfied with the pace of your lessons! My class is doing C1, C2, C3, C4, S1, S2, M1, M2 before exams and starting FP1, M3 and S3 after exams so there's no way I'm going to be asking for extra work!

2006 - Mathematics for Physics

Q1

It would have saved you a little time to notice that $2007^2-2006^2= (2007+2006)(2007-2006)$ by the difference of two squares.

It would have saved you a little time to notice that $2007^2-2006^2= (2007+2006)(2007-2006)$ by the difference of two squares.

Aah... crap.

2009-Q.03

Perhaps an easier solution - We have a right angled triangle, where opposite = 2 (radius), adj = 4 (it goes from 0,4 to 0,0). By observation the angle is 30deg

that means that the triangle completing the angle to 90 deg (to the horizontal) will have an angle of 60 - gradient is the tan of this triangle by observation, tan60 = sqrt3.

So by observation y = +/- sqrt3 +4.

Wordier than your solution, but if you can see it, it takes no working at all.

Not to say your solution isn't good as it is, I just wanted to offer an alternative. Wouldn't want to upset a female maths lover - after all we need more of you lot around

Someone suggested that on the other thread. It's neat. I'm male

I'm so sorry. I accidentally looked at the name-tag of the poster above you