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STEP I 2012 discussion thread

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Reply 20
Original post by In One Ear
Did anyone fully do question 1? I got the first part, and then showed that the min of the sum of the distances of P and Q was whatever thjey said. But when it came to doing the same with the distances between P and Q, i couldn't get a value of m out of the dy/dx=0 to put back in the distance equation.

I then tried question 2 and failed- i managed the graph and its stationary points i think, confused my self (stupidly) about the stating part, then found the two values of a for which dy/dx=0 and d2y/dx2=0 but then decided to cut my losses and move on.

I then attempted the integration question, got the first of the three integrals and then got stuck.

Then i moved onto the differential equation question (q8?) showed the transformation worked, got a solution to the diff equation but couldn't put it into the form they wanted so i moved on and tried the same substitution for the last part which worked but since it just asked for a general solution (no particular form) i just left it in my messy form.

Then i tried the cool question about dropping the stone in the well- the scary looking show that was actually fairly easy i thought. Then had to rearrange to show D=... started it but it would have been a tedious process of squaring and moving all radicals to one side and repeating and getting it in their exact form would have been time consuming and i was running out of time. Then attempted the last part which was the estimation of the well depth which went fine.

Just before time was up i made a superficial start at the mechanics question just before that (not worth more than 1-3 marks imo).

Expecting 50-60 marks which isn't really safe enough for a 2 :frown: .

What about you guys? How did you find it?


I found m=-b/a by inspection I think but it lead to a minimum distance of 0 which is clearly not true, the point made about m = (b/a)^1/3 on the other thread seems a lot more likely.

I did the integration one too and did the first two parts but I couldn't spot the "hence" trick on the last part.

I got the differential equation in the form they wanted after a lot of attempts and then the 2nd part I didn't guess the substitution, just wondering if I had used an integrating factor on that question (it was quite standard fp2 work) would I have gained credit?

Also did the N and T hyperbola question, half of the mechanics question about connected particles, and the Tn sequence question.
Original post by james22
Can anyone tell me how to do the integral of cos(2x)ln((cosx))? I still haven't figured it out.


IBP integrating cos2x (which gives 0.5*sin2x) and differentiating ln(cosx) (which gives -tanx). Then 0.5*sin2x*tanx simplifies to sin²x (which can be expressed in terms of cos2x).

That's all IIRC :tongue:

Big regret on this paper for me was not reading all the questions properly before starting ones that looked nice (but weren't). Lesson learned. :smile:
(edited 11 years ago)
Reply 22
Original post by xxxxkrishyxxxx
Omg, I'm such an idiot, that didn't even occur to me in the exam, I was trying to find a suitable substitution >;(


Exactly the same method I used and exactly the same reaction as I had after realising how simple it was! That's probably cost me my place at uni! :frown:
Reply 23
Original post by tizza1394
Anyone got ideas on boundaries?


I'd say that paper was harder than 2008 which had 81 for S, 65 for 1, 43 for 2 and 29 for 3.
Original post by fGDu
I found m=-b/a by inspection I think but it lead to a minimum distance of 0 which is clearly not true, the point made about m = (b/a)^1/3 on the other thread seems a lot more likely.

I did the integration one too and did the first two parts but I couldn't spot the "hence" trick on the last part.

I got the differential equation in the form they wanted after a lot of attempts and then the 2nd part I didn't guess the substitution, just wondering if I had used an integrating factor on that question (it was quite standard fp2 work) would I have gained credit?

Also did the N and T hyperbola question, half of the mechanics question about connected particles, and the Tn sequence question.


Looks good, the integrating factor method (assuming no slips) is fine I would think.
Reply 25
Original post by Wahrheit
Just use parts :biggrin:.


I tried that with u=ln(cos(x)) and dv/dx=cos(2x) but it all went wrong.
Reply 26
Original post by Extricated
Looks good, the integrating factor method (assuming no slips) is fine I would think.


I don't think it's possible to use integrating factor with something of the form dy/dx +(something)/y= something?
Reply 27
Original post by Extricated
Looks good, the integrating factor method (assuming no slips) is fine I would think.


Yeah but I didn't do that :biggrin: wish I had it would have been an extra 10 marks
Original post by fGDu
I found m=-b/a by inspection I think but it lead to a minimum distance of 0 which is clearly not true, the point made about m = (b/a)^1/3 on the other thread seems a lot more likely.

I did the integration one too and did the first two parts but I couldn't spot the "hence" trick on the last part.

I got the differential equation in the form they wanted after a lot of attempts and then the 2nd part I didn't guess the substitution, just wondering if I had used an integrating factor on that question (it was quite standard fp2 work) would I have gained credit?

Also did the N and T hyperbola question, half of the mechanics question about connected particles, and the Tn sequence question.


I don't think u could use an integrating factor- i did think that might be a way to quickly "cheat" a whole question. But because it was like xydy/dx i dont think you could get it into the standard form of dydx+Py=Q where P and Q are functions of x. I think this may have been deliberate because then the question would have been far too easy. Either that or i missunderstand differential equations (quite possible since i crammed fp3 in 1.5 days ).
My quick assessment is that it looks like a pretty hard paper; I'd expect lower boundaries than usual.

[I may revise my opinion once I've had a proper look].
Original post by Extricated
Looks good, the integrating factor method (assuming no slips) is fine I would think.


Original post by Wahrheit
I don't think it's possible to use integrating factor with something of the form dy/dx +(something)/y= something?


This was also my conclusion. May be wrong though?
Original post by fGDu
I found m=-b/a by inspection I think but it lead to a minimum distance of 0 which is clearly not true, the point made about m = (b/a)^1/3 on the other thread seems a lot more likely.

I did the integration one too and did the first two parts but I couldn't spot the "hence" trick on the last part.

I got the differential equation in the form they wanted after a lot of attempts and then the 2nd part I didn't guess the substitution, just wondering if I had used an integrating factor on that question (it was quite standard fp2 work) would I have gained credit?

Also did the N and T hyperbola question, half of the mechanics question about connected particles, and the Tn sequence question.


I guessed m=-b/a too and got 0 as the min distance... :/ Reckon we'll still get marks for noticing it's a sp??
Reply 32
Original post by In One Ear
This was also my conclusion. May be wrong though?


Actually I'm pretty positive you can't use integrating factor :wink:.
This paper's pretty tricky in the sense that choosing questions is more vital than usual. Lots of algebra bashing bleugh.
Original post by Wahrheit
Actually I'm pretty positive you can't use integrating factor :wink:.


Well it would have made the whole question pretty easy :tongue: .
Original post by james22
I tried that with u=ln(cos(x)) and dv/dx=cos(2x) but it all went wrong.


U = ln(cosx) ==> du/dx = -sinx/cosx = -tanx
v' = cos2x ==> v = 1/2sin2x

I = uv - S v du/dx

==> I = ln(cosx)(1/2sin2x) - S (1/2sin2x)(-tanx)

1/2sin2x = sinxcosx ==> 1/2sin2x*-tanx = sinxcosx * -sinx/cosx = -sin^2x

==> S (1/2sin2x)(-tanx) = S -sin^2x

cos2x = 1-2sin^2x

= (cos2x - 1) / 2

Integrating = 1/4sin2x - 1/2x

==> I = 1/2sin2xln(cosx) - (1/4sin2x-1/2x)

==> I = 1/2sin2xln(cosx) - 1/4sin2x + 1/2x (+c)

Sorry about lack of Latex, bit busy
(edited 11 years ago)
Original post by In One Ear
I don't think u could use an integrating factor- i did think that might be a way to quickly "cheat" a whole question. But because it was like xydy/dx i dont think you could get it into the standard form of dydx+Py=Q where P and Q are functions of x. I think this may have been deliberate because then the question would have been far too easy. Either that or i missunderstand differential equations (quite possible since i crammed fp3 in 1.5 days ).


I think so too... had to use substitution...
Reply 37
Original post by alvina9894
Is the last part of Q1 (a^2/3 + b^2/3)^3/2 ? Anyone got the same thing?


Well, we're either both correct or both grossly wrong!

Did anyone do Question 9 (mechanics, throwing a shot put at an angle)? I thought it was a great little work through to get their answers, but I messed up somewhere and got 5(x - rt(2) - 1), where x should have been rt(6) but, for some reason, wasn't:frown:

Anyone attempt Q13? I had about 20 minutes at the end so I thought I'd sink my teeth in, managed to find out there were 3125 separate integers but got stuck up afterwards. Any one got an answer?
Anyone do 4? I got t but when i tried to find n
the algebra got so messy it was unreal, like a million of each term so i left my answer as that dno if il get any marks for not simplifying
Did no one fancy 6?

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