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STEP I, II, III 2002 Solutions

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mikru
How did you decide upon your substitution for part three? It's things like this I would have no chance finding in the actual exam. I managed to get the same result as you using the substitution x=cosecย ฮธx = \mathrm{cosec\ } \theta (which I think followed on from the previous parts of the question better), but I think yours might have reached the result quicker (although I don't really follow your working in the middle).


If you look at the limits it'll give you an idea about why that substitution was considered. It's certainly not something I spotted the first time I did this question
mikru
How did you decide upon your substitution for part three?


The clue, I suppose, is in the limits of the equation. You have, in part iii), the limits in part ii) - but in reciprocal form.

The easiest way to turn part iii)'s limits into part ii)'s is to put y = 1/x. This immediately leads to limits of √3/2 and 1, but in the wrong order. The swapping of 'dy's for 'dx's fixes this by introducing a negative sign, thereby flipping the limits. Then a bit of algebra transforms the integrand into exactly the same thing as part ii)'s.

Your cosec 2θ substitution is exactly the same thing, though, if you think about it. If we're saying y = 1/x, and part ii) works by putting x = sin 2θ, then you are, essentially, saying that y = cosec 2θ. Doing this just leads to a bit of extra work if you don't immediately think "oh, hang on a sec*, I've done this bit of working before... :holmes:".

*no pun intended.
Reply 42
Adje
The clue, I suppose, is in the limits of the equation. You have, in part iii), the limits in part ii) - but in reciprocal form.

The easiest way to turn part iii)'s limits into part ii)'s is to put y = 1/x. This immediately leads to limits of √3/2 and 1, but in the wrong order. The swapping of 'dy's for 'dx's fixes this by introducing a negative sign, thereby flipping the limits. Then a bit of algebra transforms the integrand into exactly the same thing as part ii)'s.

Your cosec 2θ substitution is exactly the same thing, though, if you think about it. If we're saying y = 1/x, and part ii) works by putting x = sin 2θ, then you are, essentially, saying that y = cosec 2θ. Doing this just leads to a bit of extra work if you don't immediately think "oh, hang on a sec*, I've done this bit of working before... :holmes:".

*no pun intended.


I see now, that does indeed make alot of sense - and does cut down the work by alot. I'm clearly not in maths-mode today as I even thought to myself "Those limits are reciprocals.. hmm.. let's try cosec!", even though by now I should be used to the final part of integration questions leading to the same result as the previous part.

Oh well. Thank you :smile:
tommm
STEP II 2002 Q9

Spoiler



tommm is there any chance you could help me out with how you got from

tan(a + da) = (h + dh)(a + b)/ab to

da(seca)^2/(1 - da(tana)) = dh(a + b)/ab

I'm pretty useless with approximations so i'm probably missing something obvious.
Hmmm... There's too much working missing to say exactly what tommm did (which means, incidentally, that there's too much working missing to get full marks, IMHO).

I would go (without having actually read the question, you understand!):

h=aba+btanโกฮฑโ€…โ€ŠโŸนโ€…โ€Šdhdฮฑ=aba+bsecโก2ฮฑh = \frac{ab}{a+b} \tan \alpha \implies \frac{dh}{d\alpha} = \frac{ab}{a+b} \sec^2 \alpha

So ฮดhโ‰ˆ(aba+bsecโก2ฮฑ)ฮดฮฑ\delta h \approx \left(\frac{ab}{a+b} \sec^2 \alpha \right) \delta \alpha
DFranklin
Hmmm... There's too much working missing to say exactly what tommm did (which means, incidentally, that there's too much working missing to get full marks, IMHO).

I would go (without having actually read the question, you understand!):

h=aba+btanโกฮฑโ€…โ€ŠโŸนโ€…โ€Šdhdฮฑ=aba+bsecโก2ฮฑh = \frac{ab}{a+b} \tan \alpha \implies \frac{dh}{d\alpha} = \frac{ab}{a+b} \sec^2 \alpha

So ฮดhโ‰ˆ(aba+bsecโก2ฮฑ)ฮดฮฑ\delta h \approx \left(\frac{ab}{a+b} \sec^2 \alpha \right) \delta \alpha


I don't think that's what tommm did because he mentions using the formula for tan(A + B) but your way looks fine to me. Thanks!
No, I don't think that it's what tommm did either. But I don't think tommm's approach was the best one anyhow, so...

Incidentally: the "missing bit" that I think you'd lose marks for in tommm's approach.. At some point, he's obviously done some approximation like "for small ฮดฮฑ,tanโกฮดฮฑโ‰ˆฮดฮฑ\delta \alpha, \tan \delta \alpha \approx \delta \alpha". Without explicitly explaining that, I think you'd be guaranteed to lose marks, even if everything written was actually correct.
Reply 47
Yeah, sorry for not typing it up fully, it was a bit late really. IIRC, it took about half a page, so DFranklin's method looks better.
Reply 48
Dadeyemi's solution to STEP III question 5:

isn't the linear co-efficient (bt-d) not (b-d)?

EDIT: not that it makes a difference to the end answer, as t=1.
Dadeyemi
Some more;

Did these quite a while ago I'm afraid some may be partial solutions.


For STEP III Question 4: Don't we also have x = 0 y = - 1 and x = 1 y = 0 as solutions? The way I did the last part was by saying x = y + n where n is a non negative integer, then you get

(3n - 2)y^2 + (3n^2 - 2n)y + n^3 - n^2 = 0

It turns out that this only has roots when n < 3. Thus the only solutions are given by x = y + 2 x = y + 1 and x = y. Plugging these in gives all the solutions.

I'm not sure if this is the method they were looking for, or even if its correct. It would be great if someone could have a look for me.
Reply 50
STEP I, Question 4

Spoiler

Reply 51
I have attached my solutions of paperIII Q3 and Q6 hope they are helpful:smile:
Reply 52
This might be a good time for me to point out that STEP II Q6 is misattributed to me (I actually posted STEP III Q6).
STEP III 2002, Question 3

Spoiler

Dadeyemi
Some attached


There was a bit of a quadratic formula fail at the end of 2. But I agree, majorly boring.
Reply 55
I got my answer to be 81/40 km don't know why it's different from yours....
tommm
STEP II 2002 Q10

Spoiler

Question 5, I'll pick up where Dadeymi left off:

(bโˆ’1)(ab2+(aโˆ’1)b+(aโˆ’1))=0 (b-1)\left(ab^2 + (a-1)b +(a-1)\right)=0

Now, a1=ฬธaโ€…โ€ŠโŸนโ€…โ€Šk=ฬธ11โˆ’aโ€…โ€ŠโŸนโ€…โ€Šb=ฬธ1 a_1 \not= a \implies k \not= \frac{1}{1-a} \implies b \not= 1

So (ab2+(aโˆ’1)b+(aโˆ’1))=0 \left(ab^2 + (a-1)b +(a-1)\right)=0

โ€…โ€ŠโŸนโ€…โ€Š(b+aโˆ’12a)2=1โˆ’aaโˆ’a2โˆ’2a+14a2=โˆ’5a2+6aโˆ’14a2 \implies \left( b + \frac{a-1}{2a} \right)^2 = \frac{1-a}{a} - \frac{a^2 - 2a +1}{4a^2} = \frac{-5a^2 + 6a - 1}{4a^2}

โ€…โ€ŠโŸนโ€…โ€Šb=1โˆ’a2aยฑโˆ’5a2+6aโˆ’12a \implies b = \frac{1-a}{2a} \pm \frac{\sqrt{-5a^2 +6a - 1}}{2a}

โ€…โ€ŠโŸนโ€…โ€Šk=12aยฑโˆ’5a2+6aโˆ’12a(1โˆ’a) \implies k = \frac{1}{2a} \pm \frac{\sqrt{-5a^2 +6a - 1}}{2a(1-a)}
Reply 57
Question 10, STEP I, 2002

Using NLR with elasyic collisions, speed of seperation = speed of approach so collision with the end of the cylinder dont change the speed. This gives vnโˆ’u=vnโˆ’1+uโ‡”vn=vnโˆ’1+2uv_n - u = v_{n-1} + u \Leftrightarrow v_n = v_{n-1} +2u, which is an arithmetic progression with initial value v and common difference 2u so vn=v+2unv_n = v +2un as required.

The diference in the distances from the piston and end of cylinder is the distance the piston moves in that time so dnโˆ’dn+1=utn d_n-d_{n+1} = ut_n

tnt_n is the time take to reach the end of the cylinder + the time taken to reach the piston afterwards.
Giving tn=dnvn+dn+1vn=dn+dn+1vn t_n = \dfrac{d_n}{v_n}+\dfrac{d_{n+1}}{v_n} = \dfrac{d_n +d_{n+1}}{v_n}

Using our previous result
dnโˆ’dn+1=utn=udn+dn+1vn d_n-d_{n+1} = ut_n = u\dfrac{d_n +d_{n+1}}{v_n}
โ‡”(v+2un)(dnโˆ’dn+1)=u(dn+dn+1)\Leftrightarrow (v+2un)(d_n-d_{n+1}) = u(d_n +d_{n+1})
โ‡”(v+u(2n+1))dn+1=(v+u(2nโˆ’1))dn\Leftrightarrow (v+u(2n+1))d_{n+1} = (v+u(2n-1))d_n
โ‡”dn+1=v+u(2nโˆ’1)v+u(2n+1)dn\Leftrightarrow d_{n+1} = \dfrac{v+u(2n-1)}{v+u(2n+1)}d_n
Which was to be shown.

dn=v+u(2nโˆ’3)v+u(2nโˆ’1)dnโˆ’1=v+u(2nโˆ’3)v+u(2nโˆ’1)ร—v+u(2nโˆ’5)v+u(2nโˆ’3)ร—โ‹ฏร—v+uv+3ud1d_n= \dfrac{v+u(2n-3)}{v+u(2n-1)}d_{n-1} =\dfrac{v+u(2n-3)}{v+u(2n-1)} \times \dfrac{v+u(2n-5)}{v+u(2n-3)} \times \cdots \times \dfrac{v+u}{v+3u} d_1
โ‡’dn=v+uv+u(2nโˆ’1)d1\Rightarrow d_n = \dfrac{v+u}{v+u(2n-1)}d_1

I the case u=v u = v we have dn=d1n d_n = \dfrac{d_1}{n}

Giving utn=dnโˆ’dn+1=d1nโˆ’d1n+1=d1n(n+1) ut_n = d_n - d_{n+1} = \dfrac{d_1}{n}-\dfrac{d_1}{n+1} = \dfrac{d_1}{n(n+1)}
Reply 58
Dadeyemi
Some more;

Did these quite a while ago I'm afraid some may be partial solutions.


A couple of comments:
For question 2, a far nicer way of doing the final part is to notice that
Unparseable latex formula:

[br]\[[br]\frac{1}{{r^2 - r + 1}}[br]\][br]

is just
Unparseable latex formula:

[br]\[[br]\frac{1}{{r^2 + r + 1}}[br]\][br]

with r replaced by r-1. The last part of the question is then trivial.

For question 4 the (unordered) pair (1,0) is also a solution.
Reply 59
I tried to do the and hence that (2s2โˆ’a2โˆ’c2)2+(2b2โˆ’a2โˆ’c2)2=4a2c2(2s^2 - a^2 - c^2)^2 + (2b^2 - a^2 - c^2)^2 = 4a^2c^2 by expanding out (s2+b2โˆ’a2)2+(s2+b2โˆ’c2)2=4s2b2(s^2 + b^2 - a^2)^2 + (s^2 + b^2 - c^2)^2 = 4s^2b^2, subtracting 4s2b24s^2b^2, and then adding 4a2c24a^2c^2 to both sides. However, this gives me the wrong amount of s^2's: I get two, and I should be getting four.

Any pointers?

EDIT: crafty little dodge showed that what you get after multiplying out (s2+b2โˆ’a2)2+(s2+b2โˆ’c2)2=4s2b2(s^2 + b^2 - a^2)^2 + (s^2 + b^2 - c^2)^2 = 4s^2b^2 and then subtracting 4s2b24s^2b^2 is zero. Hence, you can multiply by two and then add 4a2c24a^2c^2, and then factorise the resulting expression in the form given.

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