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 ฮธ (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
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... ".
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... ".
*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.
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!):
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=a+babโtanฮฑโนdฮฑdhโ=a+babโsec2ฮฑ
So ฮดhโ(a+babโsec2ฮฑ)ฮดฮฑ
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,tanฮดฮฑโฮดฮฑ". Without explicitly explaining that, I think you'd be guaranteed to lose marks, even if everything written was actually correct.
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.
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 vnโโu=vnโ1โ+uโvnโ=vnโ1โ+2u, which is an arithmetic progression with initial value v and common difference 2u so vnโ=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 dnโโdn+1โ=utnโ
tnโ is the time take to reach the end of the cylinder + the time taken to reach the piston afterwards. Giving tnโ=vnโdnโโ+vnโdn+1โโ=vnโdnโ+dn+1โโ
Using our previous result dnโโdn+1โ=utnโ=uvnโdnโ+dn+1โโ โ(v+2un)(dnโโdn+1โ)=u(dnโ+dn+1โ) โ(v+u(2n+1))dn+1โ=(v+u(2nโ1))dnโ โdn+1โ=v+u(2n+1)v+u(2nโ1)โdnโ Which was to be shown.
I tried to do the and hence that (2s2โa2โc2)2+(2b2โa2โc2)2=4a2c2 by expanding out (s2+b2โa2)2+(s2+b2โc2)2=4s2b2, subtracting 4s2b2, and then adding 4a2c2 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+b2โa2)2+(s2+b2โc2)2=4s2b2 and then subtracting 4s2b2 is zero. Hence, you can multiply by two and then add 4a2c2, and then factorise the resulting expression in the form given.