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I'm pretty sure spherical coordinates are not required... cylindrical coordinates make the problem much nicer, but they're not required to solve the problem "conceptually". Anyway you can do it also like, by using spherical coordinates but in terms of like, cones etc. Not too hard, no...

Re: 2012 HSC MX2 Marathon No its not... \int_0^d \frac{1+\sin x}{\cos x} dx = \int_0^d (\tan x + \sec x) dx = \left[-\ln |\cos x| + \ln |\tan x + \sec x| \right]_0^d = \left[ \ln \left(\frac{\tan x + \sec x}{\cos x} \right)\right]_0^d = \ln \left|\frac{\tan d + \sec d}{\cos d}\right| Then...

Re: 2012 HSC MX2 Marathon uhhhhh... how can this be true? the right hand series (the harmonic series) diverges

Re: 2012 HSC MX2 Marathon Neat question ^^ First we make the substitution u = x^2. u = x^2 $ implies $ du = 2x dx and thus I = \int\frac{x^3 e^{x^2}}{(x^2+1)^2}dx = \frac{1}{2}\int\frac{u e^u}{(u+1)^2)}du And now we note that u = (u+1) - 1 so we can split up the integral: I =...

Re: 2012 HSC MX1 Marathon You do polynomial division. x = (e^y)/(3+e^y) = 1 - 3/(3+e^y) so 3/(3 + e^y) = 1 - x so 3/(1-x) = 3+e^y so y = ln[3/(1-x) - 3]

Re: 2012 HSC MX2 Marathon It's also true because of http://en.wikipedia.org/wiki/Circles_of_Apollonius

Re: 2012 HSC MX1 Marathon Or go to sleep :P

Re: 2012 HSC MX1 Marathon Timske's answer is correct

There's a lot of total bullcrap that goes around the place about how to get into overseas unis. First, academic results do matter from school (ATAR above 99 is great) but not an enormous amount, and SAT scores really don't matter if they're above a certain point. So as long as you don't...

Now I understand for most people maths is a tool and not a discipline in its own right, but your conception of what constitutes "use" is incredibly narrow-sighted. We learn huge numbers of things throughout our entire lives and throughout our entire schooling that we won't practically put to any...

The set of skills you use in Euclidean/Circle geometry problems at school are a far more useful set of skills than learning the specifics of differential equations ever will be at least in any way they'd teach them at school. The reason is this; there's a difference between specific and general...

The difficulty of various subjects depends a lot not just on the subject but on what sort of student you are and in particular how hard you work. Now I did both Latin and Classical Greek (extensions in both as well) for my HSC so I can say something about these subjects (and hopefully languages...

Thats not so much vector calculus as just normal one variable calculus done with mysterious constants attached (the unit vectors). Also at school we did "vecotr calculus" in the sense it appeared in VCE when doing circular motion, as a way to prove the eqns of motion.

Ive got another way which I think is much faster, using real/imaginary parts and modulus/argument and de moivres: Im\left(z + \frac{1}{z}\right) = Im\left(rcis(\theta) + \frac{1}{r}cis(-\theta)\right) = rsin(\theta) + \frac{1}{r}sin(-\theta) = sin(\theta)\left(r - \frac{1}{r}\right) And thus...

I don't think its any secret those proofs aren't actually written be the people who write the questions. Instead they just find the proof in academic literature and split it into the parts required for a school question. And uh isn't vector calculus decently hard? like 2nd year uni

This is actually a really ludicrous complaint to make. Students are always going to be "advantaged" in subtle ways; for instance anyone who took the time to try out olympiad-like maths or just really enjoyed geometry and worked through lots of problems on their own would likely have seen every...

This is dubious and quite hard to prove. If anyone is interested, you need to basically know about minimal polynomials and algebraic numbers (its pretty cool). Basically, each complex (and real!) number is either "algebraic" or "transcendental". What that means is this: an algebraic number is...

Re: 2012 HSC MX1 Marathon You can do this sort of thing and make it be perfectly valid - its called "reverse reconstruction". You need to be a bit careful though about the uniqueness of points and such (nooblet94's proof isnt actually right because of things lying on different sides of lines).

Re: 2012 HSC MX1 Marathon wahhhh... A, B and D are always going to be concyclic so long as <ADB is not zero (i.e. A,D,B collinear) since any three noncollinear points are concyclic. If you meant A, B, D are on a circle centre C, thats not true; although A, B, C and the centre of the circle...

Re: 2012 HSC MX1 Marathon You dont even need to use the centre. here's a pretty simple proof: Note that the combined length of arc PAQ is the same as arc PBQ (since arcs AP = PB, BQ = QA). Hence the angles they subtend are equal, so /_PAQ = /_PBQ, but by cyclic quadrilateral interior angle sum...