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Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread If you take x to be negative, the acceleration you plug in needs to be positive. And when they said the acceleration is 4 m s-2, that is a bit misleading, since what they really were referring to was the magnitude of the...

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread As I said, in simple harmonic motion, the acceleration and displacement from centre of motion are always opposite signs (this is because the acceleration is always trying to push the object back to its centre of motion, as you...

As you said, the matrices are upper triangular. It's easy to show by induction and expanding down the first column that the determinant of an upper-triangular (in fact, any triangular) matrix is the product of the diagonal elements (realised now that you knew this as you wrote it in part of...

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread We can set the mean position to be x = 0. So we can say v^2 = n^2 (A^2 - x^2), where A is the amplitude, n is the usual n (angular frequency is what n is). (If we have centre of motion (mean position) x0 instead, the formula...

Of course another easy type of counterexample is if A = B. Then clearly AB = BA, but it's easy to see that this need not equal I.

Still no. A classic counterexample would be that if A and B are both diagonal matrices, then AB = BA, but it need not be the identity matrix. If A and B are both diagonal matrices, then it is easy to verify that AB is the diagonal matrix whose diagonal elements are the corresponding diagonal...

No. A counterexample is for instance if A is the zero matrix. However, if we are given that A and B are square matrices with AB = I, then it turns out that yes, this implies that B and A are inverses and BA = I also.

So the quality of teachers hasn't dropped too much since 50+ years ago? Or has that dropped too?

The current HSC MX2 papers seem much easier than the HSC 4U / Leaving Certificate maths papers from ~50-100 years ago (probably similar story for many other HSC subjects too). Were the teaching standards in NSW higher back then? And if so, why?

What is the main reason behind this? Shortage of teachers?

I have read that we are lagging behind in mathematical education.

Re: HSC 2016 Complex Numbers Marathon I'm guessing you've already memorised most of the ones on Wikipedia?

Re: HSC 2016 Complex Numbers Marathon There is no Parad0xica, there is only parad0xica and Paradoxica (I'm getting that effect now where if you say a word too many times it seems to lose its meaning, where the word in question is "Paradoxica"). :p (Incidentally, this effect is known as...

I edited it in using complex numbers. It is clear though that it has to be some rotation for a given fixed x (at the very least a trivial one of angle 0) from the fact that the length is preserved, which means it's constrained to the circle it started on.

Given any point x in R2, the point Qx is the point x rotated by some constant angle (what angle?) counter-clockwise about the origin. Such a matrix Q is called a rotation matrix for this reason. This fact can be deduced by either geometrical means or through complex numbers. Here's how it's...

The answer would be 23,246,692*(1.0114) = 23511704.29. Rounding to the nearest integer (since we can't have fractional people), answer is 23511704.

Oh yeah, missed that, sorry.

QED is an abbreviation of the Latin words "Quod Erat Demonstrandum" which loosely translated means "that which was to be demonstrated". It is usually placed at the end of a mathematical proof to indicate that the proof is complete. Source: http://www.mathopenref.com/qed.html.

$\noindent Here's a more typical way of doing it.$ $\noindent Note that by rearranging, the statement is equivalent to $\sin^2 a \cos ^2 b - \cos^2 a \sin^2 b - \sin^2 a + \sin^2 b = 0$. But in this statement, using factorisation and the Pythagorean Trig. Identity, $LHS = \sin^2 a...

Yes. (Sorry, forgot this was in prelim thread and you hadn't learnt calculus yet.)