IB Maths Marathon (1 Viewer)

InteGrand · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

davidgoes4wce said:
$sin (5x) = sin (4x+x)$

$sin(4x)cos(x)+sin(x)cos(4x)$

$[2 sin (2x) cos (2x)] cos x+sin x (1-2sin^{2}2x)$

$[ 2 \times 2 sin(x) cos(x) cos (2x)]cos x+sin x(1- 2 \times sin 2x \times sin 2x)$

$[ 2 \times 2 sin(x) cos(x) cos (2x)]cos x+sin x(1- 2 \times 2\ sin x \ cos x \ \times 2sinx \ cos x)$

$4 \ cos^{2} x \ sin x \ cos(2x)+sin x(1-8 sin^{2}x \ cos^{2} x)$

$4 (1-sin^{2}x) sin x (1-2sin^{2}2x)+sin \ x[1-8 sin^{2}x (1-sin^{2}x)]$

$(4-4sin^{2}x_)(sinx-2 \ sin x \ \times 2 \ sinx cos x \ 2 sin x cos x)+sin x[1-8sin^{2}x+8sin^{4}x]$

$(4-4sin^{2}x_)(sinx-8 sin^{3}x \ cos^{2} x)+sin x[1-8sin^{2}x+8sin^{4}x]$

$(4-4sin^{2}x_)[sinx-8 sin^{3}x \ (1-sin^{2}x)]+sin x[1-8sin^{2}x+8sin^{4}x]$

$(4-4sin^{2}x_)[sinx-8 sin^{3}x +8 sin^{5} x]+sin x[1-8sin^{2}x+8sin^{4}x]$

$4 sin x-32 sin^{3}x+32sin^{5}x-4sin^{3}x+32 sin^{5}x-32sin^{7}x+4sin^{3}x+sinx-8sin^{3}x+8sin^{5}x$

$-32sin^{7}x+72sin^{5}x-24sin^{3}x+5sin{x}$

Is that right?

If we sub. in x = pi/2 to the final answer, since sin(pi/2) = 1, we get -32 + 72 -24 + 5 = 21, which is outside the range of the LHS, which is sin(5x), which is between -1 and 1 always.

This is the check that parad0xica mentioned in an earlier comment.

Paradoxica · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

davidgoes4wce said:
$sin (5x) = sin (4x+x)$

$sin(4x)cos(x)+sin(x)cos(4x)$

$[2 sin (2x) cos (2x)] cos x+sin x (1-2sin^{2}2x)$

$[ 2 \times 2 sin(x) cos(x) cos (2x)]cos x+sin x(1- 2 \times sin 2x \times sin 2x)$

$[ 2 \times 2 sin(x) cos(x) cos (2x)]cos x+sin x(1- 2 \times 2\ sin x \ cos x \ \times 2sinx \ cos x)$

$4 \ cos^{2} x \ sin x \ cos(2x)+sin x(1-8 sin^{2}x \ cos^{2} x)$

$4 (1-sin^{2}x) sin x (1-2sin^{2}2x)+sin \ x[1-8 sin^{2}x (1-sin^{2}x)]$

$(4-4sin^{2}x_)(sinx-2 \ sin x \ \times 2 \ sinx cos x \ 2 sin x cos x)+sin x[1-8sin^{2}x+8sin^{4}x]$

$(4-4sin^{2}x_)(sinx-8 sin^{3}x \ cos^{2} x)+sin x[1-8sin^{2}x+8sin^{4}x]$

$(4-4sin^{2}x_)[sinx-8 sin^{3}x \ (1-sin^{2}x)]+sin x[1-8sin^{2}x+8sin^{4}x]$

$(4-4sin^{2}x_)[sinx-8 sin^{3}x +8 sin^{5} x]+sin x[1-8sin^{2}x+8sin^{4}x]$

$4 sin x-32 sin^{3}x+32sin^{5}x-4sin^{3}x+32 sin^{5}x-32sin^{7}x+4sin^{3}x+sinx-8sin^{3}x+8sin^{5}x$

$-32sin^{7}x+72sin^{5}x-24sin^{3}x+5sin{x}$

Is that right?

Not possible, sin5x is a polynomial in sinx of degree 5.

Some power in there is broken.

davidgoes4wce · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

Im not sure about this question, is my expansion and simplification right?

parad0xica · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

davidgoes4wce said:
$sin (5x) = sin (4x+x)$

$\textbf{L1. } sin(4x)cos(x)+sin(x)cos(4x)$

$\textbf{L2. } [2 sin (2x) cos (2x)] cos x+sin x (1-2sin^{2}2x)$

$\textbf{L3. } [ 2 \times 2 sin(x) cos(x) cos (2x)]cos x+sin x(1- 2 \times sin 2x \times sin 2x)$

$\textbf{L4. } [ 2 \times 2 sin(x) cos(x) cos (2x)]cos x+sin x(1- 2 \times 2\ sin x \ cos x \ \times 2sinx \ cos x)$

$\textbf{L5. } 4 \ cos^{2} x \ sin x \ cos(2x)+sin x(1-8 sin^{2}x \ cos^{2} x)$

$\textbf{L6. } 4 (1-sin^{2}x) sin x (1-2sin^{2}2x)+sin \ x[1-8 sin^{2}x (1-sin^{2}x)]$

Looking at the first term only when going from line 5 to line 6, you did this:

$\cos 2x = 1 - 2 \sin ^2 2x$

I think it was an issue of focus, not of lack of knowledge

davidgoes4wce · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

I'll try it again after watching a bit of the Euro 2016 action

davidgoes4wce · Jun 26, 2016

Re: International Baccalaureate Marathon 2016

$sin (5x) = sin (4x+x)$

$sin(4x)cos(x)+sin(x)cos(4x)$

$[2 sin (2x) cos (2x)] cos x+sin x (1-2sin^{2}2x)$

$[ 2 \times 2 sin(x) cos(x) cos (2x)]cos x+sin x(1- 2 \times sin 2x \times sin 2x)$

$[ 2 \times 2 sin(x) cos(x) cos (2x)]cos x+sin x(1- 2 \times 2\ sin x \ cos x \ \times 2sinx \ cos x)$

$4 \ cos^{2} x \ sin x \ cos(2x)+sin x(1-8 sin^{2}x \ cos^{2} x)$

$4 (1-sin^{2}x) sin x (1-2sin^{2}x)+sin \ x[1-8 sin^{2}x (1-sin^{2}x)]$

$(4-4sin^{2}x) (sin x-2sin^{3}x)+sinx(1-8sin^{2}x+8sin^{4}x)$

$4sin x-8sin^{3}x-4sin^{3}x+8sin^{5}x+sin x-8sin^{3}x+8sin^{5} x$

$16 sin^{5} x-20sin^{3}x+5 sin x$

davidgoes4wce · Jun 27, 2016

Re: International Baccalaureate Marathon 2016

Maths HL Introductory Problem:

My thinking was to integrate the circumference of a circle in order to get the area of a circle.

InteGrand · Jun 27, 2016

Re: International Baccalaureate Marathon 2016

davidgoes4wce said:
Maths HL Introductory Problem:

My thinking was to integrate the circumference of a circle in order to get the area of a circle.

Yeah that's essentially one way to do it. The typical HSC method would be to do something like integrate sqrt(r^2 - x^2) between 0 and r (this getting the area of the quarter-disk), and then multiplying by 4. The integration would be carried out using a trig. substitution (or you could use integration by parts but for that definite integral, trig. sub. is nicer).

davidgoes4wce · Jun 27, 2016

Re: International Baccalaureate Marathon 2016

Something like

$\int 2 \pi r dr = \pi {r^2} + C$

InteGrand · Jun 27, 2016

Re: International Baccalaureate Marathon 2016

davidgoes4wce said:
Something like

$\int 2 \pi r dr = \pi {r^2} + C$

Pretty much. And the area of a circle of radius 0 is 0, which makes C = 0. Not sure how rigorous you are required to be though with regards to explaining the integral.

Here's a paper discussing when the perimeter of a shape will be the derivative of its area, or area the derivative of volume: https://www.math.byu.edu/~mdorff/docs/DorffPaper07.pdf .

davidgoes4wce · Jul 4, 2016

Re: International Baccalaureate Marathon 2016

I was a bit confused on this Introductory Question in Maths HL on Differentiation page 527 of Cambridge.

Im thinking the function is

$C=\frac{12+v^2}{100}$

We have to use differentiation and set $\frac{dC}{dv}$ to some value to solve for v.

Anyone with any ideas to get a kick start to this question?

InteGrand · Jul 4, 2016

Re: International Baccalaureate Marathon 2016

davidgoes4wce said:
I was a bit confused on this Introductory Question in Maths HL on Differentiation page 527 of Cambridge.

Im thinking the function is

$C=\frac{12+v^2}{100}$

We have to use differentiation and set $\frac{dC}{dv}$ to some value to solve for v.

Anyone with any ideas to get a kick start to this question?

$\noindent If the speed travelled at is $v$, then the time taken is distance divided by speed, i.e. $\frac{50}{v}$. So the total cost (in terms of litres of fuel used) of the journey is the time multiplied by the rate of fuel usage, so it is $C = \frac{50}{v}\times \frac{12+v^2}{100}$. So this is the cost function to minimise.$

davidgoes4wce · Jul 4, 2016

Re: International Baccalaureate Marathon 2016

I totally didn't see it like that, after you explained that I got it.

davidgoes4wce · Jul 4, 2016

Re: International Baccalaureate Marathon 2016

InteGrand said:
$\noindent If the speed travelled at is $v$, then the time taken is distance divided by speed, i.e. $\frac{50}{v}$. So the total cost (in terms of litres of fuel used) of the journey is the time multiplied by the rate of fuel usage, so it is $C = \frac{50}{v}\times \frac{12+v^2}{100}$. So this is the cost function to minimise.$

I had a go at finishing off this question

$$ By simplifying terms we get, $ C=\frac{12+v^2}{2v}$

$\frac{dC}{dv}=\frac{(2v)(2v)-(12+v^2)(2)}{4v^2}$

$\frac{dC}{dv}=0$

$0=4v^2-24-2v^2$

$0=2v^2-24$

$0=2(v^2-12)$

$0=v^2-12$

$v=2 \ \sqrt{3} \ miles \ per \ hour$

davidgoes4wce · Jul 17, 2016

Re: International Baccalaureate Marathon 2016

Intro Problem Chapter 16

davidgoes4wce · Jul 18, 2016

Re: International Baccalaureate Marathon 2016

Im not much of a Physics person but just found this definition of magnetic dipole: "A magnetic dipole is the limit of either a closed loop of electric current or a pair of poles as the dimensions of the source are reduced to zero while keeping the magnetic moment constant."

Im going to say the magnetic dipole is not 2 since the average of those values is 2.000 0012

InteGrand · Jul 18, 2016

Re: International Baccalaureate Marathon 2016

davidgoes4wce said:
Im not much of a Physics person but just found this definition of magnetic dipole: "A magnetic dipole is the limit of either a closed loop of electric current or a pair of poles as the dimensions of the source are reduced to zero while keeping the magnetic moment constant."

Im going to say the magnetic dipole is not 2 since the average of those values is 2.000 0012

That's still pretty close to 2, isn't it?

davidgoes4wce · Feb 2, 2017

Re: International Baccalaureate Marathon 2016

Not sure if there are any IB markers in here but I came across a question like this tonight, which is equivalent to a Band 5 HSC question:

'A circular disc is cut into twelve sectors whose angles are in an arithmetic sequence. The angle of the largest sector is twice the angle of the smallest sector. Find the size of the angle of the smallest sector.'

The answer was 20 degrees, I used the answer in radians which was $\frac{\pi}{9}$

Would I get penalized if I left my answer in radians , even though it didn't specifically mention either?

dan964 · Feb 2, 2017

Re: International Baccalaureate Marathon

***Renaming this thread since it is 2017, and stickying it for easy access***

davidgoes4wce · Feb 3, 2017

Re: International Baccalaureate Marathon

Just confirming this was an error from TZ1 Mathematics Standard SL 2015 paper.

I learnt something new today , TZ1 is the Americas time zone.

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