HSC 2016 Maths Marathon (archive) (1 Viewer)

leehuan · Mar 18, 2016

Re: HSC 2016 2U Marathon

Drsoccerball said:
Slight correction the '=' sign is supposed to be an approximation sign as the RHS is just another way of writing Simpson's rule.

Oops. Fixed

Paradoxica · Mar 18, 2016

Re: HSC 2016 2U Marathon

leehuan said:
$\\$Use your knowledge of the basis of Simpson's Rule and integration to prove the formula$ \\ \int_{a}^{b}{f\left(x\right)\,dx}\approx\frac{b-a}{6}\left[f\left(a\right)+4f \left(\frac{a+b}{2}\right) + f\left(b\right)\right]$

The proof is soundly difficult, so hints will be provided if left unanswered.

Hence show that for quadratics and cubics, the approximation is exact.

Paradoxica · Mar 18, 2016

Re: This question is 2U level difficulty, right?

Nailgun said:
$Okay, I've given it another attempt. I think I've gotten a,b,c but I'm struggling with d and e$
$Let\quad b^{ 2 }+4ac=z\\ a)\quad P\left( \frac { b-\sqrt { z } }{ 2a } ,\frac { { b }^{ 2 }-b\sqrt { z } }{ 2a } +c \right) \quad and\quad Q\left( \frac { b+\sqrt { z } }{ 2a } ,\frac { { b }^{ 2 }+b\sqrt { z } }{ 2a } +c \right) \\ b)\quad R\left( \frac { b }{ 2a } ,\frac { { b }^{ 2 } }{ 4a } \right) \\ c)\quad { A }_{ \Delta PQR }=\frac { { b }^{ 2 }\sqrt { z } }{ 8a^{ 2 } }$

$Is\quad d)\quad \int _{ P }^{ Q }{ bx+c-a{ x }^{ 2 } } dx?$

Instead of integrating to find the area of the parabolic segment, you could instead use Simpson's Rule on the abscissae of P,Q,R to determine the exact area of the parabolic segment (on the appropriate function, of course)

leehuan · Mar 18, 2016

Re: HSC 2016 2U Marathon

Paradoxica said:
Hence show that for quadratics and cubics, the approximation is exact.

Lol I was going to do that but... Didn't want to give them a bit too much

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Bestintheworld · Mar 20, 2016

Re: HSC 2016 2U Marathon

$Find\the\ maximum\and \ minimum \ values \ of \ y=2x^3-3x^2-12x for -2 \leq x \leq 4.$

took me 5 mins to write in LaTex

Drsoccerball · Mar 20, 2016

Re: HSC 2016 2U Marathon

Bestintheworld said:
$Find\the\ maximum\and \ minimum \ values \ of \ y=2x^3-3x^2-12x for -2 \leq x \leq 4.$

took me 5 mins to write in LaTex

Write it like this (Check "Reply with Quote")

$$ Find the maximum and minimum values of $ y=2x^3 -3x^2 -12x $ for $ 2 \leq x \leq 4.$

Bestintheworld · Mar 20, 2016

Re: HSC 2016 2U Marathon

Drsoccerball said:
Write it like this (Check "Reply with Quote")

$$ Find the maximum and minimum values of $ y=2x^3 -3x^2 -12x $ for $ 2 \leq x \leq 4.$

that looks so much easier, cheers mate.

Nailgun · Mar 20, 2016

Re: HSC 2016 2U Marathon

Bestintheworld said:
$Find\the\ maximum\and \ minimum \ values \ of \ y=2x^3-3x^2-12x for -2 \leq x \leq 4.$

took me 5 mins to write in LaTex

maximum (4,32)
minimum (2,-20)

from graph lelel

Paradoxica · Mar 20, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
maximum (4,32)
minimum (2,-20)

from graph lelel

By inspection (of the graph)

Nailgun · Mar 21, 2016

Re: This question is 2U level difficulty, right?

Paradoxica said:
$$\noindent The line $y = bx + c$ intersects the parabola $y =ax^2$ in two distinct points $P$ and $Q.\\$a) Find the points of intersection in terms of $a,b,c. \\ $b) Find the point on the parabola $R$ whose $x$ value is the average of the points of intersection.$\\ $c) By using the trapezoidal rule with a single strip on the points $P,Q$, and then two strips on the points $P,Q,R$, determine the area of $\Delta PQR. \\ $d) Find the area of the parabolic segment bound by the chord $PQ$ and the parabola.$\\$e) Hence prove that the area of the parabolic segment is $\frac{4}{3}$ times the area of the inscribed triangle $PQR.$

HOLY CRAP I DID IT

only took 11 days lol

I realised I used the wrong function twice when I was working out the area of the triangle

legit gna write up the entire thing now ahahah

rip half yearly study

Nailgun · Mar 21, 2016

Re: HSC 2016 2U Marathon

fml

InteGrand · Mar 21, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
fml

Try splitting up the LaTeX code in half.

Speed6 · Mar 21, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
fml

Yeah just make multiple $[/tex.] lines and that should potentially fix that issue.$

Nailgun · Mar 21, 2016

Re: This question is 2U level difficulty, right?

$Let\quad { b }^{ 2 }+4ac=z\\ \\ a)\quad { ax }^{ 2 }=bx+c\\ ax^{ 2 }-bx-c=0\\ \therefore \quad x=\frac { b\pm \sqrt { { b }^{ 2 }+4ac } }{ 2a } \\ \\ Sub\quad into\quad y=bx+c\\ \\ Q\left( \frac { b+\sqrt { z } }{ 2a } ,\frac { b^{ 2 }+b\sqrt { z } }{ 2a } +c \right) ,P\left( \frac { b-\sqrt { z } }{ 2a } ,\frac { b^{ 2 }-b\sqrt { z } }{ 2a } +c \right) \\ \\ b)\quad { R }_{ x }=\frac { { Q }_{ x }+{ P }_{ x } }{ 2 } =\frac { b+\sqrt { z } +b-\sqrt { z } }{ 4a } =\frac { 2b }{ 4a } =\frac { b }{ 2a } \\ Sub\quad into\quad y={ ax }^{ 2 }\\ R\left( \frac { b }{ 2a } ,\frac { { b }^{ 2 } }{ 4a } \right)$

$c)\quad Let\quad f\left( x \right) =bx+c,\quad and\quad g\left( x \right) ={ ax }^{ 2 }\\ { A }_{ T }\approx \int _{ P }^{ Q }{ f\left( x \right) dx } -\int _{ R }^{ Q }{ g\left( x \right) } -\int _{ P }^{ R }{ g\left( x \right) } \\ { A }_{ T }={ A }_{ 1 }-{ A }_{ 2 }-{ A }_{ 3 }\\ \\ \int _{ P }^{ Q }{ f\left( x \right) dx } \approx \frac { { Q }_{ x }-{ P }_{ x } }{ 2 } (f\left( { Q }_{ x } \right) +f\left( { P }_{ x } \right) )\\ { A }_{ 1 }=\frac { { Q }_{ x }-{ P }_{ x } }{ 2 } (f\left( { Q }_{ x } \right) +f\left( { P }_{ x } \right) )\\ \\ =\frac { \sqrt { z } }{ 2a } \left( \frac { { b }^{ 2 } }{ a } +2c \right) \\ \\ \int _{ R }^{ Q }{ g\left( x \right) dx } \approx \frac { { Q }_{ x }-R_{ x } }{ 2 } (g\left( { Q }_{ x } \right) +g\left( { R }_{ x } \right) )\\ { A }_{ 2 }=\frac { { Q }_{ x }-R_{ x } }{ 2 } (g\left( { Q }_{ x } \right) +g\left( { R }_{ x } \right) )\\ =\frac { \sqrt { z } }{ 4a } \left( \frac { 3b^{ 2 }+4ac+2b\sqrt { z } }{ 4a } \right) \\$ $\\ \int _{ P }^{ R }{ g\left( x \right) dx } \approx \frac { { R }_{ x }-P_{ x } }{ 2 } (g\left( { R }_{ x } \right) +g\left( { P }_{ x } \right) )\\ { A }_{ 2 }=\frac { { R }_{ x }-P_{ x } }{ 2 } (g\left( { R }_{ x } \right) +g\left( { P }_{ x } \right) )\\ =\frac { \sqrt { z } }{ 4a } \left( \frac { 3b^{ 2 }+4ac-2b\sqrt { z } }{ 4a } \right)$

${ A }_{ T }={ A }_{ 1 }-{ A }_{ 2 }-{ A }_{ 3 }\\ \\ { A }_{ T }=\frac { 2\sqrt { z } }{ 4a } \left( \frac { { b }^{ 2 } }{ a } +2c \right) -\frac { \sqrt { z } }{ 4a } \left( \frac { 3b^{ 2 }+4ac+2b\sqrt { z } }{ 4a } \right) -\frac { \sqrt { z } }{ 4a } \left( \frac { 3b^{ 2 }+4ac-2b\sqrt { z } }{ 4a } \right) \\ \\ =\frac { \sqrt { z } }{ 4a } \left( \frac { { 8b }^{ 2 } }{ 4a } -\frac { 3b^{ 2 }+4ac+2b\sqrt { z } }{ 4a } -\frac { 3b^{ 2 }+4ac-2b\sqrt { z } }{ 4a } +4c \right) \\ =\frac { \sqrt { z } }{ 4a } \left( \frac { { 2b }^{ 2 }-8ac }{ 4a } +4c \right) \\ =\frac { \sqrt { z } }{ 4a } \left( \frac { { b }^{ 2 }-4ac }{ 2a } +\frac { 8ac }{ 2a } \right) \\ { A }_{ T }=\frac { z\sqrt { z } }{ 8{ a }^{ 2 } } units^{ 2 }$

$d)\quad Let\quad h\left( x \right) =bx+c-a{ x }^{ 2 }\\ { A }_{ s }=\frac { { Q }_{ x }-P_{ x } }{ 6 } \left( h\left( { Q }_{ x } \right) +h\left( P_{ x } \right) +4h\left( \frac { { Q }_{ x }+P_{ x } }{ 2 } \right) \right) \\ =\frac { \sqrt { z } }{ 6a } \left( \frac { { b }^{ 2 } }{ a } +4c \right) \\ =\frac { \sqrt { z } }{ 6a } .\frac { z }{ a } \\ { A }_{ s }=\frac { z\sqrt { z } }{ 6{ a }^{ 2 } } units^{ 2 }\\ \\ e)\quad \frac { { A }_{ s } }{ { A }_{ T } } =\frac { \left( \frac { z\sqrt { z } }{ 6{ a }^{ 2 } } \right) }{ \left( \frac { z\sqrt { z } }{ 8{ a }^{ 2 } } \right) } =\frac { 4 }{ 3 }$

Q.E.D

leehuan · Mar 21, 2016

Re: HSC 2016 2U Marathon

leehuan said:
$\\$Use your knowledge of the basis of Simpson's Rule and integration to prove the formula$ \\ \int_{a}^{b}{f\left(x\right)\,dx}\approx\frac{b-a}{6}\left[f\left(a\right)+4f \left(\frac{a+b}{2}\right) + f\left(b\right)\right]$

The proof is soundly difficult, so hints will be provided if left unanswered.

Draw any arbitrary parabola through three conveniently chosen points

Nailgun · Mar 21, 2016

Re: HSC 2016 2U Marathon

btw is there like a notation for trapezoidal rule/simpsons?

i wanted to show what area i was trying to find with which bounds but idk how to like, write that
is what i did with the integral sign incorrect? the logic is a bit dodgy in my answer lol

edit: hey my answer was my 1000th post ahahaha
not a bad post for a milestone

ownagge · Mar 27, 2016

Re: HSC 2016 2U Marathon

Sketch y=2cos(x) - sin(x)

Speed6 · Mar 27, 2016

Re: HSC 2016 2U Marathon

ownagge said:
Sketch y=2cos(x) - sin(x)

Use Wolfram!

Drsoccerball · Mar 27, 2016

Re: HSC 2016 2U Marathon

ownagge said:
Sketch y=2cos(x) - sin(x)

Auxiliary form.

Paradoxica · Mar 27, 2016

Re: HSC 2016 2U Marathon

Drsoccerball said:
Auxiliary form.

I'm not sure that's assumed knowledge for 2U...

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