HSC 2013 Maths Marathon (archive) (3 Viewers)

girlworld_club · Oct 20, 2013

Re: HSC 2013 2U Marathon

Badsiii said:
Arc length= 5pi/3 = circumference of cone
Hence radius of cone= 5/6
Radius of sector= 5 = slant height of cone
Therefore height of cone, H^2= 5^2 - 25/36
h= root(875/36)
Volume cone= 1/3pi R^2 h
= 1/3 pi 25/36 x root(875/36) ?

The answer is

125 root(35) pi / 648 cm^3

Badsiii · Oct 20, 2013

Re: HSC 2013 2U Marathon

girlworld_club said:
The answer is

125 root(35) pi / 648 cm^3

Yeh that's what I got... Mines just not simplified

girlworld_club · Oct 20, 2013

Re: HSC 2013 2U Marathon

how did you get the perpendicular height for the cone?

dw i got it.

mahmoudali · Oct 20, 2013

Re: HSC 2013 2U Marathon

$$i) Consider$ \ \ f(x) = \ln x + \ln \sqrt{x}$

$$ii) Write f(x) in the form of$ \ \ \ln (g(x))$

$iii) Find the area bounded by g(x) the y axis, y=0 and y=5 in exact form$

HeroicPandas · Oct 20, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
sin(2x) = 2sinx.cosx
sinx/2sinx.cosx
1/2cosx
1/2

This is a 3U double angle formula, so no marks awarded

Menomaths · Oct 20, 2013

Re: HSC 2013 2U Marathon

HeroicPandas said:
This is a 3U double angle formula, so no marks awarded

Can it be done with nooby 2U methods?

Youi_ · Oct 20, 2013

Re: HSC 2013 2U Marathon

mahmoudali said:
$$i) Consider$ \ \ f(x) = \ln x + \ln \sqrt{x}$

$$ii) Write f(x) in the form of$ \ \ \ln (g(x))$

$iii) Find the area bounded by g(x) the y axis, y=0 and y=5 in exact form$

How many marks is part i) lol

HeroicPandas · Oct 20, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
Can it be done with nooby 2U methods?

It can be done by 2U methods

Badsiii · Oct 20, 2013

Re: HSC 2013 2U Marathon

Differentiate:

2yln(y)
__y

HeroicPandas · Oct 20, 2013

Re: HSC 2013 2U Marathon

lebomatic said:
yeh i was wondering bout that, not sure how to do 2u method. hint please

wait..... do u know that $\lim_{x \rightarrow 0} \frac{sinx}{x}=1$ ? <--have u learnt this?

Youi_ · Oct 20, 2013

Re: HSC 2013 2U Marathon

ocatal said:
$M = 2\alpha^2$

$L^2 = (-\alpha - 2\alpha)^2 = 9\alpha^2$

$\frac{L^2}{9} = \alpha^2$

$\frac{2L^2}{9} = 2\alpha^2 = M$

idk whats wrong with me today but where did alpha come from? please explain your solution to a tired guy

RealiseNothing · Oct 20, 2013

Re: HSC 2013 2U Marathon

Youi_ said:
idk whats wrong with me today but where did alpha come from? please explain your solution to a tired guy

It's the root of the quadratic.

Menomaths · Oct 20, 2013

Re: HSC 2013 2U Marathon

Prove that:

$\frac{sin\theta\cos^2\theta-sin^3\theta}{2cos^3\theta-cos\theta} = tan\theta$

andybandy · Oct 20, 2013

Re: HSC 2013 2U Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{sin\theta&space;cos^{2}\theta&space;-sin^{3}\theta&space;}{2cos^{3}\theta&space;-cos\theta&space;}&space;=&space;\frac{sin\theta&space;(cos^{2}\theta&space;-sin^{2}\theta)&space;}{cos\theta&space;(2cos^{2}\theta&space;-1)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{sin\theta&space;cos^{2}\theta&space;-sin^{3}\theta&space;}{2cos^{3}\theta&space;-cos\theta&space;}&space;=&space;\frac{sin\theta&space;(cos^{2}\theta&space;-sin^{2}\theta)&space;}{cos\theta&space;(2cos^{2}\theta&space;-1)}" title="\frac{sin\theta cos^{2}\theta -sin^{3}\theta }{2cos^{3}\theta -cos\theta } = \frac{sin\theta (cos^{2}\theta -sin^{2}\theta) }{cos\theta (2cos^{2}\theta -1)}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{sin\theta&space;}{cos\theta&space;}.\frac{cos2\theta&space;}{cos2\theta&space;}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{sin\theta&space;}{cos\theta&space;}.\frac{cos2\theta&space;}{cos2\theta&space;}" title="= \frac{sin\theta }{cos\theta }.\frac{cos2\theta }{cos2\theta }" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;tan\theta&space;.1&space;=&space;tan\theta" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;tan\theta&space;.1&space;=&space;tan\theta" title="= tan\theta .1 = tan\theta" /></a>

Edit: Realised this is the 3 unit method, for the 2 unit method, you would change

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{sin\theta(cos^{2}\theta&space;-&space;(1&space;-&space;cos^{2}\theta))}{cos\theta(2cos^{2}\theta-1)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{sin\theta(cos^{2}\theta&space;-&space;(1&space;-&space;cos^{2}\theta))}{cos\theta(2cos^{2}\theta-1)}" title="= \frac{sin\theta(cos^{2}\theta - (1 - cos^{2}\theta))}{cos\theta(2cos^{2}\theta-1)}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{sin\theta(2cos^{2}\theta&space;-&space;1)}{cos\theta(2cos^{2}\theta-1)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{sin\theta(2cos^{2}\theta&space;-&space;1)}{cos\theta(2cos^{2}\theta-1)}" title="= \frac{sin\theta(2cos^{2}\theta - 1)}{cos\theta(2cos^{2}\theta-1)}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{sin\theta}{cos\theta}&space;=&space;tan\theta" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{sin\theta}{cos\theta}&space;=&space;tan\theta" title="= \frac{sin\theta}{cos\theta} = tan\theta" /></a>

HeroicPandas · Oct 20, 2013

Re: HSC 2013 2U Marathon

^U used 3U double angles

A hint is to never write a '1' as a '1'

andybandy · Oct 20, 2013

Re: HSC 2013 2U Marathon

HeroicPandas said:
^U used 3U double angles

A hint is to never write a '1' as a '1'

I know, i corrected it using the 2 unit method, and what do you mean?

HeroicPandas · Oct 20, 2013

Re: HSC 2013 2U Marathon

andybandy said:
I know, i corrected it using the 2 unit method, and what do you mean?

ignore everything i said, im n00b

Menomaths · Oct 20, 2013

Re: HSC 2013 2U Marathon

andybandy said:
I know, i corrected it using the 2 unit method, and what do you mean?

I think he means write '1' as '2-1'

Menomaths · Oct 20, 2013

Re: HSC 2013 2U Marathon

andybandy said:
<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{sin\theta&space;cos^{2}\theta&space;-sin^{3}\theta&space;}{2cos^{3}\theta&space;-cos\theta&space;}&space;=&space;\frac{sin\theta&space;(cos^{2}\theta&space;-sin^{2}\theta)&space;}{cos\theta&space;(2cos^{2}\theta&space;-1)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{sin\theta&space;cos^{2}\theta&space;-sin^{3}\theta&space;}{2cos^{3}\theta&space;-cos\theta&space;}&space;=&space;\frac{sin\theta&space;(cos^{2}\theta&space;-sin^{2}\theta)&space;}{cos\theta&space;(2cos^{2}\theta&space;-1)}" title="\frac{sin\theta cos^{2}\theta -sin^{3}\theta }{2cos^{3}\theta -cos\theta } = \frac{sin\theta (cos^{2}\theta -sin^{2}\theta) }{cos\theta (2cos^{2}\theta -1)}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{sin\theta&space;}{cos\theta&space;}.\frac{cos2\theta&space;}{cos2\theta&space;}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{sin\theta&space;}{cos\theta&space;}.\frac{cos2\theta&space;}{cos2\theta&space;}" title="= \frac{sin\theta }{cos\theta }.\frac{cos2\theta }{cos2\theta }" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;tan\theta&space;.1&space;=&space;tan\theta" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;tan\theta&space;.1&space;=&space;tan\theta" title="= tan\theta .1 = tan\theta" /></a>

Edit: Realised this is the 3 unit method, for the 2 unit method, you would change

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{sin\theta(cos^{2}\theta&space;-&space;(1&space;-&space;cos^{2}\theta))}{cos\theta(2cos^{2}\theta-1)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{sin\theta(cos^{2}\theta&space;-&space;(1&space;-&space;cos^{2}\theta))}{cos\theta(2cos^{2}\theta-1)}" title="= \frac{sin\theta(cos^{2}\theta - (1 - cos^{2}\theta))}{cos\theta(2cos^{2}\theta-1)}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{sin\theta(2cos^{2}\theta&space;-&space;1)}{cos\theta(2cos^{2}\theta-1)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{sin\theta(2cos^{2}\theta&space;-&space;1)}{cos\theta(2cos^{2}\theta-1)}" title="= \frac{sin\theta(2cos^{2}\theta - 1)}{cos\theta(2cos^{2}\theta-1)}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{sin\theta}{cos\theta}&space;=&space;tan\theta" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{sin\theta}{cos\theta}&space;=&space;tan\theta" title="= \frac{sin\theta}{cos\theta} = tan\theta" /></a>

It's so much easier with the 3U method, wish we could use it

For this question why can't we use $y-y_1=m(x-x_1)?$

spanky125 · Oct 20, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
It's so much easier with the 3U method, wish we could use it

For this question why can't we use $y-y_1=m(x-x_1)?$

Why bother? Just integrate it then sub in the points to find the value of C

HSC 2013 Maths Marathon (archive) (3 Viewers)

Member

Member

Member

Member

Heroic!

Exaı̸̸̸̸̸̸̸̸lted Member

Member

Heroic!

Member

Heroic!

Member

what is that?It is Cowpea

Exaı̸̸̸̸̸̸̸̸lted Member

Member

Heroic!

Member

Heroic!

Exaı̸̸̸̸̸̸̸̸lted Member

Exaı̸̸̸̸̸̸̸̸lted Member

Just waitin' for a mate

Users Who Are Viewing This Thread (Users: 0, Guests: 3)