HSC 2013 Maths Marathon (archive) (1 Viewer)

[girlworld_club]

Re: HSC 2013 2U Marathon

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]

Re: HSC 2013 2U Marathon

The answer is

125 root(35) pi / 648 cm^3

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

 

[girlworld_club]

Re: HSC 2013 2U Marathon

how did you get the perpendicular height for the cone?

dw i got it.

 

[mahmoudali]

Re: HSC 2013 2U Marathon

 

[HeroicPandas]

Re: HSC 2013 2U Marathon

sin(2x) = 2sinx.cosx
sinx/2sinx.cosx
1/2cosx
1/2

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

 

[Menomaths]

Re: HSC 2013 2U Marathon

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

Can it be done with nooby 2U methods?

 

[Youi_]

Re: HSC 2013 2U Marathon

How many marks is part i) lol

 

[HeroicPandas]

Re: HSC 2013 2U Marathon

Can it be done with nooby 2U methods?

It can be done by 2U methods

 

[Badsiii]

Re: HSC 2013 2U Marathon

Differentiate:

2yln(y)
__y

 

[HeroicPandas]

Re: HSC 2013 2U Marathon

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

wait..... do u know that ? <--have u learnt this?

 

[Youi_]

Re: HSC 2013 2U Marathon

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

 

[RealiseNothing]

Re: HSC 2013 2U Marathon

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]

Re: HSC 2013 2U Marathon

Prove that:

 

[andybandy]

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]

Re: HSC 2013 2U Marathon

^U used 3U double angles

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

 

[andybandy]

Re: HSC 2013 2U Marathon

^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]

Re: HSC 2013 2U Marathon

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

ignore everything i said, im n00b

 

[Menomaths]

Re: HSC 2013 2U Marathon

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]

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>

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

For this question why can't we use

 

[spanky125]

Re: HSC 2013 2U Marathon

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

For this question why can't we use

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