# Questions from 2007 and 2008 Paper (1 Viewer)

#### Timothy.Siu

##### Prophet 9
lol i might do them later, its late now...

but for 10b (2008) just use the fact they're similar triangles and u shud be fine =)

#### cutemouse

##### Account Closed
lol, can't be bothered downloading that PDF. Could you post the question?

#### vds700

##### Member
not sure if its right, but anyways

#### Juliaan

##### Member
and your doing your HSC in 2011; dayumm.

#### Official

##### Bring it on
and your doing your HSC in 2011; dayumm.
Ah, only for 2U maths tho lol

#### Juliaan

##### Member
When i was in year 10, i didn't even know what the HSC was ~

#### Official

##### Bring it on
fair nuff =p, HSC doesn't really come into most pplz minds anyway until yr 11 +

#### Official

##### Bring it on
not sure if its right, but anyways
thank you kind sir

edit: Oops the answer seems to be wrong grr; actual answer: 11kb^4 / 12

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#### vds700

##### Member
thank you kind sir

edit: Oops the answer seems to be wrong grr; actual answer: 11kb^4 / 12
ah damn, dunno what i did wrong. Someone else might be able to solve it

#### Timothy.Siu

##### Prophet 9
i got what vds got...i'm not sure what i put in exam though..

#### Trebla

At vds700's file, when you subbed in x = b, you forgot about the k.

$\bg_white f'(x) = k\left(b^2x-\dfrac{x^3}{3}\right)\\f(x) = \dfrac{kb^2x^2}{2}-\dfrac{kx^4}{12}+c\\f(b)=0\\\Rightarrow \dfrac{kb^4}{2}-\dfrac{kb^4}{12}+c=0\\c = - \dfrac{5kb^4}{12} which is f(0) \\Therefore the distance below the beam from the x-axis at x = 0 is \dfrac{5kb^4}{12}$

Though it still doesn't agree with the supposed "actual" answer...lol

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#### Timothy.Siu

##### Prophet 9
When you subbed in x = b, you forgot about the k.

$\bg_white f'(x) = k\left(b^2x-\dfrac{x^3}{3}\right)\\f(x) = \dfrac{kb^2x^2}{2}-\dfrac{kx^4}{12}+c\\f(b)=0\\\Rightarrow \dfrac{kb^4}{2}-\dfrac{kb^4}{12}+c=0\\c = - \dfrac{5kb^4}{12} which is f(0) \\Therefore the distance below the beam from the x-axis is \dfrac{5kb^4}{12}$

Though it still doesn't agree with the "actual" answer...lol
lol i actually got what you got =P

#### jpmeijer

##### New Member
for 2007 Q10b

part (i)

at P,
$\bg_white N_{1}= \frac{L_{1}}{x^{2}}, N_{2}= \frac{L_{2}}{(m-x)^{2}}$
$\bg_white N = N_{1} + N_{2}$
$\bg_white N = \frac{L_{1}}{x^{2}} + \frac{L_{2}}{(m-x)^{2}}$

part (ii)

now we want to minimise the noise level N, so we differentiate (with respect to x)
remember when you are doing this that L1, L2 and m are all constants, so you treat them just like you would any number like 2, 5 or 6

$\bg_white \frac{dN}{dx} = \frac{-2L_{1}}{x^{3}} + \frac{2L_{2}}{(m-x)^{3}}$
$\bg_white \frac{dN}{dx} = \frac{-2L_{1}(m-x)^{3} + 2L_{2}x^{3}}{x^{3} + (m-x)^{3}}$

now we find any stationary points:

let dN/dx = 0

$\bg_white \Rightarrow 2L_{2}x^{3} = 2L_{1}(m-x)^{3}$
$\bg_white L_{2}x^{3} = L_{1}(m-x)^{3}$

now we take the cube root of both sides:
$\bg_white \sqrt[3]{L_{2}}x = \sqrt[3]{L_{1}}(m-x)$

and now expand the RHS and make x the subject, giving

$\bg_white x = \frac{\sqrt[3]{L_{1}}m}{\sqrt[3]{L_{2}}+\sqrt[3]{L_{1}}}$

Don't know if this is enough...
You might need to keep going and show that this is in fact a minimum... ie find the 2nd derivative, sub this answer in and show that it is > 0
Anyone else know?

PS: Many apologies if there are any mistakes in there... this is my 1st time using LaTeX

#### lyounamu

##### Reborn
Uh...are you sure, Official?

I think I got what Trebla got...

(I don't remember exactly though)

#### vds700

##### Member
At vds700's file, when you subbed in x = b, you forgot about the k.

$\bg_white f'(x) = k\left(b^2x-\dfrac{x^3}{3}\right)\\f(x) = \dfrac{kb^2x^2}{2}-\dfrac{kx^4}{12}+c\\f(b)=0\\\Rightarrow \dfrac{kb^4}{2}-\dfrac{kb^4}{12}+c=0\\c = - \dfrac{5kb^4}{12} which is f(0) \\Therefore the distance below the beam from the x-axis at x = 0 is \dfrac{5kb^4}{12}$

Though it still doesn't agree with the supposed "actual" answer...lol
oops thats what i did wrong, sorry