4 Unit Revising Marathon HSC '10 (5 Viewers)

shaon0 · Feb 12, 2010

untouchablecuz said:
$\\$Let $\theta=2 \tan ^{-1} t\Rightarrow d\theta=\frac{2}{1+t^2}dt\\\int \frac{2}{3-2\sqrt{2}\cos\theta} d\theta\\=\int\frac{2}{3-2\sqrt{2}\times\frac{1-t^2}{1+t^2}} \, \frac{2}{1+t^2}dt\\=4\int\frac{dt}{(3-2\sqrt2)+t^2(3+2\sqrt2)}\\=\frac{4}{3+2\sqrt2}\int\frac{dt}{\frac{3-2\sqrt2}{3+2\sqrt2}+t^2}\\=\frac{4}{3+2\sqrt2}\int\frac{dt}{(3-2\sqrt2)^2+t^2}\\=\frac{4}{3+2\sqrt2}\left[\frac{1}{3-2\sqrt2}\tan ^{-1} \frac{t}{3+2\sqrt2}\right]+C\\=4\left[\tan ^{-1} \frac{\tan \frac{\theta}{2}}{3+2\sqrt2}\right]+C$

$\\\int_{0}^{2\pi} \frac{2}{3-2\sqrt{2}\cos\theta} d\theta \\=4\left[\tan ^{-1} \frac{\tan \frac{\theta}{2}}{3+2\sqrt2}\right]_0^a+4\left[\tan ^{-1} \frac{\tan \frac{\theta}{2}}{3+2\sqrt2}\right]_a^{2\pi}\\=8\tan ^{-1} \frac{\tan \frac{a}{2}}{3+2\sqrt2}\\$As $a\to \pi^{\pm}, \tan ^{-1} \left(\frac{\tan \frac{a}{2}}{3+2\sqrt2}\right)\to \frac{\pi}{2}\\\therefore \int_{0}^{2\pi} \frac{2}{3-2\sqrt{2}\cos\theta} d\theta=4\pi$

I think your 4th to 5th line is wrong.

untouchablecuz · Feb 12, 2010

shaon0 said:
I think your 4th to 5th line is wrong.

i dont see it

shaon0 · Feb 12, 2010

untouchablecuz said:
i dont see it

Yeah my bad lol. Sorry

cutemouse · Feb 12, 2010

A group of 7 boys and 5 girls go to watch a movie together.

(i) In how many ways can they sit in a row if the boys sit together and the girls sit together?

(ii) In how many ways can they sit together in a row if no two girls are to sit together?

(iii) After the movie the group of twelve go to a café and sit at a round table so that two particular girls sit together, five particular boys sit together, the remaining three girls sit together and the remaining two boys sit together. The four groups around the table don't mind which other group they are seated next to and within each group no one minds who they sit next to. In how many different ways can they be arranged?

Pwnage101 · Feb 12, 2010

How about we give the '10er's a go to answer/ask these Q's...

fullonoob · Feb 12, 2010

jm01 said:
A group of 7 boys and 5 girls go to watch a movie together.

(i) In how many ways can they sit in a row if the boys sit together and the girls sit together?

(ii) In how many ways can they sit together in a row if no two girls are to sit together?

(iii) After the movie the group of twelve go to a café and sit at a round table so that two particular girls sit together, five particular boys sit together, the remaining three girls sit together and the remaining two boys sit together. The four groups around the table don't mind which other group they are seated next to and within each group no one minds who they sit next to. In how many different ways can they be arranged?

i) 7P7 x 5P5 x 2 o-o
ii) 7P5 x 5P5 x2 + 5P5 x 7P4 x2
dont wna embarass myself now .. correcy me if you want xD
Just a question when do you use P and C. C = order important? P = order not?

fullonoob · Feb 12, 2010

lolman12567 said:
(i) Find the sum 1 + w + w² + w³ +... to n terms, considering the cases n=3k, n=3k+1, n=3k+2 where k is an integer.

(ii) Show that
(1 - w+w²)(1 - w² +w⁴)(1 - w² +w⁸) ... to 2n factors = 2²ⁿ

i) Sn = (w^n-1)/(w-1) or is it

addikaye03 · Feb 14, 2010

Cazic said:
Show that

$\int_{0}^{2\pi} \frac{2}{3-2\sqrt{2}\cos\theta} \, d\theta = 4\pi$

Alternative Method:

- Rationalise denominator => Partial Fractions.

Haven't checked if it works yet, but i think it would work out alright.

inedible · Feb 28, 2010

How about this one,

Prove (1+cos(x) +isin(x))^{n} = (2^{n}cos\frac{x}{2}(cos\frac{nx}{2}+isin\frac{nx}{2}))

P.S. How can I use LaTex?

adomad · Feb 28, 2010

$(1+cos(x) +isin(x))^{n} = (2^{n}cos\frac{x}{2}(cos\frac{nx}{2}+isin\frac{nx} {2}))$ .

you need to put it between $and [:/tex]... remove the :$

Rezen · Apr 3, 2010

inedible said:
How about this one,

Prove (1+cos(x) +isin(x))^{n} = (2^{n}cos\frac{x}{2}(cos\frac{nx}{2}+isin\frac{nx}{2}))

P.S. How can I use LaTex?

I think the question is wrong. $LHS=(1+cos^2(\frac{x}{2})-sin^2(\frac{x}{2})+2icos(\frac{x}{2})sin(\frac{x}{2}))^n \\=(2cos^2(\frac{x}{2})+2icos(\frac{x}{2})sin(\frac{x}{2}))^n \\ =2^ncos^n(\frac{x}{2})(cos(\frac{nx}{2})+isin(\frac{nx}{2})), by \;De \;Moivre's \;Theorem$

untouchablecuz · Apr 3, 2010

a nice way using eulers:

$\\(1+\cos x+i\sin x)^n\\=(1+e^{ix})^n\\=\left(e^{i\frac{x}{2}}(e^{-i\frac{x}{2}}+e^{i\frac{x}{2}}) \right )^n\\=(2\cos \frac{x}{2})^n(\cos \frac{x}{2}+i\sin \frac{x}{2})^n\\=2^n\cos^n \frac{x}{2}(\cos \frac{nx}{2}+i \sin \frac{nx}{2})$

cutemouse · Apr 3, 2010

Question:

A point P is moving in a circular path around a centre O. Define the angular velocity of P with respect to O at time t.

Derive expressions for the tangential and normal components of the acceleration of P at time t.

Aquawhite · Apr 3, 2010

jm01 said:
Question:

A point P is moving in a circular path around a centre O. Define the angular
velocity of P with respect to O at time t.

Derive expressions for the tangential and normal components of the acceleration of P at time t.

Aww, I havn't done mechanics yet. I don't think many schools will have :S.

untouchablecuz · Apr 3, 2010

jm01 said:
Question:

A point P is moving in a circular path around a centre O. Define the angular velocity of P with respect to O at time t.

Derive expressions for the tangential and normal components of the acceleration of P at time t.

$\\$Let $\theta$ be the angle between the vertical and radius vector at point P,$\\\ddot{x}_T=g\sin \theta $ and $\ddot{x}_R=g\cos \theta\\$By definition$, \ \omega =\frac{2\pi}{T} $ where $T $ is the period of rotation,$\\\frac{d\theta}{dt}=\frac{2\pi}{T}\\\int_{0}^{\theta} d\theta =\frac{2\pi}{T} \int_{0}^{t}dt\\\theta=\frac{2\pi}{T}t\\\therefore \ddot{x}_T=g\sin \frac{2\pi}{T}t$ and $\ddot{x}_R=g \cos \frac{2 \pi}{T}$

amiright?

cutemouse · Apr 3, 2010

untouchablecuz said:
$\\$Let $\theta$ be the angle between the vertical and radius vector at point P,$\\\ddot{x}_T=g\sin \theta $ and $\ddot{x}_R=g\cos \theta\\$By definition$, \ \omega =\frac{2\pi}{T} $ where $T $ is the period of rotation,$\\\frac{d\theta}{dt}=\frac{2\pi}{T}\\\int_{0}^{\theta} d\theta =\frac{2\pi}{T} \int_{0}^{t}dt\\\theta=\frac{2\pi}{T}t\\\therefore \ddot{x}_T=g\sin \frac{2\pi}{T}t$ and $\ddot{x}_R=g \cos \frac{2 \pi}{T}$

amiright?

No.

1. $\omega =\frac{2\pi}{T}$ is not a definition.
2. It only applies to uniform circular motion.

nikkifc · Apr 9, 2010

jm01 said:
Question:

A point P is moving in a circular path around a centre O. Define the angular velocity of P with respect to O at time t.

Derive expressions for the tangential and normal components of the acceleration of P at time t.

Let P be at the position at the time t of a point describing a circle of centre O and radius r. OX is a fixed line cutting the circle at A. Suppose that at time t the particle is at P with speed v and at time t+δt it is at Q with speed v+δv.

At Q the resolved parts of velocity parallel and perpendicular to OP are (v+δv)sinδθ and (v+δv)cosδθ respectively.

Thus, the resolved parts of increase in velocity during time δt are:

Along tangent at P: (v+δv)cosδθ-v
Along normal at P: (v+δv)sinδθ-0

When δθ is small, sinδθ =:=δθ and using cos2θ=1-2sin²θ
cosδθ=1-2sin²(1/2 δθ) =:= 1 - 2(1/2 δθ)² =:= 1 (neglecting 2nd order quantities)

Thus, (v+δv)cosδθ-v = δv
and (v+δv)sinδθ-0=vδθ+δvδθ=vδθ (neglecting 2nd order quantities)

Thus, tangential component of accel at P is: $\lim{\delta t \to 0} \frac{\deltav}{\deltat}=\frac{dv}{dt}=\frac{d}{dt}(r\omega)=r\dot{\omega}=r\ddot{\theta}$

and normal component o accel at P is: $\lim{\delta t \to 0} v\frac{\delta \theta}{\delta t}=v \frac{d\theta}{dt}=v\dot{\theta}=r\omega^2=v\omega=\frac{v^2}{r} \text{ using } v=r\omega$

shaon0 · Apr 9, 2010

nikkifc said:
Let P be at the position at the time t of a point describing a circle of centre O and radius r. OX is a fixed line cutting the circle at A. Suppose that at time t the particle is at P with speed v and at time t+δt it is at Q with speed v+δv.

At Q the resolved parts of velocity parallel and perpendicular to OP are (v+δv)sinδθ and (v+δv)cosδθ respectively.

Thus, the resolved parts of increase in velocity during time δt are:

Along tangent at P: (v+δv)cosδθ-v
Along normal at P: (v+δv)sinδθ-0

When δθ is small, sinδθ =:=δθ and using cos2θ=1-2sin²θ
cosδθ=1-2sin²(1/2 δθ) =:= 1 - 2(1/2 δθ)² =:= 1 (neglecting 2nd order quantities)

Thus, (v+δv)cosδθ-v = δv
and (v+δv)sinδθ-0=vδθ+δvδθ=vδθ (neglecting 2nd order quantities)

Thus, tangential component of accel at P is: $\lim{\delta t \to 0} \frac{\deltav}{\deltat}=\frac{dv}{dt}=\frac{d}{dt}(r\omega)=r\dot{\omega}=r\ddot{\theta}$

and normal component o accel at P is: $\lim{\delta t \to 0} v\frac{\delta \theta}{\delta t}=v \frac{d\theta}{dt}=v\dot{\theta}=r\omega^2=v\omega=\frac{v^2}{r} \text{ using } v=r\omega$

Nice, i was wondering when a 10er would answer this question. btw, it'll never be tested ie Non-uniform circular motion won't ever be tested.

nikkifc · Apr 9, 2010

shaon0 said:
Nice, i was wondering when a 10er would answer this question. btw, it'll never be tested ie Non-uniform circular motion won't ever be tested.

Check the 1981 HSC paper. I think this is from that paper. Oh and I accelerated

shaon0 · Apr 9, 2010

nikkifc said:
Check the 1981 HSC paper. I think this is from that paper. Oh and I accelerated

1981? It may have been but 2010 onwards, they'll never ask it but either way it's very easy to derive. The proofs easier using polar coordinates and a lot quicker.

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