4 Unit Revising Marathon HSC '10 (1 Viewer)

nikkifc · Apr 9, 2010

shaon0 said:
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.

Yeah, it's probably not going to be ask anytime soon but it's still in the course. Also, if you check the 1989 HSC paper, they ask for this method (but they don't ask you to derive the tangential component of acceleration), so it's good learning this way too

It's a bit like proving DeMovire's theorem using induction. It's never been examined before in the HSC, but it's still examinable...

Gussy Booo · Apr 12, 2010

the equation:
|z-1-3i|+|z-9-3i|=10 corresponds to an ellipse in the argand diagram

(i) write down the complex number corresponding to the centre

(ii) write down the range of values of arg(z) for complex numbers z, corresponding to points on the ellipse

from hsc past paper =/,

Dragonmaster262 · Apr 13, 2010

Gussy Booo said:
the equation:
|z-1-3i|+|z-9-3i|=10 corresponds to an ellipse in the argand diagram

(i) write down the complex number corresponding to the centre

(ii) write down the range of values of arg(z) for complex numbers z, corresponding to points on the ellipse

from hsc past paper =/,

Is the centre 5+3i? I think I miscalculated something.

adomad · Apr 13, 2010

center is correct

range of arg(z)

0<=arg(z)<=pi/2

Gussy Booo · Apr 13, 2010

Dragonmaster262 said:
Is the centre 5+3i? I think I miscalculated something.

Lets see your working out? [=

adomad · Apr 13, 2010

Gussy Booo · Apr 13, 2010

adomad said:
center is correct

range of arg(z)

0<=arg(z)<=pi/2

omg omg tell me how. lol. i wanted to see working out, so i could kno how to do it.

explain plzzzzzzzzzzzzzzz.

+ how u got the centre.

they're both right btw.

adomad · Apr 13, 2010

well the outer most point is 0+3i... the arg of this is pi/2
the smallest arg is the tanget to the bottom of the ellipse.. work that out .. centre is 5+3i so form a triangle with sides 5 (half of 10) and 4(the distance from the middle to the point 1+3i)
and then u see that the ellipse hits the point 5+0i... arg 5=0

adomad · Apr 13, 2010

new question!:

express $(sin\frac{\pi }{3}+icos\frac{\pi}{3})^8 $ in the form a+ib$

adomad · Apr 13, 2010

or if that is too easy...

$$ use De Moivre's theorem to find an expression for tan3\theta$

Gussy Booo · Apr 13, 2010

adomad said:
new question!:

express $(sin\frac{\pi }{3}+icos\frac{\pi}{3})^8 $ in the form a+ib$

Looks to easy =/...

Um....If we express it in mod arg form we get --> 1cis(pi/3).
Now..according to De Moivre [1cis([i/3)]^8 = 1^8cis(8pi/3).
But this doesn't lie number the principle argument.
So if we play around..it becomes --> 1^8cis(5pi/3).
This furthermore gives:

cos5pi/3+isin5pi/3.
We then punch some numbers into the calculator...which i cbf doing.

adomad · Apr 13, 2010

Gussy Booo said:
Looks to easy =/...

Um....If we express it in mod arg form we get --> 1cis(pi/3).
Now..according to De Moivre [1cis([i/3)]^8 = 1^8cis(8pi/3).
But this doesn't lie number the principle argument.
So if we play around..it becomes --> 1^8cis(5pi/3).
This furthermore gives:

cos5pi/3+isin5pi/3.
We then punch some numbers into the calculator...which i cbf doing.

twaz a trick Q

omgmynameislong · Apr 13, 2010

sin+icos

not cos+isin

(sinpi/3 + icos pi/3)^8

=(cos pi/6 + isin pi/6)^8
=cis -4pi/6

cbf a+ib

Gussy Booo · Apr 13, 2010

omgmynameislong said:
sin+icos

not cos+isin

(sinpi/3 + icos pi/3)^8

=(cos pi/6 + isin pi/6)^8
=cis -4pi/6

cbf a+ib

HAHAHAH! OMGGGGGG! thats soo funny.
i remember looking at the screen and saying...wait..why are my sin and cos the other way around. Then i just switched em around.
Good Q. >.< I would so lose a mark there.

omgmynameislong · Apr 13, 2010

adomad said:
or if that is too easy...

$$ use De Moivre's theorem to find an expression for tan3\theta$

in terms of tanx?

(sinx+icosx)^3

= (sinx)^3 + i3(sinx)^2cosx - 3sinx(cosx)^2 - i(cosx)^3

equating real and imaginary parts

sin3x= (sinx)^3 - 3sinx (cosx)^2
cos3x= 3(sinx)^2cosx - (cosx)^3

tan3x = [(sinx)^3 - 3sinx(cosx)^2]/[3(sinx)^2cosx - (cosx)^3]

divide by (cosx)^3

tan3x= [(tanx)^3 - 3tanx]/[3tanx -1]

cutemouse · Apr 13, 2010

Prove by mathematical induction that

$\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{n}\le\frac{4n+3}{6}n \text{ for all integers }n\ge1$

Gussy Booo · Apr 17, 2010

Someone please do this.

nonsenseTM · Apr 17, 2010

why is that induction a 4u question?

ninetypercent · Apr 17, 2010

not sure if this is right...

EDIT: it is wrong. RTP part/ ooopss

$\sqrt{1}+ \sqrt{2}+\sqrt{3}+....+\sqrt{n} \leq \frac{4n +1}{6}n$

For n =1
LHS = 1
RHS = (4 + 3)/6 x 1 = 7/6
LHS < RHS
therefore, true for n =1

Assume true for n = k
$\sqrt{1}+ \sqrt{2}+\sqrt{3}+....+\sqrt{k} \leq \frac{4k +1}{6}k$

RTP true for n = k + 1

$\sqrt{1}+ \sqrt{2}+\sqrt{3}+....+\sqrt{k} + \sqrt{k+1}\leq \frac{4(k+1)}{6}(k+1) \\ LHS = \sqrt{1}+ \sqrt{2}+\sqrt{3}+....+\sqrt{k} + \sqrt{k+1} \leq \frac{4k+3}{6}(k) + \sqrt{k+1} \\ = \frac{4k+4 -1}{6}k + \sqrt{k+1} \\ = k\frac{4(k+1)}{6} - \frac{k}{6} + \sqrt{k+1} \\ = k\frac{4(k+1)}{6} - \frac{k - \sqrt{k+1}}{6}\\\leq \frac{4(k+1)}{6}(k) + \frac{4(k+1)}{6}\\ =\\= \frac{4(k+1)}{6}(k+1) = RHS$

The statement is true for n = k+1, if true for n =k

therefore, by induction, the statement is true

EDIT: latex looked ugly, so i had to edit a number of times

Trebla · Apr 17, 2010

Hint: First prove that

$\sqrt{k+1}\leq \dfrac{8k+5}{6}$

4 Unit Revising Marathon HSC '10 (1 Viewer)

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