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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: man0005 on February 07, 2011, 08:21:37 pm

Title: man0005's specialist question thread :)
Post by: man0005 on February 07, 2011, 08:21:37 pm: Give parametric equations for the following:
1) (x-1)^2/9 + (y+3)^2 / 4 = 9

Determine the cartesian equation of the curve determined by the following parametric equation:
x = 1 - sec(2t)
y = 1 + tan (2t)
t : (pie/4 , 3pie/4)
Title: Re: man0005's specialist question thread :)
Post by: xZero on February 07, 2011, 08:59:27 pm: firstly
you rearrange
x = 1-sec(2t) into x-1 = -sec(2t)
y = 1+tan(2t) into y-1 =t an(2t)

then you square them so (x-1)^2 = sec^2(2t) and (y-1)^2 = tan^2(2t)

then using the trig identity: 1 + tan^2(x) = sec^2(x),
1+ tan^2(2t) = sec^2(2t)
1+ (y-1)^2 = (x-1)^2
(x-1)^2 - (y-1)^2 = 1

now since pi/4< t < 3pi/4, we can see that pi/2< 2t < 3pi/2 (multiply it by 2 to match the 2t)
the range of 2t is the third quadrant so sec(2t) will always be negative or 0 so the domain of the equation is 1<x<infinite
the range of the equation should be R <- need verify
Title: Re: man0005's specialist question thread :)
Post by: m@tty on February 07, 2011, 10:48:50 pm: 1.

$\left(\frac{x-1}{3}\right)^2+\left( \frac{y+3}{2}\right)^2=9$

$\left(\frac{x-1}{9}\right)^2+\left( \frac{y+3}{6}\right)^2=1$

$\left(\frac{x-1}{9}\right)^2=\cos^2(t)$

$x=9\cos(t)+1$

$\left(\frac{x+3}{6}\right)^2=\sin^2(t)$

$y=6\sin(t)-3$
Title: Re: man0005's specialist question thread :)
Post by: man0005 on February 08, 2011, 07:21:03 am: Does it matter which one is x and which one is y?
Title: Re: man0005's specialist question thread :)
Post by: kamil9876 on February 08, 2011, 04:18:17 pm: You mean, would x=9sint+1 and y=6cost -3 also work? yes it would

edit: However it is a good habit to associate $x$ with $cos(t)$ and $y$ with $sin(t)$ if possible. Here is an example where it is important:

parametrize $x^2+4y^2=1$ with the additional constraint that $x\geq 0$ (half of an ellipse). State the domain of any parameter used.
Title: Re: man0005's specialist question thread :)
Post by: man0005 on February 08, 2011, 04:58:02 pm: oh do you get the wrong y-intercepts if you do it the other way?
Title: Re: man0005's specialist question thread :)
Post by: kamil9876 on February 08, 2011, 05:07:25 pm: the point is, you can do it two ways:

Solution 1:

$x=cost$ $y=2sin(t)$ and the domain of $t$ is $[-\pi/2, \pi/2]$

Solution 2:

$x=sin(t)$ $y=2cos(t)$ and the domain of $t$ is $[0,\pi]$

So you see, the domains for $t$ are different if you do it differently, but solution 1 is easier to visualize because of your experience with circular functions.
Title: Re: man0005's specialist question thread :)
Post by: man0005 on February 08, 2011, 05:16:59 pm: oh okay thanks heaps :D

im stuck on this question now :/

Show that every circle that touches the x axis and the y-axis has an equation of similar form to x^2 + y^2 - 2ax - 2ay + a^2 = 0
Title: Re: man0005's specialist question thread :)
Post by: xZero on February 08, 2011, 05:23:48 pm: ok firstly the standard equation of a circle is x2 + y2 = a2 (where a = radius of the circle), for the circle to only touch the x axis and the y axis, you can move the circle by a units right and up to the first quadrant (you can move the circle to any other quadrant)

(x-a)2 + (y-a)2 = a2
x2 - 2ax +a2 + y2 - 2ay + y2 = a2
x2 + y2 - 2ax -2ay +a2 =0

you can then move the circle to other quadrant and repeat the steps above
Title: Re: man0005's specialist question thread :)
Post by: kamil9876 on February 08, 2011, 05:34:54 pm: your approach works for the first and third quadrant only.

The real question should be show that every circle that touches the x axis and the y axis has the form:

$ x^2+y^2-2ax \pm 2ay - a^2=0$
Title: Re: man0005's specialist question thread :)
Post by: xZero on February 08, 2011, 05:48:51 pm: Quote from: kamil9876 on February 08, 2011, 05:34:54 pm
your approach works for the first and third quadrant only.

The real question should be show that every circle that touches the x axis and the y axis has the form:

$ x^2+y^2-2ax \pm 2ay - a^2=0$

how come?
1st quadrant x2 + y2 - 2ax - 2ay +a2 = 0
2nd quadrant x2 + y2 + 2ax - 2ay + a2 =0
3rd quadrant x2 + y2 + 2ax + 2ay + a2 = 0
4th quadrant x2 + y2 -2ax + 2ay + a2 = 0
Title: Re: man0005's specialist question thread :)
Post by: kamil9876 on February 08, 2011, 07:46:04 pm: Firstly let me restate what I wanted to say just to remove ambiguity, every such circle has either one of these forms:

(1) $x^2+y^2 - 2ax - 2ay +a^2=0$
(2) $x^2+y^2 - 2ax + 2ay +a^2=0$

For example if it is a circle or radius r centre at (r,r) then we have form (1) with a=r. If it is a circle of radius r centred at (-r,-r) then we have form (1) with a=-r. If it is a circle or radius r centred at (r,-r) then we have form (2) with a=r, while if it is at (-r,r) then it is form (2) with a=-r.

The problem with the original problem statement is that if we have the centre at $(r,-r)$ what can we set $a$ to? we need the option of form (2) to allow for the coefficients of $x$ and $y$ to have different sign (as you have) because the way they have written it we need those coefficients to both be the same ( $2a$ )
Title: Re: man0005's specialist question thread :)
Post by: golden on February 11, 2011, 05:55:00 pm: Express the equation in the modulus–argument form, 0 < theta < pi/2:

1 + tan i

The solutions state to use the modulus argument, which gets sec(theta). But it then multiplies sec(theta) by cis(theta). How did the cis(theta) come about?

Thanks.
Title: Re: man0005's specialist question thread :)
Post by: kamil9876 on February 11, 2011, 06:22:36 pm: Are you saying that you want to write $1 + tan (\theta) i$ in polar form over that domain?. Well you must write it in the form $rcis\phi$ you have figured our $r$ correctly, now you must find the argument $\phi$ , but from a simple sketch and trig you see $\tan(\phi)=(tan\theta/1)$ and so since your complex number is in the first quadrant and $0<\theta<\pi/2$ this gives $\phi=\theta$ so:

$1+itan\theta=sec(\theta) cis\theta$
Title: Re: man0005's specialist question thread :)
Post by: man0005 on February 12, 2011, 12:17:10 pm: got a really good one for you kamil d:

(sin x + sin 3x + sin 5x)
____________________ = tan 3x

(cos x + cos 3x + cos 5x)
Title: Re: man0005's specialist question thread :)
Post by: kamil9876 on February 12, 2011, 01:05:06 pm: Don't know if there is any quicker way yet but you can always do it by brute force by just expanding sin(3x)=sin(x+2x) and likewise sin(5x)=sin(3x+2x)... same for the cosines, then just cleaning up a big mess.

I think a slightly quicker way may be to write sin(x)=sin(3x-2x) and expand that out, write sin(5x)=sin(3x+2x) and expand that out.

Here is what you get for the numerator:

$ sin(3x)cos(-2x)+cos(3x)sin(-2x) + sin(3x) + sin(3x)cos(2x) + cos(3x)sin(2x)$

$=2sin(3x)cos(2x) + sin(3x)$

$=sin(3x)(2cos(2x)+1)$

Now do a similair thing for the denominator and hopefully it will work out (that is expand out $cos(x)=cos(3x-2x)$ and $cos(5x)=cos(3x+2x)$ )
Title: Re: man0005's specialist question thread :)
Post by: evaever on February 12, 2011, 05:16:48 pm: Quote from: man0005 on February 12, 2011, 12:17:10 pm
got a really good one for you kamil d:

(sin x + sin 3x + sin 5x)
____________________ = tan 3x

(cos x + cos 3x + cos 5x)

I'm not Kamil. I'll do it anyway.

cisx+cis3x+cis5x = (cosx+cos3x+cos5x)+i(sinx+sin3x+sin5x)
kcis3x+cis3x=(cosx+cos3x+cos5x)+i(sinx+sin3x+sin5x)
(k+1)cis3x=(cosx+cos3x+cos5x)+i(sinx+sin3x+sin5x)
arg of left = arg of right
3x=tan^-1 (sinx+sin3x+sin5x)/(cosx+cos3x+cos5x)
tan3x=(sinx+sin3x+sin5x)/(cosx+cos3x+cos5x)
Title: Re: man0005's specialist question thread :)
Post by: man0005 on February 12, 2011, 05:24:28 pm: O_O
evaever are we meant to know how to do it like that in spesh D:
cause i have no idea what you did LOL
Title: Re: man0005's specialist question thread :)
Post by: kamil9876 on February 12, 2011, 05:57:19 pm: evaever did not define $k$ in the solution so I don't know what he/she did entirely.

Here is another solution using complex numbers(using the idea in my previous solution to change 5x to 3x+2x and x to 3x-2x):

let $ z=cis(x)+cis(3x)+cis(5x)=cis(3x)cis(-2x) + cis(3x) + cis(2x)cis(3x)$

$=cis(3x) (cis(-2x) + 1 + cis(2x))$

But you see that cis(-2x)+cis(2x) is a real number. Hence we have that cis(x)+cis(3x)+cis(5x) is a real number multiply of cis(3x), hence $tan(3x)=Im z / Re z$ which is the result.

edit: probably won't be expected to do it like this, brute force works and that's how most of specialist rolls.

generalisation: suppose that $a_1,a_2...,a_n$ is an arithmetic sequence with arithmetical average $a$ , then $\frac{sin(a_1)+sin(a_2)...+sin(a_n)}{cos(a_1)+....+cos(a_n)}=tan(a)$ (if it is defined)
Title: Re: man0005's specialist question thread :)
Post by: man0005 on February 15, 2011, 05:30:34 pm: oh okay, lol someone eventually googled it and there some random formula to do it. wasted all my time in the end :P

anyway i have a quick question
can't seem to get this one
Let sin x = 0.75, x is an element of pie, 3pie/2, Find correct to 2 decimal places sin 1/2 x
(We're meant to use identities to get it, not calculator)
sorry if this seems like a silly question, i'm a little tired today :P
Title: Re: man0005's specialist question thread :)
Post by: kamil9876 on February 15, 2011, 09:18:09 pm: Quote
oh okay, lol someone eventually googled it and there some random formula to do it. wasted all my time in the end

what do you mean? it is doable with just the identities you know as shown in the first post after the solution. The complex number solution was just a more slick way that you don't really need to know and we didn't use the generalisation to solve it, it is just something I noticed after looking back at my complex number solution.

Anyway for your question:

I assume that is $sin(x/2)$ ? If so let $y=x/2$ you see that:

$cos(2y)=1-2sin^2 y$

Still this would require a little bit of calculator work but only for the square roots that you need to find cos(2y) using pythagoras as well as square rooting at the end (so you can find an answer with surds basically.
Title: Re: man0005's specialist question thread :)
Post by: man0005 on March 09, 2011, 09:27:58 pm: Can someone help me with this please

The amount of pollution (p), t centuries after 1850 (ie t = 0 is 1850) is given by the rule:
P = aTan^-1(bt)

Find the percentage increase in pollution from 1900 to 2000.
in 1900, P = 2.657, in 1950, P = 4.5
Title: Re: man0005's specialist question thread :)
Post by: Mao on March 10, 2011, 01:02:35 am: Substituting (t,P)=(0.5,2.657) and (1.0,4.5), we arrive at

2.657=aTan^-1[0.5b]
4.5=aTan^-1(b)

eqn1/eqn2:
2.657Tan^-1(b)=4.5Tan^-1[0.5b]

Plot 2.657Tan^-1(b) and 4.5Tan^-1[0.5b], find that the intersection is at b=1.00082. Further substitution gives a=5.72658

If the question provided exact figures, the answer would be b=1 and a=18/pi
Title: Re: man0005's specialist question thread :)
Post by: man0005 on March 24, 2011, 10:47:18 pm: How would you do this D:
Split into partial fractions

(-x-10)
__________

(x^2 - 4x -12)
Title: Re: man0005's specialist question thread :)
Post by: Mao on March 25, 2011, 12:50:53 am: $\begin{align*} \frac{-x-10}{(x-6)(x+2)} & = \frac{a}{x-6} + \frac{b}{x+2} \\ -x-10 & = a(x+2) + b(x-6)\end{align*} $
Equating coefficients
$ \begin{align*} a+b & = -1 \\ 2a-6b & = -10 \\ \implies b & = 1 \\ \implies a & = -2 \\ \frac{-x-10}{x^2-4x-12} & = \frac{1}{x+2}-\frac{2}{x-6} \end{align*}$
Title: Re: man0005's specialist question thread :)
Post by: man0005 on March 25, 2011, 07:42:02 pm: ohh right, sorry :P
what about splitting:

4x^2 - 21x + 23
___________________
x^2 - 6x + 9
Title: Re: man0005's specialist question thread :)
Post by: pi on March 25, 2011, 08:53:31 pm: Just a hint: Try using long division first
Title: Re: man0005's specialist question thread :)
Post by: m@tty on March 25, 2011, 09:05:00 pm: Don't use long division on that..

Just recognise that there are 4 x^2 on the numerator, thus try and get 4 of the denominator separate to divide.

$\frac{4x^2-21x+23}{x^2-6x+9}= \frac{(4x^2-24x+36)+3x-13}{x^2-6x+9} = 4 + \frac{3x+13}{x^2-6x+9}$
Title: Re: man0005's specialist question thread :)
Post by: pi on March 25, 2011, 09:12:05 pm: ^^Thats the way I would have done it too, I was just giving a stock-standard approach. Of course, visualising a factor is always better and neater man0005, just that its not in the texts for some reason...
Title: Re: man0005's specialist question thread :)
Post by: man0005 on March 25, 2011, 10:09:35 pm: thanks for that guys :D
i really didnt want to use long division :P
Title: Re: man0005's specialist question thread :)
Post by: iNerd on March 25, 2011, 10:39:00 pm: Quote from: m@tty on March 25, 2011, 09:05:00 pm
Don't use long division on that..

Just recognise that there are 4 x^2 on the numerator, thus try and get 4 of the denominator separate to divide.

$\frac{4x^2-21x+23}{x^2-6x+9}= \frac{(4x^2-24x+36)+3x-13}{x^2-6x+9} = 4 + \frac{3x+13}{x^2-6x+9}$
Wtf. That is insanely sexy/eloquent.
Title: Re: man0005's specialist question thread :)
Post by: man0005 on March 25, 2011, 10:51:59 pm: hm how would you sketch something like Arg z > -pi/2
nevermind ^^

what would the domain of this be?
|z – 6| – |z + 6| = 3
Title: Re: man0005's specialist question thread :)
Post by: Mao on March 27, 2011, 02:53:51 pm: Quote from: man0005 on March 25, 2011, 10:51:59 pm
what would the domain of this be?
|z – 6| – |z + 6| = 3

This is a hyperbola. It is the left-hand side only, with a Re(z) intercept of -1.5, domain is (-infinity,-1.5]
Title: Re: man0005's specialist question thread :)
Post by: man0005 on March 27, 2011, 02:59:56 pm: Ah what about this one:
If Sec^-1 x = 2
then x is equal to
a) 1.047 b) 0.8776 c) 1.100 d) -2.403 e) 0.5
Title: Re: man0005's specialist question thread :)
Post by: luken93 on March 27, 2011, 03:14:54 pm: $sec^{-1}(x) = 2$

$x = sec(2)$

$x = \frac{1}{cos(2)}$

$x = -2.403$
Title: Re: man0005's specialist question thread :)
Post by: man0005 on March 27, 2011, 03:18:45 pm: For 2Sin^-1 (1/x+1)
why do you have to exclude zero for the implied range?
Title: Re: man0005's specialist question thread :)
Post by: enpassant on March 27, 2011, 03:23:38 pm: Quote from: luken93 on March 27, 2011, 03:14:54 pm
$sec^{-1}(x) = 2$

$cos^{-1}(x) = \frac{1}{2}$

$x = cos(\frac{1}{2})$

$x = \frac{\pi}{3}$

$x = 1.047$
disagree
Title: Re: man0005's specialist question thread :)
Post by: man0005 on March 27, 2011, 03:26:32 pm: did you get -2.403 empassant? cause thats what the answers said
but i got what luken got as well :/
Title: Re: man0005's specialist question thread :)
Post by: enpassant on March 27, 2011, 03:34:05 pm: yes
Title: Re: man0005's specialist question thread :)
Post by: man0005 on March 27, 2011, 03:34:53 pm: how?!
Title: Re: man0005's specialist question thread :)
Post by: luken93 on March 27, 2011, 03:36:19 pm: Quote from: man0005 on March 27, 2011, 03:34:53 pm
how?!
check mine now..
Title: Re: man0005's specialist question thread :)
Post by: man0005 on March 27, 2011, 03:40:49 pm: isnt there a way to do it without the use of calculator then?
Title: Re: man0005's specialist question thread :)
Post by: BubbleWrapMan on March 27, 2011, 03:59:39 pm: Quote from: man0005 on March 27, 2011, 03:18:45 pm
For 2Sin^-1 (1/x+1)
why do you have to exclude zero for the implied range?
Is the inside part 1/x+1 or 1/(x+1)?
If it's the latter (which makes sense based on your question)
1/(x+1) is a hyperbola with range R\{0}, so since it isn't possible for the hyperbola to equal 0, and sin^-1(0) = 0, then 2sin^-1(1/(x+1)) can never equal 0.
Title: Re: man0005's specialist question thread :)
Post by: VCE247 on March 27, 2011, 07:14:05 pm: How would I go about doing a question like this:
Let Sec B = b, B is an element of [pi/2 , pi ]
Find in terms of B, two values in the range [-pi, pi ] which
satisfy sec x = -b and cosec x = b
Title: Re: man0005's specialist question thread :)
Post by: luken93 on March 27, 2011, 07:29:08 pm: So basically, you are trying to find two values where

$-sec(x) = cosec(x)$

$\frac{-1}{cos(x)} = \frac{1}{sin(x)}$

$tan(x) = -1$

$x = \frac{-\pi}{4} or \frac{3\pi}{4}$

I'm not sure if this is what you're asking though?

Subbing back into the other equations, we see that
$b = -\sqrt{2}$

Hence, in terms of B:

$sec(B) = -\sqrt{2}$

$B = arcsec(-\sqrt{2})$

$B = \frac{3\pi}{4}$
Title: Re: man0005's specialist question thread :)
Post by: man0005 on March 27, 2011, 08:15:30 pm: what about this one? ><
z^3 + az^2 + bz + 10 - i
find the constants a and b given they are real, if z = (1+i) is a zero of the polynomial
Title: Re: man0005's specialist question thread :)
Post by: m@tty on March 27, 2011, 08:22:24 pm: You have (z-1-i) as a factor.

Long divide, then use the quadratic formula on the result.
Title: Re: man0005's specialist question thread :)
Post by: man0005 on March 27, 2011, 08:24:06 pm: hm if a and b are real, wouldnt another factor be (z-1+i)?
cant i do that?
Title: Re: man0005's specialist question thread :)
Post by: man0005 on March 27, 2011, 10:45:28 pm: If someone could also help me with this one, that would be great
Find the set of real values for k, k doesnt equal -1, for which the roots of the equation x^2 +4x -1 + k ( x^2 + 2x + 1 ) = 0
are:
a) real and distinct
b) real and equal
c) complex with positive part
Title: Re: man0005's specialist question thread :)
Post by: m@tty on March 27, 2011, 10:47:58 pm: Quote from: man0005 on March 27, 2011, 08:24:06 pm
hm if a and b are real, wouldnt another factor be (z-1+i)?
cant i do that?

No you can't there is an imaginary unit in the expression... right at the end. complicates it a bit.

For the conjugate root theorem to apply, ALL coefficients must be real (no i's present when simplified).

And, IF you could apply the conjugate theorem, it would be (z+(1+i)).
Title: Re: man0005's specialist question thread :)
Post by: man0005 on March 27, 2011, 10:55:04 pm: ohhh, om gosh how did i miss that ._.
thanks !

do you have any ideas for the next one?
Title: Re: man0005's specialist question thread :)
Post by: m@tty on March 27, 2011, 10:59:10 pm: $x^2 +4x -1 + k ( x^2 + 2x + 1 ) = 0$

$\iff x^2(1+k)+2x(2+k)-1+k=0$

Use discriminant

$\Delta = (4+2k)^2-4(1+k)(k-1)$

Solve for:
$\Delta = 0$ (real and equal)

$\Delta < 0$ (complex)

$\Delta > 0$ (real and distinct)
Title: Re: man0005's specialist question thread :)
Post by: xZero on March 27, 2011, 11:00:11 pm: Quote from: man0005 on March 27, 2011, 10:45:28 pm
If someone could also help me with this one, that would be great
Find the set of real values for k, k doesnt equal -1, for which the roots of the equation x^2 +4x -1 + k ( x^2 + 2x + 1 ) = 0
are:
a) real and distinct
b) real and equal
c) complex with positive part

expand k, $x^2+kx^2+4x+2kx-1+k=0$

$(k+1)x^2+(4+2k)x+(k-1)=0$

now use $b^2-4ac$ and if it =0 then its real and equal, if its >0 then real and distinct and <0 will be complex

Edit: I Got Ninja'ed oO
Title: Re: man0005's specialist question thread :)
Post by: man0005 on March 27, 2011, 11:02:54 pm: ahh thanks
got another one
Let z be a complex number with | z | = 6
Let A be the point representing z. Let B be the point representing (1+i)z
Prove that OAB is an isosceles right-angled triangle

(sorry if im annoying you with all the qs D: )
Title: Re: man0005's specialist question thread :)
Post by: VCEMan94 on March 27, 2011, 11:17:21 pm: Can someone help me with this one as well? D:
Let S = { z : | z - (2root2 + i (2root2) | < (or equal to) 2
If z belongs to S, find the maximum and minimum values of |z|
If z belongs to S, find the maximum and minimum values of Arg (z)
Title: Re: man0005's specialist question thread :)
Post by: xZero on March 27, 2011, 11:19:04 pm: change 1+i and z into polar form so $1+i=\sqrt{2}cis(\frac{\pi}{4})$ and $z=6cis(\theta)$

$(1+i)z=6\sqrt{2}cis(\frac{\pi}{4}+\theta)$

well since the new line moved by 45 degrees, then the adjacent side would be $6\sqrt{2}sin(\frac{\pi}{4})$ which will get you 6, so we have 2 sides with magnitude of 6 and a 45 degree angle between the hypotenuse and the opposite, hence it is a isosceles right-angled triangle
Title: Re: man0005's specialist question thread :)
Post by: xZero on March 27, 2011, 11:23:53 pm: Quote from: VCEMan94 on March 27, 2011, 11:17:21 pm
Can someone help me with this one as well? D:
Let S = { z : | z - (2root2 + i (2root2) | < (or equal to) 2
If z belongs to S, find the maximum and minimum values of |z|
If z belongs to S, find the maximum and minimum values of Arg (z)

since my brain is dead ill just tell you what i would do,

convert S into a Cartesian equation and graph it by letting z = x + yi.

The value of |z| is asking you what is the maximum and the minimum distance from the graph to the origin.
The Arg(z) is the angle from origin to the graph, maximum and minimum occurs when the line from the origin to the graph is a tangent, so just work out where on the graph the line will be a tangent which goes through the origin
Title: Re: man0005's specialist question thread :)
Post by: VCEMan94 on March 27, 2011, 11:28:51 pm: oh okay, but how would you find the tangent? :/

and would you know how to do this one by any chance?
Let w = cis theta and z = w + (1/w)
Show that z lies on ellipse with equation (x^2/25) + (y^2/9) = 1/4
Title: Re: man0005's specialist question thread :)
Post by: m@tty on March 27, 2011, 11:29:32 pm: Quote from: xZero on March 27, 2011, 11:19:04 pm
change 1+i and z into polar form so $1+i=\sqrt{2}cis(\frac{\pi}{4})$ and $z=6cis(\theta)$

$(1+i)z=6\sqrt{2}cis(\frac{\pi}{4}+\theta)$

well since the new line moved by 45 degrees, then the adjacent side would be $6\sqrt{2}sin(\frac{\pi}{4})$ which will get you 6, so we have 2 sides with magnitude of 6 and a 45 degree angle between the hypotenuse and the opposite, hence it is a isosceles right-angled triangle

Did you assume a right angle in your explanation?
Title: Re: man0005's specialist question thread :)
Post by: xZero on March 27, 2011, 11:37:50 pm: Quote from: m@tty on March 27, 2011, 11:29:32 pm
Quote from: xZero on March 27, 2011, 11:19:04 pm
change 1+i and z into polar form so $1+i=\sqrt{2}cis(\frac{\pi}{4})$ and $z=6cis(\theta)$

$(1+i)z=6\sqrt{2}cis(\frac{\pi}{4}+\theta)$

well since the new line moved by 45 degrees, then the adjacent side would be $6\sqrt{2}sin(\frac{\pi}{4})$ which will get you 6, so we have 2 sides with magnitude of 6 and a 45 degree angle between the hypotenuse and the opposite, hence it is a isosceles right-angled triangle

Did you assume a right angle in your explanation?

Nope, the angle between the new line and z is 45 degrees (from the pi/4 + theta part) and work out the magnitude of the line connecting the new line and z, which is 6. Then use these facts to proof it
Title: Re: man0005's specialist question thread :)
Post by: xZero on March 27, 2011, 11:39:17 pm: Quote from: VCEMan94 on March 27, 2011, 11:28:51 pm
oh okay, but how would you find the tangent? :/

and would you know how to do this one by any chance?
Let w = cis theta and z = w + (1/w)
Show that z lies on ellipse with equation (x^2/25) + (y^2/9) = 1/4

you let the line passing through the origin and the graph be y=mx, sub that into S and use discriminant = 0 to solve for m (should get 2 solutions)
Title: Re: man0005's specialist question thread :)
Post by: m@tty on March 27, 2011, 11:44:11 pm: Quote from: xZero on March 27, 2011, 11:37:50 pm
Quote from: m@tty on March 27, 2011, 11:29:32 pm
Quote from: xZero on March 27, 2011, 11:19:04 pm
change 1+i and z into polar form so $1+i=\sqrt{2}cis(\frac{\pi}{4})$ and $z=6cis(\theta)$

$(1+i)z=6\sqrt{2}cis(\frac{\pi}{4}+\theta)$

well since the new line moved by 45 degrees, then the adjacent side would be $6\sqrt{2}sin(\frac{\pi}{4})$ which will get you 6, so we have 2 sides with magnitude of 6 and a 45 degree angle between the hypotenuse and the opposite, hence it is a isosceles right-angled triangle

Did you assume a right angle in your explanation?

Nope, the angle between the new line and z is 45 degrees (from the pi/4 + theta part) and work out the magnitude of the line connecting the new line and z, which is 6. Then use these facts to proof it

How did you deduce that the magnitude of the joining line must be 6? It appears to me that you have used trig (for right angled triangles) -- sorry if I'm wrong... just a little perplexed at this time...
Title: Re: man0005's specialist question thread :)
Post by: m@tty on March 27, 2011, 11:50:09 pm: Firstly, proving that it has a right angle [Yes, should have used polar, but started with this; so I stuck with it]

$A = z = x+yi$

$B = (1+i)z=x+ix+yi+yi^2=(x-y)+(x+y)i$

Switching to vector notation:

$A = xi+yj$

$B = (x-y)i+(x+y)j$

$A\cdot B = x(x-y)+y(x+y)=x^2-xy+yx+y^2 = x^2+y^2 = 36$ (see below)

Because $|A|=6, x^2+y^2=36$ ( $A=x+yi$ )

$|(1+i)z|=|1+i| |z| = \sqrt{2} \cdot 6$

Now using the angle formula $\theta = \arccos{\left(\frac{a\cdot b}{|a|\cdot |b|}\right)$ for two vectors, a, b.

$\theta = \arccos\left(\frac{36}{6\cdot \sqrt{2} \cdot 6}\right)= \arccos(\sqrt{2}^{-1}) = \frac{\pi}{4}$

Now, using cosine rule to determine the other side length (as we haven't shown a right angle yet):

$c^2=a^2+b^2-2ab\cos(\theta)$

$c^2 = 36 +2\times 36 - 2\times 6\times \sqrt{2} 6\cos\left(\frac{\pi}{4}\right)$

$c^2 = 3\times 36 - 2 \times 36 = 6$

Now, test for pythagorian realationship:

$6^2 + 6^2 = 2\times 36$

$(\sqrt{2}\cdot 6)^2=2\times 36$

Therefore a^2+b^2=c^2 <=> right angle

And since two sides are equal, you have a right angled isosceles triangle.
Title: Re: man0005's specialist question thread :)
Post by: xZero on March 27, 2011, 11:52:32 pm: Well I probably didn't explain it well enough, what I thought was I used z as the direction and if we extend or shrink that line it will eventually form a right-angled triangle with (1+i)z.

Then I use trig to work out the distance required from the point (1+i)z to the line z if the triangle were to be a right-angled. From this we can use Pythagoras theorem to work out the magnitude of the length in the direction of z, which is 6.

Hence z doesn't need to be shortened or extended to make a isosceles right-angled triangle

Edit: @OP use m@tty's way, it seems better than mine haha

Edit 2:to below, the length from (1+i)z to a line in the direction of z must be 6 to form a right-angled triangle with (1+i)z as hypotenuse. then use pythag, (6root2)^2 = 6^2 + c^2 and solve for c, which will give you 6. and since z has the magnitude of 6, then the line from (1+i)z will meet z and form a right-angled triangle

I know its a bit confusing :S
Title: Re: man0005's specialist question thread :)
Post by: m@tty on March 27, 2011, 11:57:46 pm: Our ways are exactly the same lol, just you use polar which is smarter as it's quicker/easier to compute...

Same basic logic though.

EDIT: Wait, sorry, so how did you show that the length is six, as is required for a right angled triangle?
Title: Re: man0005's specialist question thread :)
Post by: VCEMan94 on March 28, 2011, 06:46:55 am: any idea how to do this one?
Let w = cis theta and z = w + (1/w)
Show that z lies on ellipse with equation (x^2/25) + (y^2/9) = 1/4
Title: Re: man0005's specialist question thread :)
Post by: man0005 on April 09, 2011, 04:16:24 pm: can someone please help me with these
16b) and 17)
Title: Re: man0005's specialist question thread :)
Post by: xZero on April 09, 2011, 05:18:59 pm: 16b)

$AB=CB-CA$

$AB.AB=|AB|^2$ (If a vector dot products with itself, the result will be the square of its magnitude)

now sub $AB=CB-CA$ into the dot product

$(CB-CA).(CB-CA)=|AB|^2$

$(CB.CB)-2(CB.CA)+(CA.CA)=|AB|^2$

since $(CB.CA)=|CB||CA|cos(\theta)$

$|CB|^2+|CA|^2-|CB||CA|cos(\theta) = |AB|^2$

now sub in a,b and c

$a^2+b^2-2abcos(\theta)=c^2$

as required

17)
$DB=DO+OA+AB$ , since AB = OC, $DB=-OD+OA+OC$

$CE=CO+OA+AE$ , since AE = OD, $CE=-OC+OA+OD$ , $OC=OA+OD-CE$

$\frac{1}{2}DB=\frac{1}{2}OA+\frac{1}{2}OC-\frac{1}{2}OD$

$\frac{1}{2}DB=\frac{1}{2}OA+\frac{1}{2}(OD+OA-CE)-\frac{1}{2}OD$

$\frac{1}{2}DB=OA-\frac{1}{2}CE$

$OA=\frac{1}{2}DB+\frac{1}{2}CE$

If half of DB + half of CE = OA, half of DB must share a common point with half of CE, and that point is in the middle of DB and CE, hence the diagonals bisects each other.

as required

as for acute angle, just find CE and DB and use dot product to find the angle
Title: Re: man0005's specialist question thread :)
Post by: enpassant on April 09, 2011, 08:10:36 pm: Here is a simpler proof for Q17.

Let M and N be the midpoints of DB and CE respectively.

DB=OB-OD=OA+OC-OD
OM=OD+DM=OD+1/2 DB=1/2 (OA+OC+OD)

CE=OE-OC=OA+OD-OC
ON=OC+CN=OC+1/2 CE=1/2 (OA+OC+OD)

.: OM=ON

.: M and N are the same point and hence the two diagonals bisect each other.
Title: Re: man0005's specialist question thread :)
Post by: man0005 on April 10, 2011, 08:44:10 pm: what about this one ? :S
Prove that the diagonals of a parallelogram bisect each other

sorry im really bad at proofs ><
Title: Re: man0005's specialist question thread :)
Post by: enpassant on April 11, 2011, 08:13:36 am: Label parallelogram as OABC. Let OA=a, OC=c, and let M and N be the midpoints of OB and AC respectively.

OM=1/2 OB=1/2 (a+c)
ON=1/2 (a+c)

.: OM=ON

.: M and N are the same point and hence the two diagonals bisect each other.
Title: Re: man0005's specialist question thread :)
Post by: man0005 on April 11, 2011, 11:49:01 am: How would i prove the cosine rule for any triangle? :S
Title: Re: man0005's specialist question thread :)
Post by: enpassant on April 11, 2011, 01:30:56 pm: Label triangle as ABC. Let CA=a, CB=b, and AB=c.

c = b - a
c.c = (b - a).(b - a)
c.c = b.b + a.a - 2b.a
c2 = b2 +a2 - 2bacosC
Title: Re: man0005's specialist question thread :)
Post by: man0005 on April 11, 2011, 02:01:17 pm: why do you do c.c? what does it represent?
Title: Re: man0005's specialist question thread :)
Post by: vea on April 11, 2011, 02:20:30 pm: c.c=|c|^2 which is the magnitude/length of the triangle SQUARED
Title: Re: man0005's specialist question thread :)
Post by: man0005 on April 13, 2011, 09:37:19 pm: Will examiners test us on the semi circle theorem?
Title: Re: man0005's specialist question thread :)
Post by: man0005 on April 17, 2011, 11:19:18 pm: Let a = 3i - 6j + 4k
and b = 2i + j - 2k

c is the vector component of a perpendicular to b
d is the vector resolute of c in the direction of a

hence show that | a | | d | = | c | ^2

cant seem to get it :/
Title: Re: man0005's specialist question thread :)
Post by: evaever on April 18, 2011, 12:36:07 am: the result is true for any non-parallel a, b
Title: Re: man0005's specialist question thread :)
Post by: jane1234 on April 18, 2011, 12:17:29 pm: Hmm... there has to be an easier way than this:
But basically I just worked out the two required components (or resolutes) c and d and then found the magnitudes of a, c and d and found that |a|*|d| = |c|^2

c = a - (unit(b) . a)*(unit(b)) ---> this is just the formula for finding a perpendicular component in direction of b.

c = 3i - 6j + 4k - ((2/3i + 1/3j - 2/3k) . 3i - 6j + 4k)*(2/3i + 1/3j - 2/3k)
c = 3i - 6j + 4k - (-8/3)*(2/3i + 1/3j - 2/3k)
c = 43/9i - 46/9j + 20/9k

...|c| = (sqrt(485))/3
...|c|^2 = 485/9

d = (unit(a) . c)*(unit(a)) ---> formula for finding parallel component in direction of a

d = (3/sqrt(61)i - 6/sqrt(61)j + 4/sqrt(61)k . 43/9i - 46/9j + 20/9k)*(3/sqrt(61)i - 6/sqrt(61)j + 4/sqrt(61)k)
d = ((485*sqrt(61))/549)*3/sqrt(61)i - 6/sqrt(61)j + 4/sqrt(61)k
d = 485/183i - 970/183j + 1940/549k

... |d| = (485*sqrt(61))/549
and |a| = sqrt(61)

... |a|*|d| = 485/9
= |c|^2 (as required)
Title: Re: man0005's specialist question thread :)
Post by: man0005 on April 18, 2011, 01:03:25 pm: ah yeah thats right :D silly mistake by me :P thanks :D

i have another one ><

O

A D

OAD forms a triangle like so with B the midpoint of AD and C the midpoint of OD
OA - a
OB - b
OC - c

E is a point on OA produced such that OE = 4AE. If CB = k AE, find the value of k
i think the answer may be wrong..so if anyone could tell me what they get, it would be great :D
Title: Re: man0005's specialist question thread :)
Post by: brightsky on April 18, 2011, 01:54:35 pm: 1/2 OD + 1/2 DA = 1/2(OD+DA) = CB
OD + DA = OA, so 1/2 OA = CB
since OE = 4AE and OE - AE = OA
then 4AE - AE = 3AE = OA
so 1/2 (3AE) = CB
3/2 AE = CB
Title: Re: man0005's specialist question thread :)
Post by: man0005 on April 18, 2011, 07:47:36 pm: thanks! i got that too, but the answers said 2.5 :/

i have another question
how do you prove if something is "concurrent"? :S
what does it mean
Title: Re: man0005's specialist question thread :)
Post by: man0005 on April 19, 2011, 08:43:00 pm: How would you do 1c) ?
:S