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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Chavi on December 15, 2009, 02:44:16 pm

Title: Spec '10 - Help forum
Post by: Chavi on December 15, 2009, 02:44:16 pm: Could i get some help with:
Factorize over R: z^4 + 64
and Factorize over C z^4 + 64

Thanks
Title: Re: Spec '10 - Help forum
Post by: TrueTears on December 15, 2009, 03:15:05 pm: Over $\mathbb{R} \implies \left(z^2\right)^2 + \left(\left(2\sqrt{2}\right)^2\right)^2$

Let $u = z^2$ and $x = \left(2\sqrt{2}\right)^2$ and let $y = \left(z^2\right)^2 + \left(\left(2\sqrt{2}\right)^2\right)^2$

$y=u^2+x^2$

$y = u^2+2ux+x^2-2ux$

$y = (u+x)^2-2ux$

$y = (u+x)^2 - \left(\sqrt{2ux}\right)^2$

$y = \left(u+x-\sqrt{2ux}\right)\left(u+x+\sqrt{2ux}\right)$

$y = \left(z^2+\left(2\sqrt{2}\right)^2-4z\right)\left(z^2+\left(2\sqrt{2}\right)^2+4z\right)$

$y = \left(z^2-4z+8\right)\left(z^2+4z+8\right)$

$\therefore z^4+64 = \left(z^2-4z+8\right)\left(z^2+4z+8\right)$
Title: Re: Spec '10 - Help forum
Post by: Ahmad on December 15, 2009, 03:19:54 pm: This is a special case of Sophie Germain's Identity.
Title: Re: Spec '10 - Help forum
Post by: TrueTears on December 15, 2009, 03:24:24 pm: Quote from: Ahmad on December 15, 2009, 03:19:54 pm
This is a special case of Sophie Germain's Identity.
lol I did that AIME question when I saw it in the algebra section in AnC :P
Title: Re: Spec '10 - Help forum
Post by: QuantumJG on December 15, 2009, 03:38:09 pm: Quote from: TrueTears on December 15, 2009, 03:15:05 pm
Over $\mathbb{R} \implies \left(z^2\right)^2 + \left(\left(2\sqrt{2}\right)^2\right)^2$

Let $u = z^2$ and $x = \left(2\sqrt{2}\right)^2$ and let $y = \left(z^2\right)^2 + \left(\left(2\sqrt{2}\right)^2\right)^2$

$y=u^2+x^2$

$y = u^2+2ux+x^2-2ux$

$y = (u+x)^2-2ux$

$y = (u+x)^2 - \left(\sqrt{2ux}\right)^2$

$y = \left(u+x-\sqrt{2ux}\right)\left(u+x+\sqrt{2ux}\right)$

$y = \left(z^2+\left(2\sqrt{2}\right)^2-4z\right)\left(z^2+\left(2\sqrt{2}\right)^2+4z\right)$

$y = \left(z^2-4z+8\right)\left(z^2+4z+8\right)$

$\therefore z^4+64 = \left(z^2-4z+8\right)\left(z^2+4z+8\right)$

That is pretty cool!
Title: Re: Spec '10 - Help forum
Post by: Chavi on December 15, 2009, 09:13:08 pm: Quote from: TrueTears on December 15, 2009, 03:15:05 pm
$y = u^2+2ux+x^2-2ux$

Is that the only conventional way of factorising u^2 + v^2 over C?

Thanks
Title: Re: Spec '10 - Help forum
Post by: Mao on December 15, 2009, 09:23:15 pm: I would like to clarify:

Factorization over R:

$u^2 + v^2 = u^2 + 2uv + v^2 - 2uv = (u+v)^2 - 2uv = (u + v - \sqrt{2uv})(u + v + \sqrt{2uv})$

Factorization over C:

$u^2 + v^2 = u^2 - (vi)^2 = (u + vi)(u - vi)$

TT's method was purely over R. To factorize over C:

$\begin{align*} z^4 + 64 & = z^2 - (8i)^2 \\ & = (z^2 + 8i)(z^2 - 8i) \\ & = (z^2 - (i\sqrt{8i})^2)(z^2 - (\sqrt{8i})^2) \\ & = (z + i\sqrt{8i})(z - i\sqrt{8i})(z + \sqrt{8i})(z - \sqrt{8i}) \\ & = (z + 2\sqrt{2}i^{3/2})(z - 2\sqrt{2}i^{3/2})(z + 2\sqrt{2}i^{1/2})(z - 2\sqrt{2}i^{1/2}) \\ \end{align*} $

This can be further simplified with $i^{1/2} = \frac{\sqrt{2}}{2}(1+i)$ and $i^{3/2} = \frac{\sqrt{2}}{2}(-1+i)$

$\begin{align*} & = (z -2 + 2i)(z + 2 - 2i)(z + 2 + 2i)(z - 2 - 2i) \\ & = ((z - 2)+2i)((z-2)-2i)((z+2)+2i)((z+2)-2i)) \\ & = ((z-2)^2 -4i^2)((z+2)^2 - 4i^2) \\ & = (z^2 - 4z + 4 + 4)(z^2 + 4z + 4 + 4) \\ z^4 + 64 & = (z^2 - 4z + 8)(z^2 + 4z + 8)\\ \end{align*}$

Takes much more effort.
Title: Re: Spec '10 - Help forum
Post by: Chavi on December 15, 2009, 09:30:52 pm: Quote from: Mao on December 15, 2009, 09:23:15 pm

$\begin{align*} (z + 2\sqrt{2}i^{3/2})(z - 2\sqrt{2}i^{3/2})(z + 2\sqrt{2}i^{1/2})(z - 2\sqrt{2}i^{1/2}) \\ \end{align*}$

How do you take the sqrt of i, or other indicy variants such as i^(1/4)?

Thanks
Title: Re: Spec '10 - Help forum
Post by: Mao on December 15, 2009, 09:38:54 pm: Quote from: Chavi on December 15, 2009, 09:30:52 pm
Quote from: Mao on December 15, 2009, 09:23:15 pm

$\begin{align*} (z + 2\sqrt{2}i^{3/2})(z - 2\sqrt{2}i^{3/2})(z + 2\sqrt{2}i^{1/2})(z - 2\sqrt{2}i^{1/2}) \\ \end{align*}$

How do you take the sqrt of i, or other indicy variants such as i^(1/4)?

Thanks

De Moivre's Theorem (polar form) [Note, $\mbox{cis}(\theta)$ is equivalent to $e^{i\theta}$ ]

$\begin{align*} i & = e^{ i \pi / 2} \\ i^{1/2} & = e^{i \pi / 4} \\ & = \cos \left( \frac{\pi}{4} \right) + i \cdot \sin \left(\frac{\pi}{4} \right) \\ & = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \\ \end{align*}$

And so on.
Title: Re: Spec '10 - Help forum
Post by: Chavi on December 15, 2009, 09:42:27 pm: Quote from: Mao on December 15, 2009, 09:38:54 pm

De Moivre's Theorem (polar form) [Note, $\mbox{cis}(\theta)$ is equivalent to $e^{i\theta}$ ]

$\begin{align*} i & = e^{ i \pi / 2} \\ i^{1/2} & = e^{i \pi / 4} \\ & = \cos \left( \frac{\pi}{4} \right) + i \cdot \sin \left(\frac{\pi}{4} \right) \\ & = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \\ \end{align*}$

And so on.

Thanks - it makes sense now. but what is the purpose of the "e" notation?
Title: Re: Spec '10 - Help forum
Post by: Mao on December 16, 2009, 09:03:44 am: Exponential form is a lot more useful, since you can use rules that already apply to exponentials, as opposed to using the cis notation where you have to create a new set of rules, e.g.

$w = r \mbox{cis} (\theta)$

$w^n = (r \mbox{cis} (\theta))^n = r^n \mbox{cis}(n\theta)$

compared to

$w = r\cdot e^{i \theta}$

$w^n = \left( r \cdot e^{i \theta} \right)^n = r^n \cdot e^{i (n\theta)}$

The exponential form is used extensively in complex analysis.
Title: Re: Spec '10 - Help forum
Post by: Chavi on December 20, 2009, 12:57:24 pm: Could I please get some help with: $\int sin^3(x/2) \ dx$

Thanks
Title: Re: Spec '10 - Help forum
Post by: TrueTears on December 20, 2009, 01:02:37 pm: Quote from: Chavi on December 20, 2009, 12:57:24 pm
Could I please get some help with: $\int sin^3(x/2) \ dx$

Thanks

$\sin^3\left(\frac{x}{2}\right) = \sin^2\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right) = \left(1-\cos^2\left(\frac{x}{2}\right)\right)\sin\left(\frac{x}{2}\right)$

Let $u = \cos\left(\frac{x}{2}\right)$

$\implies \frac{du}{dx} = -\frac{1}{2}\left(\sin\left(\frac{x}{2}\right)\right)$

$\implies du = -\frac{1}{2}\left(\sin\left(\frac{x}{2}\right)\right) dx$

$\int \sin^3\left(\frac{x}{2}\right) dx = \int \left(1-\cos^2\left(\frac{x}{2}\right)\right)\sin\left(\frac{x}{2}\right) dx = -2\int \left(1-u^2\right)du$

Rest is trivial.
Title: Re: Spec '10 - Help forum
Post by: Chavi on January 03, 2010, 04:29:49 pm: How would I integrate the following expression?
$\int x(x^2+1){\sqrt{x^2-1}} dx$

And also out of interest, is there a fast way to insert math expressions into your messages?

Thanks
Title: Re: Spec '10 - Help forum
Post by: TrueTears on January 03, 2010, 04:33:05 pm: Try let $u = x^2-1$

So $u+1=x^2$

$\frac{du}{dx} = 2x$

Rest is trivial.
Title: Re: Spec '10 - Help forum
Post by: Chavi on January 04, 2010, 02:41:06 pm: Could i please get some help with
$\int sin^2(x) cos^3(x) dx$
Thanks
Title: Re: Spec '10 - Help forum
Post by: dekoyl on January 04, 2010, 02:48:05 pm: $\int sin^2(x)\left(1-sin^2(x)\right)cos(x) dx$

Let $u = sin(x)$

$du = cos(x) dx$

$\int u^2(1-u^2)du$

Etc.
Title: Re: Spec '10 - Help forum
Post by: Juddinator on January 24, 2010, 05:21:17 pm: In Essentials for Spec, an example which is: tan(2x - π) = square root of 3, for x is an element of [-π, π]. Let theta = 2x - π.

The next line states therefore -2π ≤ 2x – π and thus -3 π ≤ 2x – π ≤ π and -3 π ≤ theta ≤ π.

I don't understand how they have gone about this. Is it a worry if I cannot figure this out I could be in trouble for Spec this year? :S

Thanks
Title: Re: Spec '10 - Help forum
Post by: superflya on January 24, 2010, 05:36:23 pm: $\Theta =2x-\Pi$

$\Rightarrow$

ur original restrictions are $-\Pi\leq x\leq \Pi$

multiple all by 2 $-2\Pi\leq 2x\leq 2\Pi$

then subtract the $\Pi$ which was in the original restriction

$\Rightarrow$ $-3\Pi\leq 2x-\Pi\leq \Pi$

yea u shood be able to do this

Edit: thanks matt :)
Title: Re: Spec '10 - Help forum
Post by: googoo on January 24, 2010, 08:29:55 pm: Quote from: Chavi on December 15, 2009, 02:44:16 pm
Could i get some help with:
Factorize over R: z^4 + 64
and Factorize over C z^4 + 64

Thanks
By completing squares: z^4 + 64 = z^4 + 16z^2 + 64 - 16z^2
= (z^2 + 8)^2 - (4z)^2 = (z^2 - 4z + 8)(z^2 + 4z + 8) etc.
Title: Re: Spec '10 - Help forum
Post by: m@tty on January 24, 2010, 08:40:52 pm: Quote from: superflya on January 24, 2010, 05:36:23 pm
$\Theta =2x-\Pi$

$\Rightarrow$

ur original restrictions are $-\Pi\leq x\leq \Pi$

multiple all by 2 $-\Pi\leq x\leq \Pi$

then subtract the $\Pi$ which was in the original restriction

$\Rightarrow$ $-3\Pi\leq 2x-\Pi\leq \Pi$

yea u shood be ablt to do this
You missed multiplying by two in LaTeX, and yeah, that's how to do it.
Title: Re: Spec '10 - Help forum
Post by: mandy on January 24, 2010, 10:00:10 pm: SCALAR PRODUCT OF TWO VECTORS.

Can I get help with this question, please :)

OAB is an equilateral triangle with side length 1 unit. O is the origin, OA = i and OB = xi + yj.
a. Find the values of x and y, and hence express OB in terms of i and j.
Title: Re: Spec '10 - Help forum
Post by: Mao on January 24, 2010, 10:20:11 pm: Angle of AOB - 60 degrees
$\angle AOB = 60^o$

$\implies OA \cdot OB = |OA|\cdot |OB| \cos (60^o)$

$\implies x = \cos(60^o) = \frac{1}{2}$

$|OB| = 1 \implies x^2 + y^2 = 1^2 \implies y = \pm \frac{\sqrt{3}}{2}$
Title: Re: Spec '10 - Help forum
Post by: fady_22 on January 24, 2010, 10:23:58 pm: Quote from: Mao on January 24, 2010, 10:20:11 pm
Angle of AOB - 60 degrees
$\angle AOB = 60^o$

$\implies OA \cdot OB = |OA|\cdot |OB| \cos (60^o)$

$\implies x = \cos(60^o) = \frac{1}{2}$

$|OB| = 1 \implies x^2 + y^2 = 1^2 \implies y = \frac{\sqrt{3}}{2}$

Technically, couldn't it be both positive and negative $\frac{\sqrt{3}}{2}$ ?
Title: Re: Spec '10 - Help forum
Post by: kamil9876 on January 24, 2010, 10:35:46 pm: Yep you are right, with those conditions as specified in the question you can tell without even doing any working that there will be two answers since you can reflect the triangle in the y-axis and you get another triangle that satisfies those conditions.
Title: Re: Spec '10 - Help forum
Post by: mandy on January 24, 2010, 10:38:01 pm: Thank youuuuu, I get it now :)
Title: Re: Spec '10 - Help forum
Post by: mandy on January 25, 2010, 12:45:23 pm: If a = i + 3j - 2k and b = -2i + 4j -8k, resolve a into two components, one parallel to b and the other perpendicular to b.

Help please :)
Title: Re: Spec '10 - Help forum
Post by: fady_22 on January 25, 2010, 12:50:43 pm: Parallel to b, by definition: $\frac{a.b}{b.b}*b$
Perpendicular to b: Subtract the above vector from a.

Hope that helps!
Title: Re: Spec '10 - Help forum
Post by: cipherpol on January 25, 2010, 01:04:09 pm: Quote from: mandy on January 25, 2010, 12:45:23 pm
If a = i + 3j - 2k and b = -2i + 4j -8k, resolve a into two components, one parallel to b and the other perpendicular to b.

Help please :)

Let u be the component parallel to b:

$u=(a.\hat{b})\hat{b}$

$\hat{b}=\frac{1}{2\sqrt{21}}(-2i+4j-8k)$

$a.\hat{b}=\frac{1}{2\sqrt{21}}(i+3j-2k).(-2i+4j-8k)$

$\implies \frac{1}{2\sqrt{21}} (26) = \frac{13\sqrt{21}}{21}$

$u=\frac{13\sqrt{21}}{21}*\frac{1}{2\sqrt{21}}(-2i+4j-8k)$

$\implies u=\frac {13}{42}(-2i+4j-8k)$

Can you tell me if this is the correct answer :S

EDIT: looking over it, fady's way is much faster, use that instead :P
Title: Re: Spec '10 - Help forum
Post by: mandy on January 25, 2010, 01:13:54 pm: When I did it, I got the same answer as you, but its not right :(

Quote from: cipherpol on January 25, 2010, 01:04:09 pm
Quote from: mandy on January 25, 2010, 12:45:23 pm
If a = i + 3j - 2k and b = -2i + 4j -8k, resolve a into two components, one parallel to b and the other perpendicular to b.

Help please :)

$\implies u=\frac {13}{42}(-2i+4j-8k)$

The answer to the question is $\implies u=\frac {13}{21}(-i+2j-4k) + \frac {1}{21}(34i+37j+10k)$
Title: Re: Spec '10 - Help forum
Post by: cipherpol on January 25, 2010, 01:17:42 pm: Oh, I only calculated the parallel component; the answer also includes the perpendicular component.

They factored out a 2 from the parallel component to get their answer for the first part.
Title: Re: Spec '10 - Help forum
Post by: mandy on January 25, 2010, 01:52:22 pm: Oh I got it, thank you fady_22 and cipherpol :)
Title: Re: Spec '10 - Help forum
Post by: mandy on January 27, 2010, 03:03:01 pm: The points A, B and C have position vectors -i+4j+2k, i-2j-3k, and i-j-2k respectively. Find:
a. the lengths of AB, BC and AC
|AB| = $\sqrt65$
|BC| = $\sqrt2$
|AC| = $3\sqrt5$

b. the resolved part of AB in the direction of i-j-2k

I don't know how to do part b of this question, I don't even know what it's asking for. Can someone help me :)
Title: Re: Spec '10 - Help forum
Post by: cipherpol on January 27, 2010, 03:11:50 pm: I think they want you to find the component of AB parallel to i-j-2k.

Not sure though :S
Title: Re: Spec '10 - Help forum
Post by: TrueTears on January 27, 2010, 03:14:42 pm: Yeah cipherpol is spot on.
Title: Re: Spec '10 - Help forum
Post by: mandy on January 27, 2010, 03:20:37 pm: Quote from: cipherpol on January 27, 2010, 03:11:50 pm
I think they want you to find the component of AB parallel to i-j-2k.

Not sure though :S

Oh no, I've forgotten what 'the component of AB parallel to i-j-2k' means. What am I meant to do here.
Title: Re: Spec '10 - Help forum
Post by: hyperblade01 on January 27, 2010, 03:22:52 pm: Use the formula $(a.\hat{b})\hat{b}$

Where a = AB and b = i - j - 2k

EDIT: AB can be broken up in perpendicular and parallel components. Think of a right angled triangle were the hypotenuse is AB and the other two sides are the components - you still end up in the same place you just go a different direction.
Title: Re: Spec '10 - Help forum
Post by: Chavi on January 27, 2010, 03:27:22 pm: Can I get some help with the following question (specifically the restriction placed on t)

Find the Cartesian equation.
r = (4cost)i + (5sint)j , t>0

Also, is there a quick way to use mathematical symbols in this forum?
And could somebody explain the concept behind position vectors.

Thanks
Title: Re: Spec '10 - Help forum
Post by: mandy on January 27, 2010, 03:28:18 pm: Thank you hyperblade01, TrueTears and cipherpol :D
Title: Re: Spec '10 - Help forum
Post by: TrueTears on January 27, 2010, 03:29:55 pm: Quote from: Chavi on January 27, 2010, 03:27:22 pm
Can I get some help with the following question (specifically the restriction placed on t)

Find the Cartesian equation.
r = (4cost)i + (5sint)j , t>0

Also, is there a quick way to use mathematical symbols in this forum?
And could somebody explain the concept behind position vectors.

Thanks

Let $x = 4\cos(t)$ and $y = 5\sin(t)$

$\frac{x^2}{16} + \frac{y^2}{25} = 1$

http://vcenotes.com/forum/index.php/topic,10280.0.html

Position vectors are fixed, the line segment must start from the origin. They are NOT free vectors.
Title: Re: Spec '10 - Help forum
Post by: Chavi on January 27, 2010, 03:39:40 pm: Quote from: TrueTears on January 27, 2010, 03:29:55 pm

Let $x = 4\cos(t)$ and $y = 5\sin(t)$

$\frac{x^2}{16} + \frac{y^2}{25} = 1$

How about the restriction on t in relation to x
Title: Re: Spec '10 - Help forum
Post by: TrueTears on January 27, 2010, 03:41:40 pm: domain of cartesian = range of x(t)

range of cartesian = range of y(t)
Title: Re: Spec '10 - Help forum
Post by: Chavi on January 27, 2010, 03:43:52 pm: Quote from: TrueTears on January 27, 2010, 03:41:40 pm
domain of cartesian = range of x(t)

range of cartesian = range of y(t)

i.e. |x|<4 ?
Title: Re: Spec '10 - Help forum
Post by: hyperblade01 on January 27, 2010, 03:47:50 pm: $|x| \le 4$
Title: Re: Spec '10 - Help forum
Post by: QuantumJG on January 27, 2010, 04:49:14 pm: Quote from: mandy on January 27, 2010, 03:03:01 pm
The points A, B and C have position vectors -i+4j+2k, i-2j-3k, and i-j-2k respectively. Find:
a. the lengths of AB, BC and AC
|AB| = $\sqrt65$
|BC| = $\sqrt2$
|AC| = $3\sqrt5$

b. the resolved part of AB in the direction of i-j-2k

I don't know how to do part b of this question, I don't even know what it's asking for. Can someone help me :)

let d = i-j-2k

proj(AB)_d = $\dfrac{ AB . d }{ d^{2} } d$

I don't like the term 'resolute', what you are doing is projecting one vector onto another (look at this - http://nixweb.com/critters/figs/vproj.gif). Think of w as AB and v as d, this means that vector u is proj(AB)_d.

Anyway,

proj(AB)_d = 3i-3j-6k
Title: Re: Spec '10 - Help forum
Post by: QuantumJG on January 27, 2010, 05:07:50 pm: Quote from: hyperblade01 on January 27, 2010, 03:47:50 pm
$|x| \le 4$

and $|y| \le 5$
Title: Re: Spec '10 - Help forum
Post by: Chavi on January 27, 2010, 10:41:00 pm: How would I do the following?

Find the cartesian equation [and the restriction on domain + range] of the path of a particle whose position vector at any time t is given by
$r=\frac{1-t^2}{1+t^2}\mathbf{i} + \frac{2t}{1+t^2}\mathbf{j} , t\le 0$

Many thanks
Title: Re: Spec '10 - Help forum
Post by: kamil9876 on January 27, 2010, 11:33:17 pm: $x(1+t^2)=1-t^2$

$1-x=(x+1)t^2$

$t^2=\frac{1-x}{x+1}$ if $x\neq -1$ (it never does anyway as you can check by plugging in x=-1 into the second equation)

Now find you have $t$ in terms of $x$ so plug into the equation for $y$ to get it in terms of $x$
Title: Re: Spec '10 - Help forum
Post by: Chavi on January 28, 2010, 03:17:22 pm: Can I get help with some vectors:

2 ships are observed from station O. Ship A at 1pm and ship B at 2pm. Relative to O the respective position + velocity vectors of the ships are given by:

A: r= 2i + 6j and v= 5i + 4j
B: r=11i + 6j and v= 2i + 7j

time is measured in hours. Prove that if the ships maintain their velocities they will collide, and find the time they collide.

Thanks
Title: Re: Spec '10 - Help forum
Post by: hyperblade01 on January 28, 2010, 08:27:28 pm: Bit rusty...

Alright I'm assuming those position vectors are at 1pm and 2pm for A and B respectively

Antidiff the velocity vectors:

A r(t) = 5ti + 4tj + c
B r(t) = 2ti + 7tj + c

Let t be the time past 1pm...

A r(0) = 5(0)i + 4(0)j + c
2i+6j = c

$\therefore$ r(t) = (5t+2)i + (4t+6)j

Do the same for B but this time sub in t=1 to get r(t) = (2t+9)i + (7t-1)j

Now when it says they collide, their positions must be the same, so let the i components equal and the j components equal

5t+2 = 2t+9 4t+6=7t-1

Solve both equations and you should get t = 7/3 for both, which means they will indeed be in the same position at a particular time.

So they hit 7/3hrs after 1pm..which is 3:20pm
Title: Re: Spec '10 - Help forum
Post by: Chavi on January 30, 2010, 01:44:05 pm: The adjacent sides of a parallelogram are 9cm and 11cm. One of its angles is 67º. Find the length of the longer diagonal, correct to two decimal places.
I don't understand how/what they want you to find, if all the opposite angles/sides of a parallelogram are equal.
Title: Re: Spec '10 - Help forum
Post by: fady_22 on January 30, 2010, 02:44:20 pm: Find the length of the DIAGONALS: i.e. the lines drawn from opposite vertices to make the parallelogram two triangles.

You know that there are two angles of 67 (opposite angles) and two angles of 113 (again, opposite angles) so to find the length of the diagonals, use the cos rule:
$c^{2}=a^{2}+b^{2}-2bc*cos(c)$
Sub in the known values:
$c^{2}=9^{2}+11^{2}-2*11*9*cos(67)$
$c=11.16 cm$
Do this again, but use 113 as the angle:
$c=16.71cm$
Therefore, the longer diagonal is 16.71 cm long.
Title: Re: Spec '10 - Help forum
Post by: Juddinator on January 30, 2010, 06:44:15 pm: Sketch the graph of: x^2 +y^2 +3x-4y=6

I can get the equation which is (x+3/2)^2 + (y-2)^2 = 49/4 but I cannot get the x and y intercepts even when I let both equal 0 :|
Title: Re: Spec '10 - Help forum
Post by: superflya on January 31, 2010, 04:35:01 pm: Quote from: Juddinator on January 30, 2010, 06:44:15 pm
Sketch the graph of: x^2 +y^2 +3x-4y=6

I can get the equation which is (x+3/2)^2 + (y-2)^2 = 49/4 but I cannot get the x and y intercepts even when I let both equal 0 :|

there are no axial intercepts.
Title: Re: Spec '10 - Help forum
Post by: Juddinator on January 31, 2010, 04:39:29 pm: Quote from: superflya on January 31, 2010, 04:35:01 pm
Quote from: Juddinator on January 30, 2010, 06:44:15 pm
Sketch the graph of: x^2 +y^2 +3x-4y=6

I can get the equation which is (x+3/2)^2 + (y-2)^2 = 49/4 but I cannot get the x and y intercepts even when I let both equal 0 :|
there are no axial intercepts.
ahh touché... thanks man :)
Title: Re: Spec '10 - Help forum
Post by: superflya on January 31, 2010, 04:41:32 pm: no prob, whenever u cant solve for y or x after letting one of the variables equal 0, usually means there are no intercepts.
Title: Re: Spec '10 - Help forum
Post by: brightsky on January 31, 2010, 04:42:59 pm: Quote from: superflya on January 31, 2010, 04:35:01 pm

there are no axial intercepts.

Not true. Assuming that I'm interpreting stuff correctly.

Let's work with $x^2 + y^2 + 3x - 4y = 6$ to make matters simpler.

To find y-intercepts, substitute $x = 0$ .

$0^2 + y^2 + 3(0) - 4y = 6$

$y^2 - 4y - 6 = 0$

By the quadratic formula:

$y = \frac{4 + \sqrt{16 - (-24)}}{2}$ or $y = \frac{4 - \sqrt{16 - (-24)}}{2}$

So your y-intercepts are:

$(0, 2 + \sqrt{10})$ and $(0, 2- \sqrt{10})$

Do that for the x-intercepts, and you'll find that the x-intercepts are $\left(-\frac{3+\sqrt{33}}{2}, 0\right)$ and $\left(\frac{\sqrt{33} - 3}{2}, 0 \right)$
Title: Re: Spec '10 - Help forum
Post by: superflya on January 31, 2010, 04:46:20 pm: Quote from: brightsky on January 31, 2010, 04:42:59 pm
Quote from: superflya on January 31, 2010, 04:35:01 pm

there are no axial intercepts.

Not true. Assuming that I'm interpreting stuff correctly.

Let's work with $x^2 + y^2 + 3x - 4y = 6$ to make matters simpler.

To find y-intercepts, substitute $x = 0$ .

$0^2 + y^2 + 3(0) - 4y = 6$

$y^2 - 4y - 6 = 0$

By the quadratic formula:

$y = \frac{4 + \sqrt{16 - (-24)}}{2}$ or $y = \frac{4 - \sqrt{16 - (-24)}}{2}$

So your y-intercepts are:

$(0, 2 + \sqrt{10})$ and $(0, 2- \sqrt{10})$

Do that for the x-intercepts.

LOL i didnt work it out -.- tis correct brightsky :)
Title: Re: Spec '10 - Help forum
Post by: superflya on January 31, 2010, 04:50:41 pm: for the x ints

$\Rightarrow x^{2}+3x-6=0$

$\frac{-3\pm \sqrt{9+24}}{2}$

so your 2 x ints are $\frac{-3+\sqrt{33}}{2}$ & $\frac{-3-\sqrt{33}}{2}$
Title: Re: Spec '10 - Help forum
Post by: Juddinator on January 31, 2010, 05:44:24 pm: Sweet thanks superfly and brightsky. It makes so much more sense to use the quadratic formula when you have your a, b and c. Cheers (Y)
Title: Re: Spec '10 - Help forum
Post by: Chavi on February 06, 2010, 10:20:52 pm: Could I please get some help with a few questions in regards to complex numbers/graph sketching?

1. Find the solutions to the equation $z^4 - 2z^2 + 4 = 0$ in polar form. (can this be done *without* solving algebraically first?)

2. Find the square roots of $1 + i$ by cartesian methods and hence find exact values for $cos(\pi/8)$ and $sin(\pi/8)$

3. If, with an Argand diagram with origin O, the point P represent $z$ and Q represents $\frac{1}{\bar{z}}$ , prove that O,P and Q are collinear and find the ratio of OP : OQ in terms of $|z|$ .

4. Find the coordinates of the point of intersection of the following loci: $4x^2 + 4y^2 - 60x - 76y + 536 = 0$ and $x^2 + y^2 - 10x - 14y + 49 = 0$

Thanks in advance.
Title: Re: Spec '10 - Help forum
Post by: TrueTears on February 06, 2010, 10:23:53 pm: 1. Let u =z^2

2. z^2 = 1+i, solve using de moivers theorem and then change into cartesian form, rest is trivial.

3. let z = x+yi and z conjugate = x-yi, sub it in, simplify and u get an obvious result.
Title: Re: Spec '10 - Help forum
Post by: Chavi on February 06, 2010, 11:16:36 pm: 1. is there a way to do this directly into polar form, without solving for z, and then converting each answer?

2. the question asked to solve without using de moivers theorem - how would this be done algebraically?

3. I tried that - z = x + iy, and 1/z conju = (x + iy)/(x^2 + y^2) - but how do i prove they're collinear, and get the ratio?
Title: Re: Spec '10 - Help forum
Post by: TrueTears on February 06, 2010, 11:20:19 pm: 1. yes sub in u = z^2, factorise then use de moivres.

2. z^2-1-i = 0

quadratic formula?

3. z = x+yi 1/z = z/(x^2+y^2)

so the ratio is 1/(x^2+y^2)
Title: Re: Spec '10 - Help forum
Post by: Mao on February 06, 2010, 11:21:24 pm: Quote from: Chavi on February 06, 2010, 11:16:36 pm
1. is there a way to do this directly into polar form, without solving for z, and then converting each answer?

2. the question asked to solve without using de moivers theorem - how would this be done algebraically?

3. I tried that - z = x + iy, and 1/z conju = (x + iy)/(x^2 + y^2) - but how do i prove they're collinear, and get the ratio?

1. no, that would get extremely over-complicated. You will need to perform the substitution $u = z^2$

2. let $z=x+yi,\; \implies z^2 =(x+yi)(x+yi) = (x^2-y^2) + 2xyi = 1+i$ , now you have a set of simultaneous equations $x^2 - y^2 = 1$ and $2xy = 1$

3. colinear: $\vec{OP} = k \vec{OQ}$ , where k is some constant. In this case, $k = \frac{OP}{OQ} = \frac{x + yi}{(x+yi)/(x^2 + y^2)}=x^2 + y^2 = |z|^2$ , which is a constant. The ratio OP:OQ is $1:\frac{1}{k} \implies 1:\frac{1}{|z|^2}$

@TT: using the quadratic formula would land him back in pretty much the same spot: $z = \frac{\pm 2\sqrt{1 + i}}{2}$
Title: Re: Spec '10 - Help forum
Post by: TrueTears on February 06, 2010, 11:26:02 pm: expand $\sqrt{1+i}$
Title: Re: Spec '10 - Help forum
Post by: Mao on February 06, 2010, 11:28:38 pm: Quote from: TrueTears on February 06, 2010, 11:26:02 pm
expand $\sqrt{1+i}$

I doubt many specialist students know how to do that, considering it's not really covered in the course.

[Assuming you are referring to $\sqrt{x+yi} = \sqrt{\frac{r+x}{2}} + \frac{y}{\sqrt{2(r+x)}}i$ ]
Title: Re: Spec '10 - Help forum
Post by: TrueTears on February 06, 2010, 11:29:23 pm: tis covered in methods, binomial theorem
Title: Re: Spec '10 - Help forum
Post by: kamil9876 on February 06, 2010, 11:30:10 pm: That binomial theorem is A LOT different.
Title: Re: Spec '10 - Help forum
Post by: TrueTears on February 06, 2010, 11:30:53 pm: same principle, just extend it.
Title: Re: Spec '10 - Help forum
Post by: kamil9876 on February 06, 2010, 11:32:08 pm: oh right, so you mean that you have a proof of the generalized binomial using only the "same principle".
Title: Re: Spec '10 - Help forum
Post by: TrueTears on February 06, 2010, 11:33:07 pm: it is just extended to C

maybe u can enlighten me with the proof, but i am happy to just apply it for now, will look at the proof once i know how to utilise it.
Title: Re: Spec '10 - Help forum
Post by: kamil9876 on February 06, 2010, 11:34:18 pm: the word "just" doesn't do the difficulty of that any justice.

edit: how about Fermat's last theorem, can I "just" extend that so that x,y,z are any real numbers.
Title: Re: Spec '10 - Help forum
Post by: TrueTears on February 06, 2010, 11:34:56 pm: sometimes it is good to prove after you apply. after u get the confidence of seeing how it works in action
rmb?

edit: i am talking about applying it, u said urself u didnt even know why it applies over C.
Title: Re: Spec '10 - Help forum
Post by: Mao on February 06, 2010, 11:36:33 pm: Quote from: TrueTears on February 06, 2010, 11:33:07 pm
it is just extended to C

maybe u can enlighten me with the proof, but i am happy to just apply it for now, will look at the proof once i know how to utilise it.

Care to show us how it is applied? The only binomial theorem generalised over C that I know of involves an infinite series, which will give the right answer in a very round-about way.
Title: Re: Spec '10 - Help forum
Post by: TrueTears on February 07, 2010, 12:00:29 am: (1+i)^(1/2) = 1 + 1/2(i) -1/8 i^2 + 1/16i^3 -5/128i^4 +7/256i^5 - 21/1024i^6 +33/2048i^7... = 1+1/2(i) +1/8 -1/16(i) -5/128 + 7/256(i)+ 21/1024 - 33/2048(i)...

the non i terms form a series with sum rt[2(rt[2]+1)]/2

the terms with i form a series with sum rt[2(rt[2]-1)]/2(i)

Since both are converging series, their sum is also converging rt[2(rt[2]+1)]/2+rt[2(rt[2]-1)]/2(i) = (1+i)^(1/2)
Title: Re: Spec '10 - Help forum
Post by: Mao on February 07, 2010, 02:17:06 am: Quote from: TrueTears on February 07, 2010, 12:00:29 am
(1+i)^(1/2) = 1 + 1/2(i) -1/8 i^2 + 1/16i^3 -5/128i^4 +7/256i^5 - 21/1024i^6 +33/2048i^7... = 1+1/2(i) +1/8 -1/16(i) -5/128 + 7/256(i)+ 21/1024 - 33/2048(i)...

the non i terms form a series with sum rt[2(rt[2]+1)]/2

the terms with i form a series with sum rt[2(rt[2]-1)]/2(i)

Since both are converging series, their sum is also converging rt[2(rt[2]+1)]/2+rt[2(rt[2]-1)]/2(i) = (1+i)^(1/2)

A part of me wants to slap you. This has been in no way helpful to students.
Title: Re: Spec '10 - Help forum
Post by: /0 on February 07, 2010, 02:41:49 am: Quote from: TrueTears on February 06, 2010, 11:33:07 pm
it is just extended to C

maybe u can enlighten me with the proof, but i am happy to just apply it for now, will look at the proof once i know how to utilise it.

spoken like a true physicist :P

(at least that's what some people in /math/ keep telling me)
Title: Re: Spec '10 - Help forum
Post by: TrueTears on February 07, 2010, 11:36:11 am: doesn't hurt to learn something new these days
Title: Re: Spec '10 - Help forum
Post by: Chavi on February 07, 2010, 05:17:43 pm: How would i figure out the intersection of two circles?
4. Find the coordinates of the point of intersection of the following loci: $4x^2 + 4y^2 - 60x - 76y + 536 = 0$ and $x^2 + y^2 - 10x - 14y + 49 = 0$

Thanks
Title: Re: Spec '10 - Help forum
Post by: brightsky on February 07, 2010, 06:06:30 pm: $4x^2 + 4y^2 - 60x - 76y + 536 = 0$ (1)

$x^2 + y^2 - 10x - 14y + 49 = 0$ (2)

From (1),

$4(x^2 + y^2 - 15x - 19y + 134) = 0$

$x^2 + y^2 - 15x - 19y + 134 = 0$ (3)

From (2),

$x^2 - 10x + y^2 - 14y + 49 = 0$

$10x - x^2 = y^2 - 14y + 49$

$10x - x^2 = (y-7)^2$

$y - 7 = \pm \sqrt{10x - x^2}$

$y = 7 \pm \sqrt{10x - x^2}$ (4)

Substitute (4) into (3):

$x^2 + (7 \pm \sqrt{10x - x^2}^2 - 15x - 19(7 \pm \sqrt{10x - x^2}) + 134 = 0$

$x^2 + 49 \pm 14\sqrt{10x - x^2} + (10x - x^2) - 15x - 133 \pm -19\sqrt{10x - x^2} +134 = 0$

When $y = 7 + \sqrt{10x - x^2}$

$-5 \cdot \sqrt{10x - x^2} - 5x + 50 = 0$

$-5(\sqrt{10x - x^2} + x - 10) = 0$

$\sqrt{10x - x^2} + x - 10 = 0$

$\sqrt{10x - x^2} = 10 -x$

$10x - x^2 = (10 - x)^2 = 100 - 20x + x^2$

$100 - 30x + 2x^2 = 0$

$x^2 - 15x + 50 = 0$

Using quadratic formula,

$x = 5$ or $x = 10$

Do the same for when $y = 7 - \sqrt{10x - x^2}$
Title: Re: Spec '10 - Help forum
Post by: the.watchman on February 07, 2010, 06:52:10 pm: Wouldn't it have been a whole lot easier to just take eqns 2 and 3 and let them equal one another

Then you would have gotten $x=17-y$ , you can sub this into the originals then.

Or have I made a mistake?
Title: Re: Spec '10 - Help forum
Post by: brightsky on February 07, 2010, 06:55:24 pm: Quote from: the.watchman on February 07, 2010, 06:52:10 pm
Wouldn't it have been a whole lot easier to just take eqns 2 and 3 and let them equal one another

Then you would have gotten $x=17-y$ , you can sub this into the originals then.

Or have I made a mistake?

Yeah, I think you have. That would be incorrect logic as by doing that you've solved for both equations, and substituting that into both equations, you would get a weird result like $17 = 17$ . Haven't tried it yet though, so I may be wrong..
Title: Re: Spec '10 - Help forum
Post by: Mao on February 07, 2010, 07:11:00 pm: Quote from: brightsky on February 07, 2010, 06:55:24 pm
Quote from: the.watchman on February 07, 2010, 06:52:10 pm
Wouldn't it have been a whole lot easier to just take eqns 2 and 3 and let them equal one another

Then you would have gotten $x=17-y$ , you can sub this into the originals then.

Or have I made a mistake?

Yeah, I think you have. That would be incorrect logic as by doing that you've solved for both equations, and substituting that into both equations, you would get a weird result like $17 = 17$ . Haven't tried it yet though, so I may be wrong..

In this case, it works.

A trivial result would be obtained if a substitution is made in the equation it was derived from. However, this substitution is not derived independently from either equations, hence the substitution is valid.

However, you are correct in in being careful. If not done right, this method can give trivial results.
Title: Re: Spec '10 - Help forum
Post by: the.watchman on February 07, 2010, 07:17:42 pm: Whew! Lucky again! :D
Title: Re: Spec '10 - Help forum
Post by: Chavi on February 15, 2010, 09:19:31 pm: Can I please get some help with this question - according to the answer, the function randomly stops between [-1, 1]. Can somebody please explain? Thanks.

16a. I $\bar{z}$ denoted the complex conjugate of the number $z = x + iy$ ,
Sketch on an argand diagram: $\left \{ (1+i)z + (1-i)\bar{z} = -2, \arg(z) \leq \frac{\pi}{2} \right \}$
Title: Re: Spec '10 - Help forum
Post by: fady_22 on February 15, 2010, 09:58:05 pm: The function "stops" because in these values of re(z), the Arg(z)(with capital A, not lower case as you said it was-- this makes a big difference) is greater than pi/2.
Title: Re: Spec '10 - Help forum
Post by: luken93 on February 17, 2010, 08:48:03 pm: hello all, in yr 11 and just wondering if you could give us a hand with this q.
i know it will be easy for all you geniuses but i'm having trouble with them.

Solve each of the following pairs of simultaneous equations for x and y
(a + b)x + cy = bc
(b + c)y + ax = -ab

Thanks, and btw I'm not that great on latex hence the normal type.
Title: Re: Spec '10 - Help forum
Post by: m@tty on February 17, 2010, 09:20:24 pm: Those two would've just worked if you placed "tex" and "/tex", each enclosed in square brackets i.e. [...].

$\begin{align} (a+b)x+cy&=bc \ [1] \\ (b+c)y+ax&=-ab \ [2] \end{align}$

From [1]
$(a+b)x=bc-cy \implies x=\frac{c(b-y)}{a+b} \ [3]$

Sub [3] into [2]

$\begin{align} (b+c)y+a\cdot \frac{c(b-y)}{a+b}&=-ab(a+b) \\ (a+b)(b+c)y+abc-acy&=-ab(a+b) \\ y((a+b)(b+c)-ac)&=-(a^2b+ab^2+abc) \end{align}$

$\begin{align} y&=-\frac{ab(a+b+c)}{((a+b)(b+c)-ac)} \\ &=-\frac{ab(a+b+c)}{ab+b^2+bc+ac-ac} \\ &=-\frac{ab(a+b+c)}{b(a+b+c)} \\ \therefore y&= -a \end{align}$

Sub that in to [2]

$-a(b+c)+ax=-ab \implies x=\frac{-ab+ab+ac}{a}=c$

You should always check the solutions you found, by subbing both in to an equation.

$\begin{align} sub \ y=-a, \ x=c \ &into [1] \\ (a+b)c +c(-a)&=bc \\ ac+bc-ac&=bc \\ bc=bc& \end{align}$

Therefore your solutions are correct.

These questions are just a pain.
Title: Re: Spec '10 - Help forum
Post by: TrueTears on February 17, 2010, 09:22:17 pm: What you could do is, you can set up a matrix and use guass-jordon elimination =)
Title: Re: Spec '10 - Help forum
Post by: brightsky on February 17, 2010, 09:46:34 pm: Soz for hijacking this thread, but I've been musing over this question for quite some time, without any success.

Prove that $\frac{1}{m} + \frac{1}{m+1} + \frac{1}{m+2} + \frac{1}{m+3} + \frac{1}{m+4} + \cdots + \frac{1}{m+n}$ is not an integer, where $m,n \in \mathbb{N}$ .

Any clues?

Thanks. :D
Title: Re: Spec '10 - Help forum
Post by: luken93 on February 17, 2010, 10:50:49 pm: Quote from: TrueTears on February 17, 2010, 09:22:17 pm
What you could do is, you can set up a matrix and use guass-jordon elimination =)
What is that???

Quote from: m@tty on February 17, 2010, 09:20:24 pm
Those two would've just worked if you placed "tex" and "/tex", each enclosed in square brackets i.e. [...].
I'm gueesing you are talking about latex?

and finally, thanks but i worked it out after i posted it haha
Title: Re: Spec '10 - Help forum
Post by: the.watchman on February 17, 2010, 11:00:36 pm: Quote from: luken93 on February 17, 2010, 10:50:49 pm
Quote from: TrueTears on February 17, 2010, 09:22:17 pm
What you could do is, you can set up a matrix and use guass-jordon elimination =)
What is that???

It's the process of setting up a matrix equation, then using the row echelon form of the matrix, apply row operations etc.etc.
Read the wiki page on it :P
Title: Re: Spec '10 - Help forum
Post by: TrueTears on February 17, 2010, 11:19:30 pm: dw not in spesh, but it's just a systematic way of solving it.
Title: Re: Spec '10 - Help forum
Post by: TrueTears on February 17, 2010, 11:21:00 pm: Quote from: brightsky on February 17, 2010, 09:46:34 pm
Soz for hijacking this thread, but I've been musing over this question for quite some time, without any success.

Prove that $\frac{1}{m} + \frac{1}{m+1} + \frac{1}{m+2} + \frac{1}{m+3} + \frac{1}{m+4} + \cdots + \frac{1}{m+n}$ is not an integer, where $m,n \in \mathbb{N}$ .

Any clues?

Thanks. :D
What have you tried?

Try proof by contradiction. I remember doing a Q exactly like this before but with a tiny bit different series.

Also look at a specific case. When m = 1, you see its the harmonic series, does that tell you something?

What about when m = 2? that's just the harmonic series - 1

what about when m = 3?

Do you notice a pattern?
Title: Re: Spec '10 - Help forum
Post by: /0 on February 17, 2010, 11:21:18 pm: You could also use Cramer's rule or find the transpose cofactor coefficient matrix divided by the determinant of the coefficient matrix multiplied by the RHS matrix. both of which involve monstrous overkill and which would probably be less efficient than the method you have used
Title: Re: Spec '10 - Help forum
Post by: the.watchman on February 18, 2010, 06:29:14 am: Hey what about an induction proof?
They work nicely :D
(contradictions are great too)
Title: Re: Spec '10 - Help forum
Post by: TrueTears on February 18, 2010, 02:35:19 pm: there's actually no need to prove by contradiction or induction if you can see a pattern i fink , u just have to prove the harmonic series is never an integer, try comparison
Title: Re: Spec '10 - Help forum
Post by: the.watchman on February 18, 2010, 03:52:07 pm: Quote from: TrueTears on February 18, 2010, 02:35:19 pm
there's actually no need to prove by contradiction or induction if you can see a pattern i fink , u just have to prove the harmonic series is never an integer, try comparison

oh ok ;)
Title: Re: Spec '10 - Help forum
Post by: Chavi on February 25, 2010, 08:04:13 pm: Hey, can i please get some help with this tricky vector question?

A and B are defined by the position vectors: $a = 2i - 2j - k$ and $b = 3i + 4k$
Find the unit vector which bisects triangle AOB

Thanks
Title: Re: Spec '10 - Help forum
Post by: /0 on February 25, 2010, 09:16:04 pm: First make the vectors unit vectors:

$\mathbf{\hat{a}} = \frac{1}{\sqrt{5}} (2\mathbf{i} - 2\mathbf{j} - \mathbf{k})$

$\mathbf{\hat{b}} = \frac{1}{5}(3\mathbf{i}+4\mathbf{k})$

The bisecting vector is $\mathbf{\hat{a}}+\mathbf{\hat{b}}$ (verify this by drawing a diagram of the vector addition). The reason is that the diagonals of a rhombus bisect the corner angles.

Then make that vector unit length.
Title: Re: Spec '10 - Help forum
Post by: mandy on February 28, 2010, 04:20:26 pm: Find the square roots, in both polar and Cartesian form, of the following:

4 - 3 $i$ .

I'm not so sure how to get the angle bit, cos I keep getting it wrong. Can you help me?
Title: Re: Spec '10 - Help forum
Post by: brightsky on February 28, 2010, 08:39:25 pm: For the angle, just use trigonometry. Let the angle be $\theta$ , then $\tan(\theta) = \frac{3}{4}$ and work from there.
Title: Re: Spec '10 - Help forum
Post by: superflya on February 28, 2010, 08:42:57 pm: $\frac{-3}{4}$ ?
Title: Re: Spec '10 - Help forum
Post by: TrueTears on February 28, 2010, 08:43:59 pm: nah u can use 3/4 and then draw the unit circle and get the actual angle.

i hate just remembering tan theta = imaginary/real, because tan theta is not universal, you should really sketch the graph and work it out graphically.
Title: Re: Spec '10 - Help forum
Post by: superflya on February 28, 2010, 08:45:45 pm: oh okay :)
Title: Re: Spec '10 - Help forum
Post by: mandy on February 28, 2010, 09:21:05 pm: Alright, thanks guys.
Title: Re: Spec '10 - Help forum
Post by: Chavi on March 05, 2010, 06:17:40 pm: Can I please get some help with the following questions:

1. A circle with centre O and radius, r, is inscribed in a square ABCD. P is any point on the circumference of the circle.
a) Find $\overrightarrow{A P} . \overrightarrow{A P}$
b) Hence find $\mathbf{AP}^2 + \mathbf{BP}^2 + \mathbf{CP}^2 + \mathbf{DP}^2$ in terms of r.
Title: Re: Spec '10 - Help forum
Post by: Martoman on March 05, 2010, 11:55:10 pm: Quote from: Chavi on March 05, 2010, 06:17:40 pm
Can I please get some help with the following questions:

1. A circle with centre O and radius, r, is inscribed in a square ABCD. P is any point on the circumference of the circle.
a) Find $\overrightarrow{A P} . \overrightarrow{A P}$
b) Hence find $\mathbf{AP}^2 + \mathbf{BP}^2 + \mathbf{CP}^2 + \mathbf{DP}^2$ in terms of r.

This is a very_nice_question. This requires a visual proof. Go to this link to see my explanation: http://img169.imageshack.us/img169/5426/proofofsquarecircle.jpg

~~I'll leave you to do b) now you know where the first part comes from.~~

EDIT: what the hell, here is the second part, it actually involves a tiny bit of thought so I might as well give it to you.

Link: http://img256.imageshack.us/img256/9340/partb.jpg
Title: Re: Spec '10 - Help forum
Post by: mandy on March 07, 2010, 03:16:46 pm: Calculate the fourth roots of 100.

So I have :
$z^4 - 100 = 0$
and then I have no idea what to do.
Title: Re: Spec '10 - Help forum
Post by: TrueTears on March 07, 2010, 03:18:20 pm: z^4 = 100

Use de moives theorem

Let z = rcis theta

let 100 = 100cis(0)
Title: Re: Spec '10 - Help forum
Post by: mandy on March 07, 2010, 03:20:08 pm: Quote from: TrueTears on March 07, 2010, 03:18:20 pm
z^4 = 100

Use de moives theorem

Let z = rcis theta

let 100 = 100cis(0)

Yeah, is there another way to do it without using De Moivre's?
Title: Re: Spec '10 - Help forum
Post by: TrueTears on March 07, 2010, 03:21:45 pm: You can let z = x+yi

(x+yi)^4 = 100+0i

then equate coefficients

Or factorise

(z^2)^2 - (( $\sqrt{10}$ )^2)^2 use DOPS
Title: Re: Spec '10 - Help forum
Post by: mandy on March 07, 2010, 03:44:07 pm: Okay, so I used the factorising way, and this is what I got:

$z^4 = 100$
$z^4 - 100 = 0$
$z^4 - (\sqrt10 )^4 = 0$
$(z - \sqrt10)(z + \sqrt10)(z + \sqrt10)(z - \sqrt10) = 0$
$(z - \sqrt10)(z - \sqrt10)(z - \sqrt10i^2)(z - \sqrt10i^2) = 0$

So then I get $z = \sqrt10, z = -\sqrt10 i$

The answer is meant to be: $z = \pm\sqrt10, z = \pm\sqrt10 i$ .
What am I doing wrong here?
Title: Re: Spec '10 - Help forum
Post by: cipherpol on March 07, 2010, 03:58:46 pm: $z^4-100=0$

$(z^2)^2-(10)^2=0$

$(z^2-10)(z^2+10)=0$

$(z+\sqrt{10})(z-\sqrt{10})(z+\sqrt{10}i)(z-\sqrt{10}i)=0$

$\therefore z = \pm\sqrt10, z = \pm\sqrt10 i$
Title: Re: Spec '10 - Help forum
Post by: mandy on March 07, 2010, 04:02:52 pm: Thanks a lot guys :]
Title: Re: Spec '10 - Help forum
Post by: TrueTears on March 07, 2010, 04:07:46 pm: You're welcome xD