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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: dejan91 on February 05, 2009, 09:33:49 pm

Title: dejan91's questions
Post by: dejan91 on February 05, 2009, 09:33:49 pm: OK well the question:
The sides of triangle ABC are represented by the vectors A>B = a, B>C = b, and C>A = c, such that a + b = -c (sorry i have no idea how to type in vectors, so A>B is AB with arrow an at the top like this →).

Prove the cosine rule, |c|² = |a|² + |b|² - 2|a||b|cos B.

Any help would be much appreciated.
Title: Re: Vector Proofs
Post by: Mao on February 06, 2009, 12:26:13 am: let a and b be two position vectors from O. Let $c=a-b$ , thus forming a triangle.

$a-b=c$

$(a-b)\cdot (a-b) = c\cdot c$

$a\cdot a + b\cdot b - 2 (a\cdot b) = c\cdot c$

$|a|^2 + |b|^2 - 2|a||b|\cos (B) = |c|^2$

QED
Title: Re: Vector Proofs
Post by: BlueYoHo on February 06, 2009, 03:53:39 pm: oooooooh yeah now i see. wow... what a coinidence. I'd just gotten stuck on this question and decided to take a break/look here.

so a.a for example is the same as |a|^2 ?
Title: Re: Vector Proofs
Post by: Mao on February 06, 2009, 04:38:56 pm: $a\cdot a = |a| \cdot |a| \cdot \cos (0) = |a|^2$
Title: Re: Vector Proofs
Post by: monokekie on February 06, 2009, 06:37:02 pm: haha, i am stuck on this questions too..

thanks mao lol
Title: Re: Vector Proofs
Post by: dejan91 on February 06, 2009, 11:21:35 pm: Thanks for help man. However, I asked my teacher today, and this is what he said to do:

c.c = |c|²
=a.a + a.b + a.b + b.b
=a.a + b.b + 2a.b
=a.a + b.b + 2|a||b|cos B

Now, find the positive supplementary angle.
|c|² = a.a + b.b + 2|a||b|cos(180 - B)

Since 180 - B is in second quadrant, cos(180 - B) will be negative. Therefore,
|c|² = a.a + b.b - 2|a||b|cosB
= |a|² + |b|² - 2|a||b|cosB.

Of course, correct me if I'm wrong :P
Title: Re: Vector Proofs
Post by: Mao on February 06, 2009, 11:30:19 pm: ahhh, silly me. you are right. I completely forgot to take into account the direction of a and b.

However, changing angles can get messy. It is easier to define a and b to be pointing outwards from, hence c=a-b. I have edited my proof above.
Title: Re: Vector Proofs
Post by: dejan91 on February 08, 2009, 09:41:36 pm: Thanks Mao fo help anyway (Y)

Also, this isn't on vector proofs, but cbf starting new thread. Had this question that I just left ages ago till now..

If a = 2i - 3j + k and b = -2i + 3j + k,
find a unit vector perpendicular to both a and b.
Title: Re: Vector Proofs
Post by: /0 on February 08, 2009, 10:23:23 pm: Let the vector perpendicular to both $\mbox{a}$ and $\mbox{b}$ be $\mbox{c} = x\mbox{i}+y\mbox{j}+z\mbox{k}$

Then, by the dot product:

$2x-3y+z=0$

$-2x+3y+z=0$

Adding the two equations, we find that $z=0$ and so $x=\frac{3}{2}y$ .

So a possible vector could be $\mbox{c} = \frac{3}{2}\mbox{i}+\mbox{j}$

And the unit vector would be $\hat{\mbox{c}}=\frac{2\sqrt{13}}{13}\left(\frac{3}{2}\mbox{i}+\mbox{j}\right)$
Title: Re: Vector Proofs
Post by: dejan91 on February 18, 2009, 10:19:19 pm: /0 thanks, however your answer is slightly off the books answer. I don't have the book with me now, but I will post it up later.

Also, this isn't a vector proof, and may not even be in the specialist maths course, but I am interested in knowing wether this actually works or not (note this isn't my work, and I had no idea where else to post this):

Prove that if x = 0.9999...
x = 1.

10x = 9.9999...

10x - x = 9x = 9

Therefore, x = 1,
and x = 0.9999...

Is this correct??
Title: Re: Vector Proofs
Post by: /0 on February 18, 2009, 10:26:36 pm: I don't know how valid that proof is, but I prefer using geometric series:

0.999... = 0.9 + 0.09 + 0.009 + ...

a = 0.9, r = 0.1

$S_{\infty} = \frac{a}{1-r}=\frac{0.9}{1-0.1} = \frac{0.9}{0.9}=1$
Title: Re: Vector Proofs
Post by: TrueTears on February 18, 2009, 10:52:55 pm: Quote from: dejan91 on February 18, 2009, 10:19:19 pm
/0 thanks, however your answer is slightly off the books answer. I don't have the book with me now, but I will post it up later.

Also, this isn't a vector proof, and may not even be in the specialist maths course, but I am interested in knowing wether this actually works or not (note this isn't my work, and I had no idea where else to post this):

Prove that if x = 0.9999...
x = 1.

10x = 9.9999...

10x - x = 9x = 9

Therefore, x = 1,
and x = 0.9999...

Is this correct??
yes that proof is correct, and it is perfectly valid.
Title: Re: Vector Proofs
Post by: kamil9876 on February 19, 2009, 10:29:08 pm: Quote from: TrueTears on February 18, 2009, 10:52:55 pm
Quote from: dejan91 on February 18, 2009, 10:19:19 pm
/0 thanks, however your answer is slightly off the books answer. I don't have the book with me now, but I will post it up later.

Also, this isn't a vector proof, and may not even be in the specialist maths course, but I am interested in knowing wether this actually works or not (note this isn't my work, and I had no idea where else to post this):

Prove that if x = 0.9999...
x = 1.

10x = 9.9999...

10x - x = 9x = 9

Therefore, x = 1,
and x = 0.9999...

Is this correct??
yes that proof is correct, and it is perfectly valid.

The geometric series method is "more" valid since it defined the number 0.9999... as the limit of a sum. This 'proof' merely treats the number as an infinite string of 9's after the decimal rather than treating the number in terms of a quantity. Using such syntax based reasoning can lead incorrect results:

p=2*2*2*2...
2p=2(2*2*2*2...)
2p=p
2p-p=0
hence p=0
and 2*2*2*2*2...=0

However sometimes a correct result is yielded:

Let $x=\sqrt{2+\sqrt{2+\sqrt{2...}}}$

${x^2}=2+\sqrt{2+\sqrt{2...}}$

${x^2}=2+x$

and so x=2

So 2 is the answer. Both proofs used the same method; perform an operation on both sides and use the infinite repitition of the process to get the value of x or p in some some solvable equation. However one is wrong, one is right and limits fill this void, just like the limit based approach of the geometric series does. However the proof involving 0.999... can easily be made more formal and it is in fact analogous to the derivation of the sum of a geometric series (ask for the details if u want, i feel my post is already long enough LOL)
Title: Re: Vector Proofs
Post by: Mao on February 20, 2009, 09:17:49 pm: $\frac{1}{3} = 0.33333\cdots$

$\implies 0.99999 = 3\times \frac{1}{3} = \frac{3}{3} = 1$
Title: Re: Vector Proofs
Post by: kamil9876 on February 20, 2009, 09:55:23 pm: Quote from: Mao on February 20, 2009, 09:17:49 pm
$\frac{1}{3} = 0.33333\cdots$

Wouldn't the proof of this premise also use geometric series? Hence 0.333..*3=0.999 would be synonymous to 3*((3/10)+(3/100)+(3/1000)...)=(9/10)+(9/100)+(9/100)...

And so we would have to some how work out the sum on the right, which is the limit of the sum of a geometric series, or use some other method which would probably just be the same thing in disguise.
Title: Re: dejan91's questions
Post by: dejan91 on March 29, 2009, 12:24:02 pm: Help please? :)

Find the cartesian equation of the path of an object whose position is given by the vector equation $r=\frac{t}{2}i + (t^{2}+1)j$ .
Title: Re: dejan91's questions
Post by: TrueTears on March 29, 2009, 12:29:13 pm: $x = \frac{t}{2}$ ...[1]

$y = t^2+1$ ...[2]

from [1], t = 2x sub into 2

$y = (2x)^2 + 1$

$y = 4x^2 + 1$
Title: Re: dejan91's questions
Post by: kurrymuncher on March 29, 2009, 01:27:09 pm: remember to restrict the domain if your asked to graph it
Title: Re: dejan91's questions
Post by: dejan91 on March 29, 2009, 02:02:26 pm: ohk yup thanks will remember
Title: Re: dejan91's questions
Post by: dejan91 on March 29, 2009, 03:50:02 pm: Quote from: /0 on February 08, 2009, 10:23:23 pm
Quote from: dejan91 on February 08, 2009, 09:41:36 pm
Thanks Mao fo help anyway (Y)

Also, this isn't on vector proofs, but cbf starting new thread. Had this question that I just left ages ago till now..

If a = 2i - 3j + k and b = -2i + 3j + k,
find a unit vector perpendicular to both a and b.
Let the vector perpendicular to both $\mbox{a}$ and $\mbox{b}$ be $\mbox{c} = x\mbox{i}+y\mbox{j}+z\mbox{k}$

Then, by the dot product:

$2x-3y+z=0$

$-2x+3y+z=0$

Adding the two equations, we find that $z=0$ and so $x=\frac{3}{2}y$ .

So a possible vector could be $\mbox{c} = \frac{3}{2}\mbox{i}+\mbox{j}$

And the unit vector would be $\hat{\mbox{c}}=\frac{2\sqrt{13}}{13}\left(\frac{3}{2}\mbox{i}+\mbox{j}\right)$

Forgot about this one lol...

Answer to it is actually $\pm\frac{1}{\sqrt{394}}\left(9i + 8j + 14k)$

How do you get this?
Title: Re: dejan91's questions
Post by: dcc on March 29, 2009, 04:26:30 pm: If a = 2i - 3j + k and b = -2i + 3j + k,
find a unit vector perpendicular to both a and b.

$c = a \times b = \left| \begin{array}{ccc} i & j & k\\ 2 & -3 & 1\\ -2 & 3 & 1\\ \end{array} \right| = \left| \begin{array}{ccc} i & j & k\\ 2 & -3 & 1\\ 0 & 0 & 2\\ \end{array} \right| = 2 \left| \begin{array}{cc} i & j\\ 2 & -3\\ \end{array} \right| = -6i - 4j$

Which is of course, a scalar multiple of the answer you arrived at doing much the same thing, hence you are correct, and the book is talking crap, as the vector they have given is not perpendicular to a or b.

Then of course you have the negative of this vector, and the unit vector of these vectors.
Title: Re: dejan91's questions
Post by: dejan91 on March 29, 2009, 09:45:51 pm: Hmmmm, not sure what that was ^^ matrices? But ok, so the book is wrong?
Title: Re: dejan91's questions
Post by: dekoyl on March 29, 2009, 09:48:37 pm: Quote from: dejan91 on March 29, 2009, 09:45:51 pm
Hmmmm, not sure what that was ^^ matrices? But ok, so the book is wrong?
That was using the cross product as the cross product of two vectors is perpendicular to the two vectors. However it's not in the specialist course so unfortunately you can't use it in the exam. :P
Title: Re: dejan91's questions
Post by: dejan91 on March 30, 2009, 08:17:33 pm: Quote from: dekoyl on March 29, 2009, 09:48:37 pm
Quote from: dejan91 on March 29, 2009, 09:45:51 pm
Hmmmm, not sure what that was ^^ matrices? But ok, so the book is wrong?
That was using the cross product as the cross product of two vectors is perpendicular to the two vectors. However it's not in the specialist course so unfortunately you can't use it in the exam. :P

Ahhh no wonder I didn't quite understand it :P

Ok, this one is on one of TT's vector worksheets.

(http://i44.tinypic.com/257k9s4.jpg)

I got:
$\tilde{AC} = (-\lambda - 2)i + (5\lambda - 3)j$ for i.

$\left | \tilde{AC} \right\vert = \sqrt{26\lambda^2 - 26\lambda +13}$ for ii.

Is this right?

I'm stuck on b. Not sure how to aproach this exactly.
Title: Re: dejan91's questions
Post by: Over9000 on March 30, 2009, 08:27:52 pm: Quote from: dejan91 on March 30, 2009, 08:17:33 pm
Quote from: dekoyl on March 29, 2009, 09:48:37 pm
Quote from: dejan91 on March 29, 2009, 09:45:51 pm
Hmmmm, not sure what that was ^^ matrices? But ok, so the book is wrong?
That was using the cross product as the cross product of two vectors is perpendicular to the two vectors. However it's not in the specialist course so unfortunately you can't use it in the exam. :P

Ahhh no wonder I didn't quite understand it :P

Ok, this one is on one of TT's vector worksheets.

(http://i44.tinypic.com/257k9s4.jpg)

I got:
$\tilde{AC} = (-\lambda - 2)i + (5\lambda - 3)j$ for i.

$\left | \tilde{AC} \right\vert = \sqrt{26\lambda^2 - 26\lambda +13}$ for ii.

Is this right?

I'm stuck on b. Not sure how to aproach this exactly.
Yes, youre answers to 9.a) are all good, I got same answers
Title: Re: dejan91's questions
Post by: Over9000 on March 30, 2009, 08:34:15 pm: For b
$AB= 10i + 2j -(2i + 3j)$
$AB= 8i - j$
$|AB|= \sqrt{8^{2}+1}$
$= \sqrt{65}$
|AB|=|AC|
from previous question
$\sqrt{65}= \sqrt{26x^{2} - 26x +13}$
square both side
$65= 26x^{2} - 26x +13$
$26x^{2} - 26x = 52$
$x^{2} - x = 2$
$x^{2}-x-2=0$
i just made x = lamda, so sub that back in
Title: Re: dejan91's questions
Post by: dejan91 on March 30, 2009, 08:45:11 pm: Ok cool as. My method was right all along it seems I just used AB = 10i+ 2j lol. Thanks dude (Y)
Title: Re: dejan91's questions
Post by: Over9000 on March 30, 2009, 09:01:33 pm: No probs :)
Title: Re: dejan91's questions
Post by: dejan91 on March 31, 2009, 07:01:51 pm: Anyone done this question? It's on truetears' question sheet 3. If you have, are you able to post up answers just so i can check if mine are right?

(http://i42.tinypic.com/2jds9ol.jpg)
Title: Re: dejan91's questions
Post by: dejan91 on April 10, 2009, 04:39:13 pm: Bit stuck on this:
Find the intersection of the two relations $x^2 + 4y^2 - 4x - 8y + 4 = 0$ and $4x^2 + y^2 - 6y + 5 = 0$ . Thanks.
Title: Re: dejan91's questions
Post by: Over9000 on April 11, 2009, 12:11:45 am: First reaarange so that x and y are seperated
$x^{2} +4y^{2} -4x -8y +4 = 4x^{2} + y^{2} -6y +5$
$3x^{2} -3y^{2} + 2y +4x +1 =0$
$3y^{2} -2y =3x^{2} +4x +1$
divide both sides by 3 to have $x^{2}$ coefficient as 1
$y^{2} - \frac{2}{3}y= x^{2} + \frac{4}{3}x + \frac{1}{3}$
Complete the square
$(y-\frac{1}{3})^{2} - \frac{1}{9} = (x+ \frca{2}{3})^{2} - \frac{1}{9}$
$(y-\frac{1}{3})^{2} = (x+ \frca{2}{3})^{2}$
$y- \frac{1}{3} = \pm(x+\frac{2}{3})$
$y= x+1$ and $y=-x - \frac{1}{3}$

Now:
$x^{2} +4y^{2} -4x -8y +4 =0$ ......1
$4x^{2} + y^{2} -6y +5=0$ ...............2
However lets be sneaky and sub in y=x+1 and $y=-x-\frac{1}{3}$ as we have worked them out

$x^{2} + 4(x+1)^{2} - 4x - 8(x+1) + 4 = 0$
$x^{2} + 4x^{2} + 8x +4 -4x -8x -8 +4= 0$
$5x^{2} -4x =0$
$x(5x-4)=0$
$x=0$ and $x=4/5$
$y=x+1$
therefore y=1 and $y=\frac{9}{5}$
So points of intersection are (0,1) and $(\frac{4}{5}, \frac{9}{5})$

you can check, by subbing y=x+1 into equation 2 if u want
Title: Re: dejan91's questions
Post by: TrueTears on April 11, 2009, 12:14:39 am: ^^^^^^^^ SO F***KING PRO, I was stuck on this Q for like ages. pro pro legendary man.
Title: Re: dejan91's questions
Post by: dejan91 on April 11, 2009, 01:49:44 am: Thanks heaps Over9000 :)

Might I add, explanation was A++
Title: Re: dejan91's questions
Post by: Over9000 on April 11, 2009, 01:54:58 am: Quote from: dejan91 on April 11, 2009, 01:49:44 am
Thanks heaps Over900 :)

Might I add, explanation was A++
No probs, glad to help ;)
Title: Re: dejan91's questions
Post by: dcc on April 11, 2009, 11:00:15 am: Quote from: dekoyl on March 29, 2009, 09:48:37 pm
That was using the cross product as the cross product of two vectors is perpendicular to the two vectors. However it's not in the specialist course so unfortunately you can't use it in the exam. :P

Incorrect. Any maths assessor worth their weight in salt will know and understand the cross product (god, even the physics assessors would).
Title: Re: dejan91's questions
Post by: Mao on April 11, 2009, 06:22:26 pm: Quote from: dcc on April 11, 2009, 11:00:15 am
Quote from: dekoyl on March 29, 2009, 09:48:37 pm
That was using the cross product as the cross product of two vectors is perpendicular to the two vectors. However it's not in the specialist course so unfortunately you can't use it in the exam. :P

Incorrect. Any maths assessor worth their weight in salt will know and understand the cross product (god, even the physics assessors would).

I should add that, unless specified by the question, any method that is logically coherent and rigorous will be marked as correct. (But they design the questions with this in mind, so that CAS kids and UMEP/MUEP kids aren't at an advantage)

However, personally I would stay on the safe side and stay within the syllabus.
Title: Re: dejan91's questions
Post by: dekoyl on April 11, 2009, 08:15:10 pm: Quote from: dcc on April 11, 2009, 11:00:15 am
Quote from: dekoyl on March 29, 2009, 09:48:37 pm
That was using the cross product as the cross product of two vectors is perpendicular to the two vectors. However it's not in the specialist course so unfortunately you can't use it in the exam. :P

Incorrect. Any maths assessor worth their weight in salt will know and understand the cross product (god, even the physics assessors would).
Cool. I didn't know that. I made a big assumption then as I asked my teacher if the cross product could be used and he said no.
I stand corrected.
Title: Re: dejan91's questions
Post by: dcc on April 11, 2009, 08:19:48 pm: Quote from: dekoyl on April 11, 2009, 08:15:10 pm
Cool. I didn't know that. I made a big assumption then as I asked my teacher if the cross product could be used and he said no.
I stand corrected.

If your teacher says that the cross product doesn't exist, then you shouldn't use it when you are doing maths for your teacher. (Its a parallel of price discrimination. Maths discrimination: doing maths based on who will be reading it)
Title: Re: dejan91's questions
Post by: ed_saifa on April 11, 2009, 08:25:43 pm: When using maths that isn't on the syllabus, you won't get working marks if you get the answer wrong, or so i've been told.
Title: Re: dejan91's questions
Post by: dejan91 on April 13, 2009, 09:39:16 pm: EDIT: Sorry, accidently clicked post before i finished.
Question was: Solve for x where $x \in (0,\pi)$ :

$cos(2x) = -cot(2x)$
$= -\frac{1}{tan2x}$

$cos(2x)tan(2x) = -1$

$\therefore sin(2x) = -1$

Let $2x = u$
$sinu = -1$
$\therefore u = \frac{3\pi}{2}$
and $x = \frac{3\pi}{4}$

I used this but only got one solution out of a possibl four. What did I do wrong? Or haven't I gone far enough?
Title: Re: dejan91's questions
Post by: TrueTears on April 13, 2009, 09:45:03 pm: $cos(2x) = -\frac{cos(2x)}{sin(2x)}$

$cos(2x)sin(2x) + cos(2x) = 0$

$cos(2x)(sin(2x)+1) = 0$

Null factor law $cos(2x) = 0$ or $sin(2x) + 1 = 0$

Can you do it from here?
Title: Re: dejan91's questions
Post by: dejan91 on April 13, 2009, 09:51:42 pm: Cool thanks TrueTears, do you know if my way was wrong?
Title: Re: dejan91's questions
Post by: TrueTears on April 13, 2009, 09:53:44 pm: Your mistake was that you cancelled out a possible solution when you simplified by cancelling out the cos(2x)
Title: Re: dejan91's questions
Post by: dejan91 on April 13, 2009, 09:56:06 pm: Hmm I see. Ok then thanks for pointing that out.
Title: Re: dejan91's questions
Post by: TrueTears on April 13, 2009, 10:00:27 pm: Say you have sinxcosx = cosx

if you cancel cosx from both sides, you are in fact losing a solution

this is the same thing as saying

$2x^2 = x$

you wouldn't cancel a x on both sides, because it will lose a solution, so that's why you move to the LHS and do null factor.
Title: Re: dejan91's questions
Post by: dejan91 on May 08, 2009, 07:36:10 pm: If $F(x) = \int \frac{e^{2x}}{e^x + 4} dx$ and $F(0) = 3$ , find $F(x)$ .

Thanks :)
Title: Re: dejan91's questions
Post by: humph on May 08, 2009, 07:47:32 pm: Let $u = e^x + 4$ , so that $du = e^x dx$ , and so
$F(u) = \int \frac{u-4}{u}\: du = \int 1 - \frac{4}{u} \: du = u - 4 \ln u + c$ for some $c \in \mathbb{R}$ .
Thus $F(x) = e^x + 4 - 4 \ln(e^x + 4) + c$ .
Then $F(0) = 1 + 4 - 4 \ln 5 + c = 3$ , so we must have that $c = 4 \ln 5 - 2$ , and hence $F(x) = e^x + 4 \ln\left(\frac{5}{e^x + 4}\right) + 2$ .
Title: Re: dejan91's questions
Post by: dejan91 on May 09, 2009, 12:31:50 pm: Ahhhk ok thanks I know what I did wrong.
Title: Re: dejan91's questions
Post by: dejan91 on May 13, 2009, 10:09:03 pm: Am I on the right track at all?

$\int 2cos^3(3x)sin^7(3x) dx$

$= 2\int (cos(3x)sin(3x))^3sin^4(3x) dx$
$= 2\int (\frac {1}{2}sin(6x))^3sin^4(3x) dx$

I got up to this part and I'm not sure what to do from here, or even if it's correct.
Title: Re: dejan91's questions
Post by: Mao on May 13, 2009, 10:46:48 pm: no need to:

$\int 2\cos^3 (3x) \sin^7 (3x)\; dx = \int 2\cos(3x)(1-\sin^2 (3x)) \sin^7(dx)\; dx$

now substitute $u=\sin (3x)$
Title: Re: dejan91's questions
Post by: dejan91 on May 13, 2009, 11:07:59 pm: Thanks yeah I thought my way was too long.
Title: Re: dejan91's questions
Post by: dejan91 on May 29, 2009, 04:55:31 pm: 1) Find $a$ and $b$ , such that $z = a + bi$ , and $\frac{z+i}{z+2} = i$

2) If $z = x + yi$ , find $x$ and $y$ such that $z = \sqrt{3+4i}$

Need to see if what I got was correct or not :) Thanks.
Title: Re: dejan91's questions
Post by: Damo17 on May 29, 2009, 05:35:17 pm: Quote from: dejan91 on May 29, 2009, 04:55:31 pm
1) Find $a$ and $b$ , such that $z = a + bi$ , and $\frac{z+i}{z+2} = i$

2) If $z = x + yi$ , find $x$ and $y$ such that $z = \sqrt{3+4i}$

Need to see if what I got was correct or not :) Thanks.

1) $z+ i=zi+2i$
as $z=a+bi$
$a+(b+1)i=ai+bi^2+2i$
   $= -b+(a+2)i$

$\therefore$ $a=-b$ ------1
$b+1=a+2$ ------2

sub in $a=-b$ into 2
$b+1=-b+2$
   $b=\frac{1}{2}$

sub in $b=\frac{1}{2}$ into 1
$a=-\frac{1}{2}$

2) $z=\sqrt{3+4i}$ $\therefore$ $z^2=3+4i$

using the square root of z formulas:

$x^2=\frac{3+\sqrt{3^2+4^2}}{2}$
$x=\pm 2$

$y^2=x^2-a$
   $= 1$
$y=\pm 1$

therefore $z_1=2+i$ , $z_2 =-2-i$
Title: Re: dejan91's questions
Post by: dejan91 on May 29, 2009, 05:47:40 pm: Great :) Exactly what I got...albeit using a different method.
Title: Re: dejan91's questions
Post by: dejan91 on May 30, 2009, 12:14:05 pm: I get the red area, but the book gets the red AND green area. The explanation at the back says that "The region 'Arg(z) is equal to or smaller than Arg(w)' is the region where a point z has 'an angle' less than $\frac{\pi}{3}$ " Why use $\frac{\pi}{3}$ ?

(http://i43.tinypic.com/kefdl5.jpg)
Title: Re: dejan91's questions
Post by: dcc on May 30, 2009, 08:35:23 pm: Quote from: dejan91 on May 30, 2009, 12:14:05 pm
I get the red area, but the book gets the red AND green area. The explanation at the back says that "The region 'Arg(z) is equal to or smaller than Arg(w)' is the region where a point z has 'an angle' less than $\frac{\pi}{3}$ " Why use $\frac{\pi}{3}$ ?

$Arg(w) = -\frac{2\pi}{3}$ (this is obvious from the diagram you have there). So that seems like the red area (however not as large as you have drawn it, as $|z| \leq 1$ ).
Title: Re: dejan91's questions
Post by: dejan91 on May 31, 2009, 01:10:07 am: Ah woops was rushing on photoshop... but yes, it was meant to be $|z| \leq 1$

yeah well that's what I thought it would be (the red area) but apparently it's not.... According to checkpoints.
Title: Re: dejan91's questions
Post by: kamil9876 on May 31, 2009, 10:54:14 am: Checkpoints do make mistakes, I've found a few once. They should have the exam it was taken from at the bottom so if you really want to know might as well check vcaa(if they still have it there).
Title: Re: dejan91's questions
Post by: dejan91 on June 15, 2009, 11:23:44 pm: To find the area underneath, do I have to find the inverse function first?
$\int_\frac{1}{2}^2log_e(2x) .dx$
Title: Re: dejan91's questions
Post by: TrueTears on June 15, 2009, 11:26:11 pm: Since you can not integrate $log_e(2x)$ [have to use integration by parts]

You should find the inverse and then integrate with respect to y [notice limits will change, they will be the respective y values for $\frac{1}{2}$ and 2]. Then use the whole 'rectangle' area minus the area you got from the inverse.
Title: Re: dejan91's questions
Post by: dejan91 on June 15, 2009, 11:33:36 pm: Yeah knew integration by parts was the other way. Just thought there was a different method.

$\int_0^{log_e(4)} \frac{1}{2}e^y .dy$ , and area of rectangle is $\frac{3}{2}log_e(4)$ . Is that right?

Is it worth learning integration by parts by the way?
Title: Re: dejan91's questions
Post by: TrueTears on June 15, 2009, 11:37:20 pm: Yeah, it's not very hard to learn, a quick wikipedia can help :) [Good knowledge to have especially when doing integration, even though it's not in spesh course.]

Area of the rectangle is $2log_e(4)$

You integral for the inverse is fine, all you gotta do now is $2log_e(4) - \frac{3}{2}$ (Result of the integral of the inverse) which yields the area required.
Title: Re: dejan91's questions
Post by: dejan91 on June 15, 2009, 11:42:13 pm: Oh, crap. No wonder I got it wrong. This is what I was getting confused about: what values do you take for the side lengths of the rectangle? The maximum for both the x and y limits (in this case 2 and log_e(4)? What if the lower limit was x=1 or something?

Hmmm, might have a look into that integration by parts.
Title: Re: dejan91's questions
Post by: TrueTears on June 15, 2009, 11:53:47 pm: Doesn't matter what the lower limit is, you are right, it is just the upper limit and its respective y co ordinate.

To understand it graphically, when you integrate with respect to y, you are working out the area bounded by the graph and the y axis. Hence you use the rectangle minus that area to get the area required.

Here's are good links to learn integration by parts: (This is how I learnt it)

1. http://www.youtube.com/watch?v=WRKMunWZqU4

2. http://en.wikipedia.org/wiki/Integration_by_parts
Title: Re: dejan91's questions
Post by: dejan91 on June 17, 2009, 06:48:21 pm: The region between the curves $y=cos(2x)$ , $y=sin(x)$ and the lines $x=0$ and $x= \frac{\pi}{2}$ is rotated about the x-axis. Find the volume of the solid formed.

I've got the volume as: $\pi \int_0^\frac{\pi}{6}(cos^2(2x)-sin^2(x))dx + \pi \int_\frac{\pi}{6}^\frac{\pi}{2}(sin^2(x)-cos^2(2x))dx$

Not sure what to do with the $cos^2(2x)$ though. Help please?

EDIT: karma 22, posts 222. LOL.
Title: Re: dejan91's questions
Post by: TrueTears on June 17, 2009, 06:50:29 pm: $cos(2x) = 2cos^2x - 1$

sub in 2x for x.

Then solve for $cos^2(2x)$
Title: Re: dejan91's questions
Post by: dejan91 on June 17, 2009, 06:56:18 pm: Do you mean to do $cos^2(2x) = (2cos^2x-1)^2$ ?
Title: Re: dejan91's questions
Post by: TrueTears on June 17, 2009, 07:00:51 pm: I mean this

$cos(2(2x)) = 2cos^2(2x) - 1$

$\frac{1}{2}[cos(4x) + 1] = cos^2(2x)$

Sub that back in :)
Title: Re: dejan91's questions
Post by: dejan91 on June 17, 2009, 07:04:17 pm: Ahhh of course. Forgot you could do that! Thank you!
Title: Re: dejan91's questions
Post by: dejan91 on July 03, 2009, 12:14:09 am: A conical flask with a raduis of 5cm and a height of 12cm is being filled with powder at a constant rate of $\pi cm^3/min$ . Find the rate, in terms of height, at which the height of the powder is changing.
Title: Re: dejan91's questions
Post by: zzdfa on July 03, 2009, 12:34:19 am: dv/dt = pi

volume of the cone is:

$V=\frac{1}{3}\pi r^2 h$
but oh wait, its being filled from the pointy end, so

$V_{of powder}=V_{of whole container} - \frac{1}{3}\pi r^2 (h_0-h)$ , where $h$ is the height of the powder measured from the bottom of the cone. $h_0$ is the height of the flask. <- draw a diagram

find the ratio for $h_0-h$ to $r$ , (draw a diagram)
hence get an equation for h in terms of r
put this into the above equation,
find dh/dv,
chain rule it.
Title: Re: dejan91's questions
Post by: dejan91 on July 03, 2009, 09:27:28 am: Thanks :)
Title: Re: dejan91's questions
Post by: dejan91 on October 31, 2009, 09:34:40 pm: The curve formed by the graph of $f(x) = 0.5cos^{-1}(2x+1)$ is rotated around the y-axis from $y=0$ to $y=h$ where $h$ is a real constant, $0<h<\frac{\pi}{2}$ , to form the shape of a funnel. $x$ and $y$ are measured in cm. Find,in exact terms, the volume of the funnel for $h<\frac{\pi}{4}$ .
Title: Re: dejan91's questions
Post by: TonyHem on October 31, 2009, 09:43:22 pm: Does the answer come out nicely? I wrote a few things up and it looks like it's gonna come out with heaps of terms :/
Title: Re: dejan91's questions
Post by: dejan91 on October 31, 2009, 09:49:10 pm: Well, answer is: $\frac{\pi}{4}(\frac{3\pi}{8}-1) cm^3$ . Is that what you got?
Title: Re: dejan91's questions
Post by: TonyHem on October 31, 2009, 10:03:55 pm: ~~How do you work out H? are u meant to assume that since the graph touches the y-axis at pi/4, it "makes a funnel" there?~~
Nvm stupid question

If I sub in h = pi/4 into my equation I get $\frac{\pi}{2}(\frac{3\pi}{8}-1})$
So nope, I haven't got it :/

edit: forget that top as well lol, i took out the 1/2 before squaring it.
Title: Re: dejan91's questions
Post by: Damo17 on October 31, 2009, 10:10:19 pm: Quote from: dejan91 on October 31, 2009, 09:34:40 pm
The curve formed by the graph of $f(x) = 0.5cos^{-1}(2x+1)$ is rotated around the y-axis from $y=0$ to $y=h$ where $h$ is a real constant, $0<h<\frac{\pi}{2}$ , to form the shape of a funnel. $x$ and $y$ are measured in cm. Find,in exact terms, the volume of the funnel for $h<\frac{\pi}{4}$ .

$V= \frac{\pi}{4} \int_0^h (cos(2y)-1)^2 dy$

$V= \frac{\pi}{4}[\frac{sin(2y)cos(2y)}{4}-sin(2y)+\frac{3y}{2}]_0^h$

$h=\frac{\pi}{4}$

$\therefore V= \frac{\pi}{4}(\frac{3\pi}{8}-1)cm^3$
Title: Re: dejan91's questions
Post by: TrueTears on October 31, 2009, 10:11:00 pm: Quote from: dejan91 on October 31, 2009, 09:34:40 pm
The curve formed by the graph of $f(x) = 0.5cos^{-1}(2x+1)$ is rotated around the y-axis from $y=0$ to $y=h$ where $h$ is a real constant, $0<h<\frac{\pi}{2}$ , to form the shape of a funnel. $x$ and $y$ are measured in cm. Find,in exact terms, the volume of the funnel for $h<\frac{\pi}{4}$ .
$y = \frac{1}{2}\cos^{-1}(2x+1)$

Volume required $= \pi \int_0^{\frac{\pi}{4}} x^2 dy = \pi \int_0^{\frac{\pi}{4}} \left(\frac{\cos(2y)-1}{2}\right)^2 dy$

EDIT: damn I got beaten.
Title: Re: dejan91's questions
Post by: Damo17 on October 31, 2009, 10:13:44 pm: Quote from: TonyHem on October 31, 2009, 10:03:55 pm
How do you work out H? are u meant to assume that since the graph touches the y-axis at pi/4, it "makes a funnel" there?
If I sub in h = pi/4 into my equation I get $\frac{\pi}{2}(\frac{3\pi}{8}-1})$
So nope, I haven't got it :/

You most likely took out the $\frac{1}{2}$ from $x=\frac{1}{2}cos(2y)-\frac{1}{2}$ when setting up the integral for volume and forgetting to square it as well. So you would get $\frac{\pi}{2}$ out the front instead of $\frac{\pi}{4}$ .
Title: Re: dejan91's questions
Post by: TonyHem on October 31, 2009, 10:14:29 pm: Quote from: Damo17 on October 31, 2009, 10:13:44 pm
Quote from: TonyHem on October 31, 2009, 10:03:55 pm
How do you work out H? are u meant to assume that since the graph touches the y-axis at pi/4, it "makes a funnel" there?
If I sub in h = pi/4 into my equation I get $\frac{\pi}{2}(\frac{3\pi}{8}-1})$
So nope, I haven't got it :/

You most likely took out the $\frac{1}{2}$ from $x=\frac{1}{2}cos(2y)-\frac{1}{2}$ when setting up the integral for volume and forgetting to square it as well. So you would get $\frac{\pi}{2}$ out the front instead of $\frac{\pi}{4}$ .

haha just noticed after msging u
oh wells ;{
Title: Re: dejan91's questions
Post by: dejan91 on October 31, 2009, 10:14:59 pm: Why do you let $h = \frac{\pi}{4}$ though?
Title: Re: dejan91's questions
Post by: dejan91 on October 31, 2009, 10:16:17 pm: Do you just assume that h is at it's maximum value?
Title: Re: dejan91's questions
Post by: TonyHem on October 31, 2009, 10:17:09 pm: It wants the volume for h<pi/4, and h> 0, so the volume from 0 to pi/4
My guess is that you're thinking that it's saying find the volume for some random h that is lower than pi/4
Title: Re: dejan91's questions
Post by: TrueTears on October 31, 2009, 10:17:11 pm: Quote from: dejan91 on October 31, 2009, 10:14:59 pm
Why do you let $h = \frac{\pi}{4}$ though?
The question is very very badly worded, look it should really be when $h = \frac{\pi}{4}$

When it says $h < \frac{\pi}{4}$ you could actually put any value for h

so yes stupid question. Which exam was this? I remember this question slightly...
Title: Re: dejan91's questions
Post by: TrueTears on October 31, 2009, 10:17:50 pm: Quote from: TonyHem on October 31, 2009, 10:17:09 pm
It wants the volume for h<pi/4, and h> 0, so the volume from 0 to pi/4
My guess is that you're thinking h is an unknown value between 0 to pi/4
When it says find the volume for h < pi/4 , technically you can put any value for h between 0<h<pi/4
Title: Re: dejan91's questions
Post by: GerrySly on October 31, 2009, 10:18:53 pm: Quote from: dejan91 on October 31, 2009, 10:16:17 pm
Do you just assume that h is at it's maximum value?
It's asking for the total volume of the funnel, so the highest height the funnel can be is $\frac{\pi}{4}$ , therefore the volume of the funnel is achieved when $h=\frac{\pi}{4}$

That's the way I saw it anyway...
Title: Re: dejan91's questions
Post by: TonyHem on October 31, 2009, 10:21:59 pm: Quote from: TrueTears on October 31, 2009, 10:17:50 pm
Quote from: TonyHem on October 31, 2009, 10:17:09 pm
It wants the volume for h<pi/4, and h> 0, so the volume from 0 to pi/4
My guess is that you're thinking h is an unknown value between 0 to pi/4
When it says find the volume for h < pi/4 , technically you can put any value for h between 0<h<pi/4

Hmm... yeah now i'm getting confused.
If I imagine it said h = pi/4, then I would of integrated from 0 to pi/4.
But when I saw that question originally, I was thinking it'll be in terms of H, so when he showed the solution, I was thinking maybe it just means everything under pi/4
:S
Title: Re: dejan91's questions
Post by: dejan91 on October 31, 2009, 10:22:38 pm: This was MAV 2008 Exam 2. Yeah It was pretty dodgy wording (I shortened some parts for the forum though). So then if you can technically put in any number under $\frac{\pi}{4}$ , does that mean there is more than one solution? VCAA wouldn't do that would they?
Title: Re: dejan91's questions
Post by: TrueTears on October 31, 2009, 10:23:52 pm: Quote from: dejan91 on October 31, 2009, 10:22:38 pm
This was MAV 2008 Exam 2. Yeah It was pretty dodgy wording (I shortened some parts for the forum though). So then if you can technically put in any number under $\frac{\pi}{4}$ , does that mean there is more than one solution? VCAA wouldn't do that would they?
VCAA wouldn't be this ambiguous, MAV is just being a retard. They could have easily said "find the volume of the funnel when h = pi/4" but no...
Title: Re: dejan91's questions
Post by: dejan91 on October 31, 2009, 10:41:11 pm: Bastards!

Another one: when a question says "A force of magnitude 10 kg-wt", what does the "kg-wt" mean/stand for?
Title: Re: dejan91's questions
Post by: Mao on October 31, 2009, 11:03:22 pm: |F| = 10g

1 kg wt = 1g N
Title: Re: dejan91's questions
Post by: dejan91 on October 31, 2009, 11:44:47 pm: Thanks Mao :)
Title: Re: dejan91's questions
Post by: dejan91 on November 01, 2009, 12:16:15 am: If the vector $xi+yj-2k$ makes an angle of $\pi - cos^{-1}(\frac{1}{3})$ with the positive y-axis, then:

A. x = 0 and y = -1
B. x=6 and y=-1
C. x=2 and y=1
D. x=-2 and y=1
E x=-2, 2 and y = -1
Title: Re: dejan91's questions
Post by: TrueTears on November 01, 2009, 12:44:37 am: Let $a = xi+yj-2k$

$a.j= |a|\cos\left(\pi - \cos^{-1}\left(\frac{1}{3}\right)\right)$

You'd then have to test options to see which x and y works, it's the only way.
Title: Re: dejan91's questions
Post by: dejan91 on November 01, 2009, 12:51:24 am: Still don't understand :P I know what you mean the only way is to go down the list, but how did you get to that equation there^^^?
Title: Re: dejan91's questions
Post by: kamil9876 on November 01, 2009, 12:52:58 am: good, except you take $a.j$

$a.j=y=|a|cos(\pi - cos^{-1} (\frac{1}{3}))$
$=\sqrt{x^2+y^2+4}(\frac{-1}{3})$

which implies $y$ is negative hence looking at the options given it is -1:
$ 9=x^2+1+4$
$x=\pm 2$

and i don't think this solution is unique, but it is the only one out of those.
Title: Re: dejan91's questions
Post by: TrueTears on November 01, 2009, 03:58:39 am: Quote from: kamil9876 on November 01, 2009, 12:52:58 am
good, except you take $a.j$

$a.j=y=|a|cos(\pi - cos^{-1} (\frac{1}{3}))$
$=\sqrt{x^2+y^2+4}(\frac{-1}{3})$

which implies $y$ is negative hence looking at the options given it is -1:
$ 9=x^2+1+4$
$x=\pm 2$

and i don't think this solution is unique, but it is the only one out of those.
What you mean? I did that didn't I?