Author Topic: VCE Specialist 3/4 Question Thread! (Read 2579171 times) Tweet Share

Homer · « **Reply #1350 on:** February 25, 2013, 07:59:33 pm »

Quote from: Jayward on February 25, 2013, 07:57:28 pm

What would this graph look like? I have no clue. I vaguely think its a circle, but am not entirely sure

itd just be x^2+y^2=169

Jayward · « **Reply #1351 on:** February 25, 2013, 08:01:39 pm »

Quote from: Homer on February 25, 2013, 07:59:33 pm

itd just be x^2+y^2=169

damn i forgot! thanks

ashoni · « **Reply #1352 on:** February 25, 2013, 08:22:37 pm »

Hey guys! I was just wondering if there was some general rules/steps to tackle vector proof questions... Finding it difficult to do these kinds of questions since I haven't been exposed to them as much! Thanks!

Daenerys Targaryen · « **Reply #1353 on:** February 25, 2013, 08:46:25 pm »

Draw a diagram
Brainstorm some vector rules that could help you
Write down what you are trying to prove in mathematical terms

Homer · « **Reply #1354 on:** February 25, 2013, 09:39:41 pm »

find the particular solution for the differential equation, and state the largest domain for which the solution applies

$(y+1)^2\frac{dy}{dx}=1$ and y=2 when x=4

why is the domain R/{-5}, i though it would just be R

b^3 · « **Reply #1355 on:** February 25, 2013, 11:01:42 pm »

When you firsst start off finding the solution you will arrive at the step
$\frac{dy}{dx}=\frac{1}{(y+1)^{2}}$
But as we cannot divide by zero this implies that $y\neq-1$ .

When you finally get to the solution we have $x=\frac{1}{3}(y+1)^{3}-5$ but we still cannot have $y=5$ . So we need to find the value of $x$ for which this would happen.
$x=\frac{1}{3}(0+1)^{3}-5=-5$
So $x\neq-5$

duquesne9995 · « **Reply #1356 on:** February 27, 2013, 06:54:55 pm »

Hi guys, I have some questions about complex numbers.
I'm using the Essentials Textbook
Ex4G Q1b
Solve each of the following equations over C, and show the solutions for each on an Argand diagram.
$z^3=27i$
The textbook answers (please see attachment) have yellow lines going from the origin to the solution and also a blue circle connecting the solutions. Do we have to include the circle and lines on a SAC/exam?
Also, they write their solutions in Cartesian x+iy form but on the graph it's polar form. Is it okay to express solutions in polar form or are you only allowed to do that on the graph? And how do you know when to express answers in polar form/Cartesian if they don't tell you?
Thanks!

Stick · « **Reply #1357 on:** February 27, 2013, 07:06:54 pm »

Unfortunately my teacher rushed through complex relations on the Argand diagram so I don't get these (relatively simple) questions. Any help would be really appreciated.

(Just the first part)

Daenerys Targaryen · « **Reply #1358 on:** February 27, 2013, 08:24:15 pm »

A red light moves around a closed shape displaying a lighting display. It has a position vector given by:
r(t)=2cos(t)i+3sin(t)j ; $t\geq0$
where t represents time in seconds and the displacement components are measured in metres
find the cartesian equation of the path of the red light

Homer · « **Reply #1359 on:** February 27, 2013, 08:44:00 pm »

Quote from: Stick on February 27, 2013, 07:06:54 pm

Unfortunately my teacher rushed through complex relations on the Argand diagram so I don't get these (relatively simple) questions. Any help would be really appreciated.

(Image removed from quote.)
(Just the first part)

(Image removed from quote.)

just out of curiosity stick for the second question a) is the equation x=1

Bad Student · « **Reply #1360 on:** February 27, 2013, 08:46:33 pm »

Quote from: Stick on February 27, 2013, 07:06:54 pm

Unfortunately my teacher rushed through complex relations on the Argand diagram so I don't get these (relatively simple) questions. Any help would be really appreciated.

(Image removed from quote.)
(Just the first part)

(Image removed from quote.)

These aren't relatively simple questions. Heaps of people in my class, including me, had trouble with these.

Conic · « **Reply #1361 on:** February 27, 2013, 09:08:32 pm »

For 10a:

$|\frac{z-2}{z} |=\frac{|z-2|}{|z|}=1$

$\Rightarrow |z-2|=|z|$

let $z=x+yi$

$|(x-2)+yi|=|x+yi|$

$\sqrt{(x-2)^2+y^2}=\sqrt{x^2+y^2}$

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

$x^2-4x+4+y^2=x^2+y^2$

$4x=4$

$x=1$

or $Re(z)=1$

You can do 10b with the same method.

Stick · « **Reply #1362 on:** February 27, 2013, 09:11:20 pm »

Quote from: Homer on February 27, 2013, 08:44:00 pm

just out of curiosity stick for the second question a) is the equation x=1

Yeah. I think it's probably just an issue with my algebraic manipulation of the figures. I'm not very good at it when it gets very involved.

Quote from: Bad Student on February 27, 2013, 08:46:33 pm

These aren't relatively simple questions. Heaps of people in my class, including me, had trouble with these.

Really? Well that makes me feel better then!

Stick · « **Reply #1363 on:** February 27, 2013, 09:12:45 pm »

Quote from: Conic on February 27, 2013, 09:08:32 pm

For 10a:

$|\frac{z-2}{z} |=\frac{|z-2|}{|z|}=1$

$\Rightarrow |z-2|=|z|$

let $z=x+yi$

$|(x-2)+yi|=|x+yi|$

$\sqrt{(x-2)^2+y^2}=\sqrt{x^2+y^2}$

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

$x^2-4x+4+y^2=x^2+y^2$

$4x=4$

$x=1$

or $Re(z)=1$

You can do 10b with the same method.

Awesome! Much simpler than the method I was trying to employ. Thanks!

polar · « **Reply #1364 on:** February 27, 2013, 11:37:59 pm »

Quote from: duquesne9995 on February 27, 2013, 06:54:55 pm

Hi guys, I have some questions about complex numbers.
I'm using the Essentials Textbook
Ex4G Q1b
Solve each of the following equations over C, and show the solutions for each on an Argand diagram.
$z^3=27i$
The textbook answers (please see attachment) have yellow lines going from the origin to the solution and also a blue circle connecting the solutions. Do we have to include the circle and lines on a SAC/exam?
Also, they write their solutions in Cartesian x+iy form but on the graph it's polar form. Is it okay to express solutions in polar form or are you only allowed to do that on the graph? And how do you know when to express answers in polar form/Cartesian if they don't tell you?
Thanks!

there shouldn't even be a line connecting the origin to the point. also, in the exam they usually give an argand diagram with dotted circles on it anyway.

Quote from: HatersGonnaHate on February 27, 2013, 08:24:15 pm

A red light moves around a closed shape displaying a lighting display. It has a position vector given by:
r(t)=2cos(t)i+3sin(t)j ; $t\geq0$
where t represents time in seconds and the displacement components are measured in metres
find the cartesian equation of the path of the red light

any point on its path must have coordinates $(x,y) = (2\cos(t), 3\sin(t))$
hence, $x=2\cos(t) \implies \cos(t) = \frac{x}{2}$ where $t \in [0,\infty) \implies 2\cos(t) \in [-2,2] \implies x \in [-2,2]$
and $y=3\sin(t) \implies \sin(t) = \frac{y}{3}$ where $t \in [0,\infty) \implies 3\sin(t) \in [-3,3] \implies y \in [-3,3]$
since $\cos^2(t) + \sin^2(t) = 1$ , $\left(\frac{x}{2}\right)^2 + \left(\frac{y}{3}\right)^2 = 1 \implies \frac{x^2}{4} + \frac{y^2}{9} = 1$ where $x \in [-2,2], y \in [-3,3]$

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