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May 22, 2019, 05:29:16 am

### AuthorTopic: VCE Specialist 3/4 Question Thread!  (Read 1029124 times) Tweet Share

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#### Flight112

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« Reply #9465 on: April 25, 2019, 01:31:29 pm »
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If you sketch $y=x^2$ with domain restriction $[-1,1]$, don't the values for $y$ become restricted to being between 0 and 1?

Ah yes, thank you. I missed that notion.

#### f0od ##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9466 on: April 27, 2019, 10:37:43 pm »
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Hi! I was directed here from the methods question thread as this question isn't in the methods course (yet is still part of my sac ,_,) I'm not entirely sure if it part of the spesh course, so apologies if it isn't related!

Is it possible to find the y asymptote through the division of ordinates, or is it only possible to find the x asymptote using this method?

Many thanks! class of 2019

#### DBA-144

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« Reply #9467 on: April 28, 2019, 10:40:43 am »
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Hi! I was directed here from the methods question thread as this question isn't in the methods course (yet is still part of my sac ,_,) I'm not entirely sure if it part of the spesh course, so apologies if it isn't related!

Is it possible to find the y asymptote through the division of ordinates, or is it only possible to find the x asymptote using this method?

Many thanks! If you are allowed a calc for the sac, then its much easier. Just graph it on your cas and then see where the graph seems to approachs specific points. LEt's just say that y=2 is an asymptote. Then, just consider what happens as x-->inf./-inf./0 and you should be able to say where the asymptote is at.

THis is probably a good time to say that there may be oblique asymptotes. For these, you just need to use long division. Oblique asymptotes are those that are not your usual straight lines. eg. y=x^2 is an asymptote that is oblique. Also, note that this is in the methods course, under the sum, difference and product of functions section. However, to my knowledge, this stuff is rarely ever assessed in Exam 1 and has come up a couple of times in Exam 2 (referring to graphing here, not solving, etc. with these functions!)

OF course, if any discussion stems from this post relating to methods it needs to go in the appropriate thread. Some of these strategies could work for spesh as well (rational functions)

Hope this helps, sorry if anything is wrong.

#### EverySecondCounts

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« Reply #9468 on: May 19, 2019, 08:11:34 pm »
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Hi.
Just curious though but how would you interpret the polar conversion of

(8+5i)/(3+4i) on the Ti-inspire cas, cause I get this solution of e^(-i〖tan〗^(-1 (17/44))*√89/5,

I understand the e to the power, but I dont understand the tan part?
Much appreciated

Also how do I write equations in these forums, as I'm having to copy equations from word, and their coming across as not so easy to read once I create a post. Thanks

#### AlphaZero ##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9469 on: May 19, 2019, 08:49:33 pm »
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Hi.
Just curious though but how would you interpret the polar conversion of

(8+5i)/(3+4i) on the Ti-inspire cas, cause I get this solution of e^(-i〖tan〗^(-1 (17/44))*√89/5,

I understand the e to the power, but I dont understand the tan part?

For some reason, VCAA uses $\text{cis}$ notation for polar form, which literally no one beyond VCE uses. Essentially, $z=r\text{cis}(\theta)=re^{i\theta},$ where $r=|z|$, and $\theta=\arg(z)$. Note that the CAS will always give the principal argument of $z$.

Thus, when your CAS gives $u=\frac{8+5i}{3+4i}=\frac{\sqrt{89}}{5}e^{-i\arctan(17/44)},$ this means that $|u|=\frac{\sqrt{89}}{5}\quad \text{and}\quad \text{Arg}(u)=-\arctan\left(\frac{17}{44}\right).$

Also how do I write equations in these forums, as I'm having to copy equations from word, and their coming across as not so easy to read once I create a post. Thanks

You can copy equations from Microsoft Word. It will produce some code in LaTeX, which is used to display maths on pages.

You'll need to copy your code and put  $\texttt{$}$ before it, and $\texttt{$}$  after it.

For example, the code  $\texttt{\[\int_{-\infty}^\infty e^{-x^2}dx=\sqrt{\pi}$}\]will produce  $\int_{-\infty}^\infty e^{-x^2}dx=\sqrt{\pi}$ Alternatively, you can check out Rui's amazing $\TeX$ guide to learn how to write the code yourself (which is a lot faster).
2015$-$2017:  VCE
2018$-$2021:  Bachelor of Biomedicine and Concurrent Diploma in Mathematical Sciences, University of Melbourne

#### EverySecondCounts

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« Reply #9470 on: May 20, 2019, 06:32:45 pm »
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Hi, How would one find the short distance between the equation and the point:

{z:|z=6=12i|=|z+9=6i|} and point 8cis(-(-5π)/2)

From what I'm aware the Cartesian equation is:
y=1/2 x-63/12
And that the short distance would be be when the point is perpendicular to the line, but I'm not sure on how to go about creating two simultaneous equations that are of the same form?
Appreciate the help. Thanks - I'll get down to looking throughly at the syntax as soon as I can

#### AlphaZero ##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9471 on: May 21, 2019, 11:11:25 am »
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{z:|z=6=12i|=|z+9=6i|} and point 8cis(-(-5π)/2)

Could you check the equation of the line? You have equals signs within modulus brackets.

Did you mean $\{z:|z+6+12i|=|z+9+6i|\}$ ?
2015$-$2017:  VCE
2018$-$2021:  Bachelor of Biomedicine and Concurrent Diploma in Mathematical Sciences, University of Melbourne

#### EverySecondCounts

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« Reply #9472 on: May 21, 2019, 12:14:31 pm »
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Yep, sorry about that I didn't realise the + had turned into =.
Thankyou, though what would you recommend to solve distance between the polar point and line.
« Last Edit: May 21, 2019, 05:24:04 pm by EverySecondCounts »

#### AlphaZero ##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9473 on: May 21, 2019, 08:07:05 pm »
0
Yep, sorry about that I didn't realise the + had turned into =.
Thankyou, though what would you recommend to solve distance between the polar point and line.

There are a few ways to go about the problem. I'll present the most intuitive way:

Step 1: Convert the question into a problem in the cartesian plane.

Working
Where $z=x+yi$, the corresponding cartesian equation of the line is given by \begin{align*}&L:\ \sqrt{(x+6)^2+(y+12)^2}=\sqrt{(x+9)^2+(y+6)^2} \\
&\implies x^2+12x+36+y^2+24y+144=x^2+18x+81+y^2+12y+36\\
&\implies \boxed{y=\frac{x}{2}-\frac{21}{4}}.\end{align*} Then, we have $u=8\text{cis}\left(\frac{5\pi}{2}\right)=8(0+i)=8i\leadsto (0,\,8).$

Step 2: Find the equation of the line of the line segment that gives the shortest distance.

Working
The length of the line segment joining the point $u$ and the line $L$ will be minimum if it is perpendicular to $L$. So, it has the equation $y-8=-2(x-0)\implies \boxed{y=-2x+8}.$

Step 3: Find the point on the line $L$ where the line segment intersects.

Working
$\frac{x}{2}-\frac{21}{4}=-2x+8\implies x=\frac{53}{10}\\ \implies y=-\frac{13}{5}$

Step 4: Find the distance between the critical points.

Working
$d=\sqrt{\left(\frac{53}{10}-0\right)^2+\left(-\frac{13}{5}-8\right)^2}=\frac{53\sqrt{5}}{10}\,\text{units}.$
2015$-$2017:  VCE
2018$-$2021:  Bachelor of Biomedicine and Concurrent Diploma in Mathematical Sciences, University of Melbourne

#### EverySecondCounts

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« Reply #9474 on: May 21, 2019, 09:48:09 pm »
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Thankyou, I just realised, but another another negative slipped into the point. It should've been: $8cis\left( \frac{-5π}{2} ) \right$, though everything makes sense.
Much appreciated

By any chance, could someone highlight my mistake in using LATEX.Thankyou

#### RuiAce

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« Reply #9475 on: May 21, 2019, 09:49:37 pm »
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By any chance, could someone highlight my mistake in using LATEX.Thankyou
In regards to this, your \right doesn't have a bracket to go with it. Should be \right)

#### EverySecondCounts

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« Reply #9476 on: May 21, 2019, 09:59:30 pm »
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$8cis\left( \frac{-5π}{2} \right)$

Thankyou