July 24, 2019, 06:41:40 am

### AuthorTopic: VCE Specialist 3/4 Question Thread!  (Read 1065842 times) Tweet Share

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#### Flight112

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##### Re: VCE Specialist 3/4 Question Thread!
« 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

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##### 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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##### Re: VCE Specialist 3/4 Question Thread!
« 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.

#### AlphaZero

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9468 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).
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#### AlphaZero

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9469 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 $\{z:|z+6+12i|=|z+9+6i|\}$ ?
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#### AlphaZero

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9470 on: May 21, 2019, 08:07:05 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.

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
$d=\sqrt{\left(\frac{53}{10}-0\right)^2+\left(-\frac{13}{5}-8\right)^2}=\frac{53\sqrt{5}}{10}\,\text{units}.$
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#### EverySecondCounts

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9471 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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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9472 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)

#### Jackson.Sprigg

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9473 on: May 23, 2019, 04:39:50 pm »
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Hey guys just wondering what exactly this notation is saying?

0 ≤ y ≤ f(x)

Why do they use both f(x) and y?

#### AlphaZero

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9474 on: May 23, 2019, 04:51:31 pm »
+1
Hey guys just wondering what exactly this notation is saying?

0 ≤ y ≤ f(x)

Why do they use both f(x) and y?

What you can infer from this statement depends on the context of the problem.

All those inequations are just saying that $y$ is bounded below by zero and bounded above by some function of $x$.

Graphically, draw the line $y=0$, and draw some curve $y=f(x)$ (which obviously has to be above or equal to the $x$-axis). Then shade between the graphs - that's what the inequations represent.
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#### Jackson.Sprigg

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9475 on: May 24, 2019, 08:59:35 pm »
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Thank you!  I'm not even sure what I didn't understand about it before tbh.

Anywho...

For the attached question in part d. I seem to have to integrate  xe^x ?? Should I know how to do this by hand? Or am I missing an easier alternative?

#### AlphaZero

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9476 on: May 24, 2019, 10:28:43 pm »
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Thank you!  I'm not even sure what I didn't understand about it before tbh.

Anywho...

For the attached question in part d. I seem to have to integrate  xe^x ?? Should I know how to do this by hand? Or am I missing an easier alternative?

Indeed, to find the area of the region, you will have find an antiderivative of  $xe^x$.

You are not expected to know how to do this from scratch (which involves integration by parts), however in this question, you can just use integration by recognition.

In part a you would have found that $f'(x)=e^x+xe^x.$ From this, we can deduce that \begin{align*}\int \Big(e^x+xe^x\Big)dx&=xe^x\\
\implies \int e^xdx +\int xe^xdx&=xe^x\\
\implies \int xe^xdx&=xe^x-\int e^xdx\\ \end{align*}$\implies \boxed{\int xe^xdx=xe^x-e^x+c\,}$
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#### Tatlidil

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9477 on: May 28, 2019, 10:32:21 pm »
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Hey guys, how do you find the "closest" distance of a particle from the origin and the "furthest" distance of a particle from the origin? (Vector calculus)
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#### AlphaZero

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9478 on: May 28, 2019, 11:04:37 pm »
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Hey guys, how do you find the "closest" distance of a particle from the origin and the "furthest" distance of a particle from the origin? (Vector calculus)

Suppose we have a particle which has the position vector $\mathbf{r}(t)$ at time $t$ relative to an origin $O$.

The distance from the origin to the particle is given by $|\mathbf{r}(t)|$ which should be quite easy to minimise/maximise using knowledge from Mathematical Methods.
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#### Tatlidil

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9479 on: June 02, 2019, 05:45:53 pm »
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Suppose we have a particle which has the position vector $\mathbf{r}(t)$ at time $t$ relative to an origin $O$.

The distance from the origin to the particle is given by $|\mathbf{r}(t)|$ which should be quite easy to minimise/maximise using knowledge from Mathematical Methods.
Ohh so we just get the magnitude and use the equation to find the min, max?
Lets say after finding |r(t)| its sqrt 2sin(x), would you get sqrt 2 or is it something else?
« Last Edit: June 02, 2019, 05:49:14 pm by Tatlidil »
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