Author Topic: 4U Maths Question Thread (Read 665491 times) Tweet Share

RuiAce · « **Reply #1950 on:** July 25, 2018, 09:12:09 am »

Quote from: envisagator on July 24, 2018, 08:00:49 pm

Hi Rui, for the inequality proof question posted as an image below, I've been looking at it for a while, just cant see the actually logical reason to why the part highlighted in the box is true. Plus, when i do some questions, is it 'normal' just to sorta guess how some of your facts are bigger or smaller than another fact such that when you combine them you can get easily get to the desired result. I dont know whether that makes sense, also is it necessary to write equality iff - bla bla bla?? Thank You in advance, sorry for it being a mouthful to take in

\begin{align*}a^2+2ab+b^2 &\leq 2a^2+2b^2\\ \frac{a^2+2ab+b^2}{4}&\leq \frac{a^2+b^2}{2}\\ \left( \frac{a+b}{2} \right)^2 &\leq \frac{a^2+b^2}{2}\end{align*}
Note that $ 2ab \leq a^2+b^2$ comes from rearranging $ \frac{a+b}{2} \geq \sqrt{ab} $.

I'm not sure if the lecture handout is uploaded yet but the basic facts that you can assume for inequality proofs are all there. There really isn't that many (surprisingly), but people tend to struggle because they don't know what they are, which is what I tried to address.

cthulu · « **Reply #1951 on:** July 26, 2018, 01:54:28 pm »

Hello, having trouble with this one integration question.

$\int \frac{5sec^{2}x}{16tan^2x+25}dx$

The solution is:

$\frac{1}{4}tan^{-1}(\frac{4}{5}tanx)+C$

But I am getting:

$\frac{5}{16}x - tan^{-1}(\frac{4}{5}tanx)+C$

RuiAce · « **Reply #1952 on:** July 26, 2018, 02:04:57 pm »

Quote from: cthulu on July 26, 2018, 01:54:28 pm

Hello, having trouble with this one integration question.

$\int \frac{5sec^{2}x}{16tan^2x+25}dx$

The solution is:

$\frac{1}{4}tan^{-1}(\frac{4}{5}tanx)+C$

But I am getting:

$\frac{5}{16}x - tan^{-1}(\frac{4}{5}tanx)+C$

$\text{When you sub }u=\tan x \implies du = \sec^2\,dx\text{ you should get}\\ \int \frac{5}{25+16u^2}\,du\\ \text{which can be shown to integrate to}\\ \frac14 \tan^{-1} \frac{4u}{5} + C$
$\text{Which becomes their given answer.}\\ \text{Where are you getting an }x\text{ from? I can't find a need to use integration by parts.}$

cthulu · « **Reply #1953 on:** July 26, 2018, 02:27:45 pm »

Quote from: RuiAce on July 26, 2018, 02:04:57 pm

$\text{When you sub }u=\tan x \implies du = \sec^2\,dx\text{ you should get}\\ \int \frac{5}{25+16u^2}\,du\\ \text{which can be shown to integrate to}\\ \frac14 \tan^{-1} \frac{4u}{5} + C$
$\text{Which becomes their given answer.}\\ \text{Where are you getting an }x\text{ from? I can't find a need to use integration by parts.}$

Wow thanks, I did it a stupid way that is why I was getting a constant.

Another question:

$\text{The complex number }z \text{ is a function of the real number }t, \text{ given that }z = \frac{t-i}{t+i} \text{ for } 0\leq t\leq 1. \text{Describe the locus of } z \text{ in the argand diagram.}$

RuiAce · « **Reply #1954 on:** July 26, 2018, 03:27:12 pm »

Quote from: cthulu on July 26, 2018, 02:27:45 pm

Wow thanks, I did it a stupid way that is why I was getting a constant.

Another question:

$\text{The complex number }z \text{ is a function of the real number }t, \text{ given that }z = \frac{t-i}{t+i} \text{ for } 0\leq t\leq 1. \text{Describe the locus of } z \text{ in the argand diagram.}$

Whilst the question is doable by considering $x+iy = \frac{t^2-1}{t^2+1} - \frac{2ti}{t^2+1}$, I find that it's only really viable for the equation of the locus, and becomes kinda useless when dealing with the restriction on $t$. More or less because optimising $ f(t) = \frac{t^2-1}{t^2+1} $ and $g(t) = \frac{-2t}{t^2+1}$ over $0\leq t \leq 1$ is really handwavy unless we use calculus, but that's not just one, but two disgusting quotient rule bashes.
$\text{Let }z=x+iy\text{ so that we have}\\ \begin{align*}x+iy &= \frac{t-i}{t+i}\\ (x+iy)(t+i) &= t-i\\(xt-y) + i(yt+x) &= t-i \end{align*}$
$\text{So upon equating the two parts,}\\ \begin{align*}xt-y &= t\tag{1}\\ yt+x &= -1 \tag{2} \end{align*}$
$\text{If we assume that }y\neq 0\text{ temporarily,}\\ \text{(2) rearranges to }\boxed{t = -\frac{x+1}{y}}\\ \text{which upon subbing into }(1)\text{ gives us}\\ \begin{align*} -\frac{x(x+1)}{y} - y &= -\frac{x+1}{y}\\ (x^2+x) + y^2 &= x+1\\ x^2+y^2&=1 \end{align*}$
________________________________________________
$\text{So the locus of }z\text{ is the unit circle.}\\ \text{But what parts of the unit circle are actually covered?}$
$\text{By default, }-1\leq x\leq 1\text{ and }-1 \leq y \leq 1.\\ \textbf{However,}\text{ if we assume that }y > 0\\ \text{then }t = -\frac{x+1}{y}\text{ will always be a negative number.}$
$\text{Except this is false, as we've established that }t\geq 0.\\ \text{So }\boxed{y < 0}\\ \text{and in particular, }\boxed{-1 \leq y < 0}$
$\text{Note that therefore }-y > 0\text{ and }-y\leq 1.\\ \text{Now using the fact that }t \leq 1\text{ we have}\\ \begin{align*} -\frac{x+1}{y} & \leq 1\\ \therefore (x+1)&\leq -y \leq 1 \end{align*}$
$\text{Which proves that }x\leq 0\text{ as well.}\\ \text{So we've narrowed our focus down to the third quadrant,}\\ \text{under the assumption that }y \neq 0.$
________________________________________________
$\text{We now fill in the gap of when }y = 0.$
$\text{If }y=0\text{, equation (2) tells us that }x=-1.\\ \text{Subbing that into (1) gives }-t = t\text{ which implies }t=0.\\ \text{This is perfectly viable, so }(-1,0)\text{ belongs to the solution as well.}$
$\text{So the conclusion is that the locus is the unit circle }x^2+y^2=1\\ \text{in the third quadrant, including the boundary point }(-1,0)$
Note that $ (0,-1) $ is also included in the locus. But it was actually taken care of during the messier part of the question; it didn't need to be handled separately. May be a good idea to restate it as well just in case.

cthulu · « **Reply #1955 on:** July 26, 2018, 04:28:31 pm »

Quote from: RuiAce on July 26, 2018, 03:27:12 pm

Thanks for the reply, I was doing this a little differently with less working out:

Find |z| which is 1 so the locus lies on the unit circle.
Rationalise z and split it up so it is in the form z = x + iy
Use 0<=t<=1 with x and y to determine domain and arg(z) and then draw the locus. However when trying to find the arg I got 0<=arg(z)<= arctan(-2/0). Here was where I was stuck.

Your response is very good, really helps explain everything!

Another question if you don't mind:

Ali_Abbas · « **Reply #1956 on:** July 26, 2018, 07:10:18 pm »

Quote

Another question:

$\text{The complex number }z \text{ is a function of the real number }t, \text{ given that }z = \frac{t-i}{t+i} \text{ for } 0\leq t\leq 1. \text{Describe the locus of } z \text{ in the argand diagram.}$

Rui's solution is a slight overkill, so I recommend the following solution as an (easier) alternative.

$\begin{align*} z &= \frac{t-i}{t+i} \\ \Longrightarrow |z| &= |\frac{t-i}{t+i}| \\ &= \frac{|t-i|}{|t+i|} \end{align*}$
$\text{But } t+i = \overline{t-i}, \text{ so } |t-i| = |t+i|. \text{ Therefore}$
$|z| = 1$
$\text{To determine the range of values of Arg}(z), \text{ observe that}$
$\begin{align*} \text{Arg}(z) &= \text{Arg}(\frac{t-i}{t+i}) \\ &= \text{Arg}(t-i) - \text{Arg}(\overline{t-i}) \\ &= 2\text{Arg}(t-i) \end{align*}$
$\text{From here, simply draw up the first quadrant of the Argand plane (since } 0 \leq t \leq 1), \text{ and identify the vector } t-i \\ \text{(this will be the vector whose tail is at } i \text{ and whose tip is at } t), \text{ then compute Arg}(z) \text{ for } t = 0 \text{ and } t = 1 \\ \text{realizing that the remaining possible values of Arg}(z) \text{ will lie between these two values. You should get}$
$2(-\frac{\pi}{2}) \leq \text{Arg}(z) \leq 2(-\frac{\pi}{4}) \\ \Longrightarrow -\pi \leq \text{Arg}(z) \leq -\frac{\pi}{2}$
$\text{Therefore, the locus of } z \text{ is the arc of the unit circle that lies in the third quadrant, including} \\ \text{the points } (-1, 0) \text{ and } (0, -1).$

RuiAce · « **Reply #1957 on:** July 26, 2018, 08:35:10 pm »

The main reason why I avoid the arg when handling such problems is because arg properties are just hard to use and we cannot assume anything about atan2 in HSC maths. Use of arg relies exclusively on properties that don't always necessarily hold.

Plus, the concept of the principal argument isn't really well defined in the HSC. There's no real distinction between arg(z) and Arg(z) as there should be. That makes the properties of the argument more dangerous to me.

It is definitely at the expense of overkill, which is why I'm not necessarily the biggest fan of my own method. But I'm way more sure of what I'm doing without it.

Of course, as for the modulus, it's definitely easier to just compute $|z|$ directly. I just didn't anticipate that it would actually be a unit circle by inspection; I was expecting something more bizarre.

Quote from: cthulu on July 26, 2018, 04:28:31 pm

Thanks for the reply, I was doing this a little differently with less working out:
Find |z| which is 1 so the locus lies on the unit circle.
Rationalise z and split it up so it is in the form z = x + iy
Use 0<=t<=1 with x and y to determine domain and arg(z) and then draw the locus. However when trying to find the arg I got 0<=arg(z)<= arctan(-2/0). Here was where I was stuck.

Your response is very good, really helps explain everything!

Another question if you don't mind:

(Image removed from quote.)

I'll take a closer look at this soon but in the meantime is there a diagram attached to this? My hangry brain cannot visualise it right now

Edit: Actually, more importantly, are there any assumptions on the normal reaction forces of the particles? Or the masses of the particles? There's no indication that the particles actually travel with the same velocity here.

cthulu · « **Reply #1958 on:** July 27, 2018, 09:39:55 am »

Quote from: RuiAce on July 26, 2018, 08:35:10 pm

Removed

Oh yes, sorry I should have shown the first part too, my bad.

hassrax · « **Reply #1959 on:** July 27, 2018, 09:55:37 am »

Hi I need help with this question below on how to approach it and the logic behind the steps. Thank you
Sketch y=f(x)=(x^n)e^-x for x>0, n>1
Thank you

RuiAce · « **Reply #1960 on:** July 27, 2018, 11:21:31 am »

Quote from: cthulu on July 27, 2018, 09:39:55 am

Oh yes, sorry I should have shown the first part too, my bad.

(Image removed from quote.)

This is just a brainstorm here:

Assume that the angular displacement goes on forever, i.e. after it reaches $2\pi$ it'll keep going to $4\pi$, $6\pi$ and so on.

Using $v = r\omega$ you can deduce that $\omega = \sqrt{ \frac{g\tan \theta}{R} } $. Using $ \theta = \omega t$ you can then deduce that for each particle, $ \theta = t \sqrt{ \frac{g\tan \theta}{R} } $. Note that since the particles start side by side, we just naturally assume that their initial angular displacement is $\theta = 0$ for convenience.

By subbing in $g=10$ along with both corresponding values of $\tan\theta$ and $R$, you can deduce which of the two particles is travelling faster.

Then, the first time the particles are next to each other should be when $ \boxed{ \theta_{faster} - \theta_{slower} = 2\pi}$. In fact, the second time they are next to each other should also be when this difference equals to 4*pi. Essentially, at each point where their angular displacements differ by some integer multiple of $2\pi$, they are next to each other.

Quote from: hassrax on July 27, 2018, 09:55:37 am

Hi I need help with this question below on how to approach it and the logic behind the steps. Thank you
Sketch y=f(x)=(x^n)e^-x for x>0, n>1
Thank you

This question is simply screaming "multiplication of $y$-coordinates" and that's all there is to it.

Are you uncomfortable with the whole process of multiplying ordinates and require that logic to be expanded for your example?

scienceislife · « **Reply #1961 on:** August 18, 2018, 04:52:21 pm »

Find the volume bounded by y = 10 and y = x + 16/x about x = −2

RuiAce · « **Reply #1962 on:** August 18, 2018, 06:38:34 pm »

Quote from: scienceislife on August 18, 2018, 04:52:21 pm

Find the volume bounded by y = 10 and y = x + 16/x about x = −2

Do you want to use cylindrical shells or washers (slicing)?

scienceislife · « **Reply #1963 on:** August 18, 2018, 06:55:03 pm »

Either method would be great, because I've tried both but cannot seem to get a reasonable answer.

RuiAce · « **Reply #1964 on:** August 18, 2018, 07:10:51 pm »

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