Author Topic: Mathematics Question Thread (Read 1313690 times) Tweet Share

_____ · « **Reply #2505 on:** August 24, 2017, 09:51:43 pm »

Quote from: RuiAce on August 24, 2017, 09:19:28 pm

$\text{A lot of breaking down the question and 'altering' it is required}\\ \text{for this particular question. Substitution is just tediousness.}\\ \text{Note that we wish to prove that }x=\frac{r^2+r\sqrt{r^2+8}}{4}\textbf{ maximises }\ell$
That second derivative is FAR too ridiculous. So we will consider testing both sides instead.
___________________________
$\text{Now, we wish to study the }\textbf{sign}\text{ of }\frac{d\ell}{dx} = \frac{(x-1)(2x^2-r^2x-r^2)}{\sqrt{r^2-x^2}\sqrt{1-2x+r^2}\left(\sqrt{r^2-x^2}+x\sqrt{1-2x+r^2}\right)}$
$\textbf{But},\text{ clearly that long denominator is }\textbf{always positive}\\ \text{So the sign of }\frac{d\ell}{dx}\text{ is the same}\\ \text{as the sign of }(x-1)(2x^2-r^2x-r^2)$
$\textbf{But also,}\text{ we know that }0 < r < 1\text{, so by consequence }0 < x < 1\\ \text{Hence }x-1 < 0\text{, i.e. is negative.}\\ \text{Hence the sign of }\frac{d\ell}{dx}\\ \text{is now always the }\textbf{negative}\text{ of the sign of }2x^2-r^2x-r^2$
___________________________
$\text{It falls now to study the sign of the quadratic }y=2x^2-r^2x-r^2\\ \text{Here is the plot.}$
(Image removed from quote.)
$\text{It appears that if we go a bit to the right of }x=\frac{r^2+r\sqrt{r^2+8}}{4}\\ 2x^2-r^2x-r^2\text{ is positive, and hence }\boxed{\frac{d\ell}{dx}\text{ is negative}}$
$\text{It also appears that if we go a bit to the left of }x=\frac{r^2+r\sqrt{r^2+8}}{4}\\ 2x^2-r^2x-r^2\text{ is negative, and hence }\boxed{\frac{d\ell}{dx}\text{ is positive}}$
$\text{The boxed conditions are }\textbf{exactly}\text{ what we want}\\ \text{to prove that }x=\frac{r^2+r\sqrt{r^2+8}}{4}\text{ maximises }\ell$
Note that in general, there may not always be a nice approach. It varies from question to question how you could approach them. These ones are some of the worst types in 2U.

This will be added to the compilation thread.

Thanks a bunch.

I think I understand what you've done but there's no way I would have thought of doing that.

Just so I can understand your process for approaching different types of questions, may I ask how you'd prove a min TP with the question below (2003 HSC 9c). It's probably more typical of the difficulty level they tend to go for as the value to prove is simpler (V = 3/2U).

RuiAce · « **Reply #2506 on:** August 24, 2017, 10:05:09 pm »

Quote from: _____ on August 24, 2017, 09:51:43 pm

Thanks a bunch.

I think I understand what you've done but there's no way I would have thought of doing that.

Just so I can understand your process for approaching different types of questions, may I ask how you'd prove a min TP with the question below (2003 HSC 9c). It's probably more typical of the difficulty level they tend to go for as the value to prove is simpler (V = 3/2U).

(Image removed from quote.)

$\text{You could }\textit{potentially}\text{ find the second derivative here as it's a tiny bit better}\\ \text{But given that a fraction is there, testing both sides is still preferable.}$
$\text{The difference is that this time we can test both sides directly.}$
In the future, please provide relevant progress. Here, the derivative could've helped
$\text{We know that }v=\frac{3u}{2}\text{ minimises }E\\ \text{For clarity, we can write this as a decimal }v=1.5u$
$\text{Then, a bit to the left could be }1.4u\text{, and a bit to the right could be }1.6u\\ \text{Plugging }v=1.4u\text{ we see}\\ \begin{align*}\frac{dE}{dv}&= \frac{(1.4u-u)(3aL \times 1.4^2 u^2)-aL \times 1.4^3 u^3}{(1.4u-u)^2}\\ &= \frac{2.352aLu^3 - 2.744 aLu^3}{0.16u^2}\\ &= -2.45 aLu\\ &< 0 \end{align*}$
$\text{We can say this as we know }u > 0\text{ and }L > 0\\ a > 0\text{ as well but it wasn't really explicit; it was more implied.}$
$\text{Similarly, when }v=1.6u\text{ you can check that }\frac{dE}{dv} > 0\\ \text{So we have the required conditions to prove we have a min}$

Leah_Mer · « **Reply #2507 on:** August 27, 2017, 06:25:57 pm »

Quote from: Opengangs on August 23, 2017, 08:08:38 pm

Hey, Leah_Mer!

Your internal mark doesn't matter per se, rather your ranking and mark difference is. So, as you probably do know, the person who places first in the course will take the highest external mark to be their internal mark, and everyone will get scaled according to the gap between these respective ranks.

So, if you and the top 3 marks are relatively close, you'll receive a relative 'close' internal mark (maybe 2 - 4 mark difference). So, even if you didn't perform well internally, you can still make up for it with your external mark. Even if you have a weak cohort, placing fourth should still put you in a good position for a band 6 HSC mark.

Thankyou! That's great news to hear! Also a big thankyou to atar notes in general, forum advice and the notes I've purchased at lectures throughout the year have really made all the difference to my marks and taken so much stress away when dealing with exams! The maths topic tests have been a big help too, so thanks for everything you guys do

jakesilove · « **Reply #2508 on:** August 28, 2017, 10:49:38 am »

Quote from: Leah_Mer on August 27, 2017, 06:25:57 pm

Thankyou! That's great news to hear! Also a big thankyou to atar notes in general, forum advice and the notes I've purchased at lectures throughout the year have really made all the difference to my marks and taken so much stress away when dealing with exams! The maths topic tests have been a big help too, so thanks for everything you guys do

So glad that you've benefited from the forums and the notes! We'll be here for you for the rest of the year (and beyond!) so make sure to ask as many questions as you'd like

Daniyahasan · « **Reply #2509 on:** August 28, 2017, 12:02:26 pm »

Hey guys
Can someone help me with this question

RuiAce · « **Reply #2510 on:** August 28, 2017, 12:59:14 pm »

Quote from: Daniyahasan on August 28, 2017, 12:02:26 pm

Hey guys
Can someone help me with this question

$\textbf{These questions are fairly standard}\\ \text{and you need to get used to doing them.}$
$\text{If }C=Ae^{-0.14t}\text{ then by just differentiating and rearranging}\\ \begin{align*}\frac{dC}{dt} &= -0.14 \times Ae^{-0.14t}\\ &=-0.14 \times C\end{align*}$
_________
$\text{When }t=0, C=130\\ \text{So just substituting in}\\ 130 = Ae^0 \implies A =1 30$
_________
$C=130e^{-0.14t}\\ \text{When }t=7\\ C=130e^{-0.14\times 7}=\text{whatever the calculator says}$
_________
$\text{Since the initial concentration is 130}\\ \text{We require the time when }C=65:\\ \begin{align*}65&=130 e^{-0.14t}\\ \frac12 &= e^{-0.14t}\\ -0.14t &= \ln \frac12\\ t&=-\frac{1}{0.14}\ln \frac12\\ &= \text{whatever the calculator says}\end{align*}$

daniel.hu3 · « **Reply #2511 on:** August 28, 2017, 04:53:11 pm »

Hey Guys, can someone please help me with this question?

A box of rectangular cross-section sits on a train luggage rack as shown with the point C touching the wall. P, the point in contact with the rack, is the midpoint of AD. If D is 8 cm from the wall and P, the edge of the rack, is 20 cm from the wall, find how far A is from the wall.
Hint: Draw parallels to the wall through A, P and D.

RuiAce · « **Reply #2512 on:** August 28, 2017, 06:56:44 pm »

Quote from: daniel.hu3 on August 28, 2017, 04:53:11 pm

Hey Guys, can someone please help me with this question?

A box of rectangular cross-section sits on a train luggage rack as shown with the point C touching the wall. P, the point in contact with the rack, is the midpoint of AD. If D is 8 cm from the wall and P, the edge of the rack, is 20 cm from the wall, find how far A is from the wall.
Hint: Draw parallels to the wall through A, P and D.

$\text{The diagram is drawn so that wherever parallel and perpendicular lines are likely there}\\ \text{they certainly ARE there, and thus we have rectangles.}$

$\text{We are told }PF = 20\text{ and }DE = 8.\text{ We require }AG.$
$\text{As }SF = 8\text{ and }PF = 20, PS = 12$
$\triangle PDS ||| \triangle ADR\text{ as they are clearly equiangular}\\ \text{Hence we have the proportional sides }\frac{DP}{DA} = \frac{PS}{AR}$
$\text{But }P\text{ is the midpoint of }AD,\text{ so }\frac{DP}{DA} = \frac12\\ \therefore \frac12 = \frac{12}{AR} \implies AR = 24$
$\text{So as }RG = 8\\ AG = 24 + 8 = \boxed{32}$
Extra intuition: As everything is linear, an arithmetic progression is unsurprising.

_____ · « **Reply #2513 on:** August 29, 2017, 03:09:07 pm »

Quick question from 2005 HSC:

How do I prove theta = <DBE here for (i)? I understand how to do the rest of it from there (finding DE in terms of theta and then FE using DE).

Thanks

Opengangs · « **Reply #2514 on:** August 29, 2017, 03:33:32 pm »

Quote from: _____ on August 29, 2017, 03:09:07 pm

Quick question from 2005 HSC:

$\text{Proof: } \angle DAB = \angle FDE = \theta \text{ (corresponding angles in parallel lines, where AB || DE)}$ $\text{Now, } \angle ABD = \angle EDB \text{(alternate angles in parallel lines)}$ $\therefore \angle EBD = 90 - \angle ABD = 90 - \angle EDB = \theta$

Shadowxo · « **Reply #2515 on:** August 29, 2017, 04:22:41 pm »

Quote from: _____ on August 29, 2017, 03:09:07 pm

Quick question from 2005 HSC:

(Image removed from quote.)

How do I prove theta = <DBE here for (i)? I understand how to do the rest of it from there (finding DE in terms of theta and then FE using DE).

Thanks

Or, even easier:

$\angle ABD = 90-\theta \text{ as angles in a triangle add to 180 degrees} \\ \text{So }\angle DBE = 90 - (90-\theta ) = \theta \text{ As ABE is 90 degrees}$

fantasticbeasts3 · « **Reply #2516 on:** August 31, 2017, 10:17:39 pm »

don't know if it's me who can't read, or the question isn't worded nicely

RuiAce · « **Reply #2517 on:** August 31, 2017, 10:34:11 pm »

Quote from: fantasticbeasts3 on August 31, 2017, 10:17:39 pm

don't know if it's me who can't read, or the question isn't worded nicely

$\text{Set }A_n\text{ as the amount owing after }n\text{ months}\\ \text{The corresponding effective interest rate is 1.5 percent per month}\\ \text{Denote the monthly repayment as }M$
$\text{Since the first payment only happens in the third month, we have}\\ \begin{align*}A_1 &= 2000(1.015)\\ A_2 &= 2000(1.015)^2\\ A_3&=2000(1.015)^3 - M\end{align*}$
$\text{Using the usual recursive approach}\\ \begin{align*}A_4 &= 2000(1.015)^4 - M(1 + 1.015)\\ A_5 &= 2000(1.015)^6 - M(1+1.015+1.015^2)\\ &\vdots \\ A_n &= 2000(1.015)^n - M(1+1.015+1.015^2 + \dots + 1.015^{n-3})\\ &= 2000(1.015)^n - M\left(\frac{1.015^{n-2}-1}{0.015}\right)\end{align*}$
$\text{We require }A_{36}=0\\ \therefore 2000(1.015)^{36} - M\left(\frac{1.015^{34}-1}{0.015}\right) = 0\implies M = 77.80647704\dots \approx 77.81$
___________________________________
$\text{Since each payment is }77.806\dots\text{ and we make 34 payments}\\ \text{The amount we pay intotal is }77.806\dots \times 34 \approx 2645.42$

winstondarmawan · « **Reply #2518 on:** September 01, 2017, 06:53:45 pm »

Hey! Would appreciate help with the following:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21269494_1314379005354286_1994981189_n.jpg?oh=f219784caf6eb433c41fbdf2e1531b4a&oe=59ABAA26
TIA!

RuiAce · « **Reply #2519 on:** September 01, 2017, 07:02:38 pm »

Quote from: winstondarmawan on September 01, 2017, 06:53:45 pm

Hey! Would appreciate help with the following:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21269494_1314379005354286_1994981189_n.jpg?oh=f219784caf6eb433c41fbdf2e1531b4a&oe=59ABAA26
TIA!

$\text{This is an example of a GP nested in another GP}\\ \text{The subtleties are hard to explain without Taylor series, so take it for granted}$
$\text{Note that }\frac{2}{5^2} = \frac{1}{5^2} +\frac{1}{5^2}\\ \frac{3}{5^3}=\frac{1}{5^3}+\frac1{5^3}+\frac1{5^3}\\ \text{and so on.}$
$\text{Hence, by rearrangement of the relevant terms:}\\ \begin{align*}& \quad \frac15 + \frac{2}{5^2}+\frac3{5^3}+\frac4{5^4}+\frac5{5^5}+\dots\\ &= \left(\frac15 +\frac1{5^2}+\frac1{5^3}+\frac1{5^4}+\frac1{5^5}+\dots \right)\\ &\quad \quad \,+ \left( \frac1{5^2}+\frac1{5^3}+\frac1{5^4}+\frac1{5^5}+\dots\right)\\ & \quad \quad\quad \quad\,\,+ \left(\frac1{5^3}+\frac1{5^4}+\frac1{5^5}+\dots\right)\\ & \quad \quad\quad \quad\quad\quad\,\,\,+ \left(\frac1{5^4}+\frac1{5^5}+\dots\right)\\ & \quad\quad\quad\quad\quad\quad\quad\quad\,\,\,\, +\left(\frac{1}{5^5}+\dots\right)\\ & \quad\quad\quad\quad\quad\quad\quad\quad\,\,\,\, \,\quad\quad \ddots\end{align*}$
$\text{Summing an infinite number of GPs, we see this is equal to}\\ \begin{align*} &\quad\frac15 \left(\frac{1}{1-\frac15}\right) + \frac1{5^2}\left(\frac{1}{1-\frac15}\right)+\frac1{5^3}\left(\frac{1}{1-\frac15}\right) +\frac1{5^4}\left(\frac{1}{1-\frac15}\right)+\frac1{5^5}\left(\frac{1}{1-\frac15}\right)+\dots\\ &= \left(\frac{1}{1-\frac15}\right)\left(\frac15+\frac1{5^2}+\frac1{5^3}+\frac1{5^4}+\frac1{5^5}+\dots\right)\end{align*}$
$\text{So summing that new GP, we realise this sum is just}\\ \frac15 \left(\frac{1}{1-\frac15}\right)^2 = \frac{5}{16}$

Search the forums now!

We have moved!

Author Topic: Mathematics Question Thread (Read 1313690 times) Tweet Share

_____

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

Leah_Mer

Re: Mathematics Question Thread

jakesilove

Re: Mathematics Question Thread

Daniyahasan

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

daniel.hu3

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

_____

Re: Mathematics Question Thread

Opengangs

Re: Mathematics Question Thread

Shadowxo

Re: Mathematics Question Thread

fantasticbeasts3

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

winstondarmawan

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

Recent Posts

User:
Password: