Author Topic: 4U Maths Question Thread (Read 665227 times) Tweet Share

RuiAce · « **Reply #570 on:** October 10, 2016, 08:40:53 am »

Quote from: Justina Shehata on October 09, 2016, 11:49:10 pm

would the working out in the sample answers be a good indication to how much they require?

Maths sample answers are somewhat superior to every other course because they answer the question. But they may or may not make sense and you should use Excel Success One or MANSW published books.

If it's too late into the HSC and you don't want to spend any more money then use the answers that can be found here.

That being said, you shouldn't have to ask to know how much you need to answer. You should just be able to do the question and LEAVE it there. Mark allocations only indicate a RELATIVE quantity of working out required, not absolute. Some 3 markers will require more effort than other 3 markers.

massive · « **Reply #571 on:** October 10, 2016, 04:08:49 pm »

Quote from: RuiAce on October 09, 2016, 10:17:09 am

$\text{The total possible outcomes is just }4^6\text{ as we have 6 people and 4 rooms}$

How does this work? i've never really understood how do it and when to apply it.

jakesilove · « **Reply #572 on:** October 10, 2016, 04:10:59 pm »

Quote from: massive on October 10, 2016, 04:08:49 pm

How does this work? i've never really understood how do it and when to apply it.

Basically, you have 4 choices where to put the first person, 4 choices where to put the second person.... 4 choices where to put the sixth person, so

$4*4*4*4*4*4=4^6$

katherine123 · « **Reply #573 on:** October 10, 2016, 11:43:29 pm »

first ques: the answer assumes 0<=t<=1 for part ii) and i dont get why thats the case
2nd ques; dont know how to do part iii)

RuiAce · « **Reply #574 on:** October 10, 2016, 11:58:07 pm »

Quote from: katherine123 on October 10, 2016, 11:43:29 pm

first ques: the answer assumes 0<=t<=1 for part ii) and i dont get why thats the case
2nd ques; dont know how to do part iii)

You've already asked the second question and it was answered on the previous page.
_______________________
$\text{For the first one, the assumption }0\le t\le 1\\ \text{ was made because of the }\textbf{boundaries on the definite integral.}$
$\text{The definite integral is only for between }0\text{ and }1.\\ \text{Any other value of }t\text{( e.g. }t< 0)\text{ is useless.}$
$\text{Because we don't know the graph of }y=t^ne^{-t}\\ \text{The obvious assumption to be made is based }\textbf{off the definite integral.}$
$\text{Indeed, when }0\le t\le 1\text{, this inequality is clearly true.}\\ 0 \le t^ne^{-t} \le t^n\\ \text{Reason being obviously }0\le e^{-t} \le 1\text{ for }t\ge 0$
$\text{Because we restricted }0\le t\le 1\text{, we're still safe in putting }\int_0^1dt\text{ around it}\\ \text{Hence }\int_0^1 0dt\le \int_0^1t^ne^{-t}dt\le\int_0^1 t^n$

kiwiberry · « **Reply #575 on:** October 12, 2016, 08:45:47 pm »

how would you find the asymptotes for a parametric equation like
$x=\frac{2t+1}{t+1} , y=\frac{t^{2}}{t+1}$

RuiAce · « **Reply #576 on:** October 12, 2016, 09:29:27 pm »

Quote from: kiwiberry on October 12, 2016, 08:45:47 pm

how would you find the asymptotes for a parametric equation like
$x=\frac{2t+1}{t+1} , y=\frac{t^{2}}{t+1}$

$\textbf{Note that in general the methods are extremely complicated.}$
$\text{Clearly, }t\neq -1\\ \text{We may attempt to convert the parametric equation to a Cartesian equation}\\ \text{to study the behaviour.}$
$\text{Note that the }t+1\text{ can be cancelled out by division. Hence, attempt}\\ \frac{y}{x}=\frac{t^2}{2t+1}\\ \text{As }t\to -1, \frac{y}{x}\to -1$
$\text{This suggests that the line }y=-x\text{ is an asymptote.}$
$\text{That shows us an oblique asymptote.}\\ \text{The question then arises as to if there are vertical asymptotes.}\\ \text{Rewriting: }x=2-\frac{1}{t+1}$
$\text{That's all we need to show that }x\neq 2\\ \textit{Do you see why?}$
$\text{There may be more, but that's all I can just deduce by intuition.}$

kiwiberry · « **Reply #577 on:** October 13, 2016, 04:28:52 pm »

Quote from: RuiAce on October 12, 2016, 09:29:27 pm

$\textbf{Note that in general the methods are extremely complicated.}$
$\text{Clearly, }t\neq -1\\ \text{We may attempt to convert the parametric equation to a Cartesian equation}\\ \text{to study the behaviour.}$
$\text{Note that the }t+1\text{ can be cancelled out by division. Hence, attempt}\\ \frac{y}{x}=\frac{t^2}{2t+1}\\ \text{As }t\to -1, \frac{y}{x}\to -1$
$\text{This suggests that the line }y=-x\text{ is an asymptote.}$
$\text{That shows us an oblique asymptote.}\\ \text{The question then arises as to if there are vertical asymptotes.}\\ \text{Rewriting: }x=2-\frac{1}{t+1}$
$\text{That's all we need to show that }x\neq 2\\ \textit{Do you see why?}$
$\text{There may be more, but that's all I can just deduce by intuition.}$

Sorry could you please explain why x≠2? Is it because as t --> infinity, x --> 2?

jakesilove · « **Reply #578 on:** October 13, 2016, 04:37:35 pm »

Quote from: kiwiberry on October 13, 2016, 04:28:52 pm

Sorry could you please explain why x≠2? Is it because as t --> infinity, x --> 2?

That's exactly right! x will never QUITE reach 2, it will always be slightly less than 2

RuiAce · « **Reply #579 on:** October 13, 2016, 04:39:13 pm »

aoifera · « **Reply #580 on:** October 14, 2016, 04:18:17 pm »

Hey could I have a hand with this question? I'm not sure which formulae to use

jakesilove · « **Reply #581 on:** October 14, 2016, 04:23:44 pm »

Quote from: aoifera on October 14, 2016, 04:18:17 pm

Hey could I have a hand with this question? I'm not sure which formulae to use

If there are 10 revolutions per minute, then a particle on the circumference will move 10 circumferences per minute, or

$10*2\pi*r=10*5*2*\pi$

So, it will move 100*pi cm per minute making the answer C

RuiAce · « **Reply #582 on:** October 14, 2016, 04:37:45 pm »

Quote from: aoifera on October 14, 2016, 04:18:17 pm

Hey could I have a hand with this question? I'm not sure which formulae to use

i.e. (So this is just adding to Jake's answer)

This is your very standard mechanics formula v=rω

The only extra step is the conversion between "revolutions" to something more useful for ω. Which is standard for 4U so you must be aware of it.

nancy_cc · « **Reply #583 on:** October 15, 2016, 04:58:15 pm »

Could anyone have a look at this question I'm stuck on? Would you use slices or shells? Or is it a different method that I'm missing? Thanks

znaser · « **Reply #584 on:** October 15, 2016, 05:07:44 pm »

Hi. I can't seem to understand the working out of part (i).

Search the forums now!

We have moved!

Author Topic: 4U Maths Question Thread (Read 665227 times) Tweet Share

RuiAce

Re: 4U Maths Question Thread

massive

Re: 4U Maths Question Thread

jakesilove

Re: 4U Maths Question Thread

katherine123

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

kiwiberry

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

kiwiberry

Re: 4U Maths Question Thread

jakesilove

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

aoifera

Re: 4U Maths Question Thread

jakesilove

Re: 4U Maths Question Thread

RuiAce

4U Maths Question Thread

nancy_cc

Re: 4U Maths Question Thread

znaser

Re: 4U Maths Question Thread

Recent Posts

User:
Password: