Author Topic: 4U Maths Question Thread (Read 664983 times) Tweet Share

RuiAce · « **Reply #585 on:** October 15, 2016, 05:10:46 pm »

Quote from: nancy_cc 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

$\text{In any circumstance, you may choose which method to use.}\\ \text{As I am more of a fan of }dx\text{ than }dy\text{, I will use shells.}$
$\text{The equation of that circle is }x^2+(y-2)^2=9\\ \text{The }x\text{-intercept is at }(\sqrt{5},0)$
$\text{Since we are considering the bottom half, the equation of the semicircle is}\\ y=2-\sqrt{9-x^2}$
$\text{So by the method of cylindrical shells:}\\ \begin{align*}\delta V&=2\pi rh\, \delta x\\ &=2\pi x (2-\sqrt{9-x^2})\delta x\end{align*}\\ \text{Hence, }V=2\pi\int_\sqrt{5}^3(2x-x\sqrt{9-x^2})dx\\ \text{This is now just something annoying. Split the integral up and}\\ \text{use }u=9-x^2\text{ for }-\int_\sqrt{5}^3x\sqrt{9-x^2}dx$
__________________________________
$\text{If instead we were to use slices}\\ \text{The equation of the relevant semicircle is }x=\sqrt{9-(y-2)^2}$
$\text{Now, note that the outer radius is just }R=3\\ \text{But the inner radius is that exact equation.}\\ \text{Hence, by the method of slices we have }\\ \begin{align*}\delta V&=\pi(R^2-r^2)\delta y\\ &= \pi(9-(9-(y-2)^2)) \delta y\\ &= \pi(y-2)^2\end{align*}$
$\text{Hence the volume now is }\pi \int_0^2(y-2)^2dy\\ \text{It appears that the slicing method produced a much neater integral here.}$

RuiAce · « **Reply #586 on:** October 15, 2016, 05:34:18 pm »

$\textbf{2012 HSC 4U}$

Quote from: znaser on October 15, 2016, 05:07:44 pm

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

$\text{It is easier to use the formula}\\ Probability=\frac{\text{Favourable Outcomes}}{\text{Total Outcomes}}$
$\text{Let's consider the first }k\text{ integers first.}$
$\text{Our total outcomes is simple. Note that there are }n\text{ different numbers to which we get }k\text{ from}\\ \text{This can be done in }n^k\text{ ways.}$
$\text{Our favourable outcomes now denies repetition. So we have }^nP_k\\ \text{i.e. }n(n-1)(n-2)\dots (n-k+1)$
$\text{So the probability of }k\text{ different integers is }\frac{n(n-1)\dots (n-k+1)}{n^k}$
$\text{We now need to select an integer from the }k\text{ already chosen. This is done in }\binom{k}{1}=k\text{ ways}\\ \text{The probability of getting a specific integer is }\frac{1}{n}\\ \text{So here we have }\frac{k}{n}$
$\text{Multiply that in, and we arrive at the first line of BOSTES' working out.}$
Note: Consider an alternate source of answers to BOSTES' ones.
$\text{It now falls to do battle with algebra.}\\ \text{Fortunately there's not too much algebra needed.}\\ \text{We already have }\frac{(n-1)(n-2)\dots(n-k+1)k}{n^k}$
$\text{It should be clear that upon multiplying by }\frac{(n-k)!}{(n-k)!}\text{ we are done.}$

znaser · « **Reply #587 on:** October 15, 2016, 05:40:24 pm »

Quote from: RuiAce on October 15, 2016, 05:34:18 pm

$\textbf{2012 HSC 4U}$ $\text{It is easier to use the formula}\\ Probability=\frac{\text{Favourable Outcomes}}{\text{Total Outcomes}}$
$\text{Let's consider the first }k\text{ integers first.}$
$\text{Our total outcomes is simple. Note that there are }n\text{ different numbers to which we get }k\text{ from}\\ \text{This can be done in }n^k\text{ ways.}$
$\text{Our favourable outcomes now denies repetition. So we have }^nP_k\\ \text{i.e. }n(n-1)(n-2)\dots (n-k+1)$
$\text{So the probability of }k\text{ different integers is }\frac{n(n-1)\dots (n-k+1)}{n^k}$
$\text{We now need to select an integer from the }k\text{ already chosen. This is done in }\binom{k}{1}=k\text{ ways}\\ \text{The probability of getting a specific integer is }\frac{1}{n}\\ \text{So here we have }\frac{k}{n}$
$\text{Multiply that in, and we arrive at the first line of BOSTES' working out.}$
Note: Consider an alternate source of answers to BOSTES' ones.
$\text{It now falls to do battle with algebra.}\\ \text{Fortunately there's not too much algebra needed.}\\ \text{We already have }\frac{(n-1)(n-2)\dots(n-k+1)k}{n^k}$
$\text{It should be clear that upon multiplying by }\frac{(n-k)!}{(n-k)!}\text{ we are done.}$

Sweet! Thanks!

nancy_cc · « **Reply #588 on:** October 15, 2016, 05:46:38 pm »

Quote from: RuiAce on October 15, 2016, 05:10:46 pm

$\text{In any circumstance, you may choose which method to use.}\\ \text{As I am more of a fan of }dx\text{ than }dy\text{, I will use shells.}$
$\text{The equation of that circle is }x^2+(y-2)^2=9\\ \text{The }x\text{-intercept is at }(\sqrt{5},0)$
$\text{Since we are considering the bottom half, the equation of the semicircle is}\\ y=2-\sqrt{9-x^2}$
$\text{So by the method of cylindrical shells:}\\ \begin{align*}\delta V&=2\pi rh\, \delta x\\ &=2\pi x (2-\sqrt{9-x^2})\delta x\end{align*}\\ \text{Hence, }V=2\pi\int_\sqrt{5}^3(2x-x\sqrt{9-x^2})dx\\ \text{This is now just something annoying. Split the integral up and}\\ \text{use }u=9-x^2\text{ for }-\int_\sqrt{5}^3x\sqrt{9-x^2}dx$
__________________________________
$\text{If instead we were to use slices}\\ \text{The equation of the relevant semicircle is }x=\sqrt{9-(y-2)^2}$
$\text{Now, note that the outer radius is just }R=3\\ \text{But the inner radius is that exact equation.}\\ \text{Hence, by the method of slices we have }\\ \begin{align*}\delta V&=\pi(R^2-r^2)\delta y\\ &= \pi(9-(9-(y-2)^2)) \delta y\\ &= \pi(y-2)^2\end{align*}$
$\text{Hence the volume now is }\pi \int_0^2(y-2)^2dy\\ \text{It appears that the slicing method produced a much neater integral here.}$

Wow I didn't realise to use the bottom half and kept getting stuck! Thank you that really helped!

MarkThor · « **Reply #589 on:** October 15, 2016, 07:47:33 pm »

Hi guys,
Could you guys help me out with this probability multiple choice question
I always seem to get the harder probability questions wrong, does anyone have any advice for this topic that could help?

RuiAce · « **Reply #590 on:** October 15, 2016, 07:56:45 pm »

$\textbf{2013 HSC MX2}$

Quote from: MarkThor on October 15, 2016, 07:47:33 pm

Hi guys,
Could you guys help me out with this probability multiple choice question
I always seem to get the harder probability questions wrong, does anyone have any advice for this topic that could help?

$\text{6 people must be put into 4 rooms.}\\ \text{There are just far too many cases.}$
$\text{In times like this, we consider the complementary event.}\\ \text{The complementary event is either one of the following happening:}\\ \text{1. 5 people in a room}\\ \text{2. 6 people in a room.}$
$\text{Total outcomes: }4^6$
$\text{If 6 people are all in one room, we just need to choose a room. This is done in }\binom{4}{1}\text{ ways}$
$\text{If 5 people are all in one room, we choose that room first: }\binom{4}{1}\\ \text{Then, we put 6 of the 5 people in there: }\binom{6}{5}\\ \text{And then the last guy has to be chucked in one of the remaining rooms: }\binom{3}{1}\\ \text{So the number of ways 5 are in a room is given by }\binom{4}{1}\binom{6}{5}\binom{3}{1}=72$
$\text{Hence we require }4^6 - (4 + 72) = 4020$
_________________

Well this is the most annoying topic ever in the entire 4U course. Every question you can be hit by may be unique. Always break it down and choose a fastest route.

MarkThor · « **Reply #591 on:** October 15, 2016, 08:24:48 pm »

Cool thanks for the breakdown Rui
I absolutely dread probability

RuiAce · « **Reply #592 on:** October 15, 2016, 08:27:26 pm »

Quote from: MarkThor on October 15, 2016, 08:24:48 pm

Cool thanks for the breakdown Rui
I absolutely dread probability

You should see me. I'm dying with my first year uni combinatorics.

MarkThor · « **Reply #593 on:** October 16, 2016, 08:47:07 am »

That sounds way too intense

massive · « **Reply #594 on:** October 16, 2016, 10:37:46 am »

Hey guys how do you part c, the answer is: Incorrect, n=10²⁰ is one counter example

wth does this mean, could someone explain the q and answer please!

Thankss

RuiAce · « **Reply #595 on:** October 16, 2016, 10:52:47 am »

Quote from: massive on October 16, 2016, 10:37:46 am

Hey guys how do you part c, the answer is: Incorrect, n=10²⁰ is one counter example

wth does this mean, could someone explain the q and answer please!

Thankss

$\text{To prove a "for all" statement is incorrect}\\ \text{We need to give a concrete, numeric counterexample.}$
$\text{From part b), we know that }\\ 1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}+\dots+\frac{1}{\sqrt{N}}=\sum_{k=1}^N\frac{1}{\sqrt{k}} > 2(\sqrt{N+1}-1)\\ \text{Swapping sides we have}\\ 2(\sqrt{N+1}-1) < \sum_{k=1}^N\frac{1}{\sqrt{k}}$
$\text{Suppose, by way of contradiction}\\ \sum_{k=1}^N\frac{1}{\sqrt{k}} < 10^{10} \quad \text{ for all positive integers }N\\ \text{Then, combining inequalities, we have}\\ 2(\sqrt{N+1}-1) < 10^{10}\quad \text{for all positive integers }N$
$\text{Because the statement holds for ALL integers }N\\ \text{If we put }N=10^{20}\text{ it should still hold.}\\ \text{Let us test this proposition.}$
$\begin{align*}2(\sqrt{10^{20}+1}-1)&\approx 2\times 10^{10}\tag{calculator}\\ & \ge 10^{10}\end{align*}$
$\text{Hence we arrive at a contradiction.}\\ \text{So since all of our steps were mathematically correct, the original assumption was wrong.}$

massive · « **Reply #596 on:** October 16, 2016, 11:23:08 am »

Quote from: RuiAce on October 16, 2016, 10:52:47 am

$\text{To prove a "for all" statement is incorrect}\\ \text{We need to give a concrete, numeric counterexample.}$
$\text{From part b), we know that }\\ 1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}+\dots+\frac{1}{\sqrt{N}}=\sum_{k=1}^N\frac{1}{\sqrt{k}} > 2(\sqrt{N+1}-1)\\ \text{Swapping sides we have}\\ 2(\sqrt{N+1}-1) < \sum_{k=1}^N\frac{1}{\sqrt{k}}$
$\text{Suppose, by way of contradiction}\\ \sum_{k=1}^N\frac{1}{\sqrt{k}} < 10^{10} \quad \text{ for all positive integers }N\\ \text{Then, combining inequalities, we have}\\ 2(\sqrt{N+1}-1) < 10^{10}\quad \text{for all positive integers }N$
$\text{Because the statement holds for ALL integers }N\\ \text{If we put }N=10^{20}\text{ it should still hold.}\\ \text{Let us test this proposition.}$
$\begin{align*}2(\sqrt{10^{20}+1}-1)&\approx 2\times 10^{10}\tag{calculator}\\ & \ge 10^{10}\end{align*}$
$\text{Hence we arrive at a contradiction.}\\ \text{So since all of our steps were mathematically correct, the original assumption was wrong.}$

OHHH that makes more sense now, thankyou. But in an exam situation, how do you come up with a number like 10²⁰ to prove by contradiction

RuiAce · « **Reply #597 on:** October 16, 2016, 11:26:25 am »

Quote from: massive on October 16, 2016, 11:23:08 am

OHHH that makes more sense now, thankyou. But in an exam situation, how do you come up with a number like 10²⁰ to prove by contradiction

For that particular question, the inspiration was that

If N was very large, then sqrt(N+1)-1 is about equal to sqrt(N).

And sqrt(10²⁰) = 10¹⁰

But then because it's a proof, we always need to look at what we're trying to prove. We want to prove that it is greater than 10¹⁰. Quite fortunately enough, we actually had 2x10¹⁰ by our intuition.

With proof by contradiction, you need to look ahead. You need to try to think about how you would fail it. You firstly decide wait, does this make sense or not, before you try finding a counterexample.
Note that, for that question though, you could've picked say 10¹⁰⁰ and you would still have a valid counterexample.

nancy_cc · « **Reply #598 on:** October 16, 2016, 01:15:05 pm »

Hey could someone help with this question please!

RuiAce · « **Reply #599 on:** October 16, 2016, 01:18:30 pm »

Quote from: nancy_cc on October 16, 2016, 01:15:05 pm

Hey could someone help with this question please!

Part a) was done recently here in post #533

Use a similar analysis to try part b) and c) first. Come back with your progress

Search the forums now!

We have moved!

Author Topic: 4U Maths Question Thread (Read 664983 times) Tweet Share

RuiAce

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

znaser

Re: 4U Maths Question Thread

nancy_cc

Re: 4U Maths Question Thread

MarkThor

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

MarkThor

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

MarkThor

Re: 4U Maths Question Thread

massive

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

massive

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

nancy_cc

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

Recent Posts

User:
Password: