Author Topic: 4U Maths Question Thread (Read 667728 times) Tweet Share

bluecookie · « **Reply #1200 on:** May 27, 2017, 12:59:00 pm »

Quote from: RuiAce on May 27, 2017, 09:05:25 am

In essence, we will have a pair of intersecting lines and not just a single line.

Consider what happens as the eccentricity is increased. As the eccentricity increases, the foci become further distanced and the directrices approaches the y-axis. When the eccentricity becomes infinite, the foci will be infinitely away and the directrices coincide on the y-axis.

This also makes intuitive sense, as PS/PM = e. As e->inf, we have either PS -> inf, or PM -> 0.

It may be worth considering that this case is the opposite of the circle, where the foci coincide at the origin to become the centre, and the directrices are infinitely far away

Oh, thanks

beau77bro · « **Reply #1201 on:** May 29, 2017, 09:11:32 am »

Please help question 3

beau77bro · « **Reply #1202 on:** May 30, 2017, 09:18:59 pm »

Hey guys can I grab some help with this whole question I'm seriously struggling

to get it

RuiAce · « **Reply #1203 on:** June 01, 2017, 07:28:10 pm »

Quote from: beau77bro on May 29, 2017, 09:11:32 am

(Image removed from quote.)

Please help question 3

Honestly this question appears nonsensical.

The question says that the particle accelerates, until it reaches the maximum velocity. Then, once it is at the maximum velocity it starts travelling the distance S that it's supposed to.

And then they say $T_1$ is the time taken for the car travel the full distance at maximum velocity. But it doesn't travel the full distance at maximum velocity?

If the question meant something else then it was worded horribly.

RuiAce · « **Reply #1204 on:** June 01, 2017, 07:37:54 pm »

Quote from: beau77bro on May 30, 2017, 09:18:59 pm

Hey guys can I grab some help with this whole question I'm seriously struggling (Image removed from quote.)to get it

The significance is in that it means the force acts against the direction of the position/velocity respectively.
$\begin{align*}\Sigma F = m\ddot{y}& = -2my - 3m\dot{y}\\ \implies \ddot{y} + 3\dot{y} + 2y &= 0\end{align*}$
In the future, where possible please avoid the sideways photos. This request is a bit selfish but it stems from the fact that I personally hate mechanics and get put off from doing it quite easily......

kiwiberry · « **Reply #1205 on:** June 01, 2017, 09:16:09 pm »

Hey, can someone please explain these perms and combs questions, I'm at a loss

1) Four sets of twins go to a party. In how many ways can these children be arranged in pairs so that no child is with his twin?

2) How many ways can three integers be chosen which are in arithmetic progression from the numbers 1 to 101?

RuiAce · « **Reply #1206 on:** June 01, 2017, 10:11:54 pm »

Quote from: kiwiberry on June 01, 2017, 09:16:09 pm

Hey, can someone please explain these perms and combs questions, I'm at a loss

1) Four sets of twins go to a party. In how many ways can these children be arranged in pairs so that no child is with his twin?

2) How many ways can three integers be chosen which are in arithmetic progression from the numbers 1 to 101?

$\text{Illustration of a few cases for the common difference}\\ \text{reveals a pattern that may be exploited for Q2.}$
$\text{When the common difference is 1, we may pick the ordered triplets}\\ (1,2,3),(2,3,4),\dots,(99,100,101)\\ \text{When the common difference is 2, we may pick the ordered triplets}\\ (1,3,5),(2,4,6),\dots,(97,99,101)\\ \text{When the common difference is 3, we may pick the ordered triplets}\\ (1,4,7),(2,5,8),\dots,(95,98,101)$
$\text{This continues up to a common difference of 49}\\ \text{when we may pick the ordered triplets }(1,50,99),(2,51,100),(3,52,101)\\ \text{And lastly for a common difference of 50}\\ \text{we may pick the ordered triplet }(1,51,101)$
__________________________________________________________________
$\text{Of particular interest is the first term in each triplet}\\ \text{Note that the number of triplets we may give is what we desire}\\ \text{The first term gives us a reference point on how many triplets we have}\\ \text{for each common difference.}$
Note: We could have used the second or third term as well. Provided the triplets are ordered, we have a reference point.
$\text{For a common difference of 1, the triplets may start from any number between 1 and 97}\\ \text{Hence, there are 97 suitable choices}\\ \text{For a common difference of 2, the triplets may start from any number between 1 and 95}\\ \text{Hence, there are 95 suitable choices}\\ \text{For a common difference of 3, the triplets may start from any number between 1 and 93}\\ \text{Hence, there are 93 suitable choices}$
$\text{The pattern continues, and eventually for a common difference of 50}\\ \text{the triplets may only start from 1, so we have 1 suitable choice}$
$\text{Recollecting all the cases, the total number of triplets we may collect is}\\ 97+95+93+\dots+1\\ \text{This is nothing more than an AP with first term 1 and last term 97}$
$\text{By considering the common difference, }97 = 1 + 2(n-1) \implies n = 49\\ \text{From here, simply substitute into the formula.}$
I'm not confident with my working out for the first one. Can I please have the answer?

kiwiberry · « **Reply #1207 on:** June 01, 2017, 10:31:59 pm »

Quote from: RuiAce on June 01, 2017, 10:11:54 pm

I'm not confident with my working out for the first one. Can I please have the answer?

Thanks so much omg. The first one is 60

beau77bro · « **Reply #1208 on:** June 02, 2017, 01:15:20 pm »

Quote from: RuiAce on June 01, 2017, 07:28:10 pm

Honestly this question appears nonsensical.

The question says that the particle accelerates, until it reaches the maximum velocity. Then, once it is at the maximum velocity it starts travelling the distance S that it's supposed to.

And then they say $T_1$ is the time taken for the car travel the full distance at maximum velocity. But it doesn't travel the full distance at maximum velocity?

If the question meant something else then it was worded horribly.

Quote from: RuiAce on June 01, 2017, 07:37:54 pm

The significance is in that it means the force acts against the direction of the position/velocity respectively.
$\begin{align*}\Sigma F = m\ddot{y}& = -2my - 3m\dot{y}\\ \implies \ddot{y} + 3\dot{y} + 2y &= 0\end{align*}$
In the future, where possible please avoid the sideways photos. This request is a bit selfish but it stems from the fact that I personally hate mechanics and get put off from doing it quite easily......(Image removed from quote.)(Image removed from quote.)(Image removed from quote.)

thankyou soo much rui, i worked out the first question if you want me to post it? thankyou soo much.

RuiAce · « **Reply #1209 on:** June 02, 2017, 01:17:48 pm »

Quote from: beau77bro on June 02, 2017, 01:15:20 pm

thankyou soo much rui, i worked out the first question if you want me to post it? thankyou soo much.

Up to you. Good idea though for the benefit of the others

RuiAce · « **Reply #1210 on:** June 02, 2017, 11:28:17 pm »

Quote from: kiwiberry on June 01, 2017, 09:16:09 pm

Hey, can someone please explain these perms and combs questions, I'm at a loss

1) Four sets of twins go to a party. In how many ways can these children be arranged in pairs so that no child is with his twin?

2) How many ways can three integers be chosen which are in arithmetic progression from the numbers 1 to 101?

This Q1 took me ages to do and I had to call in my second year math friends to help me out. It dumbfounded me big time until I realised that I had mucked up the very last piece of my analysis...
$\text{The problem with tackling the approach directly is mainly in how the subsequent cases follow}\\ \text{Once the first two are picked for pair 1, the arrangements for the remaining groups}\\ \text{is actually dependent on which twins }(A,B,C\text{ or }D)\text{ we put into the first group}\\ \text{This branches out into far too many cases and overcomplicates the analysis.}$
$\text{As the direct approach is too messy, we turn to the complement}\\ \text{However, the complement is much harder to analyse as it is }\textit{full of repetition}\\ \text{"Double counting is the bane of combinatorics"}$
__________________________________
$\textit{Introducing the inclusion-exclusion principle}\\ \text{In set notation, for two sets it is written as this: }|A \cup B| = |A| + |B| - |A \cap B|$
$\text{Translation: If you have two different things someone can do, the amount of people who do one }\textbf{or}\text{ the other}\\ \text{Will be the sum of the amount who do at least the first task }A\text{, plus the amount who do task }B\\ \text{however }\textbf{less}\text{ the amount that do both tasks }A\textbf{ and }B$
$\text{The reason for this subtraction is to eliminate the error caused by double-counting}\\ \text{It is very much possible for there to be an }\textbf{overlap}\text{ in what someone does}\\ \text{When we consider those who do at least }A\text{, we bluntly ignore whether or not they do }B\\ \text{and vice versa.}$
$\text{Hence, to avoid double counting, we basically throw it out.}$
___________________________________
$\text{As a demonstration for 3 sets}\\ |A \cup B \cup C| = (|A| + |B| + |C|) - (|A \cap B| + |A \cap C| + |B \cap C|) + |A\cap B \cap C|\\ \text{Observe how the signs alternate: Plus, Minus, Plus}$
$\text{The short explanation for this involves using the same logic as above}\\ \text{We first consider the amount that do at least one of three tasks}\\ \text{but then reject those that do two of three tasks}$
$\text{Unfortunately, when we reject two of three tasks, we double count and subtract twice those that do }\textbf{all three}\\ \text{So instead of subtracting again, we have to }\textbf{add it back}$
$\text{However, this question will involve the inclusion-exclusion principle with }\textbf{four}\text{ sets}$
__________________________________________________________________________________________________
$\text{We return to the question.}$
$\text{First, we keep in mind that the total number of possible outcomes is the number of ways}$ $\text{we simply put 8 people into 4 groups of 2}.$
$\text{We choose 2 out of 8 to put into the first pair, 2 out of 6 to put into the second pair}\\ \text{and keep going. This yields}\\ \binom82\binom62\binom42\binom22$
$\text{However, the groups are all of the same size (2 people in each pair)}\\ \text{Since we have 4 groups of the same size, we have a repetition of }4!\\ \text{which we must divide out}$
$\text{Hence the total possible outcomes is }\binom82\binom62\binom42\binom22\frac{1}{4!} = 105$
___________________________________
$\text{In considering the complement}\\ \text{The unfavourable possible outcomes is when one or more twins are paired together}\\ \text{We use inclusion/exclusion to assist us.}$
$\text{For the number of ways in which at least 1 pair of twins can be paired together}\\ \text{Choose 1 pair of twins to be paired: }\binom41\\ \text{Then, arrange the rest just like above (keeping in mind the repetition: }\binom62\binom42\binom22\frac{1}{3!}\\ \text{So for the case of one set, we have }\binom43\binom62\binom42\binom22\frac{1}{3!}=60$
$\text{For the number of ways in which at least 2 pair of twins can be paired together}\\ \text{Choose 2 pairs of twins to be paired: }\binom42\\ \text{Then similarly we have }\binom42\binom22 \frac{1}{2!}\\ \text{So for two sets we have }18$
$\text{As for three sets, it's easier as whilst there's still }\binom43\text{ ways of choosing groups}\\ \text{There's only going to be }\binom22 = 1\text{ way of putting the last two in a group}\\ \text{So for three sets we have }4$
$\text{It's the easiest for four sets, as there's literally only 1 way (alternatively }\binom44)\\ \text{of forcing all 4 twins to be paired up}$
___________________________________
$\text{Recollecting, by the inclusion-exclusion principle the number of unfavourable outcomes is}\\ 60-18+4-1 = 45$
$\text{So the number of favourable outcomes is }105 - 45 = 60$

kiwiberry · « **Reply #1211 on:** June 03, 2017, 11:37:04 am »

Quote from: RuiAce on June 02, 2017, 11:28:17 pm

This Q1 took me ages to do and I had to call in my second year math friends to help me out. It dumbfounded me big time until I realised that I had mucked up the very last piece of my analysis...

Oh wow. Thanks Rui, that question was much more complicated than I anticipated :O

RuiAce · « **Reply #1212 on:** June 03, 2017, 03:22:03 pm »

That being said, the post was excessively long because Ext 2 students aren't expected to just know the inclusion-exclusion principle. I had to define it first before I could use it

Wales · « **Reply #1213 on:** June 04, 2017, 02:02:42 pm »

Having a bit of trouble with conics. Part 2. Would be appreciated if you could guide me in it right direction.

Cheers, Wales

RuiAce · « **Reply #1214 on:** June 04, 2017, 03:07:29 pm »

Quote from: Wales on June 04, 2017, 02:02:42 pm

Having a bit of trouble with conics. Part 2. Would be appreciated if you could guide me in it right direction.

Cheers, Wales

ST should be easy to find because T is on the x-axis so the distance is horizontal.

Hint: Did you consider the statement PS = ePM (or equivalently PS/PM = e, the focus-directrix definition)?

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