Author Topic: 3U Maths Question Thread (Read 1520967 times) Tweet Share

winstondarmawan · « **Reply #2865 on:** October 12, 2017, 09:02:01 am »

Would appreciate help with part (iii).
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22414754_1346898165435703_21769541_n.png?oh=00926eb7cbe7a08f284db25f57318094&oe=59E00EB2
TIA

teapancakes08 · « **Reply #2866 on:** October 12, 2017, 11:29:11 am »

Need help on Question 14(c)(iii) from MX1 2012

RuiAce · « **Reply #2867 on:** October 12, 2017, 05:45:34 pm »

Quote from: teapancakes08 on October 12, 2017, 11:29:11 am

Need help on Question 14(c)(iii) from MX1 2012

$\text{There's a slight subtlety in that the question gave you the rate}\\ \text{in kilometres per }\textbf{hour}\\ \text{yet they want the rate of change 5 }\textbf{minutes}\text{ after take-off}$
Which means this question can probably be done in multiple ways.
$\frac{dr}{du} = \frac{2u-\cos \alpha}{2\sqrt{1 + u^2 - u\cos \alpha}}$
$\text{Suppose we need the rate of change in kilometres per }\textbf{hour}.\\ \begin{align*}\frac{dr}{dt}&=\frac{dr}{du}\times \frac{du}{dt}\\ &= \frac{2u-\cos \alpha}{2\sqrt{1+u^2-u\cos \alpha}} \times 360\end{align*}$
$\text{Note that since }\frac{du}{dt} = 360\text{ we have }u = 360t\text{ where }t\text{ is the time in hours}\\ \text{So when }\boxed{t=\frac1{12}}\text{ (because 5 minutes = 1/12 hours)}, u = 30$
Then just plug $u=30$ back into the formula above. Again, a reminder that this gives the rate in kilometres per hour. If we wanted kilometres per minute, we would have something else.

Dragomistress · « **Reply #2868 on:** October 14, 2017, 07:52:09 pm »

Do you need to remember the locus shortcuts or is it better just to derive them?

RuiAce · « **Reply #2869 on:** October 14, 2017, 07:53:56 pm »

Quote from: Dragomistress on October 14, 2017, 07:52:09 pm

Do you need to remember the locus shortcuts or is it better just to derive them?

All the 2U things are provided on the formula sheet.

For the 3U parametrics stuff, such as the equation of the chord, tangent, ...
Technically, you need to know how to do both. If they want you to derive the equation of the tangent, you should feel very comfortable doing so. But now that it is also on your reference sheet, if you need the equation of the tangent but aren't told explicitly to derive it, you can just quote off your sheet.

fireives1967 · « **Reply #2870 on:** October 14, 2017, 08:01:06 pm »

16. twelve people are to be seated around a table.
(a) in how many ways can they be seated?
(b) in how many ways can they be seated if 2 particular people are not to be put together?
(c) Find the probability that 2 friends will be seated together.

How do you answer b & c?, answers are 32 659 200 for b, and 2/11 for c.
Thanks!

RuiAce · « **Reply #2871 on:** October 14, 2017, 08:02:29 pm »

Quote from: fireives1967 on October 14, 2017, 08:01:06 pm

16. twelve people are to be seated around a table.
(a) in how many ways can they be seated?
(b) in how many ways can they be seated if 2 particular people are not to be put together?
(c) Find the probability that 2 friends will be seated together.

How do you answer b & c?
Thanks!

Lacking some information here. Do we assume a round table or just an ordinary one where people sit in a row?

fireives1967 · « **Reply #2872 on:** October 14, 2017, 08:08:46 pm »

Quote from: RuiAce on October 14, 2017, 08:02:29 pm

Lacking some information here. Do we assume a round table or just an ordinary one where people sit in a row?

Just a round one!

RuiAce · « **Reply #2873 on:** October 14, 2017, 08:15:36 pm »

$\text{Assuming that we have a round table}\\ \text{We recall that in a) we had to }\textbf{fix}\text{ someone's position}\\ \text{before arranging the rest.}\\ \text{This gave the formula }(n-1)! = 11!.$
$\text{In part b), we }\textbf{deal with all restrictions first}\\ \text{In the two that can't sit together, say, persons A and B}\\ \text{We fix the position of one of them.}$
$\text{We fix person A's position}\\ \text{Then person B only has 9 possible seats remaining}\\ \text{to ensure they don't sit with person A.}$
$\text{The remaining 10 people can still be arranged anywhere}\\ \text{This will still have 10! arrangements.}$
$\therefore\text{ Answer} = \boxed{9\times 10!}$
______________________________________
$\text{For the favourable outcomes, we will }\textbf{address the restriction first}\text{ again.}\\ \text{We first count the number of arrangements for the two friends}\\ \text{There will only be }2!\text{ arrangements.}$
This is because either friend 1 sits to the left of friend 2, or other way around.
$\text{Now that the two friends have been arranged}\\ \text{we group them and }\textbf{treat them as one person}\\ \text{Now that we've done that, we are essentially arranging}\\ \text{only }\textbf{11}\text{ people in a circle}\\ \text{which will involve }10!\text{ arrangements}$
$\text{Hence the number of favourable outcomes is }2\times 10!$
The sample space (all possibilities) involves 11! outcomes in total, from part a).
$\text{So the probability we require is }\boxed{\frac{2\times 10!}{11!} = \frac{2}{11}}$

Dragomistress · « **Reply #2874 on:** October 14, 2017, 09:05:22 pm »

How am I meant to effectively find out the area of the graph?

Happy Physics Land · « **Reply #2875 on:** October 14, 2017, 11:24:00 pm »

Quote from: Dragomistress on October 14, 2017, 09:05:22 pm

How am I meant to effectively find out the area of the graph?

"Remember to draw a diagram" are the words on every mathematician's tombstone.

RuiAce · « **Reply #2876 on:** October 14, 2017, 11:29:11 pm »

Quote from: Happy Physics Land on October 14, 2017, 11:24:00 pm

"Remember to draw a diagram" are the words on every mathematician's tombstone.

(Image removed from quote.)
(Image removed from quote.)

Mate, you overkilled the area

Quote from: Dragomistress on October 14, 2017, 09:05:22 pm

How am I meant to effectively find out the area of the graph?

$\text{Using }A=\frac12 bh$

$\text{The base is the distance from }Y\text{ to }Z\\ \text{which is just }\frac{5}{4}$
$\text{The height is the distance from }X\text{ to the }x\text{-axis}\\ \text{which will be }2.$
$\text{Plugging it in, }A=\frac{5}{4}$

Happy Physics Land · « **Reply #2877 on:** October 14, 2017, 11:32:08 pm »

Quote from: RuiAce on October 14, 2017, 11:29:11 pm

Mate, you overkilled the area $\text{Using }A=\frac12 bh$
(Image removed from quote.)
$\text{The base is the distance from }Y\text{ to }Z\\ \text{which is just }\frac{5}{4}$
$\text{The height is the distance from }X\text{ to the }x\text{-axis}\\ \text{which will be }2.$
$\text{Plugging it in, }A=\frac{5}{4}$

When advanced linear algebra makes doubt simple mathematics

RuiAce · « **Reply #2878 on:** October 14, 2017, 11:40:56 pm »

lol

Extracurricular - Linear algebra method

$\text{Projecting the region onto the }x-y\text{ plane}\\ \text{in the }x-y-z\text{ hyperplane}$
$\text{We have the vectors}\\ \begin{align*}\overrightarrow{YZ} &= \begin{pmatrix}5/4\\0\\ 0\end{pmatrix}\\ \overrightarrow{YX}&= \begin{pmatrix}4\\2\\0\end{pmatrix}\end{align*}$
$\text{The parallelogram spanned by these two vectors will have area}\\ \begin{align*}\text{Area}&= \left|\overrightarrow{YZ}\times\overrightarrow{XZ} \right|\\ &= \left|\begin{pmatrix}0\\0\\5/2\end{pmatrix} \right|\\ &= \frac{5}{2}\end{align*}$
$\text{Since the area of the triangle is exactly half}\\ \text{the area of the parallelogram}\\ \text{our required area is still }\frac54$

Dragomistress · « **Reply #2879 on:** October 15, 2017, 08:17:19 am »

May I know why you obtained the distance between x and the x-axis as the height?

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