Author Topic: 4U Maths Question Thread (Read 665269 times) Tweet Share

RuiAce · « **Reply #915 on:** February 28, 2017, 09:11:33 pm »

Quote from: michaelalt on February 28, 2017, 09:05:44 pm

I have no idea how to do part b. and c. of this question. I've tried watching videos but the examples are all different, and theres no pattern to them so I'm having a hard time working them out...

$\text{b) is one you need to know}\\ \text{We say that the line is the }\textit{perpendicular bisector}\text{ of the points}\\ (-2,2)\text{ and }(4,0)$
$\text{For the complex numbers}a+bi\text{ and }c+di\\ \text{the perpendicular bisector ALWAYS has formula}\\ |z-(a+bi)|=|z-(c+di)|$
$\text{The reason is best explained by intuition}\\ |z|\text{ is the distance from the complex number }0+0i\text{ on the Argand diagram}\\ \text{But instead, }|z-(a+bi)|\text{ is the distance from the number }a+bi$
$\text{Hence, we want every complex number that is }\textit{equidistant}\\ \text{between }a+bi\text{ and }c+di$

$\text{Note then that the answer is going to be}\\ |z - 4i| = |z - (-2+2i)|$

RuiAce · « **Reply #916 on:** February 28, 2017, 09:13:36 pm »

cutiepie30 · « **Reply #917 on:** February 28, 2017, 10:41:52 pm »

Quote from: RuiAce on February 28, 2017, 07:03:34 pm

$\text{This trick is standard for 4U and the proof needs to be memorised}\\ \text{In the RHS, perform the substitution }u=a-x$
$\begin{align*}\int_0^a f(a-x)dx &= \int_a^0 f(u)(-du)\\ &= \int_0^a f(u)du\\ &= \int_0^a f(x)dx\end{align*}$
$\text{The substitution flips the boundaries but the negative sign flips it back}\\ \text{Then, the rule that }\int_a^bf(x)dx = \int_a^bf(u)du\text{ should make intuitive sense}\\ \text{When performing definite integration, the variables vanish, leaving the same answer.}$

Thankyou RuiAce

!! Makes sense now !!

Kle123 · « **Reply #918 on:** March 01, 2017, 09:10:25 pm »

Hey Rui, could you (or anybody) help me with this question. Thank you

RuiAce · « **Reply #919 on:** March 01, 2017, 09:19:50 pm »

Quote from: Kle123 on March 01, 2017, 09:10:25 pm

Hey Rui, could you (or anybody) help me with this question. Thank you

I'm trying to look for any shortcuts right now. I'll edit this post later.

Chances are that this is an annoying brute-force question.
_____________________

Edit: Can't see anything, so here goes.

$\text{Let }T\text{ be the point }(x_0,y_0)\\ \text{Then, note that }PQ\text{ is just the chord of contact from }T\\ \text{Hence, it can be shown that }PQ\text{ has equation }\frac{xx_0}{a^2}+\frac{yy_0}{b^2}=1\\ \text{This proof is left as your exercise. Come back if you're stuck.}$
$\text{Then, noting that }m_{OT}=\frac{y_0}{x_0}\\ \text{The equation of the line }OT\text{ is just }y=\frac{y_0}{x_0}x$
$\text{To find }S\text{, sub this equation into the chord of contact:}\\ \begin{align*}\frac{xx_0}{a^2}+\frac{xy_0^2}{x_0b^2}&=1\\ x&=\frac{1}{\frac{x_0}{a^2}+\frac{y_0^2}{x_0b^2}}\\ &= \frac{a^2b^2x_0}{b^2x_0^2+a^2y_0^2}\\ \text{a}&\text{nd}\\ \frac{yx_0^2}{y_0a^2}+\frac{yy_0}{b^2}&=1\\ y&= \frac{1}{\frac{x_0^2}{y_0a^2}+\frac{y_0}{b^2}}\\ &= \frac{a^2b^2y_0}{b^2x_0^2+a^2y_0^2} \end{align*}\\ \text{So }S\text{ is the point }\left(\frac{a^2b^2}{b^2x_0^2+a^2y_0^2}x_0 , \frac{a^2b^2}{b^2x_0^2+a^2y_0^2} y_0\right)$
$\text{Now suppose, w.l.o.g}\\ x_0 > 0\text{ and }y_0>0\text{ as in the diagram}\\ \text{i.e. }T\text{ is the first quadrant.}\\ \text{To find }R,\text{ solve }y=\frac{y_0}{x_0}x\text{ with the equation of the ellipse}$
$\begin{align*}\frac{x^2}{a^2}+\frac{x^2y_0^2}{x_0^2b^2}&=1\\ x^2 &= \frac{1}{\frac{1}{a^2}+\frac{y_0^2}{x_0^2b^2}}\\ &= \frac{a^2b^2x_0^2}{b^2x_0^2+a^2y_0^2}\\ x&=\frac{abx_0}{\sqrt{b^2x_0^2+a^2y_0^2}}\end{align*}\\ \text{having taken positive square roots because of the assumption }T\text{ in first quadrant} \\\text{and something similar for }y$
Try chucking all of that into the distance formula. Hopefully something good appears. I also worry that I made some algebra mistakes along the way.

Kle123 · « **Reply #920 on:** March 02, 2017, 08:48:44 pm »

Haha Rui, i stopped about the same place where you stopped too, where it was about to get really messy. Frankly, I think this is a question is a waste of time. Thanks again for your help Rui, i'll just leave this question incomplete because i think its a waste of time.

RuiAce · « **Reply #921 on:** March 02, 2017, 08:56:04 pm »

I'm with you. Huge waste of time tbh

lsong · « **Reply #922 on:** March 02, 2017, 09:44:41 pm »

Hi,
I'm having trouble with taking the limits of an equation to find where the curve approaches. For example, in the question below, I know I could take points but I would still like to know how to take the limits of this question just for future reference.
Thanks in advanced!

RuiAce · « **Reply #923 on:** March 02, 2017, 09:51:45 pm »

Quote from: lsong on March 02, 2017, 09:44:41 pm

Hi,
I'm having trouble with taking the limits of an equation to find where the curve approaches. For example, in the question below, I know I could take points but I would still like to know how to take the limits of this question just for future reference.
Thanks in advanced!

$\text{This is just a graphing question}\\ \text{It is clear that there is a vertical asymptote }x=-1$
$\text{And we turn to limits for horizontal asymptotes}\\ \text{Note that }\lim_{x\to \infty}\frac{1}{x+1}=\lim_{x\to \infty}\frac{\frac1x}{1+\frac1x}=0\\ \text{So as }x\to \infty,\\ y\to x-1+0 = x-1\\\text{ revealing an oblique asymptote }y=x-1$
Your question is not clear. What exactly are you asking? I didn't test points here.

lsong · « **Reply #924 on:** March 02, 2017, 09:58:09 pm »

Ah, sorry about that.
Basically, when you get the asymptotes, how do you find out which region the curve lies in (i.e. where it approaches the asymptotes)?
Hopefully I was clearer this time .-.

RuiAce · « **Reply #925 on:** March 02, 2017, 10:05:18 pm »

Methods of intuition take heaps of practice. I started sketching curves on GeoGebra and in my head randomly to develop the intuition needed to just "SEE" how exactly the asymptote is approached, i.e. from above or below.

Techniques I employ include splitting each region between vertical asymptotes up, and observing the presence of both x-intercepts and stationary points. Horizontal/Oblique asymptotes are used for any region not bound between two vertical asymptotes.

E.g. for something like y=1/(x^2-1), I know about the stationary point at x=0, and I also know that between the vertical asymptotes x=-1, x=1, there are no x-intercepts. So the stationary point (which happens to be a local min) is going to, in a way, deflect the curve back down, so that both asymptotes are approached from below.

I also note the general shape of some things. Quadratic/Quadratic and constant/quadratic look the same, but linear/quadratic looks different.

Again, HEAPS of practice needed to develop intuition.

lsong · « **Reply #926 on:** March 02, 2017, 10:19:17 pm »

Hmm, ok, thank you!
It seems I'll just have to practise more.
Btw, for equations which use two equations, like log [sin x], how would you recommend sketching them, because I've been considering asymptotes and mainly subbing in values. Surely there is a faster way to approach this?
Sorry I seem to be asking really basic questions.

RuiAce · « **Reply #927 on:** March 02, 2017, 10:21:23 pm »

Quote from: lsong on March 02, 2017, 10:19:17 pm

Hmm, ok, thank you!
It seems I'll just have to practise more.
Btw, for equations which use two equations, like log [sin x], how would you recommend sketching them, because I've been considering asymptotes and mainly subbing in values. Surely there is a faster way to approach this?
Sorry I seem to be asking really basic questions.

Sketch y=sin(x) first.

Read off the y-coordinates of that graph. What happens when you log, say 1. What happens when you "try" to log 0. What happens if sin(x) was negative the whole time?

And obviously you never have to consider when it's below -1 or above 1 because that's just defying the range of y=sin(x)

Compose the graph using the two graphs that build it.

Note - Inevitably you're going to "sub values". You're not supposed to get out of it. The important thing is realising WHICH values to sub in, and where possible just subbing in your head.

lsong · « **Reply #928 on:** March 03, 2017, 07:20:14 am »

Aah ok, it really looks like intuitive thinking. Thanks for the pointers!

Wales · « **Reply #929 on:** March 03, 2017, 04:39:10 pm »

I've just started the Graphs topic and this question seems completely foreign to me. Could anyone please guide me through it? I've had a look at the solutions but nothing makes too much sense.

Cheers, Wales

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