Higher Maths Paper 2 tip[s]

kaimera

paper 2..solid hint:

alright...after a session tonight, turns out he cant remember much about the circles on the line question

sorry about that BUT least it sorted out proving 3 points are collinear

anway...

for Section B-Q8.

For those with the Mathematices Revision for LC Higher Level [paper2] - green book....

it's the question on page 203.

ambulance at the point a, has to reach point c on the road bc.

It travels 80Km\Hr between a and any point on the road.

Once on the road, it travels at 100Km\Hr.

If |bc| = 9Km, |ba| = 3Km & |bp| = xKm,

calculate value for x for which the time is a minimum and calculate this minimum time.






Something similar is to appear as Q8 c

*************
Ratio Test possibly

Difference Equations Theorem

possible proof on Circle [eqn of tangent] or Line [angle between 2 lines OR perpendicular distance]

DSMe

damn thats a good tip , any ideas on what type of question it the rest would be ?
Any thing else u think might come up?

eirebhoy

I've searched through the circle chapter of my book and I can't find this proof anywhere. How do you prove it?

pfitz

i think its:
if the slope of the line joining the centres of the circles is the same then they are collinear?

Man U babe

Using the 3 centres as the 3 points of a triangle, show that the area of said triangle is 0.

Celtic Tiger

Theres a third way: let points equal p,q and r and just show that |pq|+|qr|=|pr|

aine

maybe this sounds silly...but its just an idea....how about ye get off the internet and actually get some sleep before ye're exam....usually works for me!

but just so ye know, from someone who has done lc higher math....I'm pretty sure Celtic Tiger is right...would have to check book though...its been a while!

Celtic Tiger

Bah sleep! Will catch up at the weekend. Easy day 2moz anyway

PHB

Celtix tiger, you didnt prove that they were coliner there, you just proved that the distances were equal.

Celtic Tiger

Yes but the distances are only going to be equal if points are all on the same line.

PHB

Nevermind you're right

Sev

three circles have their centres on a line. Prove the centres are collinear

Em.. I dont know about you guys, but when I see that I read.. "three circles have their centres on a line. Prove they have their centres on a line."

Sev

Or more appropriately "three points are collinear, prove they're collinear"

DSMe

i read this the same as sev.... and if its a straight line where does the triangle come from?

sikes

it will not be worded as that, in fact if they are really gay and annoying they will say something like this

What do you notice about the centre of circles A, B, C?

Celtic Tiger

If you are unsure what collinear means check the bottom of this page

Sev

I cant see that ever being a question, it just sounds stupid. Its tantamount to saying "Triangle abc is a triangle. Prove it's a triangle"

Sev

The only manifestation of such a question I can imagine, is if you were given the equation of three separate circles. And then asked to show that the centers of them are collinear. Or perhaps told that a particular line is a tangent to 3 identical circles, and asked to prove the centers or collinear. Something like that maybe, something that actually makes sense.

pfitz

so how exactly do you prove that three points are collinear?

PHB

You prove they are on a straight line basically by one of the following methods :

if the slope of the line joining the centres of the circles is the same then they are collinear

Using the 3 centres as the 3 points of a triangle, show that the area of said triangle is 0.

let points equal p,q and r and just show that |pq|+|qr|=|pr|

Go DSme, Man U Babe, and Celtic Tiger

kaimera

alright alright...my english may not be great :P

I'll find out for defo tonight n post up again.

original post was off the top of my head from Wednesday night

sikes

Kaimera

was your original message coded cos you said something is to be left out of above equation or something like that.

btw i am not taking the piss.

good luck on monday. Any other tips? i reckon proving cos rule

pfitz

could u get the equation of the line between two of the points and then substite the third point in to the equation to see if it is on the line

kaimera

original post edited slash updated

Sev

Where the hell you get your |bp| from in that question?

Sev

Right.. I take it p is the point where the car meets the road. Hmmm... you sure theres not a little more information to go with that question? Cos I can see where its coming from, but it doesnt seem to work without having an angle between [ba] and [ac]. Havn't put a lot of thought into this yet tho. But is that all exactly quoted from the book? With no pictures?

kaimera

I'll scan in the page 2m morning.
too late now to be arsing around with a scanner atm.

a,b,p make a right angled triangle

|ab|=3
|bp| = x
|ap| = sq root 9+x2

|pc| = (9-x)

Time = Distance/Speed

T = |ap|/80 + |pc|/100

working T out...(9+x2)^1/2/80 + 9/100 - X/100

Use Chain Rule to find dT/dX

Get T

T = D/S

Celtic Tiger

Thanks for sharing the tip Kaimera

Sev

Yeh thats how you do it.

It's simple then, just differentiate. Find the value of x when dT/dx is 0. Then throw that value of x into your equation for T.

Of course not knowing that abp formed a right triangle, meant that I had a very nasty looking expression for the final x value, in terms of cos(angle abp) which had to be found using the cosine rule on the triangle abp. It made it a more interesting question tho.

kaimera

there is a diagram with the question so should be one on the paper.

just fill in the values.

this question isn't in the Maths book I've been using and I'm not sure if it's only in revision books.

Never been asked on a paper iirc.

they seem to be moving away from volume etc on Q8

Sev

If you have any more.. keep em coming, if I cant drag myself away from the pc to actually do some study myself, then I'll get it done on the pc. Thanks for the tip.