* Maths paper 2 (Ordinary) tips / discussion / aftermath * (1 thread only please)

MathsManiac

BmxNoob wrote: »

Any help with this question ? :

p(3, 0) is a point.
t and s are two distinct points on the y-axis and |pt| = |ps| = 5.
(i) Find the co-ordinates of t and the co-ordinates of s.

Draw a diagram. Mark in what you know, and use Pythagoras' theorem.

mc3ac

just done that today, you have two right angle triangles and the 5 is the hypotenuse - the only way to understand it is to draw it out first then use pythagoras to find out co-ordinates

nitromaster

For probability...if you mess it up..and can't figure out it's 4 x 3 x 2 or whatever...can you just do it out manually and still get the marks?

BmxNoob

Still cant understand that question ... I tried drawing the triangle but the only point i have is (3,0) and the value 5 but i dont know what that is ... Can anyone help ?

Ficheall

BmxNoob wrote: »

Any help with this question ? :

p(3, 0) is a point.
t and s are two distinct points on the y-axis and |pt| = |ps| = 5.
(i) Find the co-ordinates of t and the co-ordinates of s.

Well, (3,0) is a point on the x-axis.
The other two points are on the y-axis from the question.

Draw it, it should be much clearer. You know the length between |pt| and (3,0) is 5. The distance between the origin and (3,0) is 3.
So by Pythagoras theorem the distance of the other side of the triangle is 4. This is directly above the origin (as it is on the y-axis), so the point is (0,4).
Correspondingly the other point is (0,-4).

Drawing it is the simplest option to understand it, I suspect.

SDTimeout

Anybody think something mad like no Simpsons Rule could happen after the 1st Principals incident ?

kt.alice

Is anyone else doing the theorems for Q4-b)?

Just wondering are you supposed to do the diagram in pencil? Cause I always did em in pen, but just realised this could be an issue. :rolleyes:

BmxNoob

Ficheall wrote: »

Well, (3,0) is a point on the x-axis.
The other two points are on the y-axis from the question.

Draw it, it should be much clearer. You know the length between |pt| and (3,0) is 5. The distance between the origin and (3,0) is 3.
So by Pythagoras theorem the distance of the other side of the triangle is 4. This is directly above the origin (as it is on the y-axis), so the point is (0,4).
Correspondingly the other point is (0,-4).

Drawing it is the simplest option to understand it, I suspect.

Thanks for trying to help but i still dont understand it ... Is there any website i can get the instructions to the question ?

I cant understand how im meant to draw the diagram ... I have the point (3,0) but i dont know how to get the coords of p and s ... Sorry for the trouble

Ave Sodalis

Does anyone here do Project Maths?

r0cks0l1dd

SDTimeout wrote: »

Anybody think something mad like no Simpsons Rule could happen after the 1st Principals incident ?

Was first principles on every year besides this year?

I checked my book and the only thing they could really ask in place of Simpsons Rule is Area/Volume Irregular Prisms but I'm sure that it will be there

r0cks0l1dd

sup_dude wrote: »

Does anyone here do Project Maths?

Download the one second from the top. Podcast about PM from Countdown to 806

http://www.rte.ie/radio1/podcast/podcast_countdownto806.xml

J_E

r0cks0l1dd wrote: »

Was first principles on every year besides this year?

I checked my book and the only thing they could really ask in place of Simpsons Rule is Area/Volume Irregular Prisms but I'm sure that it will be there

Simpsons Rule has been on every year from at least 1998 so I really doubt they'd do something that drastic. I have a feeling they could do either an awkward probability or awkward linear programming though.

Ficheall

BmxNoob wrote: »

Thanks for trying to help but i still dont understand it ... Is there any website i can get the instructions to the question ?

I cant understand how im meant to draw the diagram ... I have the point (3,0) but i dont know how to get the coords of p and s ... Sorry for the trouble

You know that they're on the y-axis. It's given in the question.
All you have to figure out is how far up/down on the y-axis they are.

Look
> Draw your x-axis and y-axis

> Mark the point (3,0). It's on the x-axis.

> Your other two points are on the y-axis, given from the question. K?
We don't know where on the y-axis yet though.

> Given that they are the same distance from the point (3,0), which is on the x-axis, the points ps and pt must be symmetric about the origin, though you don't know how far up and down the y-axis they are.

> You can find out though! Focus only on one half of the triangle, say the half above the x-axis. 5 is the length of the hypotenuse as given in the question (the distance between (3,0) and ps. The distance between (3,0) and the origin is 3. By Pythagoras' theorem, the length of the other side of the triangle must be 4.

> So the point is on the y-axis, as given in the question, and is 4 away from the origin, ie. (0,4).
> The same applies for the triangle underneath the x-axis, and the point turns out to be (0,-4).

Any better?

BmxNoob

Ficheall wrote: »

Well, (3,0) is a point on the x-axis.
The other two points are on the y-axis from the question.

Draw it, it should be much clearer. You know the length between |pt| and (3,0) is 5. The distance between the origin and (3,0) is 3.
So by Pythagoras theorem the distance of the other side of the triangle is 4. This is directly above the origin (as it is on the y-axis), so the point is (0,4).
Correspondingly the other point is (0,-4).

Drawing it is the simplest option to understand it, I suspect.

I've got it now Ficheall thanks ... I thought it was a difficult question !

r0cks0l1dd

Cydoniac wrote: »

Simpsons Rule has been on every year from at least 1998 so I really doubt they'd do something that drastic. I have a feeling they could do either an awkward probability or awkward linear programming though.

Well we didn't do probability (we were but our teacher was out for over 3 weeks once and the sub kept doing algebra so we didn't have enough time out of time) so that doesn't really affect me but the linear programming will be worrying if they do make it a bit odd, same with Trigonometry. I'm locked at 6 questions with 1,2,3,5,7 and 11 so I hope they don't mix it up to much.Oddly enough I find the (a) and Linear programming harder than (b).

SDTimeout

I don't think it won't be there, just preparing myself. I was really hoping 1st principals came up in a question 6 the other day as it's usually my weakest of all 6 , 7 and 8. Easy 20 marks and all that but shockingly it was probably the one i felt most confident answering during the test.

FlawedGenius

Quick question: How do you know when to use Pythagorass Thereom and what is the formula again??

soccymonster

Gonna do so bad tomorrow. And I went from honours into a special class made for honours drop outs. And I seemed to be the only person who hadn't covered vectors in that class so I don't understand them at all for the optional question. FML. Hopefully my paper 1 will save me from the trainwreck that'll occur tomorrow.

r0cks0l1dd

soccymonster wrote: »

Gonna do so bad tomorrow. And I went from honours into a special class made for honours drop outs. And I seemed to be the only person who hadn't covered vectors in that class so I don't understand them at all for the optional question. FML. Hopefully my paper 1 will save me from the trainwreck that'll occur tomorrow.

If you still have your book or anything, try to look over and understand Linear Programming Q11

Ficheall

FlawedGenius wrote: »

Quick question: How do you know when to use Pythagorass Thereom and what is the formula again??

If you have a right-angled triangle and know the length of two of its sides, you can use Pythagoras' theorem to find the length of the third side.

PT: The length of the hypotenuse (the longest side) squared = the sum of the squares of the lengths of the other two sides.

ie. if the sides are a,b,h, where h is the hypotenuse, then
h^2=a^2+b^2

soccymonster

r0cks0l1dd wrote: »

If you still have your book or anything, try to look over and understand Linear Programming Q11

is it difficult to understand/grasp? Like I'm 100% sure i'll get 0/50 in vectors coz i just dont get it so any marks in the option would be nice.

r0cks0l1dd

soccymonster wrote: »

is it difficult to understand/grasp? Like I'm 100% sure i'll get 0/50 in vectors coz i just dont get it so any marks in the option would be nice.

To be honest, the first 2 parts of (b) are nearly the same question every year just with different figures but you'd need to know how to do it or else you'll be lost. Try and look through the solutions in the link below. Although it's a bit cluttered, try and make some sense out of it before you go into the exam. This is the only I know how to do and it's what most people will be doing tomorrow so try to pick up some marks. It's one of those things where it's hard to explain but easy to do once you grasp it.

http://www.studentxpress.ie/ordinary/ordlinprog.htm

irishgirl10

I can never grasp the a part of Linear programming

dodgybarnet

is there any formulas that you need to know that arent in the tables?

soccymonster

Kinda grasped the part a of linear programming in the past hour. Inequalities are wrecking my head though. Also, anyone wanna simplify how to get a vertex (usually part ii, it seems) which according to my book, you have to do using simultaneous equations.. But where do i get these equations?

Ficheall

soccymonster wrote: »

Kinda grasped the part a of linear programming in the past hour. Inequalities are wrecking my head though. Also, anyone wanna simplify how to get a vertex (usually part ii, it seems) which according to my book, you have to do using simultaneous equations.. But where do i get these equations?

Ain't writing this out again:

http://www.boards.ie/vbulletin/showpost.php?p=72720592&postcount=8

If you're having difficulty with a particular question from a particular year, please post it. These general questions are more difficult to answer...

edit: you need only look at part b for the question you've asked. Ignore the a).

J_E

Quick question about circle, 2006 question

Circle C has equation x^2+y^2=25
Line L is a tangent to C at the point (-3,4)

How do I find the slope of L with only two points? I don't know how to find the centre of the circle, I'm having a silly moment, am I missing something simple here?
Do I always assume the centre is (0,0)?

Edit: Ah yes, forgot the formula always applies to a circle with centre (0,0). Sorry guys

dodgybarnet

Ficheall, could you please explain to me what has to be done at Q2 (b) (iii) 2010

Ficheall

L is 3x-4y-15=0.

To draw this line you just need to find two points on it and draw the line joining them. The points are values of x and y for which the equation is satisfied eg. (5,0) and (1,-3) (since they've already given us that one, we may as well use it!)
Draw the line connecting those two points - that's L.

Now, for K we have one point on the line (1,-3) again as given. So we just need to find another point.
The reason they tell you that K is perpendicular, is so that you can find the slope of K.
If two slopes are perpendicular their product is -1.

The slope of a line is given by (y2-y1)/(x2-x1)
ie. (0-(-3))/(1-5)= -3/4
so the slope of the line perpendicular to this is 4/3. (4/3 * -3/4 = -1)

Now we have the slope of K.
We also have a point on K.
If we have a point and the slope of a line we can write the equation as
y-y1=m(x-x1)
ie.
y-(-3)=(4/3)(x-1)
y+3=4x/3-4/3
y=4x/3-13/3
3y=4x-13

So now we have the equation of K, but still want to find that other point on the line. So just sub in values into 3y=4x-13 until you find one that you deem convenient enough to sketch, though any point that's on the line will do. Eq. (0,-13/3) (though that one may be slightly more difficult to graph accurately.

Then you have two points on the line K: (0,-13/3) and (1,-3).
Draw the line connecting those two points.

When drawing these lines, it is probably best to indicate that you know they are continuous lines and do not stop once they reach the points that you have chosen - continue drawing them to the edge of the page if you wish!

dodgybarnet

Cheers Ficheall, its all coming back to me. Paper 2 is actually a lot easier than 1