Post by: ankitd_1994 on May 29, 2010, 11:40:16 am
Post by: theaguia on May 29, 2010, 12:09:38 pm
hey can anyone who is an ex add math student or doing add math this yr, post up a list of formuale and short notes on all chapters for add math?? ???
i am doing, i havent studied for any other exam except add math, the hardest exam of em all, an i have eco on the same day. its bad, By the way just post doubts i reqally cant have notes for maths
Post by: J.Darren on May 29, 2010, 05:39:56 pm
Sector area : 1/2 x theta (in radian) x radius^2
Post by: jellybeans on May 29, 2010, 05:41:39 pm
Post by: adrian1993 on May 29, 2010, 06:48:34 pm
Post by: jellybeans on May 30, 2010, 02:24:46 am
Relative velocity seems to be the hardest imo.
i agreeeeeee :'( :'( :'( hahah.
Post by: theaguia on May 30, 2010, 07:50:42 am
heyeyyyy, does anybody have notes for bearings & stuff? :( like finding the.. velocity .. vectors.. and.. stuff? :/
just post a question
Post by: jellybeans on May 30, 2010, 08:30:19 am
just post a question
HIIII, lol. could you see what i did .. wrong please? :/
Post by: theaguia on May 30, 2010, 09:28:33 am
HIIII, lol. could you see what i did .. wrong please? :/
from first impression it looks like ur diagram is wrong, but i will do the question and get back to you
Post by: jellybeans on May 30, 2010, 09:30:04 am
from first impression it looks like ur diagram is wrong, but i will do the question and get back to youLOL hahahahahahaha, aite XD
Post by: J.Darren on May 30, 2010, 10:37:31 am
x^n = nx^(n-1)
e^2x = 2e^2x
ln x = 1 / x
sin x = cos x
cos x = -sin x
tan x = sec^2 x
Chain rule (Composite function):
y = ln (2x + 1)
y = ln (u)
u = 2x + 1
dy/du = 1/u
du/dx = 2
dy/du x du/dx = dy/dx
1/(2x + 1) * 2 = 2/(2x + 1)
Quotient rule :
y = u / v
dy/dx = (vu' - v'u) / v^2
Product rule (Multiple functions):
y = u x v
dy/dx = v'u + u'v
y = 3x^2 x e^x
v = 3x^2
u = e^x
v' = 6x
u' = e^x
dy/dx = 6xe^x + 3x^2e^x
Post by: J.Darren on May 30, 2010, 11:05:23 am
If smaller than 0 - Maximum
If equals to 0 - Point of inflexion
If greater than 0 - Minimum
Rules of Integration :
e^2x = 1/2 x e^2x
cos ax = 1/a x sin x
sin ax = 1/a x -cos x
Add constant c when integrating an indefinite integral
x^n = [x^(n+1)]/(n+1)
(ax+b)^n = [(ax+b)^n+1]/[a*(n+1)]
1/(ax+b) = [ln(ax+b)]/a
Post by: J.Darren on May 30, 2010, 11:18:18 am
AX = C
A–1 AX = A–1C
IX = A–1 C
X = A–1 C
Finding the inverse of a 2 x 2 matrix :
(http://www.mathwords.com/i/i_assets/inverse%20of%20a%20matrix%20formula%202.gif)
(http://www.mathwords.com/i/i_assets/inverse%20of%20a%20matrix%20formula%202b.gif)
(http://www.mathwords.com/i/i_assets/inverse%20of%20a%20matrix%20example%202.gif)
Post by: cooldude on May 30, 2010, 11:36:18 am
HIIII, lol. could you see what i did .. wrong please? :/
see the question is basically asking what is the time taken for the boat to travel the dist AC, dist BC=54 because the speed of the liner is 36 km/h and the time it travels is 1.5 hrs thus the dist=1.5*36=54, the dist AC can be found out by the cosine rule, x=angle ABC
A^2=B^2+C^2-2ABcosx
A^2=90^2+54^2-(2*54*90*cos 135)
A=AC=133.75 km
therefore the speed of the boat = 133.75/1.5=89.2 km/h
Post by: Ghost Of Highbury on May 30, 2010, 11:54:41 am
Post by: cooldude on May 30, 2010, 11:58:23 am
isnt it 28.9 km/h?
what part did i get wrong, cud u post ur working?
Post by: J.Darren on May 30, 2010, 12:06:35 pm
Am I correct ???
Post by: Ghost Of Highbury on May 30, 2010, 12:38:17 pm
Post by: Ghost Of Highbury on May 30, 2010, 12:45:59 pm
HIIII, lol. could you see what i did .. wrong please? :/
check the above post.
ur answer is correct, whats the matter? ms trouble?
Post by: J.Darren on May 30, 2010, 01:10:06 pm
check the above post.Right that means I have made a mistake somewhere, mind pointing it out for me? :)
ur answer is correct, whats the matter? ms trouble?
Post by: holtadit on May 30, 2010, 01:12:35 pm
Post by: Ghost Of Highbury on May 30, 2010, 01:21:35 pm
Right that means I have made a mistake somewhere, mind pointing it out for me? :)
lemme check...from what i saw, ur diagram is incorrect..considering 315 as the bearing the angle b/w the liner and lifeboat path shud be 135 degree which is obtuse, its acute in ur diagram...gotta extend the 90km line till the angle made by the third line is obtuse to the 54 km line
Post by: Ghost Of Highbury on May 30, 2010, 01:23:37 pm
Post by: holtadit on May 30, 2010, 01:34:41 pm
Is that right ?
Post by: J.Darren on May 30, 2010, 01:37:56 pm
and how did u get 45 degree in ur diagram?? it shud be 19.9 , check mine ..The ship is travelling at a bearing of 090, and it is 90km away from the lifeboat at the time of the call of distress. The liner is also on a bearing of 315 from the ship. When you draw the diagram, you can mark down 45 degrees twice - one is between the 54 km and the 90 km, and one is between the 90km and the normal at that point. I am guessing that you have made an assumption that the ship is at a bearing of 315 from the lifeboat AFTER it has moved 54 km ...
Post by: holtadit on May 30, 2010, 01:39:09 pm
The ship is travelling at a bearing of 090, and it is 90km away from the lifeboat at the time of the call of distress. The liner is also on a bearing of 315 from the ship. When you draw the diagram, you can mark down 45 degrees twice - one is between the 54 km and the 90 km, and one is between the 90km and the normal at that point. I am guessing that you have made an assumption that the ship is at a bearing of 315 from the lifeboat AFTER it has moved 54 km ...
Darren which year is this from so we can check our answer ?
Post by: J.Darren on May 30, 2010, 01:41:44 pm
Darren which year is this from so we can check our answer ?Honeslty I have no idea LOL, perhaps you should ask the guy who posted the question :D
Post by: jellybeans on May 30, 2010, 02:13:00 pm
check the above post.
ur answer is correct, whats the matter? ms trouble?
:O guess so! thats what it says in the ms
|
v
;)
Honeslty I have no idea LOL, perhaps you should ask the guy who posted the question :D
AND I AM NOT A GUY.
... so is the MS right? D:
Post by: theharrovian on May 30, 2010, 02:16:48 pm
cooldude, you've just made a computation error - everything is correct up to the bit when you somehow say that A is 133.75 km.
Post by: J.Darren on May 30, 2010, 02:20:31 pm
:O guess so! thats what it says in the msI do apologise :-[
|
v
;)
AND I AM NOT A GUY.
... so is the MS right? D:
Post by: jellybeans on May 30, 2010, 02:21:02 pm
(http://img33.imageshack.us/img33/6976/relativevelocityquestio.png)
Am I correct ???
heyyyeyy. how did you get the 54? ;D Lol
OH and np. hahaha, :P
Post by: theharrovian on May 30, 2010, 02:24:43 pm
Post by: jellybeans on May 30, 2010, 02:42:09 pm
I do apologise :-[
...how did you get 54? for the distance i mean. :(
:) :)
Post by: J.Darren on May 30, 2010, 02:42:59 pm
heyyyeyy. how did you get the 54? ;D Lol36 * 1.5 (0600 to 0730 - 90 minutes / 1.5 hours) = 54
OH and np. hahaha, :P
Post by: cooldude on May 30, 2010, 02:50:20 pm
Post by: jellybeans on May 30, 2010, 02:52:43 pm
36 * 1.5 (0600 to 0730 - 90 minutes / 1.5 hours) = 54
THANK YOU. ;)
Post by: Ghost Of Highbury on May 30, 2010, 03:11:28 pm
plus check the diagram i have attached, it shows how the angle is 135 too..
i have a gce-o lvel past papers book and i just saw this question is in that book as a sample question paper, there , the answer says - 28.9 as i have exactly explained..
plz correct me if im rong
Post by: jellybeans on May 30, 2010, 03:22:38 pm
The ms says 135 or 45 depending on the diagram. im pretty sure the ms is incorrect...first, in darren's diagram, the 90km line is less than the 54 km line, u have to at least roughly make it longer...
plus check the diagram i have attached, it shows how the angle is 145 too..
i have a gce-o lvel past papers book and i just saw this question is in that book as a sample question paper, there , the answer says - 28.9 as i have exactly explained..
:o so it has two answers? ..whaaaa?
Post by: theharrovian on May 30, 2010, 03:35:17 pm
The ms says 135 or 45 depending on the diagram...first, in darren's diagram, the 90km line is less than the 54 km line, u have to at least roughly make it longer...
plus check the diagram i have attached, it shows how the angle is 135 too..
i have a gce-o lvel past papers book and i just saw this question is in that book as a sample question paper, there , the answer says - 28.9 as i have exactly explained..
plz correct me if im rong
OK, I will correct you because you are wrong. Read the question - it says the ship is at a bearing of 315 from the lifeboat at 0600 hours. Your diagram shows that the ship is at a bearing of 315 from the lifeboat at 0730 hours instead of at 0600 hours.
Post by: cooldude on May 30, 2010, 03:37:27 pm
OK, I will correct you because you are wrong. Read the question - it says the ship is at a bearing of 315 from the lifeboat at 0600 hours. Your diagram shows that the ship is at a bearing of 315 from the lifeboat at 0730 hours instead of at 0600 hours.
now do u mind correcting me, thanks
Post by: Ghost Of Highbury on May 30, 2010, 03:43:44 pm
and By the way, if u zoom the angle 315 has been marked correctly, the 45 shudnt have been there :P
Post by: theharrovian on May 30, 2010, 03:45:38 pm
now do u mind correcting me, thanks
Your line AB is in the wrong direction - the question states "from the lifeboat", not "to the lifeboat". Otherwise your diagram is completely correct - you've just misread/misinterpreted the question and taken the lifeboat to be on a bearing of 315 from the liner.
Post by: Ghost Of Highbury on May 30, 2010, 03:47:22 pm
see the question is basically asking what is the time taken for the boat to travel the dist AC, dist BC=54 because the speed of the liner is 36 km/h and the time it travels is 1.5 hrs thus the dist=1.5*36=54, the dist AC can be found out by the cosine rule, x=angle ABC
A^2=B^2+C^2-2ABcosx
A^2=90^2+54^2-(2*54*90*cos 135)
A=AC=133.75 km
therefore the speed of the boat = 133.75/1.5=89.2 km/h
The 315 degree bearing has been interpreted incorrectly, it shud be clockwise...bering B from A should be 315, in you case it isnt..
Post by: cooldude on May 30, 2010, 04:01:58 pm
Post by: jellybeans on May 30, 2010, 04:20:32 pm
help please ;D
Post by: J.Darren on May 30, 2010, 04:44:39 pm
... D:(http://img689.imageshack.us/img689/2581/velocityvectorquestion.png)
help please ;D
Post by: J.Darren on May 30, 2010, 04:46:15 pm
Post by: Ghost Of Highbury on May 30, 2010, 04:50:34 pm
ii) 47 minutes
are these answers correct? if so, ill explain how..
Post by: jellybeans on May 30, 2010, 04:53:43 pm
Post by: J.Darren on May 30, 2010, 04:57:21 pm
Post by: Ghost Of Highbury on May 30, 2010, 05:01:49 pm
Post by: J.Darren on May 30, 2010, 05:03:52 pm
ok heres my explanation thenPerfect, except that it should be -110 instead of -100 :D
Post by: Ghost Of Highbury on May 30, 2010, 05:05:17 pm
Perfect, except that it should be -110 instead of -100 :D
oh yeah , -110. sorry.
DAMN!!!
Post by: J.Darren on May 30, 2010, 05:08:17 pm
oh yeah , -110. sorry.By the way Adi would you be kind enough to go through the forumla sheet on differentiation and integration that I have posted on the first page of this thread ? Thanks :D
DAMN!!!
Post by: Ghost Of Highbury on May 30, 2010, 05:12:08 pm
By the way Adi would you be kind enough to go through the forumla sheet on differentiation and integration that I have posted on the first page of this thread ? Thanks :D
I did, i cudnt find any errors. however the product rule and quotient rule, this is easire to remember
dy/dx (uv) = v(du) + u(dv)
dy/dx(u/v) = (vdu - udv)v^2
Post by: jellybeans on May 30, 2010, 05:21:04 pm
oh yeah , -110. sorry.
DAMN!!!
ok heres my explanation then
Thaaaanks, ;D one more question.. um how are we supposed to know where P & Q are? :o
Post by: Ghost Of Highbury on May 30, 2010, 05:27:19 pm
Q - again, from the origin , 50 to the right and -70 downwards, however this time u gotta make a parallel line to that one joining the OP ...this is to complete the triangle...
Post by: jellybeans on May 30, 2010, 05:30:27 pm
P - locate it roughly from the origin , using 280 and -40. like roughly 280 units to the right, and 40 down(-)
Q - again, from the origin , 50 to the right and -70 downwards, however this time u gotta make a parallel line to that one joining the OP ...this is to complete the triangle...
Q ..from the origin? or from P..?
Post by: Ghost Of Highbury on May 30, 2010, 05:33:43 pm
Q ..from the origin? or from P..?
its actually from P, but just FYI, how it comes dere, it shud be taken from O, but to complete the vector triangle , a parallel line is taken from P.
" however this time u gotta make a parallel line to that one joining the OP ...this is to complete the triangle..."
Post by: jellybeans on May 30, 2010, 05:35:16 pm
its actually from P, but just FYI, how it comes dere, it shud be taken from O, but to complete the vector triangle , a parallel line is taken from P.:o OHHH. okay, Danke :D
" however this time u gotta make a parallel line to that one joining the OP ...this is to complete the triangle..."
Post by: J.Darren on May 30, 2010, 06:02:34 pm
logc b = loga b / loga c
Post by: destructor on May 30, 2010, 06:48:06 pm
i take add math too
u guys should focus on vectors in the cartesian form (xi+yj)
cuz i feel thats gonna show up on ppr 2 this year
someone in the previous pages was askin for a formula
the important formula for this
is position vector + time(velocity vector)= new position vector
Post by: jellybeans on May 30, 2010, 06:52:36 pm
hey guys
i take add math too
u guys should focus on vectors in the cartesian form (xi+yj)
cuz i feel thats gonna show up on ppr 2 this year
someone in the previous pages was askin for a formula
the important formula for this
is position vector + time(velocity vector)= new position vector
THANK YOU. ;D
Post by: ankitd_1994 on May 31, 2010, 10:32:58 am
Changing the base of a logarithm :hey Thanks for tht formula cn u also give an eg. of it with nos.?
logc b = loga b / loga c
Post by: Ghost Of Highbury on May 31, 2010, 10:41:06 am
hey Thanks for tht formula cn u also give an eg. of it with nos.?
Post by: Dibss on May 31, 2010, 12:41:07 pm
hey can anyone who is an ex add math student or doing add math this yr, post up a list of formuale and short notes on all chapters for add math?? ???
Hope this helps and all the best for Add Math guys :)
Post by: ankitd_1994 on May 31, 2010, 02:21:47 pm
q. 9 in the june 2009 paper 1 (are there variants for add maths too?)
Post by: ankitd_1994 on May 31, 2010, 02:27:27 pm
hey guys
i take add math too
u guys should focus on vectors in the cartesian form (xi+yj)
cuz i feel thats gonna show up on ppr 2 this year
someone in the previous pages was askin for a formula
the important formula for this
is position vector + time(velocity vector)= new position vector
is velocity vector the same as magnitude?
Post by: ankitd_1994 on May 31, 2010, 03:06:26 pm
heyy anyone there?? ??? ??? pls answer the ques., i have posted it this time: :-\
9) At 10 00 hours, a ship P leaves a point A with position vector (– 4i + 8j) km relative to an origin O,
where i is a unit vector due East and j is a unit vector due North. The ship sails north-east with a speed
of 10?2 km h–1. Find
(i) the velocity vector of P, [2]
(ii) the position vector of P at 12 00 hours. [2]
At 12 00 hours, a second ship Q leaves a point B with position vector (19i + 34j) km travelling with
velocity vector (8i + 6j) km h–1.
(iii) Find the velocity of P relative to Q. [2]
(iv) Hence, or otherwise, find the time at which P and Q meet and the position vector of the point
where this happens. [3]
Post by: ankitd_1994 on May 31, 2010, 03:11:43 pm
Post by: J.Darren on May 31, 2010, 03:30:00 pm
There you go. Not sure about part iii though ...
Post by: ankitd_1994 on May 31, 2010, 03:52:53 pm
(http://img571.imageshack.us/img571/4262/velocityquestion.png)k thx but how did u get (i+j)/ sqrt 2 in the first part n (31,43) in the 4th part??
There you go. Not sure about part iii though ...
Post by: J.Darren on May 31, 2010, 04:14:26 pm
k thx but how did u get (i+j)/ sqrt 2 in the first part n (31,43) in the 4th part??Velocity vector = Direction vector / Discriminant of the direction vection x Speed. The boat sails north-east, hence 1 i and 1 j.
Post by: ankitd_1994 on May 31, 2010, 06:08:40 pm
Velocity vector = Direction vector / Discriminant of the direction vection x Speed. The boat sails north-east, hence 1 i and 1 j.thx +rep for the sum didnt noe the formula either
Post by: J.Darren on June 01, 2010, 01:05:54 pm
thx +rep for the sum didnt noe the formula eitherLOL I actually complicated the whole matter, in the question it was stated that the boat sails with unit vector north and east, unit vector = (1, 0) if it is i. (0, 1) if it is j ... :(
Post by: jellybeans on June 02, 2010, 11:26:47 am
Post by: Ghost Of Highbury on June 02, 2010, 01:06:31 pm
Question, do we get full marks on a question if we got the right answer but the working is different? :-\ like.. completely different.
depends on the ms, plus, if ur method is correct, ull get the marks.