• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Maths, Addmaths and Statistics: Post your doubts here!

Mayedah

Mayedah

Messages
435
Reaction score
238
Points
53
ashiqbal said:
ok guys i am really sorry, i should have given the method that i use for transformation matices over here earlier. though i didnt know how to explain it!
This is more of a sort of a lesson than an actual mnemonic, with tips to memorise the matrices step by step and easily. There might be other mnemonics, but you will mix those all up during the exam and then you are dooomed!
View attachment 11839
so basically, what you need to learn at first that those with the yellow go about the pattern(the small circles are zeroes! :)):
(x.....0
0.....y)
and those with red go with the pattern:
(0......x
y......0)
These two samples are important to learn before you proceed forward! Remember that yellow ones resemble an identity matrix, while red doesnt. (a small mnemonic here: yellow chicks (choozay) are identical)
Those in Red will be Reflection along y=x and y=-x and Rotation of anticlockwise 90 degrees and clockwise 90 degrees
Rotating REXY is Red (where ReXY means REflection along Y=X)
now that we know which are red, lets learn the method of memorising these four matrices:

REFLECTION ON Y=X
this is simple, all positive, so (0 1 / 1 0) the slash means next line. the x coordinate becomes y coordinate and vice versa

REFLECTION ON Y=-X
now there is a negative sign here, which means that we will put a negative sign with all the members (0 -1 / -1 0) here, the x coordinate becomes negative y coordinate and the y coordinate becomes negative x coordinate.

ROTATION OF 90 degrees CLOCKWISE:
This is Clockwise rotation, right? so remember that we cant go in the opposite direction, so the negative sign on the left shows the restriction of rotating left(anticlockwise) and the positive 1 shows that we can rotate right(clockwise) (0 1 / -1 0)

ROTAION OF 90 degrees ANTICLOCKWISE:
you can guess that this time we are allowed to rotate left, so there will be a positive sign on the left(anticlockwise) and a negative sign on the right(clockwise). Guess the matrix now before looking up on the page to see it!

YOU MUST REMEMBER THAT WE CANNOT MOVE IN DIRECTION OF NEGATIVE SIGN IN ROTATION!

All the others follow the Yellow pattern.
the thing about those following the yellow pattern is that they are like Identity matrix, with only one member changed. if the number that is changed is the first member eg (x 0 / 0 1) then it means that the value of x is changed and if the number changed is last ie (1 0 / 0 y) then it means that the value of y has changed. so basically what u have to see is which value is changing.

In Reflection along the axes, the values are not actually changed, but their signs are changed.

Reflection Along X AXIS:
In this, the value of x remains the same, while y changes, right? when you flip the whole image upside down, the Y becomes -Y and the -Y becomes Y. so actually to change the sign of y, what will we do? yup we will change value of Identity matrix so that the last one becomes negative. (1 0 / 0 -1). Multiply any coordinate with this and the value of only y will become negative.
Reflection Along Y AXIS:
again the same thing happens, with value of x changing as we invert the image laterally(right to left). the sign of X changing means that the first member will change its sign. (-1 0 / 0 1)

Rotation of 180 DEGREES:
this is not actually a rotation, it is a reflection of both x and y coordinates at the same time. Therefore, both the above conditions would be applied and the matrix would become (-1 0 / 0 -1)

STRETCH Parallel to X AXIS:
in Stretch, the value of x changes while the value of y remains the same (IT is PARALLEL to X AXIS) so what wil we do to the identity matrix this time? Surely, because we have to change the value of x, we will change the first member, but what will we replace it with? This time there is SHEAR Factor involved, and that will take place of the x axis. if shear factor is k, then the matrix will become (k 0 / 0 1) (remember that only the first member is changed)
STRETCH Parallel to Y AXIS:
What is your guess for the transformation matrix? i will let you work this out on your own. Check your matrix with the one given in the pic.

ENLARGEMENT:
This is the simplest of all! You just multiply the identity matrix with the enlargement factor, k(1 0 / 0 1) so it becomes
(k 0/0 k)

those given in green, the shear matrices are quite difficult to understand so i suggest that you just memorise those two, that is what i have done.

the method of learning the matrices is quite enjoyable, and even if you forget the matrices, you can just work them out in the examination hall in seconds! and the thing is, once you understand this method, it becomes very difficult to forget!

I am tagging some of the people i remember now, if u know others, please tag them as well!
Rimsa, a.abid Spidey21 syed1995 MustafaMotani khizer shahab SalmanPakRocks scouserlfc Mayedah asd
Click to expand...
THANX ALOT =D
 
X

xxvvyy

Messages
53
Reaction score
2
Points
18
HELP PLEASE! The sets, P, Q and R satisfy the conditions Q subset P, n(Q U R)=O and n(P intersection R)=n(R). Draw a Venn diagram to illustrate the sets.
 
X

xxvvyy

Messages
53
Reaction score
2
Points
18
HELP PLEASE! The sets, P, Q and R satisfy the conditions Q subset P, n(Q U R)=O and n(P intersection R)=n(R). Draw a Venn diagram to illustrate the sets.
 
X

xxvvyy

Messages
53
Reaction score
2
Points
18
HELP PLEASE! The sets, P, Q and R satisfy the conditions Q subset P, n(Q U R)=O and n(P intersection R)=n(R). Draw a Venn diagram to illustrate the sets.
 
syed1995

syed1995

Messages
1,824
Reaction score
949
Points
123
AlphaBeta said:
10 min pass too slow u will surely hav problem in timing:cool:
Click to expand...

haha .. i had other questions to solve as well ... plus this question was difficult took around 15 minutes :p but eventually i solved it .. let me check the answers first.
 
syed1995

syed1995

Messages
1,824
Reaction score
949
Points
123
AlphaBeta said:
10 min pass too slow u will surely hav problem in timing:cool:
Click to expand...

Typing it all takes time :p

okay so which part do you need help in?

a and b starting parts are pretty easy .. just use the s=d/t formula .. A to J Taj = 10/x
J to P speed is x+0.5 so Tjp = 15/(x+0.5)

c now this part was the thing ..

7 seconds time taken to reach from A to P. in which J took 2 seconds (to catch and throw).
so total time taken in traveling will be 7-2=5.

now.

(10/x)+(15/(x+0.5)=5 (as that's the time it took in travelling)

guess you can solve the above equation^ .. it will become 5x^2-22.5x+5=0

divide it by 2.5 to get the equation given in the question.

(5x^2-22.5x+5)/2.5=0/2.5

2x^2-9x+2=0 Your Answer.

d Next part is pretty easy .. just use the quadratic formula .. but show all parts working .. as its 4 marks! i did it in 3 lines though .. lol answers are x=4.71 and -0.2 .. reject -0.2 as speed cannot be in negative.!

e i Avg speed as ball travels from J to P. the equation was x+0.5 .. 4.71+0.5=5.21

e ii Time took for AJ = 10/4.71 = 2.12
Time took for JP = 15/5.21=2.88

Time Difference =2.88-2.12=0.76 Answer.
 
AlphaBeta

AlphaBeta

Messages
886
Reaction score
1,252
Points
153
syed1995 said:
Typing it all takes time :p

okay so which part do you need help in?

a and b starting parts are pretty easy .. just use the s=d/t formula .. A to J Taj = 10/x
J to P speed is x+0.5 so Tjp = 15/(x+0.5)

c now this part was the thing ..

7 seconds time taken to reach from A to P. in which J took 2 seconds (to catch and throw).
so total time taken in traveling will be 7-2=5.

now.

(10/x)+(15/(x+0.5)=5 (as that's the time it took in travelling)

guess you can solve the above equation^ .. it will become 5x^2-22.5x+5=0

divide it by 2.5 to get the equation given in the question.

(5x^2-22.5x+5)/2.5=0/2.5

2x^2-9x+2=0 Your Answer.

d Next part is pretty easy .. just use the quadratic formula .. but show all parts working .. as its 4 marks! i did it in 3 lines though .. lol answers are x=4.71 and -0.2 .. reject -0.2 as speed cannot be in negative.!

e i Avg speed as ball travels from J to P. the equation was x+0.5 .. 4.71+0.5=5.21

e ii Time took for AJ = 10/4.71 = 2.12
Time took for JP = 15/5.21=2.88

Time Difference =2.88-2.12=0.76 Answer.
Click to expand...
u doing wrong paper its oct/nov 2011 p22 q8:LOL:
 
You must log in or register to reply here.
Top