Maths (Paper 3)

[ahmed t]

ok wat about the difference between ordinates and intervals in the trapezium rule
and also can we solve any modulus inequality by squaring both sides

 

[robotic94]

can anybody explain question 7 part ii) for me please ,I have trouble with vector =__=
link here: http://www.xtremepapers.me/CIE/International A And AS Level/9709 - Mathematics/9709_w10_qp_31.pdf

ditto doubt ^

 

[robotic94]

ok wat about the difference between ordinates and intervals in the trapezium rule
and also can we solve any modulus inequality by squaring both sides

for ur 2nd question, YES you can
but beware square the whole thing~ eg http://www.xtremepapers.me/CIE/Internat ... _qp_31.pdf question1 ! DONT FORGET TO SQUARE THE 2
hope that helps !

and can i know what does mukesh mean by min and max argument (a related question from pastpaper) i never came across this term :S THANX

 

[zeebujha]

From wiki:
In mathematics, ordinate refers to that element of an ordered pair which represents the distance traveled parallel to the vertical axis (y-axis) of a two-dimensional Cartesian coordinate system, as opposed to the abscissa. It is the second of the two terms (often labelled x and y, but not always) which define the location of a point in such a coordinate system


For 3 intervals you will have four ordinates

 

[zeebujha]

ok wat about the difference between ordinates and intervals in the trapezium rule
and also can we solve any modulus inequality by squaring both sides

For the second question:
NO!!!!!!!!!!
You can only square both sides if both sides are always positive
If |x-3|=|x-5| both sides can be squred
BUT
If x-3=|x-5|, both sides SHOULD NOT be squared because we cannot be sure that x-3 is positive for all values of x

 

[zeebujha]

Guys shoot your questions, only 19 hours to go!

 

[ahmed t]

so intervals is the "segments" but ordinates are the imaginary lines we draw parallel to the y axis when splitting the curve into segments?
as for zeegujha
i think u can but u must chek results after u get them, i have done such questions and mostly u only get one anser, right?

 

[zeebujha]

so intervals is the "segments" but ordinates are the imaginary lines we draw parallel to the y axis when splitting the curve into segments?
as for zeegujha
i think u can but u must chek results after u get them, i have done such questions and mostly u only get one anser, right?

yes
As for the second part:
Definitely, if we square both sides and get two answers, we can remove the wrong one by checking. However, why take the topsy turvy path on the first place

 

[MukeshG93]

Numerical way of calculating the maximum and minimum argument of a complex locus

Alright, so the locus we usually get is in the form of a circle. Wherever the circle is, it will have a set of complex numbers that follow its requirements. Similarly, the locus has a point with min argument and max argument. Geometrically, the points with min and max arguments are where lines from the origin are tangent to the circle. Remember that a tangent to a circle will always make an angle of 90 degrees with the radius of the circle at that point. The locus usually follows the format |z - u| < a, where a is the radius and u is the centre. We can construct a right-angled triangle using the radius of the circle at the point with min/max argument, the line from the origin to the centre of the circle (it's length = |u|) and the line from the origin tangent to the circle at the min/max point. We also know the angle that the line from the origin to the centre of the circle makes with the x-axis (arg(u)). So, we have the length of two sides (the radius and the line from the origin to the centre) and a right-angle to work with. Trigonometry, much? use sin(theta) = opp/hyp and cos(theta) = adj/hyp and you should get the angles you need to find the argument in each case. Just try, if you don't get it, add me on yahoo messenger and we'll doodle it out.

Image of relevance:

 

[fluffycloud]

Show that the equation of the tangent to the curve at the point with parameter t is
x sin t + y cos t = a sin t cos t. [3]
(iii) Hence show that, if this tangent meets the x-axis at X and the y-axis at Y, then the length of XY
is always equal to a.


URGENT THANKS :d

 

[MukeshG93]

Show that the equation of the tangent to the curve at the point with parameter t is
x sin t + y cos t = a sin t cos t. [3]
(iii) Hence show that, if this tangent meets the x-axis at X and the y-axis at Y, then the length of XY
is always equal to a.


URGENT THANKS :d

Incomplete question, much? What curve? What point?

 

[WellWIshER]

CHEERS MUKESH!! :Bravo: :Bravo: :friends: :friends: :friends: :beer: :beer: :beer: :beer: :beer: :beer:

can someone please SOLVE JUNE 2010 Q9 1 ??????

how do we get da normal to da curve????

 

[zeebujha]

Show that the equation of the tangent to the curve at the point with parameter t is
x sin t + y cos t = a sin t cos t. [3]
(iii) Hence show that, if this tangent meets the x-axis at X and the y-axis at Y, then the length of XY
is always equal to a.


URGENT THANKS :d

dude u have missed the first half of the question!

 

[MukeshG93]

CHEERS MUKESH!! :Bravo: :Bravo: :friends: :friends: :friends: :beer: :beer: :beer: :beer: :beer: :beer:

can someone please SOLVE JUNE 2010 Q9 1 ??????

how do we get da normal to da curve????

differentiate the expression using the quotient rule. if dy/dx = gradient of tangent then -dx/dy = gradient of normal

So, take its negative reciprocal and proceed from there

 

[ks136]

List of integrals plzzzzzzzzzzz

 

[deathvalley]

aaarh one more thing to ask, in some question, the mark scheme said we must sketch the bisector perpendicular line in Argand diagram,for example question 3ii M/J/10/31, for |z-1|<|z-i| can you explain

 

[MukeshG93]

aaarh one more thing to ask, in some question, the mark scheme said we must sketch the bisector perpendicular line in Argand diagram,for example question 3ii M/J/10/31, for |z-1|<|z-i| can you explain

|z - 1|<|z - i| means the distance between z and 1 is less than the distance between z and i. Now, taking the case in which the distance between them is equal, we draw a line that perpendicularly bisects the line that goes through 1 and i. The shaded region will be on the side closer to 1 because the distance between z and 1 is lesser than the distance between z and i.

 

[fluffycloud]

may june 2009 q6

 

[ks136]

can someone tell me integral of cot 2x

 

[Madiha :)]

can anyone please explain why only -0.5 is the answer ..