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deathvalley said: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
ahmed t said: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:ahmed t said: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
yesahmed t said: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?
MukeshG93 said: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.
fluffycloud said: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!fluffycloud said: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
WellWIshER said: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????
deathvalley said: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
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