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[ToO LaZy ^*]

i can't seem to get this one out..

 

[xiao1985]

awwwwwwwwwww i just LOVE maths...

EVERY ONE!!! DO B SCI (ADV MATHS)@ USYD!!!!!!!!!!!!!!

 

[Affinity]

well do the substitution

dx = -1/u^2 du

you get integrate { sqrt(3), 1/sqrt(3) } -ln(1/u) / [( 1 + (1/u)^2 ) * ( u^2)] du

simplyfy you get I = -I
so.. I = 0

 

[ToO LaZy ^*]

ooo..icic
i know where i went wrong..differentiated wrong -_-"...

thanks

 

[ToO LaZy ^*]

got another question..hmm..

 

[+Po1ntDeXt3r+]

got another question..hmm..

1)if BXC is a right angle
then CB is a diameter of a circle with pts C,X,B on circumference
if CM=MB
then M is the centre of this circle.
hence M is equidistant to C,X,B

XM=CM=MB
then
MXB = XBM - (angles opp equal sides are equal in isosceles triangle, MXB)
QED

2) MXB = CAP (angle subtended by equal arcs on circle)
MXB = XBM (from 1)
MXC = pi/2 - MXB = pi/2-XBM (angle sum from BXC is right angle n equality)
AXN = MXC = pi/2 - XBM= pi/2 - MXB (vertically opp angles)

MNA = CAP + AXN = MXB + 90 - MXB = pi/2..
since MNA is 90deg then MN is perpendicular to AD

sorri the reasons are dodgy.. ivve just woken up :S
grr miss 4U maths :S

 

[ToO LaZy ^*]

1)if BXC is a right angle
then CB is a diameter of a circle with pts C,X,B on circumference
if CM=MB
then M is the centre of this circle.
hence M is equidistant to C,X,B

how did you get that?..is that a rule that i may have missed?

2) MXB = CAP (angle subtended by equal arcs on circle)

where is this point P you speak of?

 

[CM_Tutor]

how did you get that?..is that a rule that i may have missed?

The rule in question is the converse of the angle in a semicircle rule.

Rule: If AB is the diameter of a circle, and C is on the circumfrence, then angle ACB (the angle in the semicircle) is a right angle
Converse: If ACB is a right angle, then AB is the diameter of a semicircle through C.

 

[CM_Tutor]

where is this point P you speak of?

+Po1ntDeXt3r+ means D, not P, and the first line should be MBX, not MXB.

ie. it should start: MBX = CAD (angles at the circumfrence on the same arc are equal)

The method itself is fine.

 

[ToO LaZy ^*]

The rule in question is the converse of the angle in a semicircle rule.

Rule: If AB is the diameter of a circle, and C is on the circumfrence, then angle ACB (the angle in the semicircle) is a right angle
Converse: If ACB is a right angle, then AB is the diameter of a semicircle through C.

ohhhhh...ok.
is this done in harder 3u?...coz i swear it's not in 3u circle geometry..

+Po1ntDeXt3r+ means D, not P, and the first line should be MBX, not MXB.

ie. it should start: MBX = CAD (angles at the circumfrence on the same arc are equal)

The method itself is fine.

ohk..thanks champ

 

[+Po1ntDeXt3r+]

lolz oops.. methods good -2 marks for mistake..

look i just woke up when i did it sorri

 

[+Po1ntDeXt3r+]

thanks CM_tutor.. Diagram was too small .. should hav listened to 4U teacher.. 1/2 A4 page diagrams
med writing didnt help.. P=D.. :S

umm mabbe it is harder 3U but it wasnt tat difficult.. if its a Q8 where did u get it from??

 

[ToO LaZy ^*]

thanks CM_tutor.. Diagram was too small .. should hav listened to 4U teacher.. 1/2 A4 page diagrams
med writing didnt help.. P=D.. :S

umm mabbe it is harder 3U but it wasnt tat difficult.. if its a Q8 where did u get it from??

yeah...maybe if i had heard of the rule, i might of had a chance.... :uhhuh:
it's from the 1999 CSSA paper

 

[+Po1ntDeXt3r+]

oh .. tat wasnt hard for a CSSA was Q8 the last one? :|

umm i went thru the 4U math syllabus to get all the rules that i MUST kno..
the rest were extras i picked up in past papers..

hmm there is a hard epping boys paper.. from lik 1997 or 98..
ruse papers will definately have them.. pre-03 are pretti hard esp 4U.. but 03 trials are tame.. 3U was hard..
umm lik i always sed just get the basics right and the hard stuff should fall in to place oh experience counts too ^^
gd luck

 

[ToO LaZy ^*]

yeah q8 was the last question.

in the meantime..take a look at this question if you haev spare time
question 7 from 2000 CSSA paper...

 

[CM_Tutor]

(a) I_n = ∫ (0-->1) (x² - 1)ⁿ dx, n = 0, 1, 2, 3, ...
= [x(x² - 1)ⁿ] (0-->1) - ∫ (0-->1) x d(x² - 1)ⁿ
= [1 * (1² - 1)ⁿ - 0 * (0² - 1)ⁿ] - ∫ (0-->1) x * n(x² - 1)^n-1 * 2x dx, for n = 1, 2, 3, ...
= [0 - 0] - ∫ (0-->1) 2nx²(x² - 1)^n-1 dx
= -2n * ∫ (0-->1) (x² - 1 + 1)(x² - 1)^n-1 dx
= -2n * ∫ (0-->1) (x² - 1)(x² - 1)^n-1 dx - 2n * ∫ (0-->1) 1 * (x² - 1)^n-1 dx
= -2n * ∫ (0-->1) (x² - 1)ⁿ dx - 2n * ∫ (0-->1) (x² - 1)^n-1 dx

So, we have I_n = -2nI_n - 2nI_n-1, for n = 1, 2, 3, ...
(1 + 2n)I_n = -2nI_n-1

So, I_n = [-2n / (2n + 1)]I_n-1, for n = 1, 2, 3, ... as required

Note that I₀ = ∫ (0-->1) 1 dx = [x] (0-->1) = 1 - 0 = 1

(b) Theorem If I_n = ∫ (0-->1) (x² - 1)ⁿ dx, for n = 0, 1, 2, ..., and noting that I_n = -2nI_n-1 / (2n + 1), for n = 1, 2, 3, ..., then I_n = (-1)ⁿ2²ⁿ(n!)² / (2n + 1)!, for positive integers n.

Proof: by induction on n

A Put n = 1: LHS = I₁ = [-2(1) / (2(1) + 1)] * I₀ = (-2 / 3) * 1 = -2 / 3
RHS = (-1)¹2²⁽¹⁾(1!)² / (2(1) + 1)! = -1 * 4 * 1² / 3! = -4 / 6 = -2 / 3 = LHS

So, the result is true for n = 1.

B Let k be a value of n for which the result is true.
That is, I_k = (-1)^k2^2k(k!)² / (2k + 1)! _____ (**)
We must now prove the result for n = k + 1.
That is, we must prove that I_k+1 = (-1)^k+12^2k+2[(k + 1)!]² / (2k + 3)!

LHS = I_k+1
= -2(k + 1)I_k / (2k + 3)
= [-2(k + 1) / (2k + 3)] * (-1)^k2^2k(k!)² / (2k + 1)!, using the induction hypothesis (**)
= [(-1) * 2 * (k + 1) * (-1)^k * 2^2k * (k!)²] / [(2k + 1)! * (2k + 3)]
= [(-1) * (-1)^k * 2 * 2^2k * (k + 1) * k! * k!] / [(2k + 1)! * (2k + 3)] * [(2k + 2) / (2k + 2)]
= [(-1)^k+1 * 2^2k+1 * (k + 1) * k! * k! * 2(k + 1)] / [(2k + 1)! * (2k + 3) * (2k + 2)]
= [(-1)^k+1 * 2^2k+1 * 2 * (k + 1) * k! * k! * (k + 1)] / [(2k + 3) * (2k + 2) * (2k + 1)!]
= [(-1)^k+1 * 2^2k+2 * (k + 1)! * (k + 1)!] / [(2k + 3) * (2k + 2) * (2k + 1)!]
= [(-1)^k+1 * 2^2k+2 * ((k + 1)!)²] / (2k + 3)!
= (-1)^k+12^2k+2[(k + 1)!]² / (2k + 3)!
= RHS

So, if the result is true for n = k, then it follows that it is also true for n = k + 1.

C It follows from A and B by mathematical induction that the result is true for all positive integers n.

 

[ToO LaZy ^*]

wow thanks

i used the wrong method to start off with....-_-"
i had..

int[0->1] (x²-1)^n-1*(x²-1)ⁿ
*slaps forehead*

 

[CrashOveride]

the equation of the normal to the ellipse of standard form (a^2 > b^2) at the point P(x1,y1) is a^2y1.x - b^2x1.y = (a^2 - b^2)x1.y1

This normal meets the major axis of the ellipse at G. S is the focus of the ellipse. Show that GS = e.PS (where e is the eccentricity of the ellipse)

Ok i found the co-ords of G and doing found GS but for the PS part i went manually with the distance formula and it looks like it gets messy any easier ways to do this one/ cheers

 

[CrashOveride]

The acceleration due to gravity at a distance r from the earth's centre is directed towards the centre and equal to k/r^2 when r>R, equal to kr when r < R and equal to g when r = R. Imagine a narrow tunnel along a diameter XY of the earth and the particle is projected from X with initial velocity U towards Y. (R = radius of earth)

If U^2 < 2gR, prove that motion is oscillatory and amplitude is given by R / (1 - U^2/2gR)

Just started resisted motion, little confused.

 

[unmentionable]

i guess u can try using the definition of an ellipse ps/pm = e
where m is the perpindicular foot of p on the directrix

so then you get GS = e^2PM
and PM is just a straighht line distance