Flex your maths muscles! (2 Viewers)

[LoneShadow]

that's better. If someone tries to see how's the question done, they should be able to follow step-by-step.

btw. did u recieve my reply? Somehow I had been loged out when I clicked send.

 

[acmilan]

sup LoneShadow!

Having not done 4 unit, i dont know if this question is beyond your ability, but i'll ask anyways. I know one way to do it (beyond 4 unit level), but seeing as you're all a lot better than me in integration, maybe you know a better way

∫₀^π dx/(2-cosx)²

 

[LoneShadow]

:wave: hey V!

I'll add your question. Someone provide him with a better answer than his own. If that is actually possible

 

[LoneShadow]

Added some new questions on Conics. Some of you might have seen them before.

By the way, have you guys done Mechanics yet? Can I add questions on it?

 

[pLuvia]

I'm sure some schools have covered the course, but I haven't yet

 

[pLuvia]

20.
P(acosθ,bsinθ)
S(ae,0)
MP_x=(acosθ+ae)/2 --(1)
MP_y=bsinθ/2 --(2)
From (1)
cosθ=(2x-ae)/a
From (2)
sinθ=2y/b

(cosθ)²+(sinθ)²=(2x-ae)²/a²+4y²/b²=1
Which is in the standard form of an ellipse, hence the locus of the midpoint of PS is an ellipse with centre (ae/2,0)

 

[STx]

Q19)a)
a= √a
b²=a²(1-e²)
i.e. e²=a²-b²/a²
e=√a√a²-b²/a

Foci(±√a√a²-b²/a,0)
Directrices: x= ± a√a²-b²/a²-b²

Now, SP/PM=e, S'P/PM'=e (Ellipse Property)
i.e. ePM=SP, ePM'=S'P
and SP'+S'P = e(PM+PM')=2a=2√a
using e(PM+PM')=[√a√a²-b²/a] *[2a√a²-b²/a²-b²]
= 2√a
=2a (as required)

b) x²/a²+y²/b²=1
2yy'/b²=-2x/a²
2yy'=-2b²x/a²
y'=-b²x/a²y

At P(acosθ,bsinθ), y'=-b²(acosθ)/a²(bsinθ)
y'_Normal=a²bsinθ/b²acosθ

Therefore,
Eqn_{Normal at P}:

y-bsinθ=a²bsinθ/b²acosθ*[x-acosθ]
bcosθ(y-bsinθ)=asinθ(x-acosθ)
bycosθ-b²sinθcosθ=axsinθ-a²sinθcosθ
axsinθ-bycosθ=sinθcosθ(a²-b²)

sub y=0 for normal to ellipse at P to meet major axis at H
i.e. axsinθ=sinθcosθ(a²-b²)
x=cosθ(a²-b²)/a (as required)

i.e H(cosθ(a²-b²)/a, 0)

 

[STx]

Heres two q's i found, im not sure that their hard enough to be in this thread though. I know how to do the second one, but not the first one, because i havent done 4u integration yet.


1)

2) Let the points A₁,A₂,...,A_n represent the nth roots of unity, w₁,w₂,...,w_n, and suppose P represents any complex number z such that |z|=1.

(i) Prove that w₁,w₂,...,w_n = 0
edit:

(ii) Show that PA_i² = (z-w_i)(z'bar'-w'bar'_i) for i=1,2,...,n.

(iii) Prove that ∑_i=1 ⁿ PA_i²=2n

 

[pLuvia]

(19) The point P(a*cosθ, b*sinθ ) lies on the ellipse with equation: x2/a2 + y2/b2 = 1, with a>b.
The foci of the ellipse are S and S' and M, M' are the feet of the perpendiculars from P onto the directrices corresponding to S and S'.
The normal to the ellipse at P meets the major axis of the ellipse at H.

How would you draw this diagram? Thanks, maybe I could do the question after I draw it

 

[LoneShadow]

that's part of the question. Try sketching step by step. If still can't do it, I'll post the graph.

Just a note: The list of questions is getting a bit too long. It might reach the post length limit, if there exist one. Since I am trying to learn LaTex I'll type up the questions and if possible the answers as well.

 

[STx]

Pluvia:

edit: woops, i posted this before i read LoneShadows last post.

 

[Riviet]

23) find ∫(1 + tan²θ )e^tanθdθ

Let u=tanθ, du=sec²θ dθ

.'.∫(1 + tan²θ )e^tanθdθ=∫sec²θ.e^tanθdθ

=∫e^udu

=e^tanθ + C

Just a note: The list of questions is getting a bit too long. It might reach the post length limit, if there exist one. Since I am trying to learn LaTex I'll type up the questions and if possible the answers as well.

Don't you worry, if there is a limit, we definitely haven't reached anywhere near it... you should have a look at this. XD

 

[LoneShadow]

Please check your arithmatics. I also have added another part to that question.

 

[pLuvia]

24.
(a) I<SUB>n</SUB> = ∫<SUB>0</SUB><SUP>i</SUP>(1+x<SUP>2</SUP>)<SUP>n</SUP>dx
u=(1+x²)ⁿ -- dv=dx
du=n(1+x²)^n-12x -- dx v=x
I_n=[x(1+x²)ⁿ]{0-->i} - ∫{0-->i}xn(1+x²)^n-12x dx
=[i(1+i²)ⁿ]-[0]-2n∫{0-->i}x²(1+x²)^n-1dx
=-2n∫{0-->i}x²(1+x²)^n-1dx
=-2n∫{0-->i}(1+x²-1)(1+x²)^n-1dx
=-2n∫{0-->i}[(1+x²)ⁿ-(1+x²)^n-1]dx
=-2n[I_n-I_n-1]
=-2nI_n-2nI_n-1
I_n+2nI_n=2nI_n-1
I_n(1+2n)=2nI_n-1
I_n={2n/(1+2n)}I_n-1

(b)I₀=∫{0-->i}(1+x²)⁰dx
=∫{0-->i}dx
=[x]{0-->i}
=-i
I₁=2/3(I₀)
=2/3(-i)
I₂=4/5(I₁)
=4/5[2/3(-i)]
=8/15(-i)
I₃=6/7(I₂
=6/7[8/15(-i)]
=48/105(-i)
=16/35(-i)

 

[acullen]

Just thought I'd add some questions in to keep you school kiddies busy:

Find the following improper integrals:
28. ∫xⁿ∙sin(ax)∙dx

29. ∫dx/(cos(x)-sin(x)+1)

30. ∫(x²+1)dx/(x(x-2)(x+3))

31. ∫ln(tan(x))sec²(x)∙tan(x)∙dx

Write in cartesian form:
32. (-1+i)⁷

33. (^√2/₂ - i∙^√2/₂)¹²

Solve for z:
34. z⁴+2z³+4z²+8z+16=0

 

[pLuvia]

29.
∫dx/(cos(x)-sin(x)+1)
Let t=tan(x/2)
dt=1/2(sec²{x/2})dx
dx=dt/[1/2(1+tan²{x/2})]
dx=2dt/(1+t²)
I=∫1/[1-t²/1+t²]-[2t/1+t²]+1)*2dt/(1+t²)
=∫2dt/[1-t²-2t+1+t²]
=∫2dt/(2-2t)
=-ln(2-2t)+C
=-ln(2-2tan(x/2)+C

 

[pLuvia]

30. ∫(x²+1)dx/(x(x-2)(x+3))
By partial fractions you will obtain
x²+1=A(x-2)(x+3)+Bx(x+3)+Cx(x-2)
Let x=0
A=-1/6
Let x=2
B=1/2
Let x=-3
C=2/3
∫(x²+1)dx/(x(x-2)(x+3))
=-1/6∫dx/x+1/2∫dx/(x-2)+2/3∫dx/(x+3)
=-1/6lnx+1/2ln(x-2)+2/3ln(x+3)+C

 

[pLuvia]

32. (-1+i)⁷
Let z=(-1+i)
z=√2(cos[3pi/4]+isin[3pi/4])
z⁷=
=(√2}⁷(cos[21pi/4]+isin[21pi/4])
=8√2(-1/√2)+i8√2(-1/√2)
=-8-8i

 

[pLuvia]

33. (^√2/₂ - i∙^√2/₂)¹²
Let z=(^√2/₂ - i∙^√2/₂)
z=(cos[7pi/4]+isin[7pi/4])
z¹²=(cos[21pi]+isin[21pi]
=-1+i0
=-1

 

[STx]

32. (-1+i)⁷
Let z=(-1+i)
z=√2(cos[3pi/4]+isin[3pi/4])
z⁷=
=(√2}⁷(cos[21pi/4]+isin[21pi/4])
=8√2(-1/√2)+i8√2(-1/√2)
=-8-8i

i get 8+8i. I used arg(z)=(-pi/4)
i.e. z=√2(cos[-pi/4]+isin[-pi/4])
z⁷=(√2)⁷[cis-7pi/4]
= 8√2(cos[pi/4]+isin[pi/4]) (-7pi/4 + 2pi=pi/4 to make within principal arg)
= 8√2(1/√2+[1/√2]i)
= 8+8i