b3ta.com talk
You are not logged in. Login or Signup
Home » Talk » Message 6240719

perineum

(0800221155 Coldseal windows, Sat 20 Jun 2009, 1:03, archived)
NONCE!

(I helped save b3ta! Donkey Gums @mattcomedy, Sat 20 Jun 2009, 1:03, archived)
NONCE!

(I helped save b3ta! Donkey Gums @mattcomedy, Sat 20 Jun 2009, 1:12, archived)
DOUBLE NONCE!

(I helped save b3ta! Donkey Gums @mattcomedy, Sat 20 Jun 2009, 1:13, archived)
Daffodil.

(I helped save b3ta! JessAction Give me all your expensive brandy and hubcaps., Sat 20 Jun 2009, 1:03, archived)
WIN!
Come on Welsh contingency, yeah FTW etc.... or something.
(I helped save b3ta! The Doveston haunted by the memory of his own amnesia, Sat 20 Jun 2009, 1:07, archived)
Pints of Brains all round!

(I helped save b3ta! JessAction Give me all your expensive brandy and hubcaps., Sat 20 Jun 2009, 1:09, archived)
SA Gold.
Is what I have been drinking. Good brains, MMMmmm.
(I helped save b3ta! The Doveston haunted by the memory of his own amnesia, Sat 20 Jun 2009, 1:13, archived)
epesiotomy
sorry, is this word association?
(I helped save b3ta! Rotating Wobbly Hat That's not a banana. THIS is a banana., Sat 20 Jun 2009, 1:03, archived)
I Wiki'd that
And found the picture amusing.
(I helped save b3ta! Methylene Blue - electrohead, Sat 20 Jun 2009, 1:05, archived)
as amusing as a death threat, as I recall

(I helped save b3ta! mr horrible up yours, dickface, Sat 20 Jun 2009, 1:06, archived)
NEEDS MORE HULK HOGAN DESTROYING THE TWIN TOWERS.

(I helped save b3ta! JessAction Give me all your expensive brandy and hubcaps., Sat 20 Jun 2009, 1:08, archived)
Scruttocks

(I helped save b3ta! Methylene Blue - electrohead, Sat 20 Jun 2009, 1:04, archived)
weed

(I helped save b3ta! Captain Tripps is your taxloss lover from Liverpool, Sat 20 Jun 2009, 1:04, archived)
millenium

(I helped save b3ta! weaver., Sat 20 Jun 2009, 1:04, archived)
Willenium.

(I helped save b3ta! The Doveston haunted by the memory of his own amnesia, Sat 20 Jun 2009, 1:04, archived)
Hold please.
Will Smith's just bought the loophole.

"Will-hole". Sounds a bit gay.
(I helped save b3ta! The Mock TurtIe ™ --- Thinks you are a cunt, on, Sat 20 Jun 2009, 1:33, archived)
No.

(I helped save b3ta! The Doveston haunted by the memory of his own amnesia, Sat 20 Jun 2009, 1:04, archived)
well done with the whole "making the internet pay attention to you" thing there

(I helped save b3ta! Gilgamesh gazed in wonder as Frampton came alive, Sat 20 Jun 2009, 1:12, archived)
like anyone's got anything better to do right now

(0800221155 Coldseal windows, Sat 20 Jun 2009, 1:55, archived)

d2x/dt2 = -x. Find a general solution for x in terms of t.
Let y = dx/dt. Then dy/dt = d2x/dt2 = -x.
By the chain rule, dy/dt = (dy/dx)(dx/dt) = y(dy/dx).
Therefore y(dy/dx) = -x.
Therefore ∫y dy = ∫-x dx.
y2/2 = -x2/2 + constant.
y2 = c2 - x2 ≡ (dx/dt)2 = c2 - x2
dx/dt = √(c2 - x2)
So, ∫dx / √(c2 - x2) = ∫ dt.
Let x = csin θ. Then dx = ccos θ dθ, and √(c2 - x2) = √(c2(1 - sin2 θ)) = √(c2cos2 θ) = ccos θ.
Therefore ∫ (ccos θ / ccos θ) dθ = ∫ dθ = ∫ dt; therefore, θ = t + k.
As x = csin θ, θ = sin-1 (x/c).
So sin-1 (x/c) = t + k.
Therefore, x = csin(t + k) = c(sin t)(cos k) + c(cos t)(sin k) ≡ Asin t + Bcos t.

Not bad considering it's been about six years since I last did any kind of serious maths.
(I helped save b3ta! The Mock TurtIe ™ --- Thinks you are a cunt, on, Sat 20 Jun 2009, 1:19, archived)
NNNEEEERRRRRDDDDDDD!!!!!

(I helped save b3ta! The Doveston haunted by the memory of his own amnesia, Sat 20 Jun 2009, 1:27, archived)
And if you want proof of that last result,
Let OA, OB and OC be chords of unit length such that angle BOA = α, angle COB = β and angle COA = (α + β).
Let X be a point on the chord OB such that angle OXC = 90°; therefore |OX| = cos β and |CX| = sin β.
Let Y and Z respectively be points on the chords OB and OA such that angles CYZ = 180° and CZO = 90°. Then, |CZ| = sin(α + β) and |OZ| = cos(α + β).
The chord CZ can be split into two components, CY and YZ.
CY is the hypotenuse of right-angled triangle CXY. What is the angle CXY?
Angle OCX = 90° - β. Angle OCY = 90° - (α + β). Angle CXY = angle OCX - angle OCY = α.
As we know |CX| = sin β, and that CX is the adjacent side of right-angled triangle CXY, |CX| = |CY|cos α = sin β.
Therefore, |CY| = sin β / cos α.
YZ is the opposite side of triangle OYZ. Angle YOZ = α, so |YZ| = |OY|sin α.
As OY + YX = OX, and |OX| = cos β, |OY| = |OX| - |YX|.
YX is the opposite side of triangle CXY, so |YX| = |CY|sin α = (sin β sin α) / cos α. Therefore |OY| = cos β - (sin β sin α) / cos α.
Then |YZ| = sin α(cos β - (sin β sin α) / cos α) = sin α cos β - (sin β / cos α) sin2 α.
|CZ| = |CY| + |YZ| = sin(α + β)
= (sin β / cos α) + sin α cos β - (sin β / cos α) sin2 α
= sin α cos β + (sin β / cos α)(1 - sin2 α)
= sin α cos β + (sin β / cos α)cos2 α
therefore sin (α + β) = sin α cos β + cos α sin β
(I helped save b3ta! The Mock TurtIe ™ --- Thinks you are a cunt, on, Sat 20 Jun 2009, 2:02, archived)
Furthermore,
As |OZ| = cos(α + β), and OZ is the adjacent side of triangle OYZ, |OZ| = |OY|cos α.
= cos α(cos β - (sin β sin α) / cos α).
Therefore = cos(α + β) = cos α cos β - sin α sin β

As tan(α + β) = sin(α + β) / cos(α + β),
tan(α + β) = (sin α cos β + sin β cos α) / (cos α cos β - sin α sin β)
Dividing through by cos α cos β,
tan(α + β) = ((sin α cos β / cos α cos β) + (sin β cos α / cos α cos β)) / ((cos α cos β / cos α cos β) - (sin α sin β / cos α cos β))
= ((sin α / cos α) + (sin β / cos β)) / (1 - (sin α sin β) / (cos α cos β))
so tan(α + β) = (tan α + tan β) / (1 - tan α tan β)
(I helped save b3ta! The Mock TurtIe ™ --- Thinks you are a cunt, on, Sat 20 Jun 2009, 2:03, archived)