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This topic has moved here: Subject: Let's learn how to find the area of a rectangle!
  • Subject: Let's learn how to find the area of a rectangle!
Subject: Let's learn how to find the area of a rectangle!

"The individual has always had to struggle to keep from being overwhelmed by the tribe. If you try it, you will be lonely often, and sometimes frightened. But no price is too high to pay for the privilege of owning yourself."
-Nietzsche

First, let's consider a rectangle. On a coordinate plane, we'll say it has vertices at (0,0), (a,0), (0,b), and (a,b)

Then, an orthogonal parametrization of the boundary can be obtained by taking F(x,y) = (ax, by)

The domain of F is D = {(x,y): 0 <= x <= 1, 0 <= y <= 1}

Our general formula claims that: Area = integral (sqrt [[ Fx*Fx, Fx*Fy][Fy*Fx, Fy*Fy]] over the domain D. The * operation in this case is the vector scalar product. Since the matrix is positive definite, the root of its determinant will never be imaginary so we will always have a positive or zero real area.

The partial derivatives can be easily computed to be
Fx = (a, 0) and Fy = (0, b)

Evaluating the determinant yields a^2*b^2, so that sqrt(a^2*b^2) = a*b

Integrating over [0,1] with respect to x and over [0,1] with respect to y yields Area = (a*b)*1*1 = a*b

So we see that the area of a rectangle is, perhaps not obviously, the product of the lengths of its sides.

[Edited on 12.10.2010 12:21 PM PST]

  • 12.10.2010 12:14 PM PDT

Looks like somebody's been down here with the ugly stick.

2+2=5

Or something

  • 12.10.2010 12:15 PM PDT

Imma musician and things :3

DOWNLOAD MY MUSIC

OR

Width x height.

Hurr durr

[Edited on 12.10.2010 12:16 PM PST]

  • 12.10.2010 12:15 PM PDT

Wow. You used so many unnecessary symbols to find the area of a rectangle. Well done.

  • 12.10.2010 12:15 PM PDT

Rarity FTW

E=MC^2. Right?

  • 12.10.2010 12:16 PM PDT

"Hero's Get Remembered, But Legends Never Die. Follow Your Heart, And You Can Never Go Wrong."

___________(""""II"""" ; ;II""""")_TT______
I ---------____.`=====.-. :________\___|================[oo]
I_III___/___/_/~"""I_I_I_I'''

lolwut

  • 12.10.2010 12:17 PM PDT

Why don't you just try walking on your hands? Then you can use your feet for high fives and eating sandwiches, you know, the important stuff.

my blood hurts.

  • 12.10.2010 12:17 PM PDT

A=L*H

/thread

  • 12.10.2010 12:18 PM PDT

"The individual has always had to struggle to keep from being overwhelmed by the tribe. If you try it, you will be lonely often, and sometimes frightened. But no price is too high to pay for the privilege of owning yourself."
-Nietzsche

Posted by: DaGingaNinja
A=L*H

/thread
Prove it.

  • 12.10.2010 12:18 PM PDT


Posted by: Disambiguation
Posted by: DaGingaNinja
A=L*H

/thread
Prove it.

It doesn't need proved. Everyone over the age of 10 knows this.

  • 12.10.2010 12:19 PM PDT
  •  | 
  • Exalted Legendary Member

This signature is now diamonds.
-------------
Every time I look at my member title, all I can think is, "-blam!-, I'm old..."

You're not as cool as Mister Math.

  • 12.10.2010 12:19 PM PDT

"The individual has always had to struggle to keep from being overwhelmed by the tribe. If you try it, you will be lonely often, and sometimes frightened. But no price is too high to pay for the privilege of owning yourself."
-Nietzsche

Posted by: BSS Lazer

Posted by: Disambiguation
Posted by: DaGingaNinja
A=L*H

/thread
Prove it.

It doesn't need proved. Everyone over the age of 10 knows this.
Because some hack of a teacher with a B.A. told you so?

  • 12.10.2010 12:19 PM PDT

No signature found. Click here to change this.

we'll say it has vertices at (0,0), (0,a), (0,b), and (a,b)

vertice fail

  • 12.10.2010 12:20 PM PDT

Quartic Formula, go!

  • 12.10.2010 12:20 PM PDT

Wort Wort Wort???

Also, hilarious quote:
Posted by: Vgnut117
True. Now that I think about it having the -blam!- beat out of you with a pool ball in a tube sock by your degenerate pimp is the better choice.

.

Posted by: BSS Lazer
It doesn't need proved. Everyone over the age of 10 knows this.


Yes, and you want to know why? It's because somebody like Disambiguation, a long time ago, did that exact same math and found what Disambiguation proved - that L x W is the area.

Were it not for that proof we would not know that.

  • 12.10.2010 12:20 PM PDT
  •  | 
  • Fabled Legendary Member

They call me graland.

Posted by: Disambiguation
. . . the hell?

Why are you using derivatives and domains and matrices? Just use length times width.

I think you're trying to mathematically prove that length times width yields the area of a rectange, which is an exercise in futility, especially on the Flood.

[Edited on 12.10.2010 12:21 PM PST]

  • 12.10.2010 12:20 PM PDT

"The individual has always had to struggle to keep from being overwhelmed by the tribe. If you try it, you will be lonely often, and sometimes frightened. But no price is too high to pay for the privilege of owning yourself."
-Nietzsche

Posted by: AvengerTam1010
we'll say it has vertices at (0,0), (0,a), (0,b), and (a,b)

vertice fail
lol... Good catch. I'll just fix that real quick here...

  • 12.10.2010 12:21 PM PDT

"Give me a place to stand, and I will move the Earth."
-Archimedes

Half-Life Universe
Secular Sevens
Posted by: Soviet Revival
the only thing philosophical about this thread is mister math.

But Mr. D, what if the rectangle is no longer a rectangle, but has curved sides?

  • 12.10.2010 12:21 PM PDT


Posted by: DabilahroNinja
OR

Width x height.

Hurr durr

tru dat, now how do we find the area of 3 spheres that are coalesce?

  • 12.10.2010 12:21 PM PDT

**Devil's advocate of the Flood. My posts may or may not represent my personal opinion, I just enjoy disagreeing with people. None of my posts are representative of the official view of the Navy or any government agency.

Non Sibi Sed Patriae
Homework questions? Forget the Flood, join The Academy.
I've got a fan!


Posted by: Disambiguation
So we see that the area of a rectangle is, perhaps not obviously, the product of the lengths of its sides.


Nope, it's the lengths of two adjacent sides. But I'm just being nitpicky :)

  • 12.10.2010 12:21 PM PDT
  •  | 
  • Fabled Legendary Member

They call me graland.

Posted by: Warlord AIpha
Were it not for that proof we would not know that.

Yes, but aside from Disambiguation and Mister Math, who would care?

  • 12.10.2010 12:23 PM PDT

"The individual has always had to struggle to keep from being overwhelmed by the tribe. If you try it, you will be lonely often, and sometimes frightened. But no price is too high to pay for the privilege of owning yourself."
-Nietzsche

Posted by: Mister Math
But Mr. D, what if the rectangle is no longer a rectangle, but has curved sides?
You simply integrate over square root of the determinant of the matrix of scalar products of partial derivatives. Of course, you need an orthogonal parametrization of the boundary first. If you don't have one then this method doesn't work.

[Edited on 12.10.2010 12:24 PM PST]

  • 12.10.2010 12:23 PM PDT

This Message Will Self Destruct In 10 Seconds

Posted by: Disambiguation
First, let's consider a rectangle. On a coordinate plane, we'll say it has vertices at (0,0), (a,0), (0,b), and (a,b)

Then, an orthogonal parametrization of the boundary can be obtained by taking F(x,y) = (ax, by)

The domain of F is D = {(x,y): 0 <= x <= 1, 0 <= y <= 1}

Our general formula claims that: Area = integral (sqrt [[ Fx*Fx, Fx*Fy][Fy*Fx, Fy*Fy]] over the domain D. The * operation in this case is the vector scalar product. Since the matrix is positive definite, the root of its determinant will never be imaginary so we will always have a positive or zero real area.

The partial derivatives can be easily computed to be
Fx = (a, 0) and Fy = (0, b)

Evaluating the determinant yields a^2*b^2, so that sqrt(a^2*b^2) = a*b

Integrating over [0,1] with respect to x and over [0,1] with respect to y yields Area = (a*b)*1*1 = a*b

So we see that the area of a rectangle is, perhaps not obviously, the product of the lengths of its sides.

icwatudidthar

  • 12.10.2010 12:23 PM PDT

Bend it like Barrera

Guys, let's find the exact value of PI!

-RM-

  • 12.10.2010 12:24 PM PDT

Rarity FTW


Posted by: Real Madrid
Guys, let's find the exact value of PI!

-RM-


3.14....

Uh...you guys take the rest...

[Edited on 12.10.2010 12:25 PM PST]

  • 12.10.2010 12:25 PM PDT