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  • Subject: Let's learn how to find the area of a rectangle!
Subject: Let's learn how to find the area of a rectangle!

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Posted by: Soviet Revival
the only thing philosophical about this thread is mister math.


Posted by: Alpha Prime
Well looks like I need to learn some more maths.

Is this vector calculus? Also why did you introduce determinants? What do matrices have to do with this?
Yes, this is vector calculus.
Posted by: HS cRo55HaIr

Posted by: Mister Math

Posted by: Disambiguation
Feel free to continue this discussion, don't let it die! Math is important.
Indeed. Anyone else have any misconceptions about 0.999... = 1?

.999999999... <1

And 1.00000(insert almost infinate zeros)0001>1
FYI, there is no such thing as 1.0000....1 or .000...1. That suggests having a 1 after the last 0, but there are infinite 0s, i.e. no last 0.

As to ''almost infinite 0s,'' that is not relevant to 0.999...

[Edited on 12.10.2010 2:15 PM PST]

  • 12.10.2010 2:15 PM PDT

Good old hypocrisy, what humanity does best.

Posted by: Mister Math
Posted by: Alpha Prime
Well looks like I need to learn some more maths.

Is this vector calculus? Also why did you introduce determinants? What do matrices have to do with this?
Yes, this is vector calculus.
Hmm... I'll get to that either next semester or next year at Uni. Oh well.

  • 12.10.2010 2:21 PM PDT

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Posted by: Soviet Revival
the only thing philosophical about this thread is mister math.


Posted by: Alpha Prime
Posted by: Mister Math
Posted by: Alpha Prime
Well looks like I need to learn some more maths.

Is this vector calculus? Also why did you introduce determinants? What do matrices have to do with this?
Yes, this is vector calculus.
Hmm... I'll get to that either next semester or next year at Uni. Oh well.
I like it, it is pretty interesting.

  • 12.10.2010 2:22 PM PDT
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  • 12.10.2010 2:24 PM PDT

Good old hypocrisy, what humanity does best.

Posted by: Mister Math
I like it, it is pretty interesting.
It probably is just haven't encountered it yet and the wiki page isn't much help. I'll be doing plenty of it for physics anyway.

  • 12.10.2010 2:37 PM PDT

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Posted by: Alpha Prime
why did you introduce determinants? What do matrices have to do with this?
Matrices have everything to do with this. There's lots of proofs of the formula, I just thought this one was particularly elegant and decided to share it.

  • 12.10.2010 3:21 PM PDT

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If I was in Calculus, I would understand this.

  • 12.10.2010 3:25 PM PDT

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  • 12.10.2010 3:26 PM PDT

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  • 12.10.2010 3:29 PM PDT

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My Physics professor repeatedly tells us how easily understandable the formulas she presents to us are if we knew determinants. We don't, and she lol's at us every time, making us all sadface.

  • 02.03.2011 6:57 PM PDT
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Enlightened Ones. Join.

e^ix = cos(x) + isin(x)
Such that e is Euler's number, and i^2 = -1.

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Posted by: LazerSh0t
My Physics professor repeatedly tells us how easily understandable the formulas she presents to us are if we knew determinants. We don't, and she lol's at us every time, making us all sadface.
Why doesn't she teach you determinants?

  • 02.03.2011 6:59 PM PDT
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  • 02.03.2011 6:59 PM PDT
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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.
Posted by: Disambiguation
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.

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


[Edited on 02.03.2011 7:06 PM PST]

  • 02.03.2011 7:03 PM PDT

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Posted by: The Objectivist
Posted by: LazerSh0t
My Physics professor repeatedly tells us how easily understandable the formulas she presents to us are if we knew determinants. We don't, and she lol's at us every time, making us all sadface.
Why doesn't she teach you determinants?

Physics honours course in university = not enough time to teach us anything that isn't significantly relevant to the course. We get by fine using integrals; it's not as bad as I made it out to be.

  • 02.03.2011 7:06 PM PDT

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This thread confuses me...laymen's terms?

  • 02.03.2011 7:13 PM PDT

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  • 02.03.2011 7:15 PM PDT

Calculus calms my troubled mind

or lengthXwidth=area on any parallelogram. Most of that seemed pretty unnecessary.

  • 02.03.2011 7:15 PM PDT

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Posted by: jondoe4362
or lengthXwidth=area on any parallelogram.
Prove it.

  • 02.03.2011 7:16 PM PDT

Why go through so much trouble just to find the area of a freaking rectangle?

  • 02.03.2011 7:18 PM PDT


Posted by: Snipers Paradox
Why go through so much trouble just to find the area of a freaking rectangle?


This so fits the thread

  • 02.03.2011 7:20 PM PDT

­

L x W

*gasp*

I'm smart!

  • 02.03.2011 7:21 PM PDT

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Lol.

  • 02.03.2011 7:21 PM PDT

Posted by: XxBLU JELLOxX

Posted by: Snipers Paradox
Why go through so much trouble just to find the area of a freaking rectangle?


This so fits the thread

My sound card is broken.

  • 02.03.2011 7:23 PM PDT

The H times W

  • 02.03.2011 7:28 PM PDT
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Easy : LxW

  • 02.05.2011 10:37 AM PDT