/sci/ - Science & Math » Thread #11040583

6KiB, 831x699, Area_conversion_-_square_mm_in_a_square_cm.png

View Same Google iqdb SauceNAO

Anonymous Mon 07 Oct 23:17:27 2019 No.11040583 View Reply Original Report

Quoted By: >>11040586 >>11040596 >>11040602 >>11040614 >>11040792 >>11040838 >>11040873

Why does area and volumre calculation has to involve perpandicular lines ?

Anonymous

Anonymous Mon 07 Oct 2019 23:17:44 No.11040586 Report

Quoted By: >>11040589

>>11040583
it doesn't

Anonymous

Anonymous Mon 07 Oct 2019 23:18:53 No.11040589 Report

Quoted By: >>11040592

>>11040586
Elaborate ?

Anonymous

Anonymous Mon 07 Oct 2019 23:19:18 No.11040592 Report

Quoted By: >>11040597

>>11040589
area and volume calculation does not have to involve perpendicular lines

Anonymous

Anonymous Mon 07 Oct 2019 23:20:48 No.11040596 Report

Quoted By:

>>11040583
To generate a 3D space, you don't even need perpendicular vectors

Anonymous

View Same Google iqdb SauceNAO aid4281332-v4-728px-Find-the-Are (...).jpg, 66KiB, 728x546

Anonymous Mon 07 Oct 2019 23:21:18 No.11040597 Report

Quoted By: >>11040605 >>11040610

>>11040592
Any shape you think of , to get the area you have to draw perpendicular lines and work from there

Anonymous

Anonymous Mon 07 Oct 2019 23:22:45 No.11040602 Report

Quoted By: >>11040609

>>11040583
for instance, a cone
volume = ?*(r^2)h/3

no perpendiculars

Anonymous

Anonymous Mon 07 Oct 2019 23:23:47 No.11040605 Report

Quoted By: >>11040618

>>11040597
post how you get area and surface area of sphere using perpendicular lines.

Anonymous

Anonymous Mon 07 Oct 2019 23:25:05 No.11040609 Report

Quoted By: >>11040618

>>11040602
The H is perpendicular , and the way you derive the area of the circular base comes from the area of a polygone which involves drawing perpendicular lines

Anonymous

View Same Google iqdb SauceNAO maxresdefault.jpg, 62KiB, 1280x720

Anonymous Mon 07 Oct 2019 23:25:07 No.11040610 Report

Quoted By: >>11040618

>>11040597
No you don't

Anonymous

Anonymous Mon 07 Oct 2019 23:28:10 No.11040614 Report

Quoted By: >>11040627

>>11040583
Because the units are square.
It all comes back to square meters or cubic meters. You can find area and volume without perpendicular lines but its best to use them because the math gets much more simple.

Anonymous

Anonymous Mon 07 Oct 2019 23:29:32 No.11040618 Report

Quoted By: >>11040624

>>11040610
I don't fully understand why this is true , but dot products and cross products are perpendicular by definition
>>11040605
See this >>11040609

Anonymous

Anonymous Mon 07 Oct 2019 23:31:43 No.11040624 Report

Quoted By: >>11040636

>>11040618
>dot products
no they aren't. you can find the inner product between two polynomials, even. or between two exponential functions.
>cross products
orthogonal =/= perpendicular

Anonymous

View Same Google iqdb SauceNAO Two_rays_and_one_vertex.png, 31KiB, 1049x792

Anonymous Mon 07 Oct 2019 23:32:17 No.11040627 Report

Quoted By: >>11040631

>>11040614
Let's say i have two segments at an angle , how do i visualise the multiplication of thier length ?

Anonymous

Anonymous Mon 07 Oct 2019 23:33:53 No.11040631 Report

Quoted By: >>11040642

>>11040627
>multiplication of their length
|ray1|*|ray2|

Anonymous

Anonymous Mon 07 Oct 2019 23:34:42 No.11040636 Report

Quoted By: >>11040638

>>11040624
Well if you're dealing with well defined shape and Cartesian coordinates you're dot product is definitely going to correspond to some perpendicular line

Anonymous

Anonymous Mon 07 Oct 2019 23:35:12 No.11040638 Report

Quoted By:

>>11040636
good thing I'm not restricted to Cartesian coordinates

Anonymous

Anonymous Mon 07 Oct 2019 23:36:52 No.11040642 Report

Quoted By: >>11040645 >>11040655

>>11040631
Key word "Geometrically" i always imagined the are of a square as sliding on side over the other and i thought why i can't do that for segments at an angle

Anonymous

Anonymous Mon 07 Oct 2019 23:38:13 No.11040645 Report

Quoted By:

>>11040642
i don't know what the fuck you are talking about anymore

Anonymous

Anonymous Mon 07 Oct 2019 23:44:20 No.11040655 Report

Quoted By:

>>11040642
Think about the integral of a costant function using reiman sum

Anonymous

Anonymous Mon 07 Oct 2019 23:56:31 No.11040686 Report

Quoted By: >>11040793

Integrating using differential geometry for arbitrary Riemannian manifolds still works by using perpendicular lines, since the differentials used for "summing" are the basis vectors of the tangential space. The derivative is the linear approximation to the manifold at a certain point.

Anonymous

Anonymous Tue 08 Oct 2019 00:42:59 No.11040792 Report

Quoted By:

>>11040583
You could define and calculate an area in hyperboloid space.

Anonymous

Anonymous Tue 08 Oct 2019 00:43:05 No.11040793 Report

Quoted By: >>11040800

>>11040686
>Integrating using differential geometry for arbitrary Riemannian manifolds still works by using perpendicular lines, since the differentials used for "summing" are the basis vectors of the tangential space.
Except that you're not summing vectors, you're summing forms (recall the derivation of Stokes' theorem in eg. Baby Rudin, which doesn't require the simplicial elements to be perpendicular), and the tangent spaces play no role in the integration except, if you're being charitable with the interpretation, as a convenient coordinate space to ensure d^2=0. That's why the tangent bundle can, in general, have a nontrivial topology (different from the product topology).

Anonymous

Anonymous Tue 08 Oct 2019 00:47:08 No.11040800 Report

Quoted By:

Measure theoretic notions of volume/area have no such requirement, but the Lebesgue measure is defined by using axially oriented cubes in R^n and taking the measure generated by those, so it's nice and easy to think about it that way. Why do we do that? They have the most obvious notion of volume.
>>11040793
Thank you.

Anonymous

Anonymous Tue 08 Oct 2019 01:01:19 No.11040838 Report

Quoted By:

>>11040583
You don't. So long as you have a function that represents the shape you can find the area using integration. If you count Riemann sums as abstracted perpendicular lines then you can do Monte Carlo integration.

n-cubes also tile nicely in n-dimensional space. Pretty hard to get the volume of a cube by counting the number of spheres you can fit inside.

Anonymous

Anonymous Tue 08 Oct 2019 01:16:49 No.11040873 Report

Quoted By:

>>11040583

Using 'square units' (eg square cm, sq ft) is just a convention.

We could just as easily have decided to use triangular units (where each unit is the area of a unit triangle) circular units, etc.

You can choose whatever area you want as your basis--whether that's sq cm, a 3x2 rectangle, area of a circle diameter 1 cm, area of a football field...

Capcode	All Only User Posts Only Moderator Posts Only Admin Posts Only Developer Posts
Show Posts	All Only With Images Only Without Images
Deleted Posts	All Only Deleted Posts Only Non-Deleted Posts
Ghost Posts	All Only Ghost Posts Only Non-Ghost Posts
Post Type	All Only Sticky Threads Only Opening Posts Only Reply Posts
Results	All Grouped By Threads
Order	Latest Posts First Oldest Posts First

Your latest searches