/sci/ - Science & Math

[Anonymous No.6828940]

Hey /sci/, i recently started learning mathematics(self study), i'm learning analytic geometric at the moment and it's kinda hard in my opinion, so i was wondering how hard is college level math compared to analytic geometry.

Also, why do we use Pythagoras theorem to calculate the distance between two points?

[Anonymous No.6828964]

>Also, why do we use Pythagoras theorem to calculate the distance between two points?

Huh? What do you mean by this?

[Anonymous No.6828968]

>>6828964

using Cartesian coordinates on the plane, the distance between two points (x1, y1) and (x2, y2) is defined by the formula

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2},\!

which can be viewed as a version of the Pythagorean theorem

[Anonymous No.6828983]

>>6828968
it's a right triangle.

difference in x coordinates is one of the sides
difference in y coordinates is the other one
straight line from x to y is to be found

[Anonymous No.6829263]

>>6828940
Distance with pythagoras is only useful for euclidean distance.

[Anonymous No.6829321]

>>6828968
Because you are essentially asking the question about the hypotenuse of a right triangle

Come to Riemannian geometry and try dat shit nigur :^)

[Anonymous No.6829370]

>>6828940
>Also, why do we use Pythagoras theorem to calculate the distance between two points?

The shortest path between 2 points is a straight line connecting them. You can eventually prove it with Calculus of Variations but for now take it for face value.

For simplicity, move the origin to the first point. Now the second point represents a vector/ray from the origin with coordinates (x0=x2-x1,y0=y2-y1) in 2D. The projection onto the x axis is x0 and y0 for the y axis so the ray/vector from the origin is the hypotenuse of the right triangle of sides x0 and y0. Thus you find the distance from the origin with this length by the Pythagorean theorem. In 3D your projection is (x0,y0,z0) and you first calculate the distance in the x-y plane. Then you know the z0 line is at a right angle to projection in x-y plane and you find the distance by applying the Pythagorean theorem again

<span class="math">√( (√(x0^2+y0^2))^2 + z0^2 )=√( x0^2 + y0^2 + z0^2 )[/spoiler]

[Anonymous No.6830011]

bump

[Anonymous No.6830023]

>>6829370
you're weird

[Anonymous No.6830276]

>>6828940
The point of analytic geometry is to replicate euclidean geometry with a coordinate system. The pythagorean theorem is proposition 47 of Book 1 of Euclid's Elements, so, it is reasonable to shoe-horn in the euclidean measure into distance calculations.

[Anonymous No.6830283]

>>6828940
>astronomy
>smart

>tfw pure mathfag and your classes have about 50% women and all the creepy neckbeards fail out into physics and comp sci.