/sci/ - Science & Math

It's called line integral.

Yeah, google 'arc length formula'. Interestingly, the length of the sine function has no closed form in terms of elementary functions.

>>5745910
No it's fucking not! Line integrals are used to integrate scalar or vector fields along a curve, a point on this curve is the sum of the field at that point. What you want is the arc length

>>5745945

Well, the length is the line integral of the scalar field 1.

>>5745945
Yeah, the pages I found on that seemed a bit off

I looked up arc length and I found this page, does it do a decent job of explaining it?

http://www.mathwords.com/a/arc_length_of_a_curve.htm

Yep. In flat 2D space, if you move an infinitesimal distance in the x-direction, dx, and an infinitesimal amount in the y-direction, dy, then the total distance you've moved is given by the Pythagorean theorem:

<div class="math">ds^2=dx^2+dy^2</div>
The length of some arbitrary curve C is then an integral of ds over the curve:

<div class="math">L=\int_C ds=\int_C \sqrt{dx^2+dy^2}</div>
Then all you need to do is parametrize the curve and evaluate the integral. Here's a simple example: a basic parabola y=x^2 from x=0 to x=1. If we parametrize with respect to x and get the integral in a more useful form, it becomes:

<div class="math">L=\int_0^1 \sqrt{1+\left (\frac{dy}{dx} \right )^2}dx</div>
Now, dy/dx=2x, so we simply plug that in:
<div class="math">L=\int_0^1 \sqrt{1+4x^2}dx \approx 1.479</div>

>>5745945
and what happens if you set the scalar field to one and calculate the line integral?

>>5745945
Stop being a semantic prick.

>>5745923

>>5745958
Ooh I like that, that works

On this page about the subject it shows formulae for parametric and polar form as well

<span class="math">ds=\sqrt{\left (\frac{dx}{dt} \right )^2+\left (\frac{dy}{dt} \right )^2}dx[/spoiler]

<span class="math">ds=\sqrt{r^2+\left (\frac{dr}{d\theta} \right )^2}dx

How do they work?

(hopefully my LaTeX comes out right)[/spoiler]

>>5745958
Ooh I like that, that works

The site I'm on shows the formula for parametric and polar form as well

<span class="math">ds=\sqrt{\left (\frac{dx}{dt} \right )^2+\left (\frac{dy}{dt} \right )^2}dx[/spoiler]

<span class="math">ds=\sqrt{r^2+\left (\frac{dr}{d\theta} \right )^2}dx[/spoiler]

How do they work?

>>5746017
Ok this was also an excuse to try and get LaTeX to work for me

How come my derivatives are so small and why are the brackets different sizes in the first one?

>>5746027

Can someone explain the difference between: arc length, line integrals, surface integrals?

I think I understand volume and area integrals.

>>5746017
First one you just multiply ds by 1=dt/dt, then take the denominator inside the square root.

Second one comes from the fact that:
<div class="math">x=rcos\theta </div>
<div class="math">y=rsin\theta </div>
So, with basic product rule:
<div class="math">dx=cos\theta dr-rsin\theta d\theta </div>
<div class="math">dy=sin\theta dr+rcos\theta d\theta </div>
When you plug that into the Pythagorean theorem, everything simplifies so that you're left with:
<div class="math">ds^2=dr^2+r^2d \theta^2</div>

>>5746027
Use [*eqn][*/eqn] tags instead of [*math][*/math]

>>5746049
>First one you just multiply ds by 1=dt/dt, then take the denominator inside the square root.

Is this what physicists actually believe? It hurts to read such garbage. That's now how a pullback works.

>>5746060
>now
*not

>>5746060
If it works it works. There's no need for rigor when there's a more pedagogical explanation that yields the right answer.

>>5746049
ok this is probably a basic question but I get how

<div class="math">ds=\sqrt{1+\left (\frac{dy}{dx} \right )^2}</div>

and everything but how does the dx fit on the end when you make it into an integral?

>>5746060
Also, mathematicians seem to have a stick up their ass when it comes to hand-wavy explanations that involve multiplying or dividing by differentials, and I've never really understood why, considering that it *does* work.

>>5746080
I think it's because differentials aren't fractions.

>>5746072
<div class="math">ds=\sqrt{dx^2+dy^2}=\sqrt{1+\left (\frac{dy}{dx} \right )^2}dx</div>

>>5746085
They behave exactly like fractions for all practical purposes.

>>5746080
Most of the times it works but it's not always well-defined and you can't really prove anything with multiplying by differentials. But it's true that it's more pedagogic.

>>5746087
Oh wait yeah that's obvious, I was just dividing through by dx^2 and thinking it would be the same value for some reason

Dammit I'm an idiot, thanks for the explanations though

>>5745981
Whoa the namefag calling me a prick? Fuck off and learn some maths, idiot

>>5746108

Don't start shit.

Ok now that I get how this thing works what is it actually used for practically?

Oh and another thing I thought of

The integral squiggle is basically like capital sigma but for a continuous set of values right?

Is there a similar continuous counterpart to the total product (big pi) and would something like that even work? If so what would it be used for?