[ 3 / biz / cgl / ck / diy / fa / ic / jp / lit / sci / vr / vt ] [ index / top / reports ] [ become a patron ] [ status ]
2023-11: Warosu is now out of extended maintenance.

/sci/ - Science & Math

View post   
File: 5 KB, 360x250, image001.gif [View same] [iqdb] [saucenao] [google]
2067276 No.2067276 [Reply] [Original]

Hey /sci/. I'm talking Integral Calculus and we're working on Arc Length problems. You know, those weird graphs that you have to try and approximate. I can plug in the numbers into the formula and usually work out the right answer, but I don't understand what it *means*.

I get that we're taking smaller and smaller measurements to get a better estimate, like we've done to death with Riemann Sums, but I don't get *why*. Why is arc length of a graph, the integral of (g'(x))^2+1)^(1/2) ?? (I get that formula right?)

I've asked my teacher but he just says it's all in the book, and the thing is it *isn't*. Not even the little videos included with the book help.

so... halp? why is this the way it is?

>> No.2067302

>>2067276
The questions you ask require actual math, not the faggy engineering shit you are taking.

If you are a mathatican or physicist you will learn how to derive the formula of arc lenght through cal of variations and shit. So just wait until you take a real math course.

If you are a faggy engineer, you arent smart enough to learn real math.

\thread

>> No.2067301
File: 2 KB, 118x48, arc length formula.jpg [View same] [iqdb] [saucenao] [google]
2067301

>>2067276
>(g'(x))^2+1)^(1/2)

meant this (pic)

>> No.2067304

You're calculating the distance on the graph between two infinity close points and summing them to infinity on a piece of the graph.

The derivation is in your book. And you don't understand math. Try harder.

>> No.2067318

>>2067276
>>2067301
It just really pisses me off, because I was really good at differential calculus and now, shit, I'm thinking of dropping the class and trying later. My grades are pretty good right now, but if I don't understand this, what's the point of continuing on where they'll just build on this stuff. It feels like I'm faking my way through.

>> No.2067330

>>2067302
>engineer
>mathatican
>physicist

None of the above genius. It's just a college class I have to pass to continue on in my Computer Science.

>>2067304

>> No.2067337

>>2067318

The arc length formula itself isn't important. It's taught to hammer home the concept of the integral as an infinite sum that can be used for things other than areas and volumes.

>> No.2067342

>>2067302

hahaha fucking neckbeard aspie

>> No.2067349

you're basically using the distance formula, a^2 + b^2 = c^2 except you're using differential distances that are small enough that they create little right triangles. then you add them all up and it turns into the length of the entire line.

>> No.2067350

>>2067304
>You're calculating the distance on the graph between two infinity close points and summing them to infinity on a piece of the graph.

...I don't see how that fits with some of the example problems. Like a rocket booster falling from a main rocket, and using this stuff to track it's progress. Or something.

>And you don't understand math. Try harder.
I'm trying. fuck man.

>> No.2067362

>>2067349
Okay, I think I'm following you.

>> No.2067370
File: 102 KB, 850x1092, 389961_Dragon_ball_Z_Majin_Buu_MaujenFijeta_Rule_63_Son_Goku.jpg [View same] [iqdb] [saucenao] [google]
2067370

>>2067302
FULL OF WIN!

>> No.2067382

>>2067302
>>2067370
Sure is samefag in here.

>> No.2067384

>>2067350

You're probably given a function that relates y position to x.

Just plug the function into the formula and integrate choosing the limits of integration as necessary. Most textbook problems will simplify algebraically within the radical because it's often impossible to integrate a function within a square root by hand.

>> No.2067385

Ugh.... In calculus II the arc length and volume/areas section is exclusive, not knowing it won't hurt you for techniques of integration, although techniques of integration is a more challenging topic. Anyways, you don't need to know why the formula works in order to get through calc II.

How do people drop classes anyways? Doesn't it hurt your GPA too much to be worth it?

>> No.2067391

>>2067318
It's not hard, OP. Take an arbitrarily small segment. How long is the segment if you know dx and dy? It's just pythagorean theorem, ds^2 = dx^2 + dy^2. That's easy right?

So, what if you want to know how much arc length there is per change in x? That's ds/dx. Solve the above equation for it
ds^2/dx^2 = 1 + dy^2/dx^2
ds/dx = sqrt(1 + dy^2/dx^2)
ds/dx = sqrt(1 + (dy/dx)^2)
dy/dx is aka f'.

http://en.wikipedia.org/wiki/Arc_length

>> No.2067404

>>2067362

IT'S IN THE BOOK
LOOK AT EACH STEP CAREFULLY
AND WRITE IT DOWN AS THE TEXTBOOK DERIVES IT
IF YOU STILL DON'T UNDERSTAND YOU WILL HAVE SPECIFIC QUESTIONS THAT YOU CAN ASK YOUR PROF
HE WILL BE LIKELY TO ANSWER SPECIF QUESTIONS
THIS IS HOW MATH IS LEARNED

>> No.2067412
File: 28 KB, 350x401, 1282077028592.jpg [View same] [iqdb] [saucenao] [google]
2067412

>>2067302

the shit we have to deal with in here, ugh

>> No.2067425

>>2067385
>you don't need to know why the formula works in order to get through calc II.

I think I will. The teacher is known for giving essays on his Finals, and with the trouble people in my class have been having, it wouldn't surprise me one bit to see a question like this on a test.

>How do people drop classes anyways? Doesn't it hurt your GPA too much to be worth it?

There's 2 dates for dropping a class. Dropping before the first date means getting your money back. Dropping before the second date means you won't get a grade for the class, in effect saving your GPA from taking a hit. Though you're out that money.

Some colleges totally remove the fact you ever took the class from your transcript, and others do something like adding a "W" for Withdraw.

>> No.2067423
File: 54 KB, 521x312, 1280686396145.jpg [View same] [iqdb] [saucenao] [google]
2067423

>>2067302

>> No.2067417

>>2067301
as you can see the arc length formula closely resembles the distance formula. namely, square both sides and you have
(arc length)^2 = 1 + f'(x)^2

1 = 1^2 so we have:

(arc length)^2 = 1^2 + f'(x)^2

compare to c^2 = a^2 + b^2

create a right triangle where the hypotenuse is the arc length and the legs are 1 and f'(x). this should be able to model any graph. since f'(x) is an infinitely small amount, the curve of the graph can be represented as the straight line hypotenuse of the triangle. or something like that. honestly i'm not too sure where the 1 comes from, but it's probably a reference unit.

>> No.2067426
File: 35 KB, 389x479, harvey_milk_childrens_book.jpg [View same] [iqdb] [saucenao] [google]
2067426

>> No.2067433
File: 8 KB, 251x162, 1273739246235s.jpg [View same] [iqdb] [saucenao] [google]
2067433

Engineering? Why?

>> No.2067434

this >>2067391
is what i was trying to say with this >>2067417
i forgot that dx is 1 by definition, then dy is f'(x), so those are the legs of your triangle.

>> No.2067436
File: 48 KB, 462x700, gay-sex+-82-18.jpg [View same] [iqdb] [saucenao] [google]
2067436

>> No.2067467

>>2067349
>>2067391
>>2067417
>>2067434
So, we essentially use it, because it's more or less an interpretation of the distance formula, and we combine it (sort of) with the Pythagorean Theorem, to get a nice, clean formula?

I mean, basically?

>> No.2067492

>>2067467

The distance formula is a result of the Pythagorean thm.

But yes. Then algebra is used to simplify the result into a simple formula. Again look at the derivation in your book.

>> No.2067495

>>2067467
distance formula = pythagorean theorem

the arc length formula is basically just the distance formula plus calculus, it's used to find distances of curved things.

>> No.2067513

>>2067492
>>2067495
Ooh, okay. I'll work on that a bit. I swear the book doesn't lay it out like that. Or maybe I'm just not seeing it.

But thanks.

>> No.2067531

>>2067467
Just purely in terms of the distance formula
d = sqrt(x^2 + y^2)
you can rearrange it to say
d/x = sqrt(1 + (y/x)^2)
which becomes useful for calculus when d, x, and y are all taken to infinitesimals, because then it is
d' = sqrt(1 + y'^2)

>> No.2067533

>>2067425

No, unless your doing honours calculus II, you won't have to worry about proofs on exam. I think you might be getting "knowing when it works" and "knowing how and when to use it" mixed up.

You will probably need to do some manipulation to get it into the form in which the formula can be applied, as in other things in calculus, before you can just apply the formula.

Knowing what, when and how is enough to get you through calc 1 and 2.

>> No.2067534
File: 38 KB, 150x212, xandir_drawn_together_1.jpg [View same] [iqdb] [saucenao] [google]
2067534

>>2067436
You sure do have a lot of gay porn. I mean a LOT of gay porn. A LOOOOOOTTTTTTT of gay porn.

Compensate much?

>> No.2067555

I'm totally saving this thread btw.

>> No.2067570

I would say deriving e would be the most proofy thing you will be asked to do on an exam in calc 2.

However, improper intgrals and more specifically, the comparison test are what I thought to be most "outside the box"....

Concept is pretty easy actually, but given and integral and told to do the comparison test on it can suck major balls. (every different question requires a different method to solve, just the premise is the same)

>> No.2067959

I have no idea what's going on in this thread.

>> No.2067982

>>2067302
cov isnt interested in learning why the arc length formula is what it is, it is rather concerned with finding the extremals of functionals, and then showing through a combination of necessary conditions the sufficient condition that F_((y^')(y^')) is strongly positive to show that you have a minimum along the extremal.

>> No.2068010

>>2067982
F_y'y' *