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/sci/ - Science & Math

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7450581 No.7450581 [Reply] [Original]

Twenty years ago, this puzzle appeared on a test administered to top-tier math students from 16 countries around the world. Only 10% of test takers got it right. In the U.S., only 4% managed to provide a correct response. Can /sci/ find the “simple” solution that so many intelligent students missed?

>> No.7450589

12 cm + 4 * 4 cm = 38 cm

>> No.7450590

>>7450581
<span class="math"> l=4\sqrt{10} [/spoiler]

>> No.7450593

Is it 16sqrt2?

>> No.7450594

>>7450590
that's barely long enough to make it all the way down the cylinder without any twisting

>> No.7450597

>>7450581
its obviously 16cm

>> No.7450598

>top-tier math students
Looks more like a high-school type of question than actual math to me. "Find the right number" blabla

>> No.7450600

Unravel the string. It goes 4*4=16 cm around the rod and 12 across the rod. That means the total length is sqrt(16^2+12^2)=20 cm by pythag.

>> No.7450603

>>7450600
you don't know it goes 12 across the rod. the question only says that it goes around exactly 4 times. if the circumference is 4 cm the only logical answer you can come up with is that it's 16 cm. picture doesn't mean shit.

>> No.7450604 [DELETED] 
File: 5 KB, 188x403, gdfghfdhde.png [View same] [iqdb] [saucenao] [google]
7450604

>> No.7450605

>>7450594
Yeah I think I'm an idiot and it should be 4 times that, so <span class="math"> 16\sqrt{10} [/spoiler]

>> No.7450606

>>7450581
20 cm.

>> No.7450611

Parameterise the sting by:
(t, 4/(2 *pi) sin(t* (2 * pi * 4 / 12)), 4/(2 *pi) cos(t* (2 * pi * 4 / 12)) = (t,2/pi * sin (x*pi/1.5), 2/pi*cos(x*pi/1.5)
Then the length is the integral from 0 to 12 of the function given by the square root of the sum of the square of the derivative of each coordinate with respect to t. A simple trig identity renders the integrand constant though. Unless I fucked up the math it comes out to 12 * 5/3 = 20 cm.

>> No.7450612

>>7450603
I mean, it's implied that the string goes from top to bottom.

>> No.7450613
File: 5 KB, 185x394, nutbhtsjrtj.png [View same] [iqdb] [saucenao] [google]
7450613

>> No.7450614

>>7450603
>>7450612
Pay attention to the image. You can clearly see the thread on the right end.

>> No.7450615

>>7450611
It's clearly a geodesic so when you flatten it out it will be a straight line.

>> No.7450616

>>7450613
Oh shit that works too. No need to break out the calculus and the arc length formula. Clever.

>> No.7450619

>>7450611
>>7450606

I got 20cm as well. Drawing the cylinder as a rectangle of sides 4x12, you get 4 triangles with hypotenuse 5. 5x4 = 20

>> No.7450620

>>7450612
>>7450614

did you guys ever take a standardized test ever? Rule #1 for math problems is that you don't assume that images are drawn to scale. The question makes no mention about a particular orientation of the string except that it's symmetrical and goes around 4 times. You can't assume it runs the length of the cylinder because the question doesn't say it does.

>> No.7450622

>>7450581
It's fucking simple OP, not making your homework.

>> No.7450624

>>7450620
Scale has nothing to do with the fact that the picture shows the string the both ends of the string. Without that fact from the picture, there is obviously no unique answer...

>> No.7450625

>>7450581
16. Each loop around the tube is 4cm. There are 4 loops. 4x4 = 4^2 therefore the answer is 16. Pretty simple

>> No.7450626

>>7450615
That's true, but they don't teach the properties preserved by local isometries to high school students. At least not where I went to school.

>> No.7450628

>>7450625
Can't be exactly 16.

The string simply has to move even just a couple of millimeters from the beginning of the tube if it's wound around 4 times.

>> No.7450633

>>7450624
since the question says that the string is wrapped symmetrically around the cylinder, that means the proportion of the length of the string to the dimensions of the cylinder matters. this proportion is not given in the question and is not implied by any information within the question, therefore the picture cannot be relied upon. if the string really was end to end like you say, it would have been given in the question.

>> No.7450637

>>7450628
you're assuming the string has a width or thickness to it. if it did, there'd be no way of figuring out the answer unless it was provided in the question, so you have to assume that it has infinitesimal width/thickness

>> No.7450645

>>7450593

I think so. length L = 4sqrt(4^2 + 4^2) = 4sqrt(32) = 16sqrt(2)

>> No.7450648

>>7450637
It would not affect the calculation if the string went the whole way to the end of the tube.

So either they didn't mention that the string goes all the way or they didn't mention that it defies laws of physics because even with infinitesimal width it would still make a difference and they didn't say they don't require an exact answer.

Add in the graph and the answer is 28.

>> No.7450649

>>7450613

my thoughts exactly. that makes 16sqrt2 right?

>> No.7450650

>>7450633
I agree that the question is unanswerable if there's no picture. Obviously they include a picture to make it clear...

>> No.7450658

>>7450611
>>7450613
top kek

>> No.7450692

to those that still don't get it.
hold one end of the string against a flat surface, carefully roll the rod vertically to unravel the string. You have formed a right triangle. It should be obvious you are looking for the length of the hypotenuse as it is the string. Horizontal side is the length of the rod, And vertical is the circumference of the rod * 4. This should be obvious if you visualize the point where the rod touches the surface as you roll it down. The rest is just pythagorean theorem.

>> No.7450697

>>7450658
It's not immediate obvious to those naive of differential geometry that flattening it preserves the arc length of curves. It totally does though.

>> No.7450700

>>7450603
the picture is there for clarification, dingus

>> No.7450703
File: 12 KB, 234x55, answer.jpg [View same] [iqdb] [saucenao] [google]
7450703

>>7450581
answer is pic related (about 51.678cm)

Did it like
>>7450611
just correct

>> No.7450705

>>7450703
The circumference is 4, not 4pi. If you get rid of the pi^2 you get the right answer, 20.

>> No.7450711

>>7450581
20cm
it goes 4x4 = 16 in the 'x' axis and 12 in the Y axis
these axis are perpendicular so you use pythagorian theorem answer =squareroot(16^2+12^2) = 20.

how the fuck could you get this wrong ?.
israeli public school math .

>> No.7450713

>>7450711
The tricky part is the "these axis are perpendicular" part.
The directions of the axis are constantly changing in a way.

>> No.7450720
File: 34 KB, 527x415, rd3.png [View same] [iqdb] [saucenao] [google]
7450720

>>7450711
it's $$4\sqrt{10}$$. You have to cut the tube and will give a rectangular triangle.

>> No.7450724

>>7450720
It goes around 4 times ya dingus. Your picture has it go around once.

>> No.7450728
File: 6 KB, 388x356, ss (2015-08-08 at 12.31.20).png [View same] [iqdb] [saucenao] [google]
7450728

>>7450720
this is what it would look like if you cut the tube, each line separated by 3 cm vertically from the other

>> No.7450732

>>7450705
Dude, I fucked up. I actually took 4cm as diameter.

>> No.7450735

OKAY GUYS, STOP FOR A SECOND

Lets get this cleared out:

ASSUMING the string DID go all the way from one end of the rod to the other, and 4 times around, THEN what is the answer?

Isn't it 28?

>> No.7450738

>>7450735

>>7450611
>>7450613

>> No.7450742

>>7450738
First says 20 and second equals 28.

>> No.7450745

>>7450742
What?
4*5 = 20

>> No.7450746

>>7450742
>second equals 28.
<span class="math"> l=\sqrt{4^2+3^2}=\sqrt{25}=5[/spoiler] there are 4 rectangles.

>> No.7450761

>>7450581
>Twenty years ago, this puzzle appeared on a test administered to top-tier math students from 16 countries around the world. Only 10% of test takers got it right. In the U.S., only 4% managed to provide a correct response. Can /sci/ find the “simple” solution that so many intelligent students missed?

Unroll 4 sheets (think Riemann surface)=>diagonal of 12 by 16 sheet.
√(12^2+16^2)=4√(3^2+4^2)=4*5=20cm

>> No.7450793

>>7450745
>>7450746
Yeah nevermind, I'm stupid.

>> No.7450806

>>7450581
>Only 10% of test takers got it right. In the U.S., only 4% managed to provide a correct response
so, uh, the world is pretty fucking stupid, I guess

>> No.7450826

>>7450806
It's a little known fact that since Pi's invention average IQs have fallen by 30 points.

>> No.7450878

>>7450590
This

Holy shit, sci is stupid

>> No.7451048

I admit I'd have to use path integral.

>> No.7451150

>>7450581
Ah, i think I've got the answer. It's 20cm.

First, just examine one revolution of the string. Cut out 1/4 of the length of the cylinder. In just that quarter, the circumfrence is 4cm and the height of the cylinder is 3cm. Now, imagine that cylinder is actually a rectangle that's 3cm × 4cm, just rolled so the 3cm long ends meet. Now unroll it so it's a rectangle, and the problem becomes clear when I say "the string goes from one corner to the opposite corner on a 4 by 3 rectangle. Solve for the string." Pythagorean theorem that bitch, and you see the string, in one full revolution, is 5cm long. Now, multiply that by 4 (because we cut out 1/4 of the cylinder to begin with) and you get 20.

Im a high schooler going into Senior Year Calc, if anyone's curious, looking for a job. Don't everybody offer at once.

>> No.7451159

It's 20cm you stupid fucks.

>> No.7451161

If you seriously can't flatten the sheet to do rectangles, just do the fucking line integral.

>> No.7451173

>>7450692
this is the answer

quit all your arguing

>> No.7451174

>>7450606
>>7450619
glad some people are getting the correct answer

>>7450611
you too although you made it harder than necessary

>> No.7451179

12cm/4[winds]=3cm/[wind]
3cm length of rod with 4cm circumference can be flattened into a 4cm*3cm rectangle, with the wire spanning its diagonal.
Therefore, [length of wire] = 4[winds]*((3cm)^2+(4cm^2))^(1/2)=20cm

>> No.7451181

>>7451179
>(4cm^2)
*(4cm)^2
.

>> No.7451182

>>7450581
Isn't it just the diagonal of a 4x3 rectangle? Like imagine cutting the string along the length of the rod and unraveling the rod into a rectangle. The string would be 4 diagonals evenly spaced along the length of the rod, so the diagonal is sqrt(4^2+(12/4)^2) = 5

>> No.7451196

I got 20 cm as well as some other people in the thread by uncurling the cylinder.

>> No.7451225

>>7450581
without reading the thread for answers, I'm going to say 20cm.
>show your work
Split the rod into four sections.
String goes around each section exactly once.
Unroll each section to make a rectangle.
Rectangle is 3cm by 4cm.
String goes diagonally across rectangle.
3-4-5 right triangle, so length of string is 5cm.
Four of these rectangles so 4 * 5cm = 20cm.

>> No.7451247

4 x 4 = 16 cm

Probkem doesn't say it wraps down the whole length of the rod. Length of rod is irrelevant.

>> No.7451261

I find it annoying nobody's just gotten out a cylinder and some string to test if all these calculations are in the right area or not

>> No.7451273

>>7450590
>Show all your work.
How do we know you're right?

>> No.7451277

>>7450649
>16sqrt{2)
>not recognizing the 3-4-5 triangle
facepalm.jpg

>> No.7451282

>>7450613
Best post. Everything else is autism

>> No.7451286

>>7450692
This one is good too

>> No.7451296

>>7450581
This is false. This was not administered to jack shit. Stop lying to people and trying to convince them that a question you spent time solving is some special intelligence test. It's just a question.

>>7450590
>>7450878
That number hardly comes out to over 12 centimeters. Does that answer even make sense? Not really. And that's the first question you ask

>> No.7451297

>>7450611

You're right, but you over complicated the fuck out of it. Roll out the cylinder 4 times and use Pythagoras.

>> No.7451299

>>7450728

END OF DISCUSSION

>> No.7451303

>>7451297
no, parameterizing is the standard method. you happen to have shorthand magic, but that doesn't make it simpler, a general method is "simpler"

>> No.7451305

>>7450581
the curve is a helix. can't you just parametrize this given the constraints and use the arc length formula? seems easy

>> No.7451319

>>7450581
62.24

12 + 4(4pi)

>> No.7451322

>>7450619
>I got 20cm as well. Drawing the cylinder as a rectangle of sides 4x12, you get 4 triangles with hypotenuse 5. 5x4 = 20
but there is no assumption that the sides of the rectangles are all the same.

you can have the first round of cord very thin, then the second round very spaced from the beginning pint to the second. Here, you took what suited you

>> No.7451326

>>7451225
>Rectangle is 3cm by 4cm.
pure assumption
>>7451322

>>7451247
his guy is right too

>> No.7451327

>>7451322
It says in the image that it's wound symmetrically. Nice reading buddy.

>> No.7451329
File: 1.36 MB, 1521x1100, 1226855697464534052.png [View same] [iqdb] [saucenao] [google]
7451329

>>7451319
Googled it. Yes I'm embarrassed.

>> No.7451342

>>7450581
Was expecting some vector/parametric calculus solutions.

Instead I see these 4x circumferences or 4xPythagorean triangle pleb approximations.

Kek

>> No.7451367

>>7450613
This seemed like the obvious answer. I'm puzzled as to how so many people missed it, did something obviously wrong, or tried doing something crazy.

>> No.7451373

>>7450581
I am imagining just undoing the cylinder, like the roll at the center of paper towels.

>> No.7451377

>>7451367
Sometimes an answer is too obvious, and people who are expecting something difficult will go to great lengths to ensure that it is in fact difficult.

>> No.7451379

>>7451342
>Correct, elegant solutions are now pleb approximations

>> No.7451388

>>7451327
>wound symmetrically
means nothing

symmetrically wrt what ?

>> No.7451393

>>7451342
>I can't do basic high-school math but I know what parametric calculus is used for so therefore I'm smart

>> No.7451401 [DELETED] 

>>7451329
>>7451379
>>7451393
>The side of a cylinder is made of 4 flat rectangles
Please tell me 20 is not the real solution and you guys are trolling.

>> No.7451406

I'm not sure and I am probably wrong but, if you cut it and flattened it, wouldn't it get compressed?

Like how you cant just make a globe flat without distortion?

>> No.7451409

I feel ashamed that I didn't do this in a neat way. Maybe my college has destroyed me and I kept using integral ;_;

>> No.7451411

>>7451388
Wow you're dumb. wound symmetrically means that the lengths between the cord are equal. So it would produce 4 rectangles with the same dimensions. Do you understand what the word symmetry means?

>> No.7451419

>>7451406
>Like how you cant just make a globe flat without distortion?

It's a bit different, see Theorema Egregium. The Gaussian curvature of a surface is an invariant, since the Gaussian curvature of a cylinder is 0, it means that its isometric to a flat plane, which means it's perfectly fine to "un-roll" the cylinder in this problem. The curvature of a sphere is <span class="math"> 1/r^{2} [/spoiler] which means you can't form a flat plane out of a sphere with out some distortion.

That said I'm no expert in differential geometry.

>> No.7451420

>>7451411
symmetric is always with respect to a referrent. I understand that you are butthurt, but your definition of symmetric here is a just projecting.

>> No.7451424

Yeah, but if you did this problem in real life, you would cut a line through the middle of the cylinder and the string. If my logic is correct the string would go past the edges of the cylinder.

>> No.7451425

>>7451419

Yeah, but if you did this problem in real life, you would cut a line through the middle of the cylinder and the string. If my logic is correct the string would go past the edges of the cylinder.
(accidentally posted this twice. newfag here)

>> No.7451429

>>7451425
You know, you can delete the post.

>> No.7451432

>>7451419
Anyways I just feel the answer to this problem is too easy and there are mistakes in the way the question is asked. I'm not a teacher, I havent gone to college yet, I dont know what I am talking about, but logic is logic right?

>> No.7451433

>>7451429
teach me

>> No.7451444

>still no posts with a picture of an actual rod and a string
this is why you fucking retards are too dumb to be engineers

>> No.7451460
File: 150 KB, 700x700, rope-wrapped-curtain-rod-102139694.jpg [View same] [iqdb] [saucenao] [google]
7451460

nothing says that the string goes to the top of the rod


one day, perhaps, people will understand that this little problems are never stated rigorously

>> No.7451465

>>7451460
> rigorously
this is why you have no friends.

>> No.7451466

>>7451465
that's true

>> No.7451468

>>7451393

> I get offended by an anonymous self-proclaimed calculus expert.

>> No.7451478

Unravel the triangle. Height is length of the cylinder. Length is four times the circumference. Then use the pathagorean theorm.

>> No.7451485
File: 6 KB, 120x107, false-truthometer.gif [View same] [iqdb] [saucenao] [google]
7451485

>>7450581
>this puzzle appeared on a test administered to top-tier math students
Truth-O-Meter says "false"

>> No.7451492

>>7451420
I'm sorry you were too autistic to understand a simple math problem. My bad.

>> No.7451510

>>7450581
Didn't look in thread so here's my guess:

Parametrize the string as a spiral as r(t) = <cos(t), sin(t), 3t / (2pi)>, then do a path integral from t = 0..4*2pi = 0..8pi

>> No.7451517

>>7451510
I just read the geometric solutions, yeah that would be a bit easier. Ah well.

>> No.7451522

Here is a similar question:
You have a shoebox with a dot on one side panel and another dot on a different side panel, find the shortest line between the two dots.

>> No.7451532
File: 101 KB, 1437x869, path integral solution.png [View same] [iqdb] [saucenao] [google]
7451532

>>7451510
Ok, I just did it in Maple because I'm lazy, corrected the cos(t) and sin(t) for the correct radius and yep, the answer is 20.

>> No.7451551

>>7451522
Fold it in the 4th dimension :^)

>> No.7451569

Each lap corresponds to the hypotenuse of a triangle of height 12cm/4 (=3cm) and base 4cm, so sqrt(3^2 + 4^2) cm = 5cm. There are 4 laps, so the answer is 20cm.

Note that the proportions in the picture are highly misleading. If the rod is 12cm long and the circumference is 4cm, its diameter is only around 1.3cm, so it should be about 10 times as long as it is thick.

>> No.7451585

>>7451532
how do you find the parametrization in detail ?

>> No.7451598

>>7451585
Do you mean in x = r cos(t), what is r? Well, the problem says that the circumference is 4, so then r = d / 2 = c / pi / 2 = 4 / (2 pi) = 2 / pi. This is pretty basic though, might wanna brush up on this with Paul's Notes.

>> No.7451620
File: 29 KB, 1198x778, New Bitmap Image.png [View same] [iqdb] [saucenao] [google]
7451620

>>7450581
this should answer everyone's questions

>> No.7451623

do they teach arc length/line integrals in high school in the US?

>> No.7451661

>>7450581
OP is obviously baiting. This is just a simple computational exercise. No "Top-tier math student" would have any difficulty with this shit. Just parameterise for theta, and z. Maybe throw in a triple integral somewhere. done

>> No.7451664

>>7450611
THANK YOU.

>> No.7451740

>>7450581
<span class="math">\sqrt{(4*4)+12^2} = 20 \ \mathrm{cm}.[/spoiler]

>> No.7451794

>>7451623
nope

>> No.7451800

>>7451794
then the solution expected for the problem is the pythag one, not the line integral one
I mean, ok maybe a smart student will figure the integral one, but he'll still feel stupid for not seeing the obvious other one

>> No.7451901

>>7450611
Lol you idiot.
What a plodder.

>> No.7451920

>>7451598
>Do you mean in x = r cos(t), what is r? Well, the problem says that the circumference is 4, so then r = d / 2 = c / pi / 2 = 4 / (2 pi) = 2 / pi. This is pretty basic though, might wanna brush up on this with Paul's Notes.
but for e_z with the factor 3/2 times t/pi ?

>> No.7451929

>>7450581
What is the diameter of the string?

>> No.7451972 [DELETED] 

>>7450581
4 x 12

4 x x x
3 x x x
2 x x x
1 x x x
0 1 2 3 4 5 6 7 8 9 0 1 2

Pythagoras:
9+16=25

5 x 4=20 cm

>> No.7451985 [DELETED] 

>>7450581
4 x 12

4 x o o o x o o o x o o o
3 o o o x o o o x o o o x
2 o x o o o x o o o x o o
1 x o o o x o o o x o o o
0 1 2 3 4 5 6 7 8 9 0 1 2

Pythagoras:
9+16=25

5 x 4=20 cm

>> No.7451986

imagine a fly walking on the string.
It turns in the xy plan and walks on the z line.
Then:
x=4/(2pi) cos alpha (1)
y=4/(2Pi) sin alpha (2)
z= 8/(2pi) alpha (3)
(1) and (2) are easy.
if (3) is not ask me.
where alpha is the angle between the fly and the x line.
So we have to integrate (dxdx+dydy+dzdz)**0,5 with alpha between 0 and 8Pi(because there are four circles) and it gives 20.

>> No.7451991

>>7450581
4 x 12

4 o o o x o o o x o o o x
3 o o x o o o x o o o x o
2 o x o o o x o o o x o o
1 x o o o x o o o x o o o
0 1 2 3 4 5 6 7 8 9 0 1 2

Pythagoras:
9+16=25

5 x 4=20 cm

>> No.7451997

>>7451991
>>7451986
this has been debunked

>>7451247

>> No.7452079

>>7451997
no it hasn't, its not symmetrical in that case

>> No.7452091

>>7451991
Agree. Brute force:

r=4/(2 pi)=2/pi.
parameterized curve h(t), where 0<=t<=1
h(t)=(r cos(8 pi t), r sin(8 pi t),12 t)
h' = (-8 pi r sin(8 pi t), 8 pi r cos(8 pi t), 12)
= (-16 sin(8 pi t), 16 cos(8 pi t), 12)
|h'|^2 = (-16 sin(8 pi t))^2 + (16 cos(8 pi t))^2 + (12)^2 = 400
|h'|=20
<span class="math">\int_0^1 20\, dt = 20 [/spoiler] cm

>> No.7452108

>>7452091

Generally,

Cylinder length L, wrapped n times, radius r.
Parameterized curve h(t), where 0<=t<=1
h(t)=(r cos(2 pi n t), r sin(2 pi n t),L t)
h' = (-2 pi n r sin(2 pi n t),2 pi n r cos(2 pi n t), L)
|h'|=((2 pi n r)^2 + L^2)^(1/2)
\int_0^1 ((2 pi n r)^2 + L^2)^(1/2) dt
= ((2 pi n r)^2 + L^2)^(1/2) cm.
= ((n C)^2 + L^2)^(1/2) cm.

It's the Pythagorean theorem applied to triangle with sides L and n C, with C being the circumference.

This makes sense since the cylinder can be cut into a long thin strip following along the string and unrolled into the plane. The cylinder has Gaussian curvature zero, i.e. it's developable.

>> No.7452140

I am not good in english. Does debunked mean stupid ?

>> No.7452167

>>7452140
In general, it means "proven to be false."

>> No.7452186

>>7452140
It means "proven wrong".
I don't think that guy was "proven wrong", but the picture is included to show that his idea is not the intended problem. It would be like saying that Grothendieck solved the Hodge conjecture because he pointed out a trivial error in the way it is formulated. It is the intention that matters, not getting all autistic about the specific details of somebody's use of language or notation, especially in a problem like this.

>> No.7452196

>>7451247
>>7451997
It is implied that it wraps around, see the right end of the rod, there's a small a part of the string coming out from the back.

>> No.7452202

>>7450581
4 x 4 = 16

16 + 12/4 = 19cm

>> No.7452234
File: 51 KB, 600x490, FreeSlaps_For_WhiteMen.jpg [View same] [iqdb] [saucenao] [google]
7452234

translate in mathematics the first sentence : the one about the symmetry. you will see that you cannot make sense of it.

>> No.7452682

>>7450649
no fool, it makes 20

>> No.7452696

>>7451929
>engineer detected

>> No.7452700

>>7450600
Yeah. Although I thought of it as unraveling the cylinder, putting 4 of these unwrapped cylinders next to each other that essentially one of the cords would be the hypotenuse.

>> No.7452705

>>7450581
Just compute the line integral for the helical function that describes the string.

>> No.7452755

>>7452234

How sad does your existence have to be that this is how you spend your time?

>> No.7452787

l=sqrt(12^2+4*4) cm=12,65 cm

make a right triangle, base is 4 times the circumfere and height is same as rod's height

>> No.7452814

ok guys i built it and measured it
haha

what a fucking waste of time that was

>> No.7452833

>>7452814
lol
what answer did you get?

>> No.7452859

>>7450581
Are top-tier maths people really expected to be good at solving this type of problems though?
The simple solution seem to require one to have something more akin to artistic/mechanistic spatial intelligence rather than any mathematical refinement.

>> No.7452864

>>7451150
>>7451159
>>7451196
>>7451225
I agree

>> No.7452868

>>7451532
Thanks.

The way I got this was by examining each loop of the string as the hypotenuse of a base 3,4, triangle which is clear from the given dimensions. Then 4*5.

>> No.7453035

>>7451920
Well, the distance between two loops is 3, because the length of the cylinder is 12 and there are 4 sections. Now, t is going to put sin(t) and cos(t) though one cycle every 2pi, to if we want to move the spiral by 3 in a time of 2pi, we just divide 3 up in 2pi and multiply by t. That way, z increases linearly by 3 every cycle. Then we just let the entire thing run through 4 cycles (t=0..8pi) to get OPs pic and path integrate.

>> No.7453052

>>7450581
Easy as shit. Do people not know basic trig? It's basically 4 triangles with base 4cm, height 3cm (12/4). So 4 triangles with a 5cm hypotenuse. Ans: 20cm.

>> No.7453173

>>7450581
Unwind it. It's just a right triangle wrapped around itself. The base is 4*4 cm, height 12 cm, by the Pythagorean theorem, its hypotenuse is 20.

>> No.7453224

>>7453173
>Unwind it.
Just smoke weed if you wanna unwind, man.

http://420.moe blaze it

>> No.7453257

>>7450581
Not looking at the posts above me
I think the answer is
<span class="math">\sqrt(n^2 c^2 + l^2)[/spoiler]
where c is the circumference of the cylinder, l is its length, and n is the number of times the string winds around it
>inb4 it's easy

>> No.7453261
File: 138 KB, 800x964, likeb.jpg [View same] [iqdb] [saucenao] [google]
7453261

I'd like to congratulate on the special mathematically talented snowflakes in this thread who easily, elegantly figured the solution within a few minutes. brilliance my friends, can not be taken lightly. a toast to you all, cheers

>> No.7453280

Are you all retarded? Just imagine a 12 x 16 cm rectangle to represent the surface the string travels across and find the length of the diagonal

>> No.7453314

>>7453257
I don't think it's trivial like they were saying

>> No.7453334
File: 20 KB, 306x306, 79e0f76f22f429dc6a79ebbe8f3b484d.jpg [View same] [iqdb] [saucenao] [google]
7453334

>A HURR DURRR
>let's repeat the same fucking answer and act all smart
>am I SMRT guys?
Sage and hide

>> No.7453366

>>7453280

This was my solution too. The fact that a cylinder is involved is irrelevant; if you imagine the string traveling diagonally across a flat surface 12 cm tall and 16 cm long, all you need is the Pythagorean theorem.

>> No.7453494
File: 102 KB, 500x500, 10418388_818996094819044_8682617934594336344_n.jpg [View same] [iqdb] [saucenao] [google]
7453494

Hello There! Anyone here know something about Multipole expansion in general relativity? Maybe books, articles internet pages explaining the subject! (Picture unrelated)

>> No.7453500

>>7450611
Spotted the engineer

>> No.7453502

>>7453494
Make a separate thread for this, if you're serious

>> No.7453551

>>7453035
yes indeed, thank you

>> No.7453880

>>7450590
Literally this, and it's incredibly obvious.

>>7451296
Yes, it makes sense.

>> No.7453940

>>7453880

You're autistic

>>7453334

Nice answer you've got there, stay mad faggot

>>7450611

You're also autistic but the smart kind

>>7453261

If you could an hero that'd be great

>>7453173

Good job, you understood that the cylinder is of no consequence unlike half the fucking idiots here

>>7452202

How the fuck did you manage that answer

>> No.7453977
File: 128 KB, 728x426, 1436031849434.gif [View same] [iqdb] [saucenao] [google]
7453977

>>7453940
Wow look at this sperg

>> No.7454108

>>7450581
I got exactly 13cm

>> No.7454112

>>7454108
Nevermind I was retarded, I did this
>>7450613
Except I used 4/Pi instead of 4 because I haven't had to use my brain for 3 months

>> No.7456017

>>7450581
2 + 3 4 2 = 16 + 9 = 25, being the square root of 25 = 5. Now only 5 should be multiplied by 4 turns of rope to obtain a total rope length of 20 cm

>> No.7456247

7450581
38.2?
i just took an eigth of the whole black line and saw it as a stretched half circle. then i found the radius of it with pythagoras and a coefficient k by dividing the new radius through the original radius of the cylinder. then i just calculated a half circle and muliplied it with my coefficient...

>> No.7456305
File: 17 KB, 481x261, Untitled.jpg [View same] [iqdb] [saucenao] [google]
7456305

>top-tier math students
>10% of test takers got it right
>intelligent students

51.678cm

>> No.7456390

>>7456305
>acts like a smug cunt
>gets the question wrong

good work

>> No.7456421

>>7450581
28. am i retarded?

>> No.7456470

>>7456421
I have some bad news for you...

>> No.7456483

why don't people read the thread before answering

>> No.7456498

>>7456483
idk, but so far im the only one that's given the correct answer.

>> No.7456573

I drew it on a paper, then measured 4*5=20. Not the smartest way, but I guess still better than looking it up on the internet.

>> No.7456907

I think most people misread the question. It clearly states the circumference is 4, not the diameter. No π involved.

>> No.7457942

the answer is 20, faggots

you can "take apart" the cylinder and form a rectangle

it is wound symetrically 4 times around the cylinder, so you can divide it into 4 parts, each part being 3 cm long

so, you have the length of 2 sides of a right triangle, so the pythagorean theorem can be used.

the full length of one coil to the next is 5, and it is would around it 4 times, so the answer is 20 cm.

come on, faggots, a 9th grade geometry student could answer

>> No.7458651 [DELETED] 

>>7450581
24cm? can it be done by cutting up the rear ends of the cylinder and smooshing em up

>> No.7458843

>>7457942
ITT: People pretending it's a shit problem because they fucked it up.

>> No.7459960

>>7450711
The dimensions are even chosen so that the calculation can be done in the head (256 + 16 squared is 400, square root of 4 is 2, square root of 100 is 10, solution is 20).

>> No.7460047

>>7450589
This makes no sense no matter how you look at it.

>> No.7460067

>>7460047
Well it has to be 12 cm long to go the length of the rod, and has to be 4 circumferences to do the spin arounds

>> No.7460357

>>7451297
it's not complicated at all

>> No.7460393

I got twenty. If you unroll the string, it is the hypotenuse of a right triangle where the length is 12 and the height is 16 since you are unrolling the circumference (4) 4 times.
Then just use the Pythagorean theorem.

I'm really curious how people are ending up with an answer that isn't 20.