4:30 i paused the video and just sat there, wondering why it was never taught like this. That moment you wrote out 1+ m^2, the reason behind the formula "clicked" instantly, far better than the "memorize this it's on the exam" approach my classes took. Excellent job!

I remember getting a similar problem in class long ago about a rope around the Earth. The question was "If you start with a rope wrapped around the equator of the Earth, how much extra rope would you need to raise it 1 m above the surface?" Finding out it was only 6.28 m was pretty mind blowing at first (Just found out this question is such a classic it has its own Wiki page)

Here's a real world example for you- I work in telecommunications, and our data will often run on a fiber optic cable which is strung between poles above ground. If a garbage truck (it's always the garbage trucks!) takes down a fiber line, we can measure the length to the fiber break by sending a light pulse which reflects off of the broken end of the fiber and returns to the source. This time delay (with the speed of light in glass) gives us the length of fiber- this is called Optical Time Domain Reflectometry (OTDR) if you want to read more about it. The time measurements have a precision on the order of tens of nanoseconds, giving break locations with a precision on the order of several meters. Unfortunately, that measured distance is more like the arc length, but what we really need is the ground position where we need to send the technician with the splice equipment. Since the fiber cables hang in a catenary-like curve, we have to solve this problem every time a break takes place above ground... or get lucky, and hear from the police or a traffic report the intersection where someone has crashed and downed a pole.

For Q 1 Simple pythagorean formula 5.02 meters. For Q 2 I think it involves catenary calculations

Ships have a piece of small rope called a "tattle tale" secured to 2 points on a mooring line. When there is no tension on the mooring line, the tattle tale has a lot of slack in it. As the mooring line stretches, the 2 points move apart and the tattle tale slack is removed. This way, the ship's crew can tell when there is too much strain on the mooring line. The length of the tattle tale and how far apart to secure the ends depends on the specifcations of the mooring line.

Math was just numbers to me, but after coming across your channel. I see the beauty in it. The art. Every video you've posted has been blowing my mind.

People like you that knows how to explain maths well, they deserve a place in Heaven. It is always a strugle for me to understand a book math, but then I see videos like yours and it makes everthing so easy.

The large amount of horizontal motion needed to take up a tiny bit of slack on the rope also relates to the mechanical advantage that pulling the rope horizontally has on the tension of the rope, how taut rope needs to be to not deflect by that much.

professors never seem to explain why any of this is possible. This video makes it incredibly straight forward to understand why

for Q1 I actually guessed 5m. I didn't expect to be so close, but I knew it had to be much bigger than the difference in length. I'd always been surprised by the 5-12-13 Pythagorean triangle which shows that a tiny difference in length between two lines creates a big distance between them at an angle, so I remembered it well.

The athletics example about slack reminded me of the high jump - with a relatively stiff bar, there is still over an inch in height difference at the stanchions and in the middle of the high jump.

I dont understand why we assumed the rope makes a parabola. I always thought a uniform density line used catenaey shapesninvolving e^x and e^-x terms

It should be actually quite intuitive for anyone who has been playing with stretched string or rope. It's so easy to make large sag even if string is really tightly stretched. If you think about it for a minute, you'd realize that according to Hooke law, the elongation of the string is small, as well as the applied force.

Thank you Professor Math The World, this will do me good justice.

I just used Heron’s formula after summing 50, plus 2 times 25, plus 1 and used the base of 50m for the lower part of the triangle and the area I got of approximately 125.6m for a 5.02m drop of your friend when you add 1 extra meter of rope.

Such a well done video. Thank you !

I don’t know who you are but you are surely one of the best teachers for the modern generation.

"The simplist way is to ask your friend" "Hey!! How much did you fall off!!" "Four point threAAAAAAAAAAAAAAAAAAAAAAAAAAAHHHHHHHHHH!!!!!!" "Approximately 3 seconds"

O the slope formula explanation to derive arc length formula makes so much sense

This is a fantastic lesson! (I can adapt this into a class period in about 3 minutes…TEACH LIKE A PIRATE, ARGHH!)

As an amateur in sewing and pattern design, this has caused issues for me. A 1mm discrepancy in a 30cm seam can result in a 1cm arc! That's why I think half the battle is cutting fabric pieces as accurately as possible.

really amazing, explanation, animation and story thank you very much! subbed :)

I really liked this video, the way you explained how we get to the arc length formula was so clear.. it really blew my mind!

I work on sea going tugboats. We tow barges with a cable. Knowing the sag in the cable is important so it doesn’t touch the bottom. We call it shining the wire. The company policy is add 10 feet of depth for ever layer off the drum. The difficulties arise because every layer is a different diameter, every drum is a different size,and the hight of the barge is always different. The cables themselves are usually 1500 feet long ,but you never use the last layer on the drum because that is how it holds on. The diameter of the cable is mostly 1 31/32”. The math is daunting. We rely on experience.

So interesting. Thanks!

In structural engineering applications, reminds us of calculating the sag of a hanging power cable under its own weight. The shape the cable takes is known as a catenary. According to the problem posed here, the hyperbolic cosh function closely approximates this sag calculation. I didn’t realize this during my undergraduate years studying civil engineering. It was a difficult problem to solve with few computer programs those days.

The exact drop depends on the properties of the rope, and the weight of both the rope and the man. To simplify the problem I assumed the maximum drop possible, where the rope sections are kept straight by the weight of the man. (Rope weight is assumed to be insignificant) Assuming a right angle triangle is created , the triangle sides are hypotenuse, 25.5m, the new length, the original length 25m, and the drop in m. This gives a maximum drop of 5m.

How KZfaq got me here today I don't know. This really blew my mind. My intuition was wrong. After watching, what I was imagining was how any slack, would instantly create a arc from a huge circle. as x (delta of the chord)->0, y I wasn't doing the integration, but just some geometry. Looking at this from the arc and coord ratio, 4.3431 was my answer, between your two examples. 78.5 would be the max amount of rope could go before it no longer fits that limit of a circle, and then it would droop down.

I like to explain this to my students as "there's not a lot of length difference between a tight rope and a saggy one, and that's why it's pretty much impossible to load at long rope and keep it straight". Think high voltage lines, or laundry...

For those that play with ropes and overhead conductors, a huge change in tension can be caused by a small change in the arc length!

Interesting and my thanks. I ran into this problem when we were discussing the length of cable or wire between telephone poles in a High School Physics class or two. Turned out that there had to be some drop because of gravity. Which related to the amount of tension applied to the wire. A big deal if you think about it, and relates the the reliability of the then telephones and electricity coming out of your light sockets.

If instead of the parabolic approximation we use a circlar segment and trunkated Taylor series sin(x)=x-x^3/6, cos(x)=1-x^2/2, we get h=sqrt(153/8)=4.3732 approximately. It is almost the same as the parabolic approximation, however we can easily compute all the numbers without Wolfam Alpha.

On a related note I was on site and had to record the dimensions of an existing arched bridge. From afar it looked like a segment of a circle. I got a mate to I ball a spirit level end to end and measured the bow in the middle to where a string line would have been. Then measured along the curve thinking I had enough information to define the circle and the rise at one metre intervals. I spent hours trying to figure out the trig to get the actual resulting complete circle. I ended up doing some sort of iterative calc in excel to get the answers. Im

Considering a circunference I got the closed approx ==> DROP = sqrt(6·SLIP·LENGTH)/4 for any SLIP

I posted a comment about the similarity of this problem with the one where a (very, very long) rope is girdling the earth around the equator. Add one meter of slack, and then pinch and raise the rope so it's taut again. How high is it gonna go? A fellow commenter asked for the answer, so here goes: There's gonna be a baseline between the points where the rope goes tangent to the surface and starts to gain altitude. But we cannot use the Pythagorean theorem because the excess length (slack) is relative to the slightly curved equator, not the straight chord through the earth. We still use sine and cosine (actually, tangent and cosine) in a right triangle with corners at the Earth's center, apex of the raised rope, and tangent point. As you can guess, or you can see if you try it, one has to make use of approximations. But they are excellent and yield fairly accurate results. Speaking of, the simple version (triangle case) of the chasm problem can be tackled in an approximate fashion, even though the exact answer is ridiculously simple. The approximate answer for the sag, s, in terms of x=0.5m (half the excess length, or slack) and L=25m (half the chasm gap, or baseline) is: sag = s = sqrt(2xL) I mean, sqrt(2*x*L) = sqrt(25) = 5m The exact value is s=5.0249... Now, to better appreciate the effectiveness of the approximation, and to bridge the gap (so to say) between the original problem here and the other one with the Earth's equator, let's consider a huge chasm (Grand canyon type of chasm) of 5km, but the same amount of slack, 2x=1m The approximate formula yields: sag = s = 50 meters, while the exact value is s = 50.0025 So, two things: (i) The approximation gets better when L/x gets larger, and (ii) The sag is much greater for the same amount of slack! Now, back to the equator. The answer for the height (instead of sag) is h = 121.5 meters, or so, which is about 133 yards, or almost 400 feet. And, by the way, the baseline between the tangent points is 78.7 km, or 48.9 miles

My intuitive guess is that the first problem is so simple to calculate, we should calculate. The answer to the second problem is, "less", unless I want to do the not so simple calculation for a catenary curve.

Assuming the weight is in the middle of the rope, and it is perfectly inelastic, the absolute maximum drop should be the height of a right triangle winh hypotenuse af 25.5m and length of 25m, which is just about 5 meters. The slack variation is much harder to calculate intuitively, but given that it's a hyperbola, without appealing to major math tools (as you asked us not to) my intuition says it would be between 2 and 4 meters drop (about half the weighted distance at least)

Another place you see the underestimating of Arc Length is in sailing yacht sizes. They are classified by length. For example a 42ft yacht being much, much bigger internally than a 40ft yacht. That extra 2ft resulting in enough increased volume for additional cabins!

3:43 Just a “In case it helps comment” This slope section has different scales for the X and Y axis, this may confuse some how the values are found. Thanks for the videos, nice job!

Another example where this is *extremely* important: Slack line walking, and canyon jumps as an extension

This reminds me of the "string girdling earth" problem where an additional 6.3m of extra string allows a meter of additional height all around the earth. Very counterintuitive

If this is the method of solution you expect, make sure to specify to your students that you're seeking an _approximation_, not an exact solution. Most students delving into advanced calculus will assume they're expected to give exact solutions. To them, these questions feel more like 'invent a method for deriving the distance between the apex of an arc of length x with unknown shape, and the midpoint of the arc's terminations.' To these students; the problem is the unknown shape of the arc. This is why it's important to specify that you expect an approximation, rather than a solution.

My idea was to divide the graph into two equal triangles with the following properties ( adjacent side = x, opposite = 25 m, hypotenuse = y). In addition, I thought that the sum of both hypotenuses must be 51 m and since both are equal, I derived a formula from this. 51 m = 2root(25 m² + x²) This results in two answers: Answer 1: x = Root(101)/2 Answer 2: x = -(root(101)/2) I hope my thoughts where correct

It depends on the stretchyness of the rope. If it has no real give to it, they would drop about 5 meters (using the Pythagorean Theorem), so I would guess that it would be less if there was no weight pushing down on the rope. Then it depends on the weight of the rope and whatever the formula for a catenary is 😀.

Without any calculations i say its less than 1m cuz the 1m is spread across the gap other than that i could try to collapse the right side on the left side to get the same arch to stright diffrence across the rope except the meddle and use that to get the spread of the middle compared to the rest(aka a percentage)×1=the middle gap acrch to stright deference (i have no mathmaticla proof but it fell like the right way to do this with the least amount of calculations) Edit:welp i was waayy off

The first answer being 5 meters blew my mind the math checks out but how logically does that make sense just breaks my brain thinking about it but I guess it must come from the rest of the line as well not just the 1 meter of slack added in like you would intuitively think.

The slack plays a bit role in a ship's anchor ⚓ . Because the chain slacks it is hard to un slack it by increasing the horizontal distance if for insane there is a wave it is stable

The way the first puzzle is drawn the rope is straight. The pythagorean theorem tells us the rope falls just over 5 meters. Real world it will be less because the eope follows a catenary curve.

Made me think of bernoullis brachistochrone problem that shows up in physics mechanics courses, the shape ends up being a cycloid. I’d love to see a vid on that

Power lines come to mind first. 😊

The string-girdling-the-earth problem is just as enjoyable, or maybe a lot more. But it isn't the one mentioned in the other comments here; in fact, it's similar to this one, because the question is the same; namely, given a certain (small) amount of slack, how far can it get you? For the circumference of the earth, you add one meter of length to the rope encircling earth, and then you pluck the rope and pull up until it's taut again. Find how high above the earth does it reach. Like some giant plucking the rope above the chasm here, instead of hanging down. And the straight 50-meter-chasm is replaced by the round earth with a circumference of about 40 thousand km. So, it's gonna be geometry again, like the triangle case, but a little more nasty.

My intuitive guess is that it's a catenary and we can calculate it with hyperbolic functions (sinh, cosh, etc)

preliminary answer, haven't watched the video yet. will check later the initial and final point of the chord describe an angle theta at the center of the circle. start by finding this angle, and the radius. by geometry, r*theta = 51 m and, by trig, r*sin(theta/2) - r*sin(-theta/2) = 50 m. by simmetry, the last one is just r*sin(theta/2) = 25 m. solve for r and theta to get r ≈ 74.124998 m and theta ≈ 0.688027 rad. the height lays in correspondence with the midpoint of the angle, thus its corresponding value will be given by r - r*cos(theta/2) ≈ 4.34 m EDIT oh, it was a problem on a parabolic curve. thought it was a problem on a circle from the thumbnail. still close though.

Long bridges during hot days and cool nights. As it heats up and expands, that extra length has to go somewhere.

At first for both questions I guessed about a metre at most? Then I calculated the drop for the first (as in the video) and realised I *massively* underestimated the drop. So for the second... my gut went to 3 metres, but after last time I decided to guess 4.5 metres. I'm glad i continued to ignore my gut 😆

Why don't you use the p value associated to the parabola to try to guess the ecuation considering 4p for the horizontal distance? Is it a bad guess? Just suppose the horizontal distance is latus rectum.

Guesses without calculation: case 1 triangle, drop is maybe 6 m or so? Case 2 arc (note disclaimer at end) perhaps 4 m? Disclaimer: I learned something in college years ago about a cable hanging under its own weight does not take on the shape of a circle, but rather a shape described by the hyperbolic cosine function. (Or arccosh, I don’t recall.) So if the solution here is based on circular arclength, I think it won’t be exactly correct.

Are you demanding that the curvature of the rope forms a circular arc or the curve of a slack rope which is neither a circle nor a conic?

Wonder where my errors are adding up from. I just ran the number on my catenary model and I'm getting about 4.36m, which is awfully close but not exact. Without going into too many details, let L be the length of half of the rope and D be the distance from the end to the middle. Then, you can find your solution catenary as h(x) = cosh(ax)/a + C where h is the height, x is distance from the centre, and C doesn't matter since we can move "ground level" to wherever we would like. The parameter a winds up being a solution to sinh(aD) = aL Finding the appropriate a = 0.0138152, you can plug in L = 25.5 and D = 25 to obtain h(D) - h(0) = 4.360337.

The reason I forgot about the arc length formula and only derive it when I can use it.... is that... there are infinitely more possible arcs than can be described by simple functions. Even with functions which we can easily integrate... the formula makes it hard to integrate... and sometimes impossible to do so.

Analogous to the problem cited in this video, there is a practical example of "A cylindrical tank laid horizontal, length of the cylinder along the ground". Calculate the volume of Diesel Fuel while filling this tank at any instant. I have an engineering background but too old to figure this out. Perhaps youngster can suggest a formula for this problem.

Having seen tbe answer to how far the rope drops or sags when 1m is added to 50m, here's another "surprising" answer: suppose there was a rope running around the equator of the earth. Assuming the earth was a smooth solid sphere of radius 6,378,000,000 m how much extra rope would be needed to raise the rope 1m off the ground all the way round the equator?

For the first one, it has to be less than 1/2 meters. My guess is 1/4 m And for the case without friend it should be even less maybe 10cm So apparently I have no intuition and mistook the upper bound for the lower bound 😂 I actually derived the formula for the chain line by setting up the differential equations when I was in ~10th grade. Unfortunately I no longer have the post-it I calculated it on but as far as I remember I also calculated it by setting up the length of the line for a short distance to get the weight per small segment in x-direction

It would be interesting to iterate between the two questions. Start with the triangle, then have it be a quadrilateral, and so on. I have to admit that the use of a calculator that does the integration for you is a bit dissatisfying, because, at that point, the intuition disappears that connects that part. I think the above might help keep it all connected.

The most unintuitive question is how much longer the earth circumference compare at the ground level vs 1 meter above above it . just around 6.28 meters

5m and 4m. I really had to restrain myself from doing any math, and just go with gut feelings. That's such a huge ask. 😛

The first question is easily splved by just using the pitagorean theorem, it's about ~5.025m The second question I know hanging ropes use some kind of hyperbolic function but as an engineering in training approximating it with a parabola should be fine hahaha But you asked for intuition and not me calculating the arc length of the parabola, so I'd say it should be less than 5 cuz the load is pretty evenly distributed across the entire rope and it's the same total length so I'd say it's like 4m :3

6:54 Dos cosh(x) assume no bending stiffness (tension only), evenly distributed mass and gravity, and parabolic assumes no tension stiffness (bending only)? And if so would the "truth" not be somewhere between them?

MOre interesting is the question how strong the Rope pulls ON the fixing Points left and right depending ON the deepness of the sagging. (Antenna Problem)

What is that interactive software for functions?

Make more videos ❤❤

A old rhyme : No force, however great, can pull a thread, however fine, into a horizontal line .

Surely less than 1m... LOL. Thanks for righting my brain

i got carried away and did calculatings, the first one (should be) about 5.02 m (i may be stupid??) with the classic sqrt((51/2)^2-25^2) i dont know the second one

It's always embarrassing when the guy half your age just hops up and figures out something like this when you just wanted to go measure it.

In what way does the noise in the background make this instruction better? Noisepolution? My professor never had a DJ as assistent!

0:51 I guessed roughly 5 meters. I did Pythagorean with the hypotenuse 25.5 being half of the actual rope and 25 being one of the legs as the half the distance between the tree. It’s probably wrong but I don’t know what else to apply to this situation

sqr(50,75) is the solution of the first. When you just drop the lilne, it does not get an arc. That is physics.

Why would you go trough all the cool math and then just say 'nah, the real cosh is too hard' ? Don't pick the rope problem if you will not solve it. You could have sticked to the basketball case ! That was already cool !

What are the chances that my next chapter is on this and I see this post!

I must admit to always taking the lazy path of writing ds^2 = dx^2 + dy^2 and just pretend what follows is rigorous, where I think of ds as "just a bit of arc length" and dx and dy as "a tiny change in x and y"; if I zoom in enough each step is "approximately a right triangle". "Divide" by dx^2 ds^2/dx^2 = 1 + (dy/dx)^2. Take the square root multiply over by dx and integrate. Also works for different geometries like in special relativity where you have ds^2 = c^2 dt^2 - dx^2. The dx * sqrt(1+m^2) notion is a bit more rigorous of a take (without sacrificing intuition) that I might use the next time I explain this formula to someone.

So I solved this problem, I am going to give the answer, I dont know if is correct. I based my answer on Greco-Babylonian geometry and the basis of developing Acrs and Spans. The basis of the technique is that if you bisect a circle or inscribe a hexagon into a circle of radius r you create two known answers, In the first case you have a 2 suoerpositioned spans of two, in the second case you have 6 end on end spans of length one. In the first case you have have two superposition spans of length 2, so 2 x 2 = 4. this is the coversge of 4/2pi. We can bisect either span creating a tangent outward. Doing this we have an ouward facing triangle of 1/2 span (2 * 1/2 = 1) and the cosine of 1 is zero, so the ouward facing triangle is 1 -0 in the outer part of the busecting radial. The new span created is this 1^2 + 1^2 = s^2 of the square root. We can repeat this process for generating sides of polygons until at some point Lspan * sides = 2 pi. This is roughly 13 steps for most computers. We can do the same with the hexagon. So the ratio you give is 50/51 = 0.980392 which is close to a span 39.375 degrees 0.980438. This was derive from a series of back tables generating by using the 180° based and 120°[60°] based span tables. In these tables you can combine variables of two angles to come up with precise information for a third angle and then go about busecting the soans and creating outer triangles to creat new smaller spans. I try to keep the babylonian (360 = 2 pi() whole angles. This particular angle comes out of a theoretical 315° that splits 3 times to 39.375° As a consequence its half span is 19.6875° [Yes, indeed I have this on an Excel worksheet an no I did not make this for the puzzle, I was testing to see how far babylonians and ancients greeks could fill out a span table just using the pythagorean theorem, right now the table has the largest gap of 1.5°. I could just continue the process until the perimeter of the polygon did not change, then easily get every span to 2° by making the arguemnt that cosine of trivially small angles is indistinuishable from 1, so that if I have an angle like say 1 degree, and I divide it say 7 times to get 1/(2^7) degrees, if I just multiple the sin of 1° x 2 and reverse the division process, I can return all the parameters for 2°, but since the ancients dont have 64 bit processors that would defeat the effort]? While i could just get the cosine of 19.6875 that would be cheating in a greek, persian or babylonian scribal school. So from this we can see that the span of 39.375 is 0.67378 and the bisect to the span is 0.94158. Ok this is the answer but we (the scribe needs to fix it) need to fix it so that the construction crew understands it. You span is 50. 50 ~= 0.67378 x r. We dont know r. R = 74.2. R * 0.94158 = 68.875. Thus the dustance between the arc and the span at the bisect is 74.2 - 68.9 or 4.37 units. Or 4 and 3/8ths units. And so, 2000+ years before calculus and integrals, this problem can be solved (crossing fingers and hoping I didnt make any errors😎) Now of you are a stickler for ancient history, you should note I am using the decimal system instead of fractions,

From the Thumbnail I thought the arc was a circle segment and got an answer of 4.3431 (by recursive calculation) Then watched video and saw PARABOLA but if it was suspended it would be a catenary curve.

No weight other than the weight of the rope? That would be catenary curve. Time to pull out my Calculus of Variations textbook from 45 years ago? Definitely not an easy problem.

the friend (if they're in the middle) drops by a total of root(25.5^2 - 25^2) meters, i think

"let's take a simple case" -> shows a function that doesn't pass the horizontal line test.

my guess: with a friend: 8mm, without him 2cm. I tried to tigh a rope straigt several times and it is very hard. And 50m is really long.

First one is simple pythagorean Second one can be solved using parabolas

There is absolutely no way a normal person can solve for that parabola. For the triangle it is extremely simple that the answer is sqrt(x+1/4) which is very close to sqrt(x) for such big values of x like 25. (x+1/2)^2-x^2 = x+1/4

You lay down a rope to cover the equator, it will be roughly 40,000 kilometers, if you increase its length by ONE meter, guess how high the rope can go above the ground??

I used the pythagorean formula to get ~5m and then guessed 4m for the second part. Not bad!

I guess we're assuming that my friend is infinitely heavy.

I guessed "a meter or two" for the first one, but I was still way off!

I thought I had a quick guess remembering the 12 13 5 Pythagorean triangle. I thought that 24 25 would be about half as steep, so roughly 2.5 meters Then I conluded 25 and 25.5 would be half as steep again, so roughly 1.25 meters. Where did I go wrong? Oh wait, I now see it. 12 13 5 means 48 52 20, so I needed to multiply by 4 before dividing by 4... silly me!

I am torn between liking this for the nice explanation of the concept, or hating it for the cheating using apps. I guess it depends on the audience. Math students will need a different approach than engineering students.

0:50 Woah, intuition is so lacking here. Intuition kind of told me between 0,5 and 1m and that is just so wrong. If you approximate the problem with 2 triangles, you get more than 5m

My logic says that 50m of the length is "used up" by the width, therefore the extra 1m would comprise the drop, which would make it 0.5m

For the first one, using the Pythagorean theorem, I got that the middle of the rope should drop by sqrt(25.25) meters. I have NO clue how to do the second part!

For 1st it can't be more than 1m so .5m was my first guess. But now I'm feeling like it could be 1m too For 2nd it will be much lower so maybe .25m?

You might even say the question just doesnt strike the right chord with them 😊