The Catenary: A Vector Calculus Problem

[@bigboy4789]

This channel is criminally underrated

[@Mcolive123]

This is the best solution I have ever seen for this for this problem, keep up the amazing work!!! Greetings from Turkey ✌🏻✌🏻

[@LearnPlaySolve]

Wow, thank you for those kind words! 😃

[@runekingthor5958]

You Are Most the Underrated youtube i Ever Seen You really explained Jee advance Level Concept In Simple word You Got My Sub Bro❤ A lot Of Love from 🇮🇳India

[@LearnPlaySolve]

Thank you so much! That means a lot. I have another calculus video coming very soon. 😃

[@AlexandreRibeiroXRV7]

I've known that the shape was a catenary since learning Calculus back on my uni days, but I don't recall seeing a proof as to why that was the case until today. This was beautiful and thoroughly explained in a simple way, great video!

[@LearnPlaySolve]

Thank you so much!

[@skilz8098]

It was a bit more than that. He did not skip over any assumed substitutions. He showed every step of the process starting with the visual description of a hanging cable between two poles then composing a function of its curve based on its length. He geometrically constructed the necessary vector components through coordinate composition. Then he showed properties of lines and their angles in conjunction with properties of the triangle beyond basic geometry involving trigonometry. Then created the necessary polynomial equations in order to introduce calculus into the problem solving technique. From there he used known differentials, partial differentials, and integrals along with their antiderivatives. Through the use of U substitution, the Fundamental Theorem of Calculus, and other Calculus tricks while using both algebraic and trigonometric substitutions he was able to easily simplify it down to a well known simple function. In the process he even used and shown well known logarithmic equations and their properties to reduce them into a form that will simplify even further right into a well known hyperbolic function. He never skipped over any substitutions, factorings, combining of like terms, simplifying by reducing, and always accounted for and showing proper signed values. It is 100% rigorous by showing all the work throughout each step of the process. Yet it is 100% elegant through the demonstration and simple explanation of each step while using an excellent animation and presentation scheme allowing it to easily be understood even if someone doesn't know Algebra, Geometry, Trigonometry, Linear Algebra or Vector Calculus. As someone with a very strong background in mathematics and physics it was a great refresher course that took less than a half hour. And this in itself is quite impressive. Even if someone doesn't know the math, they could still easily follow it and understand it only if they are willing to do their own research into the topics he mentioned such as how the actual trig identities are formed, or the definition and proof of the Fundamental Theorem of Calculus and such. Not even from a mathematics, physics or even from an engineering perspective, this entire presentation by being simple, elegant, and complete makes it a work of art! Even the speech is clear and well spoken and the timing is relatively good (subjective) but very good. Too fast and you'll lose too many people, too slow and they'll get bored and don't want to feel like they're being lectured. For beginners it might be a little fast and for professionals or experts it might be a little slow. And I think that this compensates very well as a middle point to reach both target audiences. It definitely deserves more likes! I can't say that it is perfect as there maybe other methods of solving this equation that he didn't use or done in a different order. But to reduce the problem in less possible steps then gets into the notion of solving for Big O complexity of Algorithms based on how many computations or steps are involved, and that's a different ball game and that's a topic for another discussion and another day. I can say that it was put together very well, very clean, and as far as I can tell it is also mistake free all in a seamless production.

[@LearnPlaySolve]

😳Wow! Thank you so much for saying all of that. I can't tell you how much that means to me. I'm almost done with another video right now that I hope you will like for all those same reasons.

[@bradryan8071]

Love the video, I have taught every technique that you used to solve the final equation to my grade 11's over the past 20 years ( U sub, trig sub, partial fractions, separation of variables, etc ) This is by far the reason why I love teaching so much and continue to do it for the past 37 years.

[@Zero-ov6xs]

very elaborated and pleasingly done. deserves more views and recognition. i'll definitely show this video to my students. keep on the good work!

[@LearnPlaySolve]

Thank you!!! 😊

[@cleon_teunissen]

What I find interesting about this treatment is the end stage. The solution to the differential equation is a function with the following structure: (e^(x) + e^(-x))/2 So that is the sum of two exponential functions, the functions mirrored in the y-axis. We have that the exponential function has a cyclic property: the derivative of the exponential function is the exponential function: d(e^x)/dx = e^x In the case of the hyperbolic cosine the constituent exponential functions are such that you get a cyclic property too, but it is a two-stage cycle: d(cosh(x)/dx = sinh(x) d(sinh(x)/dx = cosh(x) So: the simplest differential equation that has the cosh function as solution is to state: the *second* derivative of this function is equal to the function itself: d^2x/dy^2 = y (Yeah, the above differential equation has *two* solutions; cosh and sinh) So the question then is: starting from the equation at 5:33 , what is the most efficient way to arrive at the solution of the form (e^(x) + e^(-x))/2 ? I'm concerned that the series of substitutions that you deploy is more of a Rube Goldberg derivation. I suspect the same can be achieved in fewer steps (but I don't know how, so I'm not in a position to criticize.) On my own website (in the article about the catenary), I use the following property of sinh and cosh: sinh^2 + 1 = cosh^2

[@LearnPlaySolve]

I'm not sure what you mean by "Rube Goldberg derivation". If you are saying there's another way to approach this problem, I agree. There are countless ways to solve any problem. I value the process of problem solving over the end result. My goal is to entertain and educate by demonstrating calculus concepts, using problems that are interesting and sometimes counterintuitive. Thank you for watching!

[@cleon_teunissen]

@@LearnPlaySolve What I mean is: I expect the same can be achieved with fewer intermediary steps. That said: I am in no position to criticize, since I don't know how. More generally: A comparison: in the case of a force that follows Hooke's law the resulting motion is harmonic oscillation. The solution to the differential equation is a function f(x) such that when you differentiate it twice the result is -f(x). The sin(x) and the cos(x) functions satisfy that demand: d(sin(x))/dy = cos(x) d(cos(x))/dy = -sin(x) The catenary problem has in common with Hooke's law that you get a differential equation with second derivative. The difference is: with the catenary the solution is a function f(x) such that when you differentiate it twice the result is that you are back to f(x).

[@davidemasi__]

Great video, I really appreciate the fact that you explained carefully the reasoning and all the substitutions. I think this is the way math should be taught! However, I didn't get the horizontal force, I'm not so good at physics 😅 could you please make it clearer to me?

[@SO-th8gc]

Horizontal (i.e. parallel to the X-axis) force from the poles acts tangential to each point of the cable.

[@LearnPlaySolve]

At the vertex of the catenary, the tangent is perfectly horizontal. And at that same point, one half of the cable is essentially pulling on the other half, and the force vector is horizontal. This is why, at the other end of that segment, the vector is diagonal, because the tangent there is also diagonal. Remember, in that context, we were only referring to a segment of the cable, with one end starting at the vertex, not the entire cable. The only external force acting on it is gravity, which is the vertical vector. I hope this helps.

[@ominollo]

Very nice 👍

[@dakerbal]

why is the vertical component of V equal to mass? Isn't it more complicated as we're dealing with a whole structure and the force acts on the center of mass (and would be its weight, not mass)?

[@LearnPlaySolve]

Actually, that's a great point! You are right, I should have said weight instead of mass. Thank you for pointing that out. The reason the vertical vector is equal to its weight, is because that vector just represents the amount of force pulling on the cable in that direction, and the only thing forcing it down in the vertical direction is gravity. That's how we measure the effect that gravity has on an object, by its weight. Gravity does not just act on the center of mass of something; it acts on every part of it. We are focusing on just a small portion of the cable, and the amount of force gravity has on it, to derive an equation for the whole thing.

[@shawnlowery7162]

I wanna smoke what he's smoking.

[@kruGrockz]

By minute 10 I had already forgotten what the problem was about lol

[@LearnPlaySolve]

It's about the catenary. That's why I titled it "the catenary". 😉 Thank you for watching!!

[@kruGrockz]

@@LearnPlaySolve Even though it was kind of true i really appreciated the long video and will rewatch until it gets printed on my brain

[@El0melette]

great proof, but I think some u-sub in integrals are not necesary.

[@LearnPlaySolve]

I agree. I think I went too much into unnecessary detail in this video with some of the integration. I've tried to avoid that in my other videos. Thank you so much for watching!

[@polarisalpine2638]

it was really satisfying to see the cosx dx equal to du tho

[@SimchaWaldman]

Clever video indeed!

[@swatishaw1351]

i think if we increase the length as the diagram suggests by flopping that wire as it HAS to increase its length for "a" to change so i think a=0 is true if we have the enough length... idk tbh im just a kid from class 10.. soo ;-;

[@amirkefiraltroterminator2556]

16:00 really?!

[@LearnPlaySolve]

😉

[@AJ-et3vf]

The integration part went too long because you didn't use hyperbolic identities, but if you used it, the proof would've been so much shorter than otherwise!!!!

[@LearnPlaySolve]

That is true. But my goal wasn't to do it quickly. It was to demonstrate calculus concepts and integration techniques. I appreciate your advice. In the future, I hope to make a video about the hyperbolic trigonometric functions and identities.

[@user-rm5md2do6d]

You forgot g...

[@LearnPlaySolve]

g

[@Cqrt3r]

holy fuck

[@realcygnus]

nifty