the firgure is exaggerated ...the only instance that the distance 10m from the ground to the middle of the curve is when the poles are 0m apart... by then the rope is folded halfway or 40m from the top... so that 40 + 10 = 50m, i.e. the length of each post
Absolutely correct, pretending that the cable does not create a parabolic function in the way it hangs is absurd. The drawing is false, the solution is false.
I arrived at the conclusion that the question was nonsense mentally. Assume the rope forms a triangle, a parabolic curve would be longer. With the hypotenuse having to be longer than the 40m drop at the centre, and the rope only being twice this drop in length, meant no required right angled triangle was possible.
Why didn't you put the logical equation at the beginning? Also, don't forget that you're dealing with a cable, and it's impossible to go down 40m and up 40m again unless you have a cut in between - what about that r-like curve you kept gesturing about? It made me think the guy was crazy. Тhe end - there are hyperbolic functions used in HDD calculations.
When the rope is fully stretched the poles are 80 meters apart and 50 meters from the ground at the centre. When the rope and the poles are brought together at he center yhe centres of the rope is 10 meters from the ground. St this point the distance between the poles is zero. So the answer is zero meters distance between the poles at the point the center of the rope is 10 meters from the ground.
Mathematically, when not taking into account the physical properties of the poles and the cable: The answer is 0 50 - 80/2 = 50 - 40 = 10 Meaning in order for the cable to lower its base to 10m, it needs half of its length (40m) from each pole. Making the distance between the poles 0. Physically it's impossible! As it requires the cable's diameter to be 1 unit of measurement (lets say an atom). And the poles might also have to share the same volume. I wonder how would an Amazon interviewer, react to the physical explanation.😇
This question has been solved using 4 different methods in a book "Theoretical Insights into Data-Physics and Philophysics" published in January 27, 2024 in Amazon. You will find the details of the steps in solving the problem are in pages 34 to 45 in chapter-3. Thank you for sharing us.
@Mtmonaghan See that book and you will see method 1 using calculus the same as the video with many cases imbedded to it. Method 2 solving using analytic geometry. Method 3 using Eculidean geometry and last best is using intutive philosophy based on philophysics.
So let's see the end - This hyperbolic functional calculations аrе used in engineering HDD calculations - history of meth. "Data-Physics and Philo-physics" - sound crazy...
Nice answer. Additionally you can find a description in Wikipedia (in German better than in English) but not for the actual problem. They write only "The actual shape that the rope ultimately takes on is calculated by adjusting x0, y0 und a so that the curve passes through the suspension points and has the specified length l." The different approaches to solving the problem seem to turn out different solutions according to their accuracy and the formula for the arc length used. So I and I think the other audience would welcome if you would provide the exact solutions in numbers.
No need for the solution to a hanging cable; it's pretty obvious from symmetry and the Pythagorean Theorem. Since the cable length to the middle low point is 40 meters and so is the height of the post (symmetry), and the minimum distance of a curve between the low point and the top of the post is 40 (Pythagoras), the only way this is possible is if d = 0.
The calculation of the second question (50/20/50) 80m length is overcomplicated and I got a different result. The poles are 50m high and the deepest point of the cable is 20m above the ground. If I apply a 1:10 scale, shortening the 5m-poles by 1m to 4m (no influence on r), I get the exact Cosh function shape. Then the distance r calculates as r=20*ArcCosh[4]=41.2687m instead of 45.4m. But what's about the cable length then? It calculates as l1=20*ArcLength[Cosh[t],{t,0,r/20}]= Integrate[Sqrt[1+D[Cosh[t],t]^2],t]= Integrate[Cosh[t],t]=Sinh[t]. We obtain l1=N[20*Sinh[ArcCosh[4]]]=N[20*Sqrt[4^2-1]]=77.4597m. Thus, the exact shape may be produced with a shorter cable only. But the cable should be 80m long. We calculate back from the ArcLength t=N[Cosh[ArcSinh[80/20]]]=N[Sqrt[4^2+1]]=4.12311m and we get for r=N[20*ArcSinh[4]]=2*20.9471m=41.8943m. Is that the correct result? The answer is no. If I calculate the arc length I get the 80m now. But what's about the hight at the pole 20.9471m from the centre? Including the suppressed 10m it calculates to 51.2311m. So we have to “add by saw” additional 1.2311m, otherwise the whole issue will sag in the middle to 18.7689m only. This is the situation with an exact Cosh curve. It is characterized by the fact that the force on the pole in y-direction equals the one in the x-direction. Due to the fixed arclength and the pole height we are limited to one single option: the poles have to be moved further apart to raise the center to 20m. But thus we say goodbye to the exact Cosh function and hello to elliptic integrals. Cosh is still used, but shrunk in y-direction. For that purpose we multiply the Cosh-function with a. It applies x'²=1 and y'²=Simplify[1+a*D[Cosh[t],t]^2]=1+a*Sinh[t]^2 for the arclength we get Integrate[Sqrt[1+a*Sinh[t]^2],t]=-I*EllipticE[I*t,a] EllipticE[I*t,a] is the incomplete elliptic integral of the type E see en.wikipedia.org/wiki/Elliptic_integral#Incomplete_elliptic_integral_of_the_first_kind Now, starting from the arc length, we could calculate the distance r/2 at which the compressed function has the value 4.But there are three problems. Firstly, it is no longer possible to form an inverse function for the arc length. Secondly, we do not know the factor a and thirdly, we not only reduce the value 4.1231 to 4, but also the value 1 at r=0. So we have to add the value (1-a). Fortunately, this has no effect on the arc length. We calculate the arc length as follows t3=20*(a3*ArcCosh[x3]+(1-a3))+10=50 x3=Chop[x/.FindRoot[20*ArcLength[a3*Cosh[t]+(1-a3),{t,0,ArcCosh[x]}]-80==0,{x,77}]] or better x3=Chop[x/.FindRoot[-20*I*EllipticE[I*ArcCosh[x],a3^2]-80==0,{x,77}]]=46.9897m a3=Chop[a30/.FindRoot[-20*I*EllipticE[I*ArcCosh[4],a30^2]-80==0,{a30,1}]]=0.8136 l3=N[20*ArcLength[a3*Cosh[t]+(1-a3),{t,0,ArcCosh[x3]}]]=80m y0=10(1+a3*Cosh[0]+(1-a3))=20m To solve the whole system it's necessary to sample a3. Starting with 1 you must approach the correct value maybe using a program. The whole thing is not that simple, especially if you want to use integrals. Now let's see if the solution in the movie is correct. The solution there is only an approximation. I have checked the whole thing with a3=0.8136. At a distance of 45.4m, the line would then have to be suspended at a height of 49.4263m and should only be 77.3186m long. The zero height of 20m would be correct. Obviously another value a4=0.858845 would be helpful here. Then the 45.4m, 20m and 80m would be correct. However, the suspension height should then be 50.5044m. After all, only ½m deviation without integrals is already a success.The remaining problem is, the speaker did not exactly explained his a. The exact solution for r is 46.9897m.
Every time I looked at this problem before my brain said "maths maths maths"... but for some reason this time when I looked at it, my brain picked out the 50 and 40 and 10 and I thought "no, it can't be." So I watched the video and found out that yeah, it is actually what I thought... it's kind of a trick question.
It was an interview test question for prospective amazon employees. If you figured out that the answer was zero or the poles are touching, then you were too smart for the job and didn't get it!
In countries that use commas as decimals, how do you differentiate between a decimal figure and a list of naturals? 4,2 is either 4.2 or 4 a list of 4 and 2.
Sinh is in formula which is a length of the curve. Apply integration. Cosh will be a different video. The video would be too long if I derived the catenary formula.
@@problem_analysis The catenary curve itself (for the hanging cable under gravity) is derived in a rare calculus called "the calculus of variations", which is rarely taught anywhere (it would belong in 3rd or 4th year university, or higher). "The calculus of variations" is in danger of almost disappearing, though I hope not.
My question for the case where the cable is 20 ft above the ground is how he got the equation that he solved. I mean to me the hanging rope looks like it's parabolic, so, I would think it could be solved using the equation of a parabola in cartesian coordinates. I don't know where the hyperbolic trig functions came from. I guess it's just assumed that we know they are applicable or something.
You can aproximate a catenary equation in an infinitely small surrounding of one dot with a parabola. However you can't aproximate an entire catenary with a parabola. This is an old problem, even Galileo had skin in the game. I will derive the catenary equation in another video, reason why I didn't do that in this video is because of the length. Getting that catenary equation is half an hour problem.
Estan juntos, porque para que haya 10 mt de la parte mas baja de la cuerda al piso significa que existen 40 mts a cada lado y como la longitud de la cuerda es 80 entonces la única solución posible es que esten juntos porque cada piste mide 50 mt
I first thought one would need a hyperbolic trigonometric function, but then I saw it: Move the poles together, and you will get 10 m + (80 m)/2 = 50 m.
Bottom line: It's a trick question in the sense that the diagram is misleading. It However, it demonstrates that if you just jump to the math and don't look at the initial conditions, you'll waste a lot of time (and fail the interview). Full disclosure--I, too, missed it and wouldn't have been hired!
An introductory statics text should cover it. My memory of it from my first year of engineering is that you divide the cable into small sections that are at an angle and consider the forces. I didn’t major in civil engineering so I didn’t do more. It is very useful if you design electric transmission lines because it is important to calculate the forces that the wire exerts on the tower.
You don't have 20m above the ground. From the very beginning of your problem you mentioned that the rope was only 10m above the ground. So you calculation would end up being incorrect. Review your own problem from the very start.
The picture is misleadind. When the rope is 10 m above the ground the poles must be touching. This distamce between the two poles must be zero. It does not ned any calculations. It needs common sense. Any idiot must be able to solve it.
This question is wrong. Those who failed at shoemaking & ditching stayed home and indulged in writing Math Carols. Robbery is not restricted to select rich areas.
This is so funny. You might know a lot of mathematics. The shape of the hung chain is a catenary, for which you need to know about hyperbolic sines, and stuff....horrible! Where do I begin? I think the more you know, the worse this is, until you apply a bit of common sense and actually look at the numbers. Which I didn't. Duh...
I doubt they would take this problem to enter Amazon.. Amazon is not an Engineering Company. Fist of all you need to demonstrate that the curve is a cosh(x). As I remember that needs differential calculus AND simplification considerations. Afterwards you need to set contour properties. This will take you a whole morning. You can not say just: Ah this is a Cosh.
the firgure is exaggerated ...the only instance that the distance 10m from the ground to the middle of the curve is when the poles are 0m apart... by then the rope is folded halfway or 40m from the top... so that 40 + 10 = 50m, i.e. the length of each post
That is not proof
The drawing itself is not true to scale as the posts were not together ……..Arse ! 😛😀😉
Actually, with any non-zero minimum bend radius, the 10m height will never be achieved.
Absolutely correct, pretending that the cable does not create a parabolic function in the way it hangs is absurd. The drawing is false, the solution is false.
@@joeesquire5927catenary is different from a parabola
I arrived at the conclusion that the question was nonsense mentally. Assume the rope forms a triangle, a parabolic curve would be longer. With the hypotenuse having to be longer than the 40m drop at the centre, and the rope only being twice this drop in length, meant no required right angled triangle was possible.
Why didn't you put the logical equation at the beginning? Also, don't forget that you're dealing with a cable, and it's impossible to go down 40m and up 40m again unless you have a cut in between - what about that r-like curve you kept gesturing about? It made me think the guy was crazy. Тhe end - there are hyperbolic functions used in HDD calculations.
When the rope is fully stretched the poles are 80 meters apart and 50 meters from the ground at the centre. When the rope and the poles are brought together at he center yhe centres of the rope is 10 meters from the ground. St this point the distance between the poles is zero.
So the answer is zero meters distance between the poles at the point the center of the rope is 10 meters from the ground.
Mathematically, when not taking into account the physical properties of the poles and the cable:
The answer is 0
50 - 80/2 = 50 - 40 = 10
Meaning in order for the cable to lower its base to 10m, it needs half of its length (40m) from each pole.
Making the distance between the poles 0.
Physically it's impossible!
As it requires the cable's diameter to be 1 unit of measurement (lets say an atom).
And the poles might also have to share the same volume.
I wonder how would an Amazon interviewer, react to the physical explanation.😇
I do not think amazon would ask this question to a normal employee applicant and they expect the applicant to be an expert in hyperbolic trigonometry.
This question has been solved using 4 different methods in a book "Theoretical Insights into Data-Physics and Philophysics" published in January 27, 2024 in Amazon. You will find the details of the steps in solving the problem are in pages 34 to 45 in chapter-3. Thank you for sharing us.
In can’t be solve, that is the answers
@Mtmonaghan See that book and you will see method 1 using calculus the same as the video with many cases imbedded to it. Method 2 solving using analytic geometry. Method 3 using Eculidean geometry and last best is using intutive philosophy based on philophysics.
So let's see the end - This hyperbolic functional calculations аrе used in engineering HDD calculations - history of meth. "Data-Physics and Philo-physics" - sound crazy...
@@andreasandre4756catenary is different from a parabola
Nice answer. Additionally you can find a description in Wikipedia (in German better than in English) but not for the actual problem. They write only "The actual shape that the rope ultimately takes on is calculated by adjusting x0, y0 und a so that the curve passes through the suspension points and has the specified length l." The different approaches to solving the problem seem to turn out different solutions according to their accuracy and the formula for the arc length used. So I and I think the other audience would welcome if you would provide the exact solutions in numbers.
No need for the solution to a hanging cable; it's pretty obvious from symmetry and the Pythagorean Theorem. Since the cable length to the middle low point is 40 meters and so is the height of the post (symmetry), and the minimum distance of a curve between the low point and the top of the post is 40 (Pythagoras), the only way this is possible is if d = 0.
The calculation of the second question (50/20/50) 80m length is overcomplicated and I got a different result. The poles are 50m high and the deepest point of the cable is 20m above the ground. If I apply a 1:10 scale, shortening the 5m-poles by 1m to 4m (no influence on r), I get the exact Cosh function shape. Then the distance r calculates as
r=20*ArcCosh[4]=41.2687m
instead of 45.4m. But what's about the cable length then? It calculates as l1=20*ArcLength[Cosh[t],{t,0,r/20}]=
Integrate[Sqrt[1+D[Cosh[t],t]^2],t]=
Integrate[Cosh[t],t]=Sinh[t].
We obtain
l1=N[20*Sinh[ArcCosh[4]]]=N[20*Sqrt[4^2-1]]=77.4597m.
Thus, the exact shape may be produced with a shorter cable only. But the cable should be 80m long. We calculate back from the ArcLength
t=N[Cosh[ArcSinh[80/20]]]=N[Sqrt[4^2+1]]=4.12311m
and we get for
r=N[20*ArcSinh[4]]=2*20.9471m=41.8943m.
Is that the correct result? The answer is no. If I calculate the arc length I get the 80m now. But what's about the hight at the pole 20.9471m from the centre? Including the suppressed 10m it calculates to 51.2311m. So we have to “add by saw” additional 1.2311m, otherwise the whole issue will sag in the middle to 18.7689m only. This is the situation with an exact Cosh curve. It is characterized by the fact that the force on the pole in y-direction equals the one in the x-direction. Due to the fixed arclength and the pole height we are limited to one single option: the poles have to be moved further apart to raise the center to 20m. But thus we say goodbye to the exact Cosh function and hello to elliptic integrals. Cosh is still used, but shrunk in y-direction. For that purpose we multiply the Cosh-function with a. It applies x'²=1 and
y'²=Simplify[1+a*D[Cosh[t],t]^2]=1+a*Sinh[t]^2
for the arclength we get
Integrate[Sqrt[1+a*Sinh[t]^2],t]=-I*EllipticE[I*t,a]
EllipticE[I*t,a] is the incomplete elliptic integral of the type E see
en.wikipedia.org/wiki/Elliptic_integral#Incomplete_elliptic_integral_of_the_first_kind
Now, starting from the arc length, we could calculate the distance r/2 at which the compressed function has the value 4.But there are three problems. Firstly, it is no longer possible to form an inverse function for the arc length. Secondly, we do not know the factor a and thirdly, we not only reduce the value 4.1231 to 4, but also the value 1 at r=0. So we have to add the value (1-a). Fortunately, this has no effect on the arc length.
We calculate the arc length as follows
t3=20*(a3*ArcCosh[x3]+(1-a3))+10=50
x3=Chop[x/.FindRoot[20*ArcLength[a3*Cosh[t]+(1-a3),{t,0,ArcCosh[x]}]-80==0,{x,77}]] or better
x3=Chop[x/.FindRoot[-20*I*EllipticE[I*ArcCosh[x],a3^2]-80==0,{x,77}]]=46.9897m
a3=Chop[a30/.FindRoot[-20*I*EllipticE[I*ArcCosh[4],a30^2]-80==0,{a30,1}]]=0.8136
l3=N[20*ArcLength[a3*Cosh[t]+(1-a3),{t,0,ArcCosh[x3]}]]=80m
y0=10(1+a3*Cosh[0]+(1-a3))=20m
To solve the whole system it's necessary to sample a3. Starting with 1 you must approach the correct value maybe using a program. The whole thing is not that simple, especially if you want to use integrals. Now let's see if the solution in the movie is correct. The solution there is only an approximation. I have checked the whole thing with a3=0.8136. At a distance of 45.4m, the line would then have to be suspended at a height of 49.4263m and should only be 77.3186m long. The zero height of 20m would be correct. Obviously another value a4=0.858845 would be helpful here. Then the 45.4m, 20m and 80m would be correct. However, the suspension height should then be 50.5044m. After all, only ½m deviation without integrals is already a success.The remaining problem is, the speaker did not exactly explained his a. The exact solution for r is 46.9897m.
Every time I looked at this problem before my brain said "maths maths maths"... but for some reason this time when I looked at it, my brain picked out the 50 and 40 and 10 and I thought "no, it can't be." So I watched the video and found out that yeah, it is actually what I thought... it's kind of a trick question.
It was an interview test question for prospective amazon employees. If you figured out that the answer was zero or the poles are touching, then you were too smart for the job and didn't get it!
d=0.0 This is trivial
beautiful video.
With luck and more power to you.
hoping for more videos.
In the world of mathematics, engineering, and physics we need proof not wild assumptions, speculations, or imaginations
In countries that use commas as decimals, how do you differentiate between a decimal figure and a list of naturals? 4,2 is either 4.2 or 4 a list of 4 and 2.
Obvious from the context.
A comma to denote decimals is NOT followed by a space, all other commas are.
9:00 if you multiply by a, there will be -10*70+60=0. Thats clear indicator, the way you following is not correct or there is no valid solution.
In a right triangle, the hypotenuse must be longer than the right angle side，right？so the cable's lenth is impossible only 80m. The question is wrong.
How did you come up with sinh and cosh?
Deus ex machina?
Sinh is in formula which is a length of the curve. Apply integration.
Cosh will be a different video. The video would be too long if I derived the catenary formula.
@problem_analysis why the h... Then did you not start with introducing the catena formula. The way you done it makes all your efforts indigestible.
@@problem_analysis The catenary curve itself (for the hanging cable under gravity) is derived in a rare calculus called "the calculus of variations", which is rarely taught anywhere (it would belong in 3rd or 4th year university, or higher). "The calculus of variations" is in danger of almost disappearing, though I hope not.
Impossible, unless the poles are next to each other... The distance is 0.00m
My question for the case where the cable is 20 ft above the ground is how he got the equation that he solved. I mean to me the hanging rope looks like it's parabolic, so, I would think it could be solved using the equation of a parabola in cartesian coordinates. I don't know where the hyperbolic trig functions came from. I guess it's just assumed that we know they are applicable or something.
You can aproximate a catenary equation in an infinitely small surrounding of one dot with a parabola. However you can't aproximate an entire catenary with a parabola. This is an old problem, even Galileo had skin in the game. I will derive the catenary equation in another video, reason why I didn't do that in this video is because of the length. Getting that catenary equation is half an hour problem.
Estan juntos, porque para que haya 10 mt de la parte mas baja de la cuerda al piso significa que existen 40 mts a cada lado y como la longitud de la cuerda es 80 entonces la única solución posible es que esten juntos porque cada piste mide 50 mt
Si juntas los postes, la altura del piso a a la parte más baja del cable (colgando) es 10m.
How about using an "integral"?
Also, I doubt that there is rounded law for cable while gravitating. Maybe, squared y = f(x) = x^2 or similar...
Good idea, I'm working on it but the preparation of the text will take a while. --- Ready
@@GerdPommerenke Awaiting for your resolution with honor
Written even in the video cover 10m, why you said it 20m?
Problem has two cases.
I first thought one would need a hyperbolic trigonometric function, but then I saw it: Move the poles together, and you will get
10 m + (80 m)/2 = 50 m.
Great content my fellow Mathematician ❤ keep up I have subscribed
Welcome to the channel!
With Pitagora you obtain an approx. of 45 meters on the 20mt height version
All that mathematics, when clearly the answer must be 0 if the low point of the string is at 10.
I may be old, but I never used cosine hyperbolical.
You're not old enough ;)
Distance is zero....no need calculation...
Bottom line: It's a trick question in the sense that the diagram is misleading. It However, it demonstrates that if you just jump to the math and don't look at the initial conditions, you'll waste a lot of time (and fail the interview). Full disclosure--I, too, missed it and wouldn't have been hired!
"Mind your decision"already answered this and this doesn't have a solution.
I don't think he explained what "a" is?
What does the letter (a) mean here?
LMAO had to do this almost every day at work!
The poles would be touching each other.
How did you find the equations y=cosh(x/a) -a and the others?
An introductory statics text should cover it. My memory of it from my first year of engineering is that you divide the cable into small sections that are at an angle and consider the forces. I didn’t major in civil engineering so I didn’t do more. It is very useful if you design electric transmission lines because it is important to calculate the forces that the wire exerts on the tower.
The poles are NOT "50 metres from the ground" They are 50 metres high!
Don’t be so picky, how many languages can you speak
You’d to prove how you got the second equation of sinh(r/a)🎉
Length of the curve. Just use a integral.
This is a badic problem yused in all engineering texts. Solved it more than one way. In this case the poles are 0m apart.
There is no solution, that is the answer
You are the man
You don't have 20m above the ground. From the very beginning of your problem you mentioned that the rope was only 10m above the ground.
So you calculation would end up being incorrect. Review your own problem from the very start.
Problem has two cases. First case is 10m and the second case is 20m.
If I remember good, the shape of the cable is the cosh(x) function...
Zero - the poles are touching.
The picture is misleadind. When the rope is 10 m above the ground the poles must be touching. This distamce between the two poles must be zero. It does not ned any calculations. It needs common sense. Any idiot must be able to solve it.
This question is wrong.
Those who failed at shoemaking & ditching stayed home and indulged in writing Math Carols.
Robbery is not restricted to select rich areas.
This is so funny. You might know a lot of mathematics. The shape of the hung chain is a catenary, for which you need to know about hyperbolic sines, and stuff....horrible! Where do I begin? I think the more you know, the worse this is, until you apply a bit of common sense and actually look at the numbers.
Which I didn't. Duh...
Calculus is required
nop.
catenary equation or hyperbolic is requred.
@@Carlos-im3hn not really.
No. The distance is 0 because the poles are touching. Diagram is misleading.
@@jamesanthony5681 you killed the suspense....
I doubt they would take this problem to enter Amazon..
Amazon is not an Engineering Company.
Fist of all you need to demonstrate that the curve is a cosh(x).
As I remember that needs differential calculus AND simplification considerations.
Afterwards you need to set contour properties.
This will take you a whole morning.
You can not say just: Ah this is a Cosh.
cosh is pronounced simply as 'cosh' and sinh as 'shine'.
In your brain, maybe. To me those sound like a problem waiting to happen.
I have always pronounced sinh as "cinch"
THANK YOU I HAD THE HARDER VERSION BUT I DIDNT KNOW HOW to sove
No problem brother. Welcome to the channel!
d= 45.4 m 10:45
Amazon interview question? Are they hiring math teachers? They hire cheap immigrant labour to push boxes.
Can anyone solve it with integral calculus?
Sure
I get 12.649, any other solutions?
For cases discussed in this video there are only the solutions represented in the video.
Not true on height off the ground.
You also have to indicate :
Cable material
Ambient temperature
Tension in the cables
Supports material and shape
Any pretension in the cable
Mengapa kok yang diterangkan pakai angka 20 dan bukan 10. Kalau 10 m sih d = 0, tidak usah rumit-rumit ngitungnya.
Horrible way of approaching and solving the problem, very very disappointing
Zero
80m
Amazon clowns
Are fool how u r getting 80 mts explain first that re
WaaaaaaaO
-.02m
Pure "BULLSHIT"
0.0
Lamentable !
.
idiot cable is 10m from the ground.
It is bigger than your head. Just relax. He proved it mathematically and logically.
zero