How to answer this question?  Any math help, hint or even a correct answer will be accepted.  

Why is this not in Homework ?

Hints

What is s ?

What is velocity in terms of s?

What is acceleration in terms of velocity ?

I barely remember anything about catenaries, but I thought they were hyperbolic cosine functions, not tangents. 

s is the arc length of a particle P from a fixed point P₀(S=0) on curve C.

Velocity is [math] \vec{v}(t)= \frac{d\vec{r}}{ds}\times \frac{ds}{dt}[/math]  where r⃗ (t) is the position of particle P at time t.

 Acceleration is [math] \vec{a}(t) = \frac{d\vec{v}}{dt}= \frac{d^2\vec{r}}{dt^2}[/math]

Edited September 17, 20214 yr by Dhamnekar Win,odd

56 minutes ago, Dhamnekar Win,odd said:

s is the arc length of a particle P from a fixed point P₀(S=0) on curve C.

Velocity is v⃗ (t)=dr⃗ ds×dsdt   where r⃗ (t) is the position of particle P at time t.

 Acceleration is a⃗ (t)=dv⃗ dt=d2r⃗ dt2

 

So is there a problem making the substitution ?

You have nearly done the question.

1 hour ago, exchemist said:

I barely remember anything about catenaries, but I thought they were hyperbolic cosine functions, not tangents. 

https://en.wikipedia.org/wiki/Whewell_equation

Homework. No linear algebra or group theory here.

17 minutes ago, joigus said:

https://en.wikipedia.org/wiki/Whewell_equation

Homework. No linear algebra or group theory here.

Aha. Thanks very much.

1 hour ago, studiot said:

 

So is there a problem making the substitution ?

You have nearly done the question.

[math] s= c \cdot \tan{(\psi)}, \frac{ds}{dt} = c\cdot (\tan^2{(\psi)} + 1)[/math]. Now how to compute [math] \frac{d\vec{r}}{ds}[/math] to find velocity?      

Edited September 17, 20214 yr by Dhamnekar Win,odd

1 hour ago, exchemist said:

Aha. Thanks very much.

You're welcome. I didn't know that form either. I did remember that the equation of an ellipse in polar coordinates is an ungodly mess if you try to express it in polar coordinates with the foci equidistant from the centre, but looks nice and simple with one focus sitting at the origin.

I just assumed something similar happens for the catenary. The rest was a wikipedestrian approach.

3 hours ago, exchemist said:

I barely remember anything about catenaries, but I thought they were hyperbolic cosine functions, not tangents. 

There are several ways to handle catenaries, but I often find that splitting the vertical axis into two with one section constant and the other parallel to the horizontala axis, as in the following.

 

!

Moderator Note

Moved to Homework Help.

 

On 9/17/2021 at 3:06 AM, exchemist said:

I barely remember anything about catenaries, but I thought they were hyperbolic cosine functions, not tangents. 

y as a function of x, yes, [tex]y= a cosh\left(\frac{x}{a}\right)[/tex].   But this is s, the arclength, as a function of [tex]\phi[/tex], then 

y as a function of x, yes- y= a cosh(x/a).  But this is s, the arclength, as a function of phi, the angle the graph makes with the x-axis.   It is the "Wethwell equation"- Catenary - Wikipedia

1 hour ago, Country Boy said:

y as a function of x, yes, [tex]y= a cosh\left(\frac{x}{a}\right)[/tex].   But this is s, the arclength, as a function of [tex]\phi[/tex], then 

y as a function of x, yes- y= a cosh(x/a).  But this is s, the arclength, as a function of phi, the angle the graph makes with the x-axis.   It is the "Wethwell equation"- Catenary - Wikipedia

Yes, thanks, that's what @joigus told me.

 

Example: Find the radial and transverse acceleration of a particle moving in a plane curve in Polar coordinates.

 

How can we use this example to solve our original catenary problem?

 

We know [math] \tan{\psi}=\frac{s}{a}, s=c\cdot \sinh{\frac{X}{c}},  V_0 = c , e^\psi= \cosh{\psi} + \sinh{{\psi}}[/math]

 I am able to get the magnitude of velocity [math]V = c\cdot e^{\psi=0}=c,V = V_0\cdot e^{\psi}[/math] but i am not getting magnitude of acceleration [math]\|\vec{a}\|= \frac{\sqrt{2}}{c} \cdot c^2 \cdot e^{2\psi} \cdot \cos^2{\psi}[/math] 

 

Edited October 7, 20214 yr by Dhamnekar Win,odd

 

 

Now let us replace ψ by y and solve the aforesaid differential equation.

 

Now we replace y by [math]\psi(t)[/math], we get,

[math]\psi(t)= C_2 - \displaystyle\int {\frac{1}{C_1 + t -2 \int {\psi(t)}dt}dt}[/math]

Now , how to show that velocity and acceleration of a moving particle along a catenary are equal to the given values in the question?

Edited October 9, 20214 yr by Dhamnekar Win,odd

 Now, here is the final correct answer provided to me by one great expert mathematician from UK (United Kingdom).

That's it.

Well done +1