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Vector caculus with applications(catenary)

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Win,odd Dhamnekar:
 This is the math(other science) question.

A particle moves along a catenary s = c tan ψ. The direction of acceleration at any point makes equal angles with the tangent and normal to the path at that point. If the speed at vertex (ψ = 0) be 𝑣0, show that the magnitude of velocity and acceleration at any point are given by 𝑣0eψ and [itex]\frac{\sqrt{2}}{c}[/itex]𝑣02e2ψ cos2 ψ respectively.

My answer is velocity [itex]\|\frac{ds}{d\psi}\|= \|c(\tan^2(\psi)+1)\|[/itex] and acceleration is |2c tan(ψ)(tan2+1)|

But in the question velocity and acceleration are different. How is that? Where am i wrong?

mjc123:
How many times have we been here before? We can't tell how you went wrong unless you show us your working.

Win,odd Dhamnekar:
You may go through with my following workings:

 

mjc123:
I haven't tried to work this out in detail, but I have a few concerns about what's been written.
1. The equation s = c tanψ doesn't seem to define a unique curve in terms of y vs. x. What are x and y in terms of ψ? I don't know if this matters.
2. Saying velocity is ds/dψ implies dψ/dt is constant. Is this the case?
3. You appear to be deriving the magnitude of the acceleration as the derivative of the magnitude of the velocity. But this is not valid. For example, for circular motion at constant speed, the magnitude of the velocity is unchanging, but the magnitude of the acceleration is non-zero.

Win,odd Dhamnekar:
 Here is the answer to this question provided to me by one great expert mathematician from UK(United Kingdom).

That's it.

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