Derivation of equation for catenary

[demonelite123]

I am a bit confused on one part of the derivation of the catenary equation. At one point my book says ds² = dx² + dy² and thus \frac{ds}{dx}=\sqrt{1 + {y'}^2}.

however that doesn't seem very rigorous to me and i am a little wary of accepting that explanation. i know that s = \sqrt{{x'}^2 + {y'}^2} so i tried to take the derivative of this with respect to x in order to hopefully obtain the same thing as above. so from the chain rule i have \frac{1}{2\sqrt{{x'}^2 + {y'}^2}} and now i have to take the derivative of (x'² + y'²) with respect to x and here is where i am having trouble.

i have \frac{d}{dx}{x'}^2 = 2x' \frac{d}{dx}\frac{dx}{dt} = 2x' \frac{d}{dt}\frac{dx}{dx} = 0. then i have \frac{d}{dx}{y'}^2 = 2y' \frac{d}{dt}\frac{dy}{dx} = 2y'y''x' which does not seem right to me. it seems like this way should work but i am just confusing the chain rule in this last portion. can someone help straighten this out? thanks.

 

[Jasso]

I am a bit confused on one part of the derivation of the catenary equation. At one point my book says ds² = dx² + dy² and thus \frac{ds}{dx}=\sqrt{1 + {y'}^2}. [...] i know that s = \sqrt{{x'}^2 + {y'}^2}

If you are given ds^2 = dx^2 + dy^2, then s \neq \sqrt{{x'}^2 + {y'}^2}.



Take ds^2 = dx^2 + dy^2, divide it by dx^2 and then simplify. (hint: \frac{dx}{dx} = x' = 1)