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: $\frac{dx}{dx} = x' = 1$)