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Derivation of equation for catenary

  1. Feb 17, 2012 #1
    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 [itex] \frac{ds}{dx}=\sqrt{1 + {y'}^2} [/itex].

    however that doesn't seem very rigorous to me and i am a little wary of accepting that explanation. i know that [itex] s = \sqrt{{x'}^2 + {y'}^2} [/itex] 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 [itex] \frac{1}{2\sqrt{{x'}^2 + {y'}^2}} [/itex] 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 [itex] \frac{d}{dx}{x'}^2 = 2x' \frac{d}{dx}\frac{dx}{dt} = 2x' \frac{d}{dt}\frac{dx}{dx} = 0 [/itex]. then i have [itex] \frac{d}{dx}{y'}^2 = 2y' \frac{d}{dt}\frac{dy}{dx} = 2y'y''x' [/itex] 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.
     
  2. jcsd
  3. Feb 18, 2012 #2
    If you are given [itex]ds^2 = dx^2 + dy^2[/itex], then [itex] s \neq \sqrt{{x'}^2 + {y'}^2} [/itex].



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