Undergrad Why is my derivation of the catenary wrong?

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SUMMARY

The discussion centers on the incorrect derivation of the differential equation for the catenary. The primary error identified is the misuse of the tangent approximation, specifically the assumption that the derivative of tangent is simply 1. The correct differentiation method involves applying the quotient rule to the function tan(θ), leading to a more accurate representation of the derivative. This highlights the importance of proper calculus techniques in deriving equations related to the catenary curve.

PREREQUISITES
  • Understanding of differential calculus, specifically derivatives.
  • Familiarity with trigonometric functions, particularly tangent.
  • Knowledge of the quotient rule in calculus.
  • Basic concepts of catenary curves and their mathematical representation.
NEXT STEPS
  • Study the quotient rule in calculus for differentiating complex functions.
  • Learn about the properties and applications of catenary curves in physics and engineering.
  • Explore advanced differentiation techniques for trigonometric functions.
  • Review common mistakes in calculus derivations to avoid similar errors.
USEFUL FOR

Students and professionals in mathematics, physics, and engineering who are working with calculus and catenary curves, as well as anyone looking to improve their differentiation skills.

phantomvommand
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TL;DR
I have "derived" a differential equation for the catenary, and have attached my working. It looks slightly different from the correct expression, which can be found here: https://www.math24.net/equation-catenary

Please do tell me where I made a mistake. Thank you!
Important note: I only derived the differential equation, I did not solve it.

WhatsApp Image 2021-03-04 at 1.22.37 AM.jpeg

What I think caused the mistake:
- the tangent approximation (tan(theta+dtheta) ~ tan theta + d theta
 
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Hi,

phantomvommand said:
What I think

That's what I think too :wink:
The proper way to differentiate ##\tan\theta## is not ##{d\tan\theta\over d\theta} = 1 ## but $${d\tan\theta\over d\theta} = {d\over d\theta}\Biggl ( {\sin\theta\over\cos\theta}\Biggr ) =\ ... $$
​
##\ ##​
 

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