Problem I'm having with intrinsic equations when studing differential geometry

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SUMMARY

The discussion focuses on intrinsic equations in differential geometry, specifically proving that for the curve \( y = \ln|\sec x| \), \( \frac{ds}{dx} = \sec x \) and that \( \psi = x \). The user initially struggled with the proof but later confirmed the correctness of their calculations. The intrinsic equation is derived from the relationship between arc length and the parameterization of the curve, leading to the conclusion that \( s = \int \sec \psi \, d\psi \).

PREREQUISITES
  • Understanding of differential geometry concepts such as intrinsic equations and arc length.
  • Familiarity with calculus, particularly derivatives and integrals.
  • Knowledge of trigonometric functions, specifically secant and tangent.
  • Experience with curve parameterization in Cartesian and polar coordinates.
NEXT STEPS
  • Study the derivation of intrinsic equations in differential geometry.
  • Learn about the relationship between arc length and parameterization in curves.
  • Explore the applications of secant and tangent functions in calculus.
  • Investigate advanced topics in differential geometry, such as curvature and geodesics.
USEFUL FOR

Students and educators in mathematics, particularly those focusing on differential geometry, as well as anyone seeking to deepen their understanding of intrinsic equations and their applications in calculus.

finchie_88
I've just started learning about intrinsic equations after learning about envelopes, arc length and curved surface area in cartesian, parametric and polar coordinates, and understand the principle behind intrinsic equations, but my book doesn't explain it particularly well, here is a question I'm stuck on and my attempt so far:

Question:
For the curve [tex]y= \ln|secx|[/tex] prove that [tex]\frac{ds}{dx} = secx[/tex] and that [tex]\psi = x[/tex]. Hence find the intrinsic equation.

My answer (Very incomplete):
[tex]y = \ln|secx| \therefore \frac{dy}{dx} = tanx[/tex]
[tex]\frac{ds}{dx} = \frac{ds}{dy} x \frac{dy}{dx}[/tex]
[tex]\frac{ds}{dx} = \sqrt{1 + (\frac{dx}{dy})^2} = \frac{tanx.secx}{tanx} = secx[/tex]

I don't know how to prove that [tex]x = \psi[/tex], but I think for the last part that [tex]s = \int sec \psi . d \psi[/tex], is this correct?

Can someone explain or show me if what I have so far is right, and how to complete the question? any help would be appreciated.

edit: Where it says ds/ds = ds/dy x dy/dx, the x in the middle is suppose to be a multiply sign.
 
Last edited by a moderator:
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Doesn't matter, after a lot of thought and a little scribbling I realized how to prove that [tex]\psi=x[/tex], and that my answer was correct.
 

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