I want to prove that the curve $$\alpha(t)=(r\cos t,r\sin t,bt),~b>0,r>0$$ (which is a circular helix) and $$\beta(t)=(e^s,e^{-s},\sqrt{2}s)$$ both share the property of "constant lifting". I'm not sure how to mathematically say that, my first thought was "constant curvature and torsion", which satisfies for the helix. However, the second curve does not satisfy that. Maybe with the projection onto the $XZ$ or $YZ$ plane and computing the "slope"?
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Where precisely did you find this term? – Ted Shifrin Mar 28 '23 at 23:51
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Hi @TedShifrin, first of all, your differential geometry textbook is so great, I have a physical copy :) It's on the set of exercises for my differential geometry course. – Fabrizio Gambelín Mar 29 '23 at 06:34
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2Then ask your professor what he means by the term. “Lifting” is usually used in the context of covering spaces, but constant makes no sense. Is the second curve also a generalized helix? – Ted Shifrin Mar 29 '23 at 07:10
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Not sure about the second one. Next week I'll be able to ask. Just thought about checking if its third component derivative is positive, then the third component is increasing? And in a constant way, since derivatives doesn't deppend on $t$ or $s$. – Fabrizio Gambelín Mar 29 '23 at 08:24
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1The first place to turn when trying to interpret any phrase used in a course is the course material - particularly the text book. If the textbook has an index, you can look for the phrase in it and find the page where it was first introduced. Otherwise, you can search through the sections covered by the exercises and see if a definition is given. When that doesn't work, asking the professor is definitely the next choice. My best guess is exactly as you describe: that the $z$-component is linear in the variable. But as Ted Shifrin has said, that would not be the normal use of "lifting". – Paul Sinclair Mar 29 '23 at 11:39
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1@ Fabrizio Gambelín : Guessing. Maybe "Lifting" here refers to the constant rate of "height" gained with respect to unit length of spiral arc traversed $ \frac{dz}{ds}$=const., instead of with respect to unit rotation around z-axis of symmetry as it happens in a circular helix. – Narasimham Mar 29 '23 at 13:15
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1Yeah, I think this is a language issue. The parametrized curve rises at a constant rate. – Ted Shifrin Mar 29 '23 at 14:56
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@PaulSinclair thanks for your reply. Exactly, I asked it here because i have to wait a week till I can ask a professor. I couldn't find in the material and Do Carmo's book that definition, but sure, "lifting" is the closest word I could think of from spanish. – Fabrizio Gambelín Mar 29 '23 at 15:31
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Exactly @Narasimham. – Fabrizio Gambelín Mar 29 '23 at 15:32
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Thank you all for your answers! – Fabrizio Gambelín Mar 29 '23 at 15:32
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Constant Lifting or climb rate on a cylinder is also a circular helix. Not interesting. So I computed for a unit sphere $(a=1)$ by numerically integrating.
The following differential relation gives climb Rate per unit arc length.
$$ \dfrac{dz}{ds}=\cos \phi \cos \psi=\lambda=0.0174524~ ; \dfrac{dr}{ds}=\sin \phi \cos \psi ;$$
Slope =$\phi$, arc/meridian angle= $\psi;$. $ \lambda$ must be less than 1.0;
Ordinary Differential Equation of non-linear helix:
$$ r\cdot \cos \psi = a \lambda $$
Starting helix angle at base= $1^{\circ}$
Narasimham
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Parametric Equation of non-linear helix is yet to be finalized: $ {\sqrt {a^2 - (\lambda s)^2}\cos \theta, \sqrt {a^2 - (\lambda s)^2} \sin\theta,\lambda s}$ – Narasimham Mar 29 '23 at 20:07
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