Interpreting Ambiguous HW Question on Curve Length and Surface Patch Domains

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In summary, the conversation discusses a homework question about finding the length of a curve on a sphere, but there is ambiguity about the domains and the surface being considered. The question also includes finding the angles of intersection between the curve and certain parallels. The expert suggests sticking to the given domain for the curve and possibly exploring further if desired.
  • #1
quasar987
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I got this HW question made up by the professor that I find ambiguous. It says

Consider the curve [itex]\theta (t)=\pi/2-t[/itex], [itex]\phi(t)=\log \cot(\pi/4-t/2)[/itex] on the sphere [itex]r(\theta,\phi)=(\sin\theta \cos\phi,\sin\theta\sin\phi,cos\theta)[/itex]

Find the length of the curve btw the points t=pi/6 and pi/4


He did not specify domains for either the curve nor the "surface" r. On one hand, if we take r to be a surface patch, this requires that the (maximum) domain be [itex]0 < \theta < \pi[/itex], [itex] 0 < \phi < 2\pi[/itex]. Anything bigger and the domain is not open or r is not injective. But this surface patch does not cover the whole sphere.

I could also consider two other surface patches of the form [itex]r_{2,3}(\theta,\phi)=(\sin\theta \cos\phi,\sin\theta\sin\phi,cos\theta)[/itex] with appropriate domains, that together with r above form an atlas for the unit sphere.

Any thoughts? How would you interpret this question?
 
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  • #2
I don't know what you mean. The domain is pi/6<t<pi/4. From the above relations you can get r as a function of t, and this is just some curve.
 
  • #3
There is another question after that:

Find the angles of intersection btw this curve and the parallels [itex]\theta = const.[/itex]

would you still say that the curve's domain is (pi/6,pi/4)? Or would you study it more carefully to find what is the maximum domain where the curve is defined and thus find all the possible intersectino points?
 
  • #4
I would stick to (pi/6,pi/4), at least for this question. If you're curious, keep going, but then you're doing more than what's asked.
 

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