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Homework Help: How should I interpret?

  1. Oct 27, 2006 #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 lenght 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?
     
    Last edited: Oct 27, 2006
  2. jcsd
  3. Oct 27, 2006 #2

    StatusX

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    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.
     
  4. Oct 27, 2006 #3

    quasar987

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    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?
     
  5. Oct 27, 2006 #4

    StatusX

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    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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