How is this, have the t's been
suppressed? Like is it a''(t).(T(t) X n(t))?
Yes. Calculate this and show it is is zero on the curves v=constant.
Revolving the unit speed curve c(u)=(g(u),h(u),0) about the x-axis
produces a surface which can be covered by the coordinate patch
PHI(u,v)=(g(u),h(u)cosv,h(u)sinv)
Assume the h(u)>0 for all u. Show that the u-parameter curves are all
geodesics
[Difficulty]
I'm revising for an exam and I don't know what to do!!
[Thoughts]
Ok first of all, should I be able to know how they got the coordinate
patch? Once I have it I know how to work with it but I don't know how
they got it!
I have loads of theory to work off, I'm just not sure how to apply it.
U-parameter curves can only be geodesic if unit speed so for starters
do I need to check that this is unit speed? Or make it unit speed?
Why is it only the u-parameter curves that will be geodesics?
A regular curve a(t) on a surface is a geodesic of the surface is
a''_tan=0
Looking at my notes I have the following formula
a''.(T X n) as the geodesic curvature. How is this, have the t's been
suppressed? Like is it a''(t).(T(t) X n(t))?
This doesn't really help me though does it? Since I don't have the
original curve except in terms of functions?
Sorry I'm really confused!