incl the noemal and a eodesic CURVAture asa Lunctioh oR t Hi

incl the noemal and a eodesic CURVAture asa Lunctioh oR t Hint: noemalise the tangent ve ctoR

Solution

Let ?? be a unit-speed parametrization of CC with ?(0)=p?(0)=p. The projection of ?? in TpSTpS along N(p)N(p) is given by

?(s)=?(s)+?p??(s),N(p)?N(p),?(s)=?(s)+?p??(s),N(p)?N(p),

whence ??(s)=??(s)????(s),N(p)?N(p)??(s)=??(s)????(s),N(p)?N(p) and ???(s)=???(s)?kn(??(s))N(p)??(s)=??(s)?kn(??(s))N(p). For s=0s=0 we obtain

??(0)=??(0)and???(0)=D??ds(0).??(0)=??(0)and??(0)=D??ds(0).

Computing the curvature of ?? at 00 we have

??(0)=???(0)×???(0)????(0)?3=???D??ds(0)???sin?1,??(0)=???(0)×??(0)????(0)?3=?D??ds(0)?sin??1,

where ?1=?(??(0),(D??/ds)(0))?1=?(??(0),(D??/ds)(0)). On the other hand, the geodesic curvature of ?? at pp is given by:

kg(p)=?D??ds(0),N(p)×??(0)?=???D??ds(0)????N(p)×??(0)?cos?2=???D??ds(0)???cos?2kg(p)=?D??ds(0),N(p)×??(0)?=?D??ds(0)??N(p)×??(0)?cos??2=?D??ds(0)?cos??2

since N(p)N(p) and ??(0)??(0) are orthogonal unit vectors, where ?2=?((D??/ds)(0),N(p)×??(0))?2=?((D??/ds)(0),N(p)×??(0)). Since ?1+?2=?/2?1+?2=?/2, the result follows.

Since your curve has unit speed (i.e., is parametrized by arc length), the equation

dTds=???(s)=kg(s)u^(s)+kn(s)m^(s)dTds=??(s)=kg(s)u^(s)+kn(s)m^(s)

is the decomposition of the accereration into tangential and normal components at ?(s)?(s). That is, the projection of ???(s)??(s) on the tangent plane T?(s)MT?(s)M is ???(s)??(s) minus the normal component:

???(s)?kn(s)m^(s)=kg(s)u^(s).