Question

7. Let a be a unit-speed curve in M CR?. Instead of the Frenet frame field on a, consider the Darboux frame field T, V, U—where T is the unit tangent of a, U is the surface normal restricted to a, and V = U * T (Fig. 5.34). (a) Show that T' = gV + kU V' =-gT + tU, U' = -KT - tv, 263/518 where k = S(T) · T is the normal curvature k(T) of M in the T direction and t = S(T) · V. The new function g is called the geodesic curvature of a. (b) Deduce that a is geodesic g = -0, asymptotic k = 0, principal t=0.

Answer 1

E solution :- Here we have a unit speed curve Tunit tangent of a U surface normal to a K Normal currature Let a be ang Curre on the surface 2=x(s) '2" - kûtsa, + lol km = kot rd, + M2 [2" E " _ km] - prin icipal normal to curve ô - Normal to surface The vector q=1,2, + Moda is Geodesic currature of curve. Firstly let curre be geodesic ñ = kô= km û tid, tud, Equating coebficient of dy+d, on both sides we get 120, l=0 19=0

(2) Conversely let q=0 o fa, + M0₂ 20 - 120, l=0 km = km û ñ is ll to ô Hence Curre is geodesic

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