Simple surfaces, unit normal, tangent vectors, tangent planes Assumptions. All functions are C1: Notations. @~x @~x ~x1 = @u1 ; ~x2 = @u1 spanning tangent vectors M = ~x (U) image of a simple surface TpM tangent plane at P De¯nitions. (1) A simple surface is a 1-1 C1 function ~x : U! R3, where U ½ R2 is open and ~x1 £ ~x2 6= ~0: (2) The tangent plane to a simple surface ~x : U! R3 at P = ~x (a; b) is the plane spanned by ~x1 and ~x2, i.e., the plane perpendicular to ~x1 £ ~x2: (3) The unit normal ~n (a; b) is ~x £ ~x ~n = 1 2 : j~x1 £ ~x2j (4) X~ is a tangent vector at P 2 ~x (U) if there exists a curve ~® : d~® ~ (¡"; ") ! ~x (U) such that ~® (0) = P and dt (0) = X: Results. Lemma 1.16. The set of tangent vectors at P is a vector subspace of R3: Proof. Let X~ and Y~ be tangent vectors at P and let r 2 R: Then there exist curves ~® :(¡"; ") ! ~x (U) ; ¯~ :(¡"; ") ! ~x (U) ~ d~® ~ d¯~ ~ such that ~® (0) = ¯ (0) = P , dt (0) = X; and dt (0) = Y: The lemma follows from showing that: (a) rX~ is a tangent vector at P: (b) X~ + Y~ is a tangent vector at P: " " Proof of (a). De¯ne ~´ :(¡ r ; r ) ! ~x (U) by ~´ (t) = ~® (rt) : Then by the chain rule, d~´ d~® (0) = r (0) = rX:~ dt dt Proof of (b). De¯ne functions ®1 (t) ; ®2 (t) ; ¯1 (t) ; ¯2 (t) [from (¡"; ") to R] by ¡ ¢ ¡ ¢ ~® (t) = ~x ®1 (t) ; ®2 (t) ; ¯~ (t) = ~x ¯1 (t) ; ¯2 (t) : 1 ¡ ¢ ¡ ¢ If you like, you may write ®i = ~x¡1 i ± ~® and ¯i = ~x¡1 i ± ¯~ for i = 1; 2; ³¡ ¢ ¡ ¢ ´ where ~x¡1 = ~x¡1 1 ; ~x¡1 2 : Note that since ¡ ¢ ~x (a; b) = P = ~® (0) = ~x ®1 (0) ; ®2 (0) ; ¡ ¢ ~x (a; b) = P = ¯~ (0) = ~x ¯1 (0) ; ¯2 (0) and since ~x is 1-1, we have ¡ ¢ ¡ ¢ ®1 (0) ; ®2 (0) = ¯1 (0) ; ¯2 (0) = (a; b) : We compute by the chain rule d~® d®1 d®2 X~ = (0) = ~x (a; b) (0) + ~x (a; b) (0) ; dt 1 dt 2 dt d¯~ d¯1 d¯2 Y~ = (0) = ~x (a; b) (0) + ~x (a; b) (0) : dt 1 dt 2 dt Note we ¯nd a curve corresponding to X~ + Y:~ De¯ne ~γ :(¡±; ±) ! ~x (U) by ¡ ¢ ~γ (t) = ~x γ1 (t) ; γ2 (t) ; γ1 (t) = ®1 (t) + ¯1 (t) ¡ a; γ2 (t) = ®2 (t) + ¯2 (t) ¡ b ¡ ¢ where 0 < ± · " is chosen small enough so that ®1 (t) + ¯1 (t) ¡ a; ®2 (t) + ¯2 (t) ¡ b 2 U for t 2 (¡±; ±) : Note that ~γ (0) = ~x (a; b) : We compute by the chain rule d~γ dγ1 dγ2 (0) = ~x1 (a; b) (0) + ~x2 (a; b) (0) dt µdt dt¶ µ ¶ d®1 d¯1 d®2 d¯2 = ~x (a; b) (0) + (0) + ~x (a; b) (0) + (0) 1 dt dt 2 dt dt d®1 d®2 = ~x (a; b) (0) + ~x (a; b) (0) 1 dt 2 dt d¯1 d¯2 + ~x (a; b) (0) + ~x (a; b) (0) 1 dt 2 dt = X~ + Y:~ 2 This completes the proof of Lemma 1.16. d®1 1 d®2 2 Remark. Often the notation dt (0) = X and dt (0) = X is used, so that 1 2 X~ = X ~x1 + X ~x2: Proposition 1.18. The set of tangent vectors is the same as the tangent plane. Remark. Tangent vector is de¯ned by (4) above, whereas tangent plane is de¯ned by (2) above. Proof. Recall ~x : U! R3 is a simple surface. The tangent plane is the span of f~x1; ~x2g. De¯ne the curves ~γ1 :(¡"; ") ! ~x (U) and ~γ2 :(¡"; ") ! ~x (U) by ¡ 1 2 ¢ ~γ1 (t) = ~x γ1 (t) ; γ1 (t) ; 1 γ1 (t) = a + t; 2 γ1 (t) = b 1 2 dγ1 dγ1 (so that dt (0) = 1 and dt (0) = 0) and ¡ 1 2 ¢ ~γ2 (t) = ~x γ2 (t) ; γ2 (t) ; 1 γ2 (t) = a; 2 γ2 (t) = b + t 1 2 dγ2 dγ2 (so that dt (0) = 0 and dt (0) = 1). Then d~γ dγ1 dγ2 1 (0) = ~x 1 (0) + ~x 1 (0) = ~x ; dt 1 dt 2 dt 1 d~γ dγ1 dγ2 2 (0) = ~x 2 (0) + ~x 2 (0) = ~x : dt 1 dt 2 dt 2 Thus ~x1 and ~x2 are both tangent vectors. Since the set of tangent vectors is a vector space containing ~x1 and ~x2, we conclude that the span of f~x1; ~x2g, i.e., the tangent plane, is a subset of the set of tangent vectors. Conversely, given a tangent vector, we shall show that it is a linear combination of ~x1 and ~x2: Let Let X~ be a tangent vector at P: Then there exists a curve ~® : d~® ~ (¡"; ") ! ~x (U) such that ~® (0) = P and dt (0) = X: We have previously shown d®1 d®2 X~ = (0) ~x + (0) ~x : dt 1 dt 2 This completes the proof of Proposition 1.18. 3.