Helix (or general helix), then we’ve got E, U =1 exactly where c is often a continuous, n = 0, and g = 0.- 1 n c 1 2 g(3.eight)Proof. Let the curve be a 1-type helix in E4 ; then, for a constant field U, we get 1 T, U = c that is a continuous, and differentiating this equation with respect to s, we have T ,U = 0 From the Frenet equations in EDFSK (two.two), we’ve four n N, U = 0 and it follows that four n N, U = 0 If n = 0, where four is really a continual, we can create N, U = 0 and differentiating (3.10) with respect to s, we obtain N , U = 0. Employing the Frenet equations in EDFSK and (3.9), we come C6 Ceramide site across (3.eight). Theorem three.six. Let be a curve together with the Frenet formulas in EDFSK from the Minkowski space E4 . 1 Hence, when the curve can be a 2-type helix, we have1 three 2 D, U 4 g N, U = 0 g(three.9)(three.10)(three.11)Proof. Let the curve be a 2-type helix. Then, for any constant field U, E, U = c1 is usually a continuous. Differentiating this equation with respect to s, we get E ,U = 0 and making use of the Frenet equations in EDFSK (2.2), we’ve (3.11). Theorem 3.7. Let be a curve using the Frenet formulas in EDFSK with the Minkowski space E4 . 1 Then, if the curve is really a 3-type helix, then we’ve the following equation: N, U = -1 where c2 is actually a constant, two = 0, and g = 0. g 2 three g c 1 two 4 g(3.12)Proof. Let the curve be a 3-type helix; hence, to get a constant field U, D, U = c2 (three.13)Symmetry 2021, 13,six ofis a continuous. Differentiating (3.13) with respect to s, we get D ,U = 0 Utilizing the Frenet equations in EDFSK (2.two), we are able to write- 2 two E, U = 0 gIf 2 = 0, exactly where 2 is actually a continual, we get g E, U = 0 and differentiating (3.14) with respect to s, we acquire E ,U =1 Utilizing the Frenet equations in EDFSK and (three.13), if g = 0, we uncover (3.12).(three.14)Theorem three.8. Let be a curve with Frenet formulas in EDFSK of your Minkowski space E4 . Then, 1 if the curve can be a Safranin Cancer 4-type helix, we have1 1 n T, U two g E, U =(three.15)Proof. Let the curve be a 4-type helix; then, for any constant field U, N, U = c3 can be a constant. By differentiating (three.16) with respect to s, we have N ,U = 0 Working with the Frenet equations in EDFSK (two.2), we locate (3.15). four. (k,m)-Type Slant Helices in E4 1 Within this section, we are going to define (k, m)-type slant helices in E4 , as in [7]. 1 Definition four.1. Let be a curve in E4 with EDFFK (or EDFSK) T, E, D, N is known as a 1 (k, m)-type slant helix if there exists a non-zero continuous vector field U E4 that satisfies 1 Uk , U = ck , Um , U = cm (ck , cm are constants) (or T, U = c1 , E, U = c2, D, U = c3 , N, U = c4 (c1 , c2 , c3 , c4 are constants)) for 1 k, m 4, k = m. The constant vector U is around the axis of . Theorem 4.1. When the curve is often a (1, two)-type slant helix in E4 , then we’ve 1 D, U = where c1 , c2 are constants. Proof. Let the curve be a (1, two)-type slant helix in E4 ; then, for a continual field U, we’ve got 1 T, U = c1 that is a continual, and E, U = c2 (four.two) (four.1) 1 1 c1 g 3 two g (three.16)1 1 1 c2 g g3 two n gSymmetry 2021, 13,7 ofwhich is really a continual. Differentiating (4.1) and (4.two) with respect to s, we’ve that T ,U = 0 and E ,U = 0 By utilizing the Frenet equations in EDFFK ((4.1) and (4.two)) the following equations is often obtained: two 1 c2 four n N, U = 0 (four.three) g1 – 1 1 c1 3 two D, U 4 g N, U = 0 g g(4.4)From (4.three), we’ve got that N, U = – Substituting (4.5) into (4.four), we uncover D, U = which completes the proof. Theorem 4.2. In the event the curve is a (1, three)-type slant helix in E4 , we have2 4 2 n 1 g g g2 g c2 four n(4.5)1 1 c1 g three 2 g1 1 1 c2 g g3 two n g(four.six)N, U =(4.7)Proof. Let the curve be a (1, three)-type sla.
