ASIMPTOTIC LINES OF ONE-SHEETED GIPERBOLOID

18 Danish Scientific Journal No100, 2025 MATHEMATICAL SCIENCES ASIMPTOTIC LINES OF ONE-SHEETED GIPERBOLOID Abdumajidova Sh. https://doi.org/10.5281/zenodo.17249782 Introduction Let us consider a surface 𝛷 which is given by equation around point 𝑃 𝑟=𝑟(𝑢,𝑣). If we intersect with a plane 𝛱 passing through a point 𝑃 on it, we obtain a smooth curve 𝛾 passing through point 𝑃 in the intersection, which we call such a curve a plane section. A plane section 𝛾 lies in the plane 𝛱 ,so its torsion is necessarily zero. Writing the equation of the plane section in terms of the natural parameter 𝑠 (i.e., arc length), and using the Frenet formulas for it (taking into account that the torsion is equal to zero), we write: {𝜏=𝑘𝑣 𝑣=−𝑘𝜏 Here, 𝜏 is the unit tangent vector, 𝑣 is the unit principal normal vector, and 𝑘 is the curvature of the curve 𝛾 at point 𝑃(𝑢0,𝑣0).. Then for second quadratic form we have 𝐼𝐼(𝜏,𝜏)=(𝜏,𝑛 󰇍  )=(𝑘𝑣,𝑛)=𝑘cos𝜃 Here 𝜃 is the angle between the vectors 𝑛 󰇍  and 𝑣. Now, if we define 𝛾 by equation 𝜌=𝜌(𝑡) (where t is an arbitrary parameter), then since 𝑡 is a function of 𝑠, and considering the following equalities, 𝑑𝜌 𝑑𝑡=𝜌′=𝜌𝑑𝑠 𝑑𝑡,𝜌″=𝜌″(𝑑𝑠 𝑑𝑡)2+𝜌′𝑑2𝑠 𝑑𝑡2 we have: 𝐼𝐼(𝜌′,𝜌′)=(𝜌″,𝑛 󰇍  )=(𝑑𝑠 𝑑𝑡)2(𝜌..,𝑛 󰇍  )=(𝑑𝑠 𝑑𝑡)2𝑘cos𝜃 We obtain following equality 𝑘cos𝜃=𝐼𝐼(𝜌′,𝜌′) (𝑑𝑠 𝑑𝑡)2=𝐼𝐼(𝜌′,𝜌′) 𝐼(𝜌′,𝜌′) It can be seen that the right side of this equality depends only on the vector 𝜌′. If we take another plane section 𝛾′ other than 𝛾 and they have a common tangent (i.e., they have the same direction), then the right side of equality (1) is the same for them. Now let the the plane section be parallel to the normal vector. Therefore, equality (1) becomes: 𝑘=±𝐼𝐼(𝜌′,𝜌′) 𝐼(𝜌′,𝜌′) Definition 1. The number 𝐼𝐼(𝜌 󰇍 󰇍  ′,𝜌 󰇍 󰇍  ′) 𝐼(𝜌 󰇍 󰇍  ′,𝜌 󰇍 󰇍  ′) obtained here is called the normal curvature of the surface 𝛷 at point 𝑃 in the direction 𝑎=ϱ′ and is denoted by 𝑘𝑎(𝑎) . Thus, the absolute value of the normal curvature of the surface in the direction 𝑎 is equal to the curvature of the normal section that defines the normal vector, possibly differing in sign. Definition 2. If 𝑘𝑎(𝑎)=0 in some direction 𝑎, then such a direction is called an asymptotic direction. For a given vector 𝑎=(𝑥,𝑦), it is necessary and sufficient that 𝐿𝑥2+2𝑀𝑥𝑦+𝑁𝑦=0 for the direction defining the asymptotic direction. Here, L, M, N are the coefficients of the second quadratic form. Definition 3. If a curve 𝛾 on a surface is given by the equation 𝑢=𝑢(𝑡),𝑣=𝑣(𝑡), and its tangent vector at any point defines the asymptotic direction, then such a curve is called an asymptotic curve. Naturally, if a straight line lies on a surface, it is an asymptotic line. We find the asymptotic lines of a hyperbolic paraboloid. The hyperbolic paraboloid is a surface of the second order and is given by the following second-order equation: 𝑧=𝑥2−𝑦2 We write the parametric equations of the hyperbolic paraboloid: 𝑥=𝑢,𝑦=𝑣,𝑧=𝑢2−𝑣2 To calculate the first and second quadratic forms , we need to know the vectors denoted by 𝑟𝑢 󰇍 󰇍 󰇍  ={1,0,2𝑢} 𝑟𝑣 󰇍 󰇍 󰇍  ={1,0,−2𝑣} 𝑟𝑢𝑢 󰇍 󰇍 󰇍 󰇍 󰇍  ={0,0,2} 𝑟𝑢𝑣 󰇍 󰇍 󰇍 󰇍 󰇍  ={0,0,0} 𝑟𝑣𝑣 󰇍 󰇍 󰇍 󰇍 󰇍  ={0,0,−2} The coeffisients of the first and second quadratic forms are 𝐸=1+4𝑢2,𝐹=−4𝑢𝑣,𝐺=1+4𝑣2, 𝐿= 2 √1+4𝑢2+4𝑣2,𝑀=0,𝑁= −2 √1+4𝑢2+4𝑣2 We construct the differential equation of asymptotic lines: 𝑑𝑢2−𝑑𝑣2=0 Its solutions are: 𝑢1=𝑡+𝑐1, 𝑢2=−𝑡+𝑐 Thus, the equations of the asymptotic lines of the hyperbolic paraboloid in space are: 𝛾1:{𝑥=𝑡+𝑐1 𝑦=𝑡 𝑧=2𝑐1𝑡+𝑐12 𝛾2:{𝑥=𝑡+𝑐2 𝑦=𝑡 𝑧=2𝑐2𝑡+𝑐2 2 Asymptotic lines of a one-sheet hyperboloid A one-sheet hyperboloid is a quadratic surface of second order, given by the following equation: 𝑥2 𝑎2+𝑦2 𝑎2−𝑧2 𝑎2=0 Parametric equations of the one-sheet hyperboloid 𝑥=𝑐𝑜𝑠𝑢𝑐ℎ𝑣,𝑦=𝑏𝑠𝑖𝑛𝑢𝑐ℎ𝑣,𝑧=𝑐𝑠ℎ𝑣, Or, in vector form: 𝑟={𝑎𝑐𝑜𝑠𝑢𝑐ℎ𝑣,𝑎𝑠𝑖𝑛𝑢𝑐ℎ𝑣,𝑎𝑠ℎ𝑣} To compute the first and second quadratic forms, we calculate the partial derivatives: 𝑟𝑢={−𝑎𝑠𝑖𝑛𝑢𝑐ℎ𝑣,𝑎𝑐𝑜𝑠𝑢𝑐ℎ𝑣,0}, 𝑟𝑢𝑢 ={−𝑎𝑐𝑜𝑠𝑢𝑐ℎ𝑣,−𝑎𝑠𝑖𝑛𝑢𝑐ℎ𝑣,0}, 𝑟𝑣 ={𝑎𝑐𝑜𝑠𝑢𝑠ℎ𝑣,𝑎𝑠𝑖𝑛𝑢𝑠ℎ𝑣,𝑎𝑐ℎ𝑣}, 𝑟𝑣𝑣 ={𝑎𝑐𝑜𝑠𝑢𝑐ℎ𝑣,𝑎𝑠𝑖𝑛𝑢𝑐ℎ𝑣,𝑎𝑠ℎ𝑣},𝑟𝑢𝑣 ={−𝑎𝑠𝑖𝑛𝑢𝑠ℎ𝑣,𝑎𝑐𝑜𝑠𝑢𝑠ℎ𝑣,0} The calculation of the first quadratic forms yields 𝐸=𝑟𝑢2 󰇍 󰇍 󰇍 󰇍  =𝑥𝑢 2+𝑦𝑢 2+𝑧𝑢 2𝐹=𝑟𝑢𝑣 󰇍 󰇍 󰇍 󰇍 󰇍  =𝑥𝑢𝑥𝑣+𝑦𝑢𝑦𝑣+𝑧𝑢𝑧𝑣𝐺=𝑟𝑣2 󰇍 󰇍 󰇍 󰇍  =𝑥𝑣 2+𝑦𝑣2+𝑧𝑣 2 Danish Scientific Journal No100, 2025 19 𝐸=𝑎2𝑠𝑖𝑛2𝑢𝑐ℎ2𝑣+𝑎2𝑐𝑜𝑠2𝑢𝑐ℎ2𝑣=𝑎2𝑐ℎ2𝑣,𝐹 =−𝑎2𝑠𝑖𝑛𝑢𝑐𝑜𝑠𝑢𝑐ℎ𝑣𝑠ℎ𝑣 +𝑎2𝑠𝑖𝑛𝑢𝑐𝑜𝑠𝑢𝑐ℎ𝑣𝑠ℎ𝑣=0 , 𝐺=𝑎2𝑐𝑜𝑠2𝑢𝑠ℎ2𝑣+𝑎2𝑠𝑖𝑛2𝑢𝑠ℎ2𝑣+ 𝑎2𝑐ℎ2𝑣=𝑎2𝑠ℎ2𝑣+𝑎2𝑐ℎ2𝑣=𝑎2𝑐ℎ2𝑣 The calculation of the second quadratic forms yields 𝐿= ∣ ∣ ∣ ∣ 𝑥𝑢𝑢 𝑦𝑢𝑢 𝑧𝑢𝑢 𝑥𝑢𝑦𝑢𝑧𝑢 𝑥𝑣𝑦𝑣𝑧𝑣 ∣ ∣ ∣ ∣ √𝐸𝐺−𝐹2,𝑀= ∣ ∣ ∣ ∣ 𝑥𝑢𝑣 𝑦𝑢𝑣 𝑧𝑢𝑣 𝑥𝑢𝑦𝑢𝑧𝑢 𝑥𝑣𝑦𝑣𝑧𝑣 ∣ ∣ ∣ ∣ √𝐸𝐺−𝐹2,𝑁 = ∣ ∣ ∣ ∣ 𝑥𝑣𝑣 𝑦𝑣𝑣 𝑧𝑣𝑣 𝑥𝑢𝑦𝑢𝑧𝑢 𝑥𝑣𝑦𝑣𝑧𝑣 ∣ ∣ ∣ ∣ √𝐸𝐺−𝐹2 𝐿=−𝑎3𝑐ℎ3𝑣, 𝑀=0, 𝑁=2𝑎3𝑐ℎ𝑣 The differential equation of the asymptotic lines is derived as follows 𝐿𝑑𝑢2+2𝑀𝑑𝑢𝑑𝑣+𝑁𝑑𝑣2=0 −𝑎3𝑐ℎ3𝑣𝑑𝑢2+2𝑎2𝑐ℎ𝑣𝑑𝑣2=0 𝑐ℎ2𝑣𝑑𝑢2=2𝑑𝑣2 {𝑑𝑢 𝑑𝑣}2=2 𝑐ℎ2𝑣 ∫𝑑𝑢=∫ √2 𝑐ℎ𝑣𝑑𝑣=∫ 2√2 𝑒𝑣+𝑒−𝑣𝑑𝑣=2√2∫ 𝑒𝑣𝑑𝑣 𝑒2𝑣+1 =𝑑𝑒𝑣 (𝑒𝑣)2+1 The intrinsic coordinate equations of the asymptotic lines can be expressed in the form 𝑢1=2√2𝑎𝑟𝑐𝑡𝑔𝑒𝑣 𝑢2=−2√2𝑎𝑟𝑐𝑡𝑔𝑒𝑣 The spatial equations of the asymptotic lines of the one-sheet hyperboloid can be written in the form 𝛾1:𝑥=𝑎𝑐𝑜𝑠(2√2𝑎𝑟𝑐𝑡𝑔𝑒𝑡)𝑐ℎ𝑡,𝑦 =𝑎𝑠𝑖𝑛(2√2𝑎𝑟𝑐𝑡𝑔𝑒𝑡)𝑐ℎ𝑡,𝑧=𝑎𝑠ℎ𝑡 𝛾2:𝑥=𝑎𝑐𝑜𝑠(2√2𝑎𝑟𝑐𝑡𝑔𝑒𝑡)𝑐ℎ𝑡,𝑦 =−𝑎𝑠𝑖𝑛(2√2𝑎𝑟𝑐𝑡𝑔𝑒𝑡)𝑐ℎ𝑡,𝑧 =𝑎𝑠ℎ𝑡 References: 1. A. Narmanov. Differensial geometriya va topologiya. (1), (2018). 2. M.A.Sobirov, A.Y. Yusupov. Differensial geometriya kurs., (2) (1959), 158 3. A.Ya.Narmanov, A.S.Sharipov, J.O.Arslonov. Differensial geometriya va topologiya kursidan masalalar to’plami. (2014)