- Problem 6 First Fundamental Form Of A Surface S Given A Parameterized Surface S R U V The Differential Form Ds 1 (35.02 KiB) Viewed 37 times
Please solve it completely.
Problem 6 (First Fundamental Form of a surface S). Given a parameterized surface S: r(u, v), the differential form ds² = Edu² + 2F dudv+Gdy² E = T..T. F = T..Tv G = r₁ r₂ is called the First Fundamental Form of S. This form is "fundamental" because it permits us to calculate lengths, angles, and areas on S. Prove all of the following: (a) For a curve C: u= u(t), v=v(t), 1 = [a,b], on the surface S, the formula for the length / of the curve C is given by dr dr - [√ ² 4ª - [²√² ( 4 ) ² + ²× ( 4 ) ( 4 ) + ¤ (4) ²¹ dt dt. di dt (b) Prove that the angle y between two intersecting curves, C₁; a=g(t), v=h(t) and C₂: u=p(t), v=q(t), on a surface S parameterized by r(u, v) is obtained from cosy= where a=rug'(1) +r,h'(1) and b=rup' (t)+r₁4 (1) are tangent vectors of C₁ and C₂. (c) Show that the square of the length of the normal vector n(u, v) can be written as n² = EG-F² (d) Show that for polar coordinates (u, v)= (r. e) and (x(u, v), y(u, v)) = (ucosv, u sin v), we have E = 1, F = 0, and G=u, so that ds=du²+udv=dr² +rde. Use this differential form and the results from part (c) to calculate the area of a disk of radius a.