Let us consider a profile curve to be a rational polynomial of degree n, see Figure 10.7 R(t)=[r(t),2(t) By simple implicitization of R=R(t), we get fn(r,2)=0 where n is the maximum total degree of f. Also, 10.6 Next we eliminate r from equations 10.5-10.6 by rewriting equations 10.5-10.6 as follows fn(;,2)=a0(2)rn+a1(2)r2-1+…+an(2)=0 r2+(x2+y2)=0 The resultant of eliminating r from these two equations is 之)a 1(2)an(z) 0 an-2(2)an-1(2)an(z) 0 0 and the degree of D= f(a, y, 2)=0 is 2n. An example is a torus(degree 4 algebraic (b) Implicitization of a cylindrical ruled surface igure 10.8: Cylindrical ruled surfaceLet us consider a profile curve to be a rational polynomial of degree n, see Figure 10.7 R(t) = [r(t), z(t)] By simple implicitization of R = R(t), we get: fn(r, z) = 0 (10.5) where n is the maximum total degree of f. Also, r 2 = x 2 + y 2 (10.6) Next we eliminate r from equations 10.5 - 10.6 by rewriting equations 10.5 - 10.6 as follows: fn(r, z) = a0(z)r n + a1(z)r n−1 + · · · + an(z) = 0 ⇒ −r 2 + (x 2 + y 2 ) = 0 The resultant of eliminating r from these two equations is D =                a0(z) a1(z) · · · an−1(z) an(z) 0 0 a0(z) · · · an−2(z) an−1(z) an(z) −1 0 x 2 + y 2 · · · −1 0 x 2 + y 2 . . . −1 0 x 2 + y 2                = 0 and the degree of D ≡ f(x, y, z) = 0 is 2n. An example is a torus (degree 4 algebraic surface). (b) Implicitization of a cylindrical ruled surface x y nˆ R(t) tˆ a z Figure 10.8: Cylindrical ruled surface. 14