tailieunhanh - Stewart - Calculus - Early Transcendentals 6e HQ (Thomson, 2008) Episode 12

Tham khảo tài liệu 'stewart - calculus - early transcendentals 6e hq (thomson, 2008) episode 12', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 1074 CHAPTER 16 VECTOR CALCULUS FIGURE 10 Figure 9. If x y z is a point on 5 then 3 x x y f x cos 0 z f x sin 0 Therefore we take x and 0 as parameters and regard Equations 3 as parametric equations of 5. The parameter domain is given by a x b 0 2ir. EXAMPLE 8 Find parametric equations for the surface generated by rotating the curve y sin x 0 x 2ir about the x-axis. Use these equations to graph the surface of revolution. SOLUTION From Equations 3 the parametric equations are x x y sin x cos 0 z sin x sin 0 and the parameter domain is 0 x 2tt 0 0 2tt. Using a computer to plot these equations and rotate the image we obtain the graph in Figure 10. We can adapt Equations 3 to represent a surface obtained through revolution about the y- or z-axis. See Exercise 30. TANGENT PLANES We now find the tangent plane to a parametric surface 5 traced out by a vector function r u v x u v i y u v j z u v k at a point Po with position vector r uo vo . If we keep u constant by putting u Uo then r uo v becomes a vector function of the single parameter v and defines a grid curve C1 lying on 5. See Figure 11. The tangent vector to C1 at Po is obtained by taking the partial derivative of r with respect to v bx dy dz 4 rv Uo vo i - v Uo vo j u vo k Similarly if we keep v constant by putting v vo r u vo that lies on 5 and its tangent vector at Pq is we get a grid curve C2 given by dx . dy 2 dz ru du uo vo i aU Uo vo j ău Uo vq k 5 SECTION PARAMETRIC SURFACES AND THEIR AREAS 1075 If rM X rv is not 0 then the surface S is called smooth it has no corners . For a smooth surface the tangent plane is the plane that contains the tangent vectors rM and rv and the vector rM X rv is a normal vector to the tangent plane. V EXAMPLE 9 Find the tangent plane to the surface with parametric equations x u2 y v2 z u 2v at the point 1 1 3 . Figure 12 shows the self-intersecting surface in Example 9 and its tangent plane at 1 1 3 . 1 1 3 SOLUTION We first compute the tangent vectors dx . r 7- i du dx. rv

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