Vector Functions 10. Parametric Surfaces 10.5 3 Parametric Surfaces We have looked at surfaces that are graphs of functions of two variables. Here we.

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Presentation transcript:

Vector Functions 10

Parametric Surfaces 10.5

3 Parametric Surfaces We have looked at surfaces that are graphs of functions of two variables. Here we use vector functions to discuss more general surfaces, called parametric surfaces. In much the same way that we describe a space curve by a vector function r(t) of a single parameter t, we can describe a surface by a vector function r(u, v) of two parameters u and v.

4 Parametric Surfaces We suppose that r(u, v) = x(u, v) i + y(u, v) j + z(u, v) k is a vector-valued function defined on a region D in the uv-plane. So x, y, and z, the component functions of r, are functions of the two variables u and v with domain D.

5 Parametric Surfaces The set of all points (x, y, z) in such that x = x(u, v)y = y(u, v)z = z(u, v) and (u, v) varies throughout D, is called a parametric surface S and Equations 2 are called parametric equations of S. Each choice of u and v gives a point on S ; by making all choices, we get all of S.

6 Parametric Surfaces In other words, the surface S is traced out by the tip of the position vector r(u, v) as (u, v) moves throughout the region D. (See Figure 1.) A parametric surface Figure 1

7 Example 1 Identify and sketch the surface with vector equation r(u, v) = 2 cos u i + v j + 2 sin u k Solution: The parametric equations for this surface are x = 2 cos uy = vz = 2 sin u So for any point (x, y, z) on the surface, we have x 2 + z 2 = 4 cos 2 u + 4 sin 2 u = 4 This means that vertical cross-sections parallel to the xz-plane (that is, with y constant) are all circles with radius 2.

8 Example 1 – Solution Since y = v and no restriction is placed on v, the surface is a circular cylinder with radius 2 whose axis is the y-axis (see Figure 2). cont’d Figure 2

9 Parametric Surfaces In Example 1 we placed no restrictions on the parameters u and v and so we got the entire cylinder. If, for instance, we restrict u and v by writing the parameter domain as 0  u   /2 0  v  3 then x  0, z  0, 0  y  3, and we get the quarter-cylinder with length 3 illustrated in Figure 3. Figure 3

10 Parametric Surfaces If a parametric surface S is given by a vector function r(u, v), then there are two useful families of curves that lie on S, one family with u constant and the other with v constant. These families correspond to vertical and horizontal lines in the uv-plane.

11 Parametric Surfaces If we keep u constant by putting u = u 0, then r(u 0, v) becomes a vector function of the single parameter v and defines a curve C 1 lying on S. (See Figure 4.) Figure 4

12 Parametric Surfaces Similarly, if we keep v constant by putting v = v 0, we get a curve C 2 given by r(u, v 0 ) that lies on S. We call these curves grid curves. (In Example 1, for instance, the grid curves obtained by letting u be constant are horizontal lines whereas the grid curves with v constant are circles.)

13 Surfaces of Revolution

14 Surfaces of Revolution Surfaces of revolution can be represented parametrically and thus graphed using a computer. For instance, let’s consider the surface S obtained by rotating the curve y = f (x), a  x  b, about the x-axis, where f (x)  0. Let  be the angle of rotation as shown in Figure 10. Figure 10

15 Surfaces of Revolution If (x, y, z) is a point on S, then x = x y = f (x) cos  z = f (x) sin  Therefore we take x and  as parameters and regard Equations 3 as parametric equations of S. The parameter domain is given by a  x  b, 0    2 .

16 Example 8 – Graphing a Surface of Revolution Find parametric equations for the surface generated by rotating the curve y = sin x, 0  x  2 , 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  z = sin x sin  and the parameter domain is 0  x  2 , 0    2 .

17 Example 8 – Solution Using a computer to plot these equations and rotate the image, we obtain the graph in Figure 11. cont’d Figure 11