Advanced Calculus fi..

Chapter 2 Differential Calculus of Functions of Several Variables 93Figure 2.6 Mapping and approximating linear mapping for Example 2.(see Fig. 2.6). We study this linear mapping in more detail. For dy = 0 we havedu = 3d.y and dv = d.r. so that (du, dv) follows a line of slope {; for d.x = 0we have du = -2dy and dl] = 4dy. so that (du, dv) follows a line of slope -2.Similarly, the linear mapping can be studied along other lines. In particular, we verifythat the points of the squarecorrespond to the points of the shaded parallelogram in the (du, dv) diagram. Thearea of the square is 1, and the area of the parallelogram isl(3i + j) x (-2i + 4j)l = 1; = 14 sq. units.For the given nonlinear mapping 14 = .Y' - xy. v = .ry + y' the lines y = 1 and.Y = 2 (on which dy = 0 and d .~ = 0, respectively) correspond to parabolas through(2, 3) in the uv-plane, as shown. The square 2 5 A- 5 3. 1 5 y 5 2 (the same squareas above) corresponds to a "curved parallelogram" as in the figure. Thus we see in ageometric way how the linear mapping approximates the given mapping. We observethat the linear mapping takes the line dy = 0 or the line dx = 0 to a line tangent tothe curve obtained from the nonlinear mapping.Remark.We have seen that for the square 0 5 dx 5 1.0 5 dy 5 1, the corre-sponding region for the linear mapping is a parallelogram of area 14 = 1: -:I. Asimilar calculation applies to an arbitrary square, and we find that the area is alwaysmultiplied by 14. Since arbitrary figures can be approximated by unions of squares,one concludes by a limiting process that all areas are multiplied by 14 under thelinear mapping.