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Saddle Points and Inflection Points ( ) ( ) − f k h . lim 2 f k 1 h→0 2 f (h) − f (0) Since f is continuous by hypothesis, it follows that this limit is indeed f (0). In the case when the function is a parabola, a calculation shows that, for fixed h,the area of ABP is constant even as the point a varies (see Figure 2). This geometric curiosity may well have been known to the ancient Greeks of course. What about the converse question: for which functions is Area(a, h) independent of a?Thetheorem we just proved provides an answer to that: only parabolas (and lines, in which case all the triangles are degenerate and the areas vanish). Figure 2. Corollary. Let f have a continuous non-zero second derivative on an open interval I . If, for all x in I , Area(x, h) depends only on h, then the graph of f is a parabola, and conversely. Acknowledgment. The author wishes to thank the referee for several suggested improvements. References 1. W. Rudin, Principles of Mathematical Analysis, 3rd ed., McGraw-Hill, 1976. ◦ Saddle Points and Inflection Points Felix´ Mart´ınez de la Rosa ([email protected]), Universidad de Cadiz,´ Cadiz,´ Spain. In the study of surfaces in multivariate calculus, we notice some similarities be- tween saddle points on surfaces and inflection points on curves. In this note, we make a direct connection between the two concepts. Throughout, we assume that f (x, y) has continuous second partial derivatives in an open set in the plane, and that (a, b) is a critical point in that set (that is, fx (a, b) = fy(a, b) = 0). 380 c THE MATHEMATICAL ASSOCIATION OF AMERICA Let A = fxx(a, b), B = fxy(a, b),andC = fyy(a, b). The standard test for extrema uses the discriminant D = AC − B2: f has a relative maximum at (a, b) if D > 0and A < 0, and a minimum at (a, b) if D > 0andA > 0. If D < 0, f is said to have a saddle point at (a, b). (If D = 0, the test is inconclusive.) In our analysis we find it useful to use the quadratic form F(x, y) = Ax2 + 2Bxy + Cy2. If (a, b) is a saddle point of f then F(x, y) is indefinite, this will be positive at some points and negative at others. Theorem. Let f be a function with continuous second partial derivatives in an open set U in the plane and let (a, b) be a saddle point in U. Then there exists a continu- ous function y = g(x) with g(a) = b for which the projection on the xz-plane of the intersection of the surface z = f (x, y) and the cylindrical surface y = g(x) has a inflection point at x = a. Proof. By hypothesis, there exist points (x1, y1) and (x2, y2) with x1, x2 = 0for y1 which F(x1, y1)>0andF(x2, y2)<0. Now let g1(x) = (x − a) + b and u1(x) = x1 f (x, g1(x)). Taking the second derivative, we find that 2 u (a) = fxx(a, b) + 2 fxy(a, b)g (a) + fyy(a, b)(g (a)) + fy(a, b)g (a) 1 1 1 1 = ( , ( )) = , y1 = 1 ( , )> . F 1 g1 a F 1 2 F x1 y1 0 x1 x1 Thus u (a) = 0andu(a)>0. 1 1 y2 Similarly, we let g2(x) = (x − a) + b and u2(x) = f (x, g2(x)),andthen x2 u (a) = 0andu(a)<0. 2 2 Now let g (x) if x < a g(x) = 1 . g2(x) if x ≥ a Then g is continuous at x = a and g(a) = b, and the theorem follows. As a example, we consider the function f (x, y) = y2 − x 2, which has a saddle point at (0, 0) as shown in Figure 1. We let (x1, y1) = (1, 2) and (x2, y2) = (2, 1),so F(x1, y1)>0andF(x2, y2)<0. As in the proof of the theorem, let the function < ( ) = 2x if x 0 . g x 1 ≥ 2 x if x 0 In Figure 2 we show the intersection between the surfaces z = f (x, y) and y = g(x). In Figure 3 we show this intersection from the point of view of the y-axis, and we can observe that the projection on the xz-plane has a point of inflection at x = 0 (Figure 4). The theorem gives a simple constructive method to obtain continuous functions, g(x), that allows us to connect the saddle points of functions of two variables with the inflection points of functions of one variable, showing that the behavior in both VOL. 38, NO. 5, NOVEMBER 2007 THE COLLEGE MATHEMATICS JOURNAL 381 z y x Figure 1. Figure 2. z x Figure 3. Figure 4. class of points is related. (The author has not found any references to this result in the literature.) A question about g(x). For every choice of g1 and g2 with second-order derivative ( ) = ( ) = ( ) = y1 ( ) = y2 continuous at a, verifying g1 a g2 a b, g1 a and g2 a , the func- x1 x2 tion g(x) obtained in the proof satisfies the theorem but is not differentiable at x = a ( ) = y1 = y2 = ( ) y1 = y2 1 ( , ) = because g1 a g2 a . We note that if ,then 2 F x1 y1 x1 x2 x1 x2 x1 1 ( , ) ( , ) ( , ) 2 F x2 y2 , and therefore F x1 y1 and F x2 y2 have the same sign. x2 However, we can take the function y = x − x 2, differentiable at x = 0. In Figure 5 we show the intersection between the surfaces z = y2 − x 2 and y = x − x 2. In Figure 6 we show this intersection from the point of view of the y-axis, and in Figure 7 we observe that the projection on the xz-plane has a point of inflection at x = 0. We conclude by asking whether there always exists such a function g(x) that is differentiable at x = a. We don’t know the answer, but we count on some readers to explore this question. Acknowledgment. The author would like to thank the editor and an anonymous referee whose suggestions im- proved the clarity of this paper. 382 c THE MATHEMATICAL ASSOCIATION OF AMERICA z x y x y Figure 5. Figure 6. z x Figure 7. ◦ Conic Sections from the Plane Point of View Sidney H. Kung ([email protected]), Cupertino, CA 95014 We give an analytic proof of the fact that the conic sections are obtained by cutting a cone at various angles. Our proof does not involve spheres or circles (see [1, 3, 4]), but primarily depends upon the cutting plane itself. Figure 1 shows a two-napped circular cone C which may be viewed as the result of rotating the line g (generator) about the fixed line l (z-axis) while maintaining the same angle (β) between g and l. We choose the intersection o of g and l as the origin. Let P(x, y, z) be a point on the surface of C. From Figure 1 (see also [2,p.751]),we a = c . see that r z It follows that the equation of C is x 2 + y2 = z2 tan2 β. (1) Suppose a cutting plane E that does not contain o makes an angle α with the z-axis. π − α. The angle between E and the xy-plane is 2 Let the equation of plane E be π z = tan − α y + h (h = 0). (2) 2 Substituting (2) into (1) gives the equation of the intersection of C and E: π 2 x 2 + y2 − tan2 β tan − α y + h = 0, 2 VOL. 38, NO. 5, NOVEMBER 2007 THE COLLEGE MATHEMATICS JOURNAL 383.
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