(10/7/08)Math 10C. Lecture Examples.Sections 12.1, 12.2, and 12.3. Functions of two variables†Example 1 (a) What is the domain of f(x, y) = x2+ y2? (b) What are the valuesf(2, 3) and f(−2, −3) of this function at (2,3) and (−2, −3)? (c) What isits range?Answer: (a) The domain of f is the entire xy-plane. (b) f (2, 3) = 13 • f (−2, −3) = 13. (c) The range off is the closed infinite interval [0, ∞).Example 2 Determine the shape of the surface z = x2+ y2in xyz-space by studyingits cross sections in the planes x = c perpendicular to the x-axis.Answer: The intersection of the surface z = x2+ y2with the plane x = c is a parabola that opens upward andwhose vertex is at the origin if c = 0 and is c2units a bove the xy-plane if c 6= 0 • Figure A2a • The surfacehas the bowl-like shape in Figure A2byxzx = cc2Figure A2a Figure A2b†Lecture notes to accompany Sections 12.1, 12.2, and 12.3 of Calculus by Hughes-Hallett et al.1Math 10C. Lecture Examples. (10/7/08) Sections 12.1, 12.2, and 12.3, p. 2Example 3 Determine the shape of the surface z = x2+ y2of Example 2 by studyingits cross sections in the planes y = c perpendicular to the y-axis.Answer: The intersection of the surface z = x2+ y2with the plane y = c is parabola that open s upward andwhose vertex is at the origin if c = 0 and is c2units a bove the xy-plane if c 6= 0. • Figure A3a • The surfacehas the bowl-like shape from Example 2. (Figure A3b shows the cross sections from Examples 2 and 3 tog ether.)yxzy = cc2yxzFigure A3a Figure A3bExample 4 Determine the shape of the surface z = y2− x2by studying its cross sectionsin the planes x = c perpendicular to the x-axis.Answer: The intersection of the surface z = y2− x2with the plane x = c is a parabola that o pens upward andwhose vertex is c2units below the xz-plane. • Figure A4a • The vertex is at the origin for c = 0 and dropsbelow the xy-plane as c moves away from zero. • The surface has the saddle shape in Figure A4b.yzc2Figure A4a Figure A4bSections 12.1, 12.2, and 12.3, p. 3 Math 10C. Lecture Examples. (10/7/08)Example 5 Determine the shape of the surface z = y2− x2of Example 4 by studyingits cross sections in the planes y = c perpendicular to the y-axis.Answer: The intersection of the surface z = y2− x2with the plane y = c is a parabola that opens downwardand whose vertex is c2units above the xy-plane. • Figure A5a • The vertex is at the origin for c = 0 and risesabove th e xy-plane as c moves away from zero. • The surface has the saddle shape from Example 4. (Figure A5bshows the two sets of cross sections together.)yxzy = cyxzFigure A5a Figure A5bExample 6 Use the curve z = y −112y3in the yz-plane of Figure 1 to determine theshape of the surface z = y −112y3−14x2.y2z12−2−13z = y −112y3−2FIGURE 1Answer: One solution: The cross section of the surface in the plane x = c has the shape of the curve in Figure 1 ifc = 0, is that curve moved down and forward if c > 0 and is that curve moved down and back if c < 0. •The surface has the boot-like s hape in Figure A6Another solution: The cross section in the plane y = c is a parabola that opens downward and has its vertex on thecurve in Figure 1. • The surface has the boot-like shape in Figure A6.Math 10C. Lecture Examples. (10/7/08) Sections 12.1, 12.2, and 12.3, p. 4Figure A6Example 7 Describe the level curves of the function f(x, y) = x2+ y2from Examples 2and 3.Answer: Figur e A7a shows horizontal cross sections of the graph of f and Figure A7b shows the correspondinglevel curves. • The level curve f = c is the circle of radius√c with its center at the origin if c > 0, is the originif x = 0, and is empty if c < 0. (The surface is called a “circular paraboloid.”)Figure A7a Figure A7bExample 8 Describe the level curves of g(x, y) = y2− x2from Examples 4 and 5.Answer: Figur es A8a and A8b • The level cu rve g = c is a hyperbola with the equation y2− x2= c. (Thesurface is a “hyperbolic paraboloid.”)Sections 12.1, 12.2, and 12.3, p. 5 Math 10C. Lecture Examples. (10/7/08)Level curves of g(x, y) = y2− x2Figure A8a Figure A8bExample 9 Figures 2 and 3 show horizontal cross sections o f the graph ofh(x, y) = y −112y3−14x2from Example 6 and the corresponding levelcurves. Describe how the surface can be reconstructed from the level curves.x2 6y26h = 0h = 1h = 5h = −5h = 1h = 0FIGURE 2 FIGURE 3Answer: Leave the two parts of the level curve h = 0 on the xyplane. • Raise the two parts of the curve labeledh = 1 one unit on the “toe” and “leg” of the “boot.” • Lower the curves above the upper part of h = 1 to form thesides of the “boot.” • Raise the curves below the lower part of h = 1 to form the more of the “leg” of the “boot.”Interactive ExamplesWork the following Interactive Examples on Shenk ’s web page, http//www.math.ucsd.edu/˜ashenk/:‡Section 14.1: Examples 1–6‡The chapter and section numbers on Shenk’s web site refer to his calculus manuscript and not to the chapters and sectionsof the textbook for t he
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