math237_a1 - SOLUTIONS FOR ASSIGNMENT 1 A1. Sketch typical...

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SOLUTIONS FOR ASSIGNMENT 1 A1. Sketch typical and any exceptional level curves, typical cross-sections and the surface z = f ( x, y ) in R 3 for the following functions f : R 2 R : a). f ( x, y ) = 4 x 2 y 2 ; d). f ( x, y ) = e 4 - x 2 - y 2 1 . a). The range of f is R ( f ) = R . Typical level curves are given by: 4 x 2 y 2 = C, C n = 0 which represent hyperbolas with asymptotes y = ± 2 x . The exceptional level curves correspond to C = 0 and are straight lines of equa- tions y = ± 2 x . The cross-sections are given by: z = 4 k 2 y 2 , z = 4 x 2 l 2 , k, l = const which represent parabolas. See fig 1 for sketches of level curves and cross- sections. The surface z = f ( x, y ) is a hyperbolic paraboloid (fig 2). y x C=0 C<0 C>0 C>0 C<0 C=0 l=2 l=1 k=2 k=1 k=0 l=0 x z y z Figure 1: Sketches of level curves (left) and cross-sections (centre and right). d). Typical level curves are given by: e 4 - x 2 - y 2 1 = C or 4 x 2 y 2 = ln( C + 1) , if C > 1 1
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x y z Figure 2: Sketch of the surface z = 4 x 2 y 2 . or x 2 + y 2 = 4 ln( C + 1) , if C e 4 1 . Thus the range of f is R ( f ) = ( 1 , e 4 1] . The exceptional level curve corresponds to C = e 4 1 and is the point (0 , 0) . The cross-sections are given by: z = e 4 - k 2 - y 2 1 , z = e 4 - x 2 - l 2 1 , k, l = const which represent parabolas with horizontal asymptote z = 1 . See fig 3 for sketches of level curves and cross-sections. The surface
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math237_a1 - SOLUTIONS FOR ASSIGNMENT 1 A1. Sketch typical...

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