Calculus with Analytic Geometry by edwards & Penney soln ch13

# Calculus with Analytic Geometry

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-3 -2 -1 1 2 3 -2 -1 1 2 Section 13.2 C13S02.001: Because f ( x, y ) = 4 3 x 2 y is defined for all x and y , the domain of f is the entire two-dimensional plane. C13S02.002: Because x 2 + 2 y 2 0 for all x and y , the domain of f ( x, y ) = x 2 + 2 y 2 is the entire two-dimensional plane. C13S02.003: If either x or y is nonzero, then x 2 + y 2 > 0, and so f ( x, y ) is defined—but not if x = y = 0. Hence the domain of f consists of all points ( x, y ) in the plane other than the origin. C13S02.004: If x = y then the denominator in f ( x, y ) is nonzero, and thus f ( x, y ) is defined—but not if x = y . So the domain of f consists of all those points ( x, y ) in the plane for which y = x . C13S02.005: The real number z has a unique cube root z 1 / 3 regardless of the value of z . Hence the domain of f ( x, y ) = ( y x 2 ) 1 / 3 consists of all points in the xy -plane. C13S02.006: The real number z has a unique cube root z 1 / 3 regardless of the value of z . But 2 x is real if and only if x 0. Therefore the domain of f ( x, y ) = (2 x ) 1 / 2 + (3 y ) 1 / 3 consists of all those points ( x, y ) for which x 0. C13S02.007: Because arcsin z is a real number if and only if 1 z 1, the domain of the given function f ( x, y ) = sin 1 ( x 2 + y 2 ) consists of those points ( x, y ) in the xy -plane for which x 2 + y 2 1; that is, the set of all points on and within the unit circle. C13S02.008: Because arctan z is defined for every real number z , the only obstruction to the computation of f ( x ) = arctan( y/x ) is the possibility that x = 0. This obstruction is insurmountable, and therefore the domain of f consists of all those points ( x, y ) in the xy -plane for which x = 0; that is, all points other than those on the y -axis. C13S02.009: For every real number z , exp( z ) is defined and unique. Therefore the domain of the given function f ( x, y ) = exp( x 2 y 2 ) consists of all points ( x, y ) in the entire xy -plane. C13S02.010: Because ln z is a unique real number if and only if z > 0, the domain of f ( x, y ) = ln( x 2 y 2 1) consists of those points ( x, y ) for which x 2 y 2 1 > 0; that is, for which y 2 < x 2 1. This is the region bounded by the hyperbola with equation x 2 y 2 = 1, shown shaded in the following figure; the bounding hyperbola itself is not part of the domain of f . C13S02.0011: Because ln z is a unique real number if and only if z > 0, then domain of f ( x, y ) = ln( y x ) consists of those points ( x, y ) for which y > x . This is the region above the graph of the straight line with equation y = x (the line itself is not part of the domain of f ). 1

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C13S02.012: Because z is a unique real number if and only if z 0, the domain of the given function f ( x, y ) = 4 x 2 y 2 consists of those points ( x, y ) for which x 2 + y 2 4. That is, the domain consists of all those points ( x, y ) on and within the circle with center (0 , 0) and radius 2. C13S02.013: If x and y are real numbers, then so are xy , sin xy , and 1 + sin xy . Hence the only obstruction to computation of f ( x, y ) = 1 + sin xy xy is the possibility of division by zero. So the domain of f consists of those points ( x, y ) for which xy = 0; that is, all points in the xy -plane other than those on the coordinate axes.
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