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Sec 14.2-sm.dvi copy

# Sec 14.2-sm.dvi copy - S E C T I O N 15.1 Functions of Two...

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S E C T I O N 15.1 Functions of Two or More Variables (ET Section 14.1) 625 x 2 + y 2 2 | xy | x 2 + y 2 2 | xy | Therefore, (1) gives | c | = xy x 2 + y 2 + 1 = | xy | x 2 + y 2 + 1 x 2 + y 2 2 x 2 + y 2 + 1 = 1 2 x 2 + y 2 x 2 + y 2 + 1 < 1 2 · 1 = 1 2 That is, | c | < 1 2 . It follows that for c 1 2 the level curve is empty. x y z We now show that if | c | < 1 2 the level curve is a hyperbola. x y z Let t = y and s = x - y 2 c . Then for | c | < 1 2 the equation of the level curve (2) can be rewritten as 1 4 c 2 - 1 t 2 - s 2 = 1 t 2 2 c 1 - 4 c 2 2 - s 2 = 1 This is the equation of a hyperbola in standard position in the ts -plane. The equation (2) for | c | < 1 2 is the equation of a hyperbola obtained by a change of variables for a hyperbola in standard position, in the xy -plane. For c = 0 . 5 the level curve is empty: x y z

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626 C H A P T E R 15 DIFFERENTIATION IN SEVERAL VARIABLES (ET CHAPTER 14) 15.2 Limits and Continuity in Several Variables (ET Section 14.2) Preliminary Questions 1. What is the difference between D ( P , r ) and D * ( P , r ) ? SOLUTION D ( P , r ) is the open disk of radius r and center ( a , b ) . It consists of all points distanced less than r from P , hence D ( P , r ) includes the point P . D * ( P , r ) consists of all points in D ( P , r ) other than P itself. 2. Suppose that f ( x , y ) is continuous at ( 2 , 3 ) and that f ( 2 , y ) = y 3 for y = 3. What is the value f ( 2 , 3 ) ? SOLUTION f ( x , y ) is continuous at ( 2 , 3 ) , hence the following holds: f ( 2 , 3 ) = lim ( x , y ) ( 2 , 3 ) f ( x , y ) Since the limit exists, we may compute it by approaching ( 2 , 3 ) along the vertical line x = 2. This gives f ( 2 , 3 ) = lim ( x , y ) ( 2 , 3 ) f ( x , y ) = lim y 3 f ( 2 , y ) = lim y 3 y 3 = 3 3 = 27 We conclude that f ( 2 , 3 ) = 27. 3. Suppose that Q ( x , y ) is a function such that 1 Q ( x , y ) is continuous for all ( x , y ) . Which of the following statements are true? (a) Q ( x , y ) is continuous at all ( x , y ) . (b) Q ( x , y ) is continuous for ( x , y ) = ( 0 , 0 ) . (c) Q ( x , y ) = 0 for all ( x , y ) . SOLUTION All three statements are true. Let f ( x , y ) = 1 Q ( x , y ) . Hence Q ( x , y ) = 1 f ( x , y ) . (a) Since f is continuous, Q is continuous whenever f ( x , y ) = 0. But by the definition of f it is never zero, therefore Q is continuous at all ( x , y ) . (b) Q is continuous everywhere including at ( 0 , 0 ) . (c) Since f ( x , y ) = 1 Q ( x , y ) is continuous, the denominator is never zero, that is, Q ( x , y ) = 0 for all ( x , y ) . Moreover, there are no points where Q ( x , y ) = 0. (The equality Q ( x , y ) = ( 0 , 0 ) is meaningless since the range of Q consists of real numbers.) 4. Suppose that f ( x , 0 ) = 3 for all x = 0 and f ( 0 , y ) = 5 for all y = 0. What can you conclude about lim ( x , y ) ( 0 , 0 ) f ( x , y ) ? SOLUTION We show that the limit lim ( x , y ) ( 0 , 0 ) f ( x , y ) does not exist. Indeed, if the limit exists, it may be computed by approaching ( 0 , 0 ) along the x -axis or along the y -axis. We compute these two limits: lim ( x , y ) ( 0 , 0 ) along y = 0 f ( x , y ) = lim x 0 f ( x , 0 ) = lim x 0 3 = 3 lim ( x , y ) ( 0 , 0 ) along x = 0 f ( x , y ) = lim y 0 f ( 0 , y ) = lim y 0 5 = 5 Since the limits are different, f ( x , y ) does not approach one limit as ( x , y ) ( 0 , 0 ) , hence the limit lim ( x , y ) ( 0 , 0 ) f ( x , y ) does not exist.
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