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Unformatted text preview: 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 | x y | x 2 + y 2 2 | x y | Therefore, (1) gives | c | = x y x 2 + y 2 + 1 = | x y | 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 x y-plane. For c = . 5 the level curve is empty: x y z 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 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 ) y= ( , ) . (c) Q ( x , y ) 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 ) 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 ( , ) . (c) Since f ( x , y ) = 1 Q ( x , y ) is continuous, the denominator is never zero, that is, Q ( x , y ) y= 0 for all ( x , y ) . Moreover, there are no points where Q ( x , y ) = 0. (The equality Q ( x , y ) = ( , ) is meaningless since the range of Q consists of real numbers.) 4. Suppose that f ( x , ) = 3 for all x y= and f ( , y ) = 5 for all y y= 0. What can you conclude about lim ( x , y ) ( , ) f ( x , y ) ?...
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This note was uploaded on 07/15/2009 for the course MATH 210 taught by Professor Hubscher during the Spring '08 term at University of Illinois at Urbana–Champaign.
- Spring '08