Sec 14.3-sm.dvi copy

# Sec 14.3-sm.dvi copy - 636 C H A P T E R 15 D I F F E R E N...

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636 C H A P T E R 15 DIFFERENTIATION IN SEVERAL VARIABLES (ET CHAPTER 14) - 2 y x z 0 0 0 2 1 2 1 2 - 1 - 1 - 2 - 2 The graph supports our result in part (a). For g ( x , y ) = xy x 2 + y 2 : (a) g ( x , y ) = xy x 2 + y 2 is defined for ( x , y ) = ( 0 , 0 ) . In Exercise 37 we showed that lim ( x , y ) ( 0 , 0 ) x a y b x 2 + y 2 does not exist if a + b 2. In particular the limit does not exist if a = b = 1, that is, lim ( x , y ) ( 0 , 0 ) g ( x , y ) does not exist. It follows that the domain of g ( x , y ) cannot be extended to all of R 2 to result in a continuous function. (b) The graph of g ( x , y ) confirms our observation: y x z 0.5 0.25 0 0 0 1 2 - 1 - 2 1 2 - 1 - 2 - 0.25 - 0.5 43. The function f ( x , y ) = x 2 y x 4 + y 2 provides an interesting example where the limit as ( x , y ) ( 0 , 0 ) does not exist, even though the limit along every line y = mx exists and is zero (Figure 9). (a) Show that the limit along any line y = mx exists and is equal to 0. (b) Calculate f ( x , y ) at the points ( 10 - 1 , 10 - 2 ) , ( 10 - 5 , 10 - 10 ) , ( 10 - 20 , 10 - 40 ) . Do not use a calculator. (c) Show that lim ( x , y ) ( 0 , 0 ) f ( x , y ) does not exist. Hint: Compute the limit along the parabola y = x 2 . SOLUTION (a) Substituting y = mx in f ( x , y ) = x 2 y x 4 + y 2 , we get f ( x , mx ) = x 2 · mx x 4 + ( mx ) 2 = mx 3 x 2 ( x 2 + m 2 ) = mx x 2 + m 2 We compute the limit as x 0 by substitution: lim x 0 f ( x , mx ) = lim x 0 mx x 2 + m 2 = m · 0 0 2 + m 2 = 0 (b) We compute f ( x , y ) at the given points: f ( 10 - 1 , 10 - 2 ) = 10 - 2 · 10 - 2 10 - 4 + 10 - 4 = 10 - 4 2 · 10 - 4 = 1 2 f ( 10 - 5 , 10 - 10 ) = 10 - 10 · 10 - 10 10 - 20 + 10 - 20 = 10 - 20 2 · 10 - 20 = 1 2 f ( 10 - 20 , 10 - 40 ) = 10 - 40 · 10 - 40 10 - 80 + 10 - 80 = 10 - 80 2 · 10 - 80 = 1 2

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S E C T I O N 15.3 Partial Derivatives (ET Section 14.3) 637 (c) We compute the limit as ( x , y ) approaches the origin along the parabola y = x 2 (by part (b), the limit appears to be 1 2 ). We substitute y = x 2 in the function and compute the limit as x 0. This gives lim ( x , y ) 0 along y = x 2 f ( x , y ) = lim x 0 f ( x , x 2 ) = lim x 0 x 2 · x 2 x 4 + ( x 2 ) 2 = lim x 0 x 4 2 x 4 = lim x 0 1 2 = 1 2 However, in part (a), we showed that the limit along the lines y = mx is zero. Therefore f ( x , y ) does not approach one limit as ( x , y ) ( 0 , 0 ) , so the limit lim ( x , y ) ( 0 , 0 ) f ( x , y ) does not exist. 44. Is the following function continuous? f ( x , y ) = x 2 + y 2 if x 2 + y 2 < 1 1 if x 2 + y 2 1 x z y FIGURE 9 Graph of f ( x , y ) = x 2 y x 4 + y 2 . SOLUTION f ( x , y ) is defined by a polynomial in the domain x 2 + y 2 < 1, hence f is continuous in this domain. In the domain x 2 + y 2 > 1, f is a constant function, hence f is continuous in this domain also. Thus, we must examine continuity at the points on the circle x 2 + y 2 = 1. x 0 y 1 y 2 + x 2 We express f ( x , y ) using polar coordinates: f ( r , θ ) = r 2 0 r < 1 1 r 1 Since lim r 1 - f ( r , θ ) = lim r 1 - r 2 = 1 and lim r 1 + f ( r , θ ) = lim r 1 + 1 = 1, we have lim r 1 f ( r , θ ) = 1. Therefore f ( r , θ ) is continuous at r = 1, or f ( x , y ) is continuous on x 2 + y 2 = 1. We conclude that f is continuous everywhere on R 2 . 15.3 Partial Derivatives (ET Section 14.3) Preliminary Questions 1. Patricia derived the following incorrect formula by misapplying the Product Rule: x ( x 2 y 2 ) = x 2 ( 2 y ) + y 2 ( 2 x ) What was her mistake and what is the correct calculation?
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