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HM4-soln

# HM4-soln - Homework 4 Solution 6.47 1 x2 y 2 = 1 1 2 x y >...

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Homework 4 Solution October 17, 2011 6.47 1. x 2 + y 2 6 = 1 . 1

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2. x + y > 0 . 3. 2 x - y x + y 1 . 2
6.59 (a) Continuous. The function is the summation of two continuous func- tions. (b) Discontinuous. Since lim x 0 (lim y 0 x 3 x + 5 y ) = lim x 0 1 3 = 1 3 , lim y 0 (lim x 0 x 3 x + 5 y ) = lim y 0 0 = 0 , lim ( x,y ) (0 , 0) x 3 x +5 y cannot exist. (c) Continuous. lim ( x,y ) (0 , 0) ( x 2 + y 2 ) sin 1 x 2 + y 2 = 0 . 6.63 (a) f x (0 , 0) = lim h 0 f ( h, 0) - f (0 , 0) h = lim h 0 ( h 2 - 0) / ( h + 0) h = 1 . (b) f y (0 , 0) = lim h 0 f (0 , h ) - f (0 , 0) h = lim h 0 (0 - 0) / (0 + h ) h = 0 . 6.67 (a) f x = 2( x + y ) - (2 x - y ) ( x + y ) 2 = 3 y ( x + y ) 2 , f xy = 3( x + y ) 2 - 3 y · 2( x + y ) ( x + y ) 4 = 3( x - y ) ( x + y ) 3 . f y = - 3 x ( x + y ) 2 , f yx = 3( x - y ) ( x + y ) 3 = f xy . Exceptional points x + y = 0 . Derivatives don’t exist. (b) f x = tan xy + xy sec 2 xy, f xy = sec 2 xy · x + x · sec 2 xy + xy · 2 sec xy · sec xy tan xy · x = 2 x sec 2 xy +2 x 2 y sec 2 xy tan xy. f y = x sec 2 xy · x = x 2 sec 2 xy, f yx = 2 x sec 2 xy + x 2 · 2 sec xy · sec xy tan xy · y = 2 x sec 2 xy +2 x 2 y sec 2 xy tan xy = f xy . 3

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Exceptional points xy = + 1 2 π, k = 0 , ± 1 , ± 2 , . . . . Derivatives don’t exist. (c) f x = sinh( y + cos x ) · ( - sin x ) = - sin x sinh( y + cos x ) , f xy = - sin x cosh( y + cos x ) . f y = sinh( y + cos x ) , f yx = cosh( y + cos x ) · ( - sin x ) = f xy . No exceptional points. 6.68 z x = 2( x - a ) ( x - a ) 2 + ( y - b ) 2 . z xx = 2[( x - a ) 2 + ( y - b ) 2 ] - 2( x - a ) · 2( x - a ) [( x - a ) 2 + ( y - b ) 2 ] 2 = 2[( y - b ) 2 - ( x - a ) 2 ] [( x - a ) 2 + ( y - b ) 2 ] 2 . Similarly we have z yy = 2[( x - a ) 2 - ( y - b ) 2 ] [( x - a ) 2 + ( y - b ) 2 ] 2 . It is clear that z xx + z yy = 0 . 6.69 z x = cos y x + y x sin y x - y x 2 sec 2 y x , z y = - sin y x + 1 x sec 2 y x , z xx = - y 2 x 3 cos y x + 2 y x 3 sec 2 y x + 2 y 2 x 4 tan y x sec 2 y x , z xy = y x 2 cos y x - 1 x 2 sec 2 y x - 2 y x 3 sec 2 y x tan y x , z yy = - 1 x cos y x + 1 x 2 sec y x · tan y x sec y x · 1 x x 2 z xx + 2 xyz xy + y 2 z yy = ( - y 2 x cos y x + 2 y x sec 2 y x + 2 y 2 x 2 tan y x sec 2 y x ) + ( 2 y 2 x cos y x - 2 y x sec 2 y x - 4 y 2 x 2 tan y x sec 2 y x ) + ( - y 2 x cos y x + 2 y 2 x 2 tan y x sec 2 y x ) = 0 . 4
6.71 d z = z x d x + z y d y = (3 x 2 - y )d x + ( - x + 6 y )d y Δ z = z ( x + Δ x, y + Δ y ) - z ( x, y ) .

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HM4-soln - Homework 4 Solution 6.47 1 x2 y 2 = 1 1 2 x y >...

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