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assignment3_sol

# assignment3_sol - MATH1211/10-11(2/Asol3/TNK Department of...

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MATH1211/10-11(2)/Asol3/TNK Department of Mathematics, The University of Hong Kong MATH1211 (2010-11, Second semester) Suggested solution to Assignment 3 Note : Some solutions given below are outline only. You may need to give more details in your solution. The parts that are typed in red color are remarks for you to read; they are not a part of the solution. 1. (a) For ( x,y ) 6 = (0 , 0), f ( x,y ) = x 3 y - xy 3 x 2 + y 2 . Hence f x ( x,y ) = (3 x 2 y - y 3 )( x 2 + y 2 ) - ( x 3 y - xy 3 )2 x ( x 2 + y 2 ) 2 = x 4 y + 4 x 2 y 3 - y 5 ( x 2 + y 2 ) 2 , f y ( x,y ) = ( x 3 - 3 xy 2 )( x 2 + y 2 ) - ( x 3 y - xy 3 )2 y ( x 2 + y 2 ) 2 = x 5 - 4 x 3 y 2 - xy 4 ( x 2 + y 2 ) 2 . (b) For y 6 = 0 and x = 0, by result of (a) we have f x (0 ,y ) = 0 + 0 - y 5 (0 + y 2 ) 2 = - y . To ﬁnd f x (0 , 0) we evaluate: f x (0 , 0) = lim h 0 f ( h, 0) - f (0 , 0) h = lim h 0 0 - 0 h = 0 . For x 6 = 0 and y = 0, by result of (a), we have f y ( x, 0) = x 5 - 0 - 0 ( x 2 + 0) 2 = x . To ﬁnd f y (0 , 0), we evaluate: f y (0 , 0) = lim k 0 f (0 ,k ) - f (0 , 0) k = lim k 0 0 - 0 k = 0 . (c) From result of (b), we have f xy (0 , 0) = lim k 0 f x (0 ,k ) - f x (0 , 0) k = lim k 0 - k - 0 k = - 1 , f yx (0 , 0) = lim h 0 f y ( h, 0) - f y (0 , 0) h = lim h 0 h - 0 h = 1 . Hence the mixed second order derivatives f xy (0 , 0) and f yx (0 , 0) are not equal. Here Theorem 4.3 does not apply. It is because, from (a), for ( x,y ) 6 = (0 , 0), f xy ( x,y ) = ( x 4 + 12 x 2 y 2 - 5 y 4 )( x 2 + y 2 ) 2 - ( x 4 y + 4 x 2 y 3 - y 5 )2( x 2 + y 2 )2 y ( x 2 + y 2 ) 4 = x 6 + 9 x 4 y 2 - 9 x 2 y 4 - y 6 ( x 2 + y 2 ) 3 . Along the line x = 0, we have lim y 0 f xy (0 ,y ) = lim y 0 - y 6 y 6 = - 1. Along the line y = 0, we have lim x 0 f xy ( x, 0) = lim x 0 x 6 x 6 = 1. Hence lim ( x,y ) (0 , 0) f xy ( x,y ) does not exist, which implies that f xy is not continuous at (0 , 0). Thus the hypothesis in Theorem 4.3, namely, that all ﬁrst order and second order partial derivatives are continuous, does not hold. 2. (a) f x (0 , 0) = lim h 0 f ( h, 0) - f (0 , 0) h = lim h 0 0 - 0 h = 0. f y (0 , 0) = lim h 0 f (0 ,h ) - f (0 , 0) h = lim h 0 0 - 0 h = 0. 1

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(b) f x ( t ) = f ( t,at ) = ( at 3 (1+ a 2 ) t 2 if t 6 = 0 0 if t = 0 = ( at (1+ a 2 ) if t 6 = 0 0 if t = 0 . Since ± at 1+ a 2 ² t =0 = 0, we can combine the two cases of t 6 = 0 and t = 0 together, to get f x ( t ) = at 1+ a 2 for all t R . Obviously f x is diﬀerentiable, and D ( f x )(0) = a 1+ a 2 . (c) By deﬁnition Df (0 , 0) = ³ f x (0 , 0) f y (0 , 0) ´ = ³ 0 0 ´ , and hence Df (0 , 0) D x (0) = ³ 0 0 ´ µ 1 a = 0, which is not equal to D ( f x )(0) as obtained in (b). This is because f is actually not diﬀerentiable at (0 , 0) and hence the chain rule cannot apply. To verify that
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assignment3_sol - MATH1211/10-11(2/Asol3/TNK Department of...

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