tutorial5_sol

# tutorial5_sol - MATH1211/10-11(2/Tu5sol/TNK Department of...

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MATH1211/10-11(2)/Tu5sol/TNK Department of Mathematics, The University of Hong Kong MATH1211 Multivariable Calculus (2010-11 2nd semester) Solution to Tutorial 5 Note : Some solutions given below are outline only. You may need to give more details in your solution. 1. We calculate f,f 0 ,f 00 ,f 000 as follows: f ( x ) = x, f 0 ( x ) = 1 2 x - 1 / 2 , f 00 ( x ) = - 1 4 x - 3 / 2 , f 000 ( x ) = 3 8 x - 5 / 2 ; f (1) = 1 , f 0 (1) = 1 2 , f 00 (1) = - 1 4 , f 000 (1) = 3 8 . Thus p 3 ( x ) = f (1)+ f 0 (1)( x - 1)+ f 00 (1) 2! ( x - 1) 2 + f 000 (1) 3! ( x - 1) 2 = 1+ 1 2 ( x - 1) - 1 8 ( x - 1) 2 + 1 16 ( x - 1) 3 . 2. (a) Df (1 , π 2 , 1) = ± f x f y f z ² ( x,y,z )=(1 π 2 , 1) = ± 2 x cos y - x 2 sin y + 2 yz y 2 ² ( x,y,z )=(1 , π 2 , 1) = ± 0 π - 1 π 2 4 ² . Hf (1 , π 2 , 1) = f xx f xy f xz f yx f yy f yz f zx f zy f zz ( x,y,z )=(1 , π 2 , 1) = 2 cos y - 2 x sin y 0 - 2 x sin y - x 2 cos y + 2 z 2 y 0 2 y 0 ( x,y,z )=(1 , π 2 , 1) = 0 - 2 0 - 2 2 π 0 π 0 . (b) Using result of part (a), the 2nd-order Taylor polynomial of

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tutorial5_sol - MATH1211/10-11(2/Tu5sol/TNK Department of...

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