ACJC 9649 2023 Prelim P1 Solutions

1 3d 4 d4 , 8dd xy ttt t t       22 3 6 2 2 62 2 62 2 2 3 24 2 2 2 2 d d 4 48dd 1616 32 64 1616 32 116 2 116 16 1 xy ttt t t t t t t tt t tt t t t t t                                   Surface area     4 2 6 2 6 2 1 1 5 1 4 2d 32 d 1232 32 23 1 π4 π 1 1 1ππ 6 2 6 k k k t t tt t tt tt kk                  Solving 62 π12 3 1π 38 2 4632 k k     , we get 2k  . Since 1k  , 2k  . 2 Multiply throughout by 1nn  1 14( 1) 2nn nnn x n x     Let n nnn x y , 142nnyy  where 1 1y  142ny   2 24 4 2 2ny     32 34 4 2 4 2 2ny        1 2 3 14 4 2 4 2 4 2 2n n ny           1 1 2(1 4 )4 14 n n       11 24 1 4 3 nn  

12 5 4 3 n   Hence 12 5 4 3 n n nx n    . Alternative solution 142nnyy  where 1 1y  The general solution is of the form   14n n Aky  . Let 0k  , so 2 342A AA   . So the general solution is   12 43 n ny k   . Since 1 1y  , we have 3 21 34 5k k  . Hence   125 433 n ny   . 3(a) Let y ux  d d dd y uuxxx 2 3 2 3d d yx y x x y yx           32 3 2 d d ux ux u x x x ux uxx       2 3 4 3 3 3 3 d d uu x ux x ux u xx     23d 1d uux u u ux      32 d1 d1 uu xxu u u     2 1dd( 1)( 1) u ux xuu     Let 22 11( 1)( 1) ( 1) u A B C uuu u u       2( 1) ( 1)( 1) ( 1)u A u B u u C u       Let 1u : 14 A  1 4A Let 1u : 12 C   1 2C  Let 0u : 110 42 B    1 4B  2 1 1 1 d ln4( 1) 4( 1) 2( 1) u x cuu u       1 1 1ln 1 ln 1 ln4 4 2( 1)u u x c u        1 1 1ln ln4 1 2( 1) u xcuu    

 4( 1)11ln4 1 2( 1) xu cuu    4( 1) 2ln 11 xu cuu   (where 4cc ) Sub. yu x : 4 1 2ln 11 yx x cyy xx      4() 2ln x y x x cy x y x   Since xy , 0yx , so the general solution is 4() 2ln x x y x cy x y x   3(b) 3 2 3 2 d d y x x y y x xy  Let   3 2 3 2 22f , 1 x x y y y xxy yx y x     , 0 2x  , 0 1y  . Using Euler method with step size 0.5,  1 0 0 0 2 2 0.5f , 211 0.5 1 1 2 2.375 y y x y      3(c) The approximation is an over-estimate. 4(i)           T3det det det ( 1) det det     A A A A A since A is skew symmetric Since      det det det 0  A A A . 4(ii)(a) Method 1 x y (2, 1) (2.5, 2.375)

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