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)

ax 11 22 33 ax ax ax                  2 3 3 2 3 1 1 3 1 2 2 1 a x a x a x a x a x a x    3 2 1 3 1 2 2 1 3 0 0 0 a a x a a x a a x             32 31 21 0 0 0 aa aa aa     M Method 2 1 23 32 10 0 0 a aa aa                                ai 13 2 31 0 10 0 aa a aa                               aj 12 21 3 0 0 10 aa aa a                                ak 32 31 21 0 0 0 aa aa aa     M 4(ii)(b) M is skew symmetric and so by (i)  det 0 M Hence M is not invertible. 4(ii0(c)    ker |T k k a , the set of vectors parallel to a or line through origin and parallel to a   3R( ) | 0T    v a v , the set of vectors perpendicular to a or plane through origin and perpendicular to a

5(i) cos cos 2 sin 2 cos sin sin 2 cos 2 sin cos 2 cos sin 2 sin sin 2 cos cos 2 sin cos(2 ) sin(2 ) cos sin                                         T sin cos 2 sin 2 sin cos sin 2 cos 2 cos cos 2 sin sin 2 cos sin 2 sin cos 2 cos sin(2 ) cos(2 ) sin cos                                        T The eigenvalues are 1 and 1 with eigenvector cos sin     and sin cos     respectively. 5(ii) 110 01     T R R 5(iii) π 3  13 22 31 22     T 13 3122 22 2 11 131 3 222                   T 5(iv) TT cos 2 sin 2 cos 2 sin 2 sin 2 cos 2 sin 2 cos 2                 cos 2 cos 2 sin 2 sin 2 cos 2 sin 2 sin 2 cos 2 sin 2 cos 2 cos 2 sin 2 sin 2 sin 2 cos 2 cos 2                  

cos 2( ) sin 2( ) sin 2( ) cos 2( )             which is a rotation through an angle of 2( ) 6(i) 2 1f ( ) 2xx x   2 11f (2) 2 2 024     2 18f (3) 3 2 039     Since f ( )yx is a continuous curve over the interval  2,3 with f (2) f (3) 0 , therefore there exists a root  in the interval. 6(ii) Let 1 2 f (3) 3 f (2) 91 f (3) f (2) 41x   Note that 91f0 41   . 912, 41   91 91 41 41 91 41 2 f ( ) f (2) 2.106450495f ( ) f (2)   2.11  (3.s.f) 6(iii) 2 1f ( ) 2xx x     1 2 3 21f ( ) 2 0 2xx x       for 912x 41   3 2 4 61f ( ) 2 0 4xx x       for 912x 41 2 1f ( ) 2xx x   is concave upward and its gradient negative over the interval 912x 41 ,  is an over - estimate of the root. 6(iv) 2 1f ( ) 2xx x   , 3 21f ( ) 22 x x x     Applying 1 f ( ) f ( ) n nn n xxx x    with 0 2.5x  , 0 10 0 f ( ) 1.844866f ( ) xxx x      1 3 11 21f ( ) 22 x xx     is undefined for 1 1.84486x  . Therefore the Newton-Raphson method failed.

6(v)   1 4 12n n x x   0 3x    1 4 0 12 2.012x x      2 4 1 12 2.061x x      3 4 2 12 2.055x x     is 2.06 correct to 3 s.f. Since f (2.055) 0.00227 0 and f (2.065) 0.02044 0  , 2.06  is sufficiently accurate correct to 3 significant figures. 7(i) Volume of revolution about y-axis, V 3π 12 0 2π d a xy x      3π 2 0 d2π ( sin ) (1 cos ) dd xa t t a t t t      3π 22 0 2π ( sin ) (1 cos ) da t t a t t   3π 32 2 0 2π ( sin )(1 cos ) da t t t t   (Shown) 7(ii) V 3π 3 2 22 0 2π ( 2 cos cos ) ( sin )(1 cos ) da t t t t t t t t      Let cos d sin cosI t t t t t t c    ----(1) Let 2cos dJ t t t du 1dut t   2 1 1 1 2 2 4 dv cos (1 cos 2 )d sin 2d t v t t t tt       21 1 1 1 2 4 2 2 sin 2 ( sin 2 )dJ t t t t t t     221 1 1 1 1 2 4 2 2 4 sin 2 cos 2J t t t t t c       21 1 1 1 2 2 2 4 sin 2 cos 2J t t t t c   

Let 2( sin )(1 cos ) dK t t t   23 1 3( sin )(1 cos ) d (1 cos )K t t t t c      3π 3 2 22 0 2π ( 2 cos cos ) ( sin )(1 cos ) d V a t t t t t t t t      3π 21 2 2 32 1 1 1 1 2 2 2 4 31 3 0 2( sin cos ) 2π sin 2 cos 2 (1 cos ) t t t t a t t t t t         3π 23 2 4 3 11 48 31 3 0 2 sin 2cos 2π sin 2 cos 2 (1 cos ) t t t t a t t t t       227 16 33 11 88 1 3 π 0 3π 0 2 0 2π 0 2π 0 0 aa             32 27 17 16 122π π 3πa    8(a) d d vm C kvt   1 d 1 d v C kv t m   11 ddvtC kv m    1 ln tC kv ckm    (for arbitrary constant c)  ln ktC kv kc m    C kv e kt kcm  e kt mA   (where e kcA  ) When 0t  , 0vv : 0A C kv   0 e kt mC kv C kv       0 e kt mC C kvv k   8(b) As t , e0 kt m  , so Cv k .

After a long time, the velocity of the falling body approaches the constant value C k , which is independent of the initial velocity 0v . 8(c)(i) d d vm C qt kvt     d d C qtvk vt m m  Multiply both sides by integrating factor d ee k kttmm  : de e ed kt kt kt m m m C qtvk vt m m     de ed kt kt m mv C qttm    e kt mv  e d kt m C qt tm     ee d kt kt mm C qt q tkk       e e d kt ktm mqC qt tkk     ee kt kt mm q mC qt ck k