RVHS 2023 T4 Session 2 (Soln to self-mark)

1   2023 Term 4 Timed Practice (Structured Remedial Session 2) Question 1 (N16/I/9) A stone is held on the surface of a pond and released. The ston e falls vertically through the water and the distance, x metres, that the stone has fallen in time t seconds is measured. It is given that 0x and d 0d x t  when 0t  . ( i ) The motion of the stone is modelled by the differential equation 2 2 dd 21 0dd xx tt   . (a) By substituting d d xy t , show that the differential equation can be written as d 10 2d y yt  . [1] Solution Learning Points and self mark d d xy t Differentiate with respect to t, (Use the substitution given and diff wrt t) 2 2 dd dd yx tt .  Sub 2 2 dd dd yx tt and d d xy t , (use above 2 results to now substitute into the original equation, with objective that your new equation is one that involves only y and t, free of x) 2 2 dd 21 0dd d 21 0d d 10 2 (Shown)d xx tt y yt y yt    (Need know this substitution is not the typical one, but one that helps you step down from 2 nd derivative equation to a 1st derivative equation. The concept and process is similar to the typical substitution. In a nutshell, you must follow the prompt “By substituting…., show that ….” , otherwise its impossible to solve! 1 mark to show you have the steps leading to the shown resul t

2   ( b ) Find y in terms of t and hence find x in terms of t. [6] Solution 22 2 d 10 2d 1d 110 2 d 1 d 1 d10 2 1 ln 10 22 ln 10 2 2 , 2 10 2 e e , e 5e 2 tC t C t y yt y yt yty yt c yt C Cc yA A Ay                      When 0t  , we have 0x and d 0d xy t , 005 e 2 10 A A   Therefore, 2 2 2 2 55 e d 55 ed 55 e d 55e 2 t t t t y x t x t x tD             When 0t  , we have 0x ,  0505 0 e 2 5 2 D D       Therefore, 2555e 22 txt   1 mark- do separable variables. (these 7 steps on left are standard process - pls learn) 1 mark for ln | | expression 1 m a r k – a r r i v e a t expression 1 mark – substitute initial conditions and attempt find the constant A. 1 mark- method for integrating one more time wrt t after rewriting d as d xy t . Cautious of the various negative signs! 1 mark- final answer (all terms correct)

3   ( i i ) A second model for the motion of the stone is suggested, given by the differential equation 2 2 d1 10 5sind2 x tt  . Find x in terms of t for this model. [3] Solution 2 2 d1 10 5sind2 d1 10 5sin dd2 d1 10 10cosd2 x tt x ttt x tt Et      When 0t  , we have d 0d x t  , 01 0 0 1 0 c o s 0 10 E E      2 110 10cos 10 d2 152 0 s i n 1 02 x tt t x tt t F        When 0t  , we have 0x ,   205 0 2 0 s i n 0 1 0 0 0 F F      Therefore, 2 152 0 s i n 1 0 .2x tt t  (note this is a second order DE, need do twice integrating to get to x) 1 m a r k – i n t e g r a t i n g sine function correctly. Need arbitrary constant E here. Note the negative sign. 1 mark – integrate one more time AND attempt to find the 2 arbitrary constants 1 mark – accurate answe r

4   Question 2 (i) Given that 1s i nyx , prove that d2c o sd yyx x  . By repeated differentiation of this result expand y as a series of ascending powers of x up to and including the 3x term. [5] (ii) Using the standard series from the List of Formulae (MF26), ve rify that the same result can be obtained for y, up to the 2x term, for small x. [3] (i) Method 1 1s i nyx Differentiate with respect to x, dc o s d 21 s i n d21 s i n c o sd d2 cos (Shown)d yx x x yx xx yyx x     Method 2 (preferred) 2 1s i n 1s i n yx yx   Differentiate with respect to x, (must use implicit diffn) d2 cos (Shown)d yyx x  2 2 2 22 3 22 3 23 23 Diff implicitly wrt , dd22 s i nd d dd d d d42 2 c o sdd dd d dd d62 c o sd dd x yy yxx x yy y y y yxxx xx x yy y yxx xx               When 0,x This method requires you to rewrite 1s i nx    back as y. 1 mark 1 mark- get this 2 nd derivative expression correct. (implicit diffn must be fluent. Note that Product rule applies. View the two terms as d2 and d yy x ) 1 mark - get 3rd derivative expression correct.

5    2 22 2 33 33 2 dd 121 c o s 0 ,dd 2 1d d 122 0 ,2d d 4 11 d d 162 124 d d 8 10 1 ,y yy xx yy xx yy xx                     . 2 23 3 11 1 481s i n 1 22 ! 3 ! 1 284 8 xx x x xx x              1 mark (for all values of derivatives evaluated at x=0) Refer/consult MF26 to quote Maclaurin formula. 1 (final answer) (do the above steps systematically to find the derivatives when x=0, don’t skip essential working steps although its worth 1 mark only) (ii) Using the MF26 sin x series, observe that sin x  x . (Alternate, By small angle approximation, since sin x  x)  1 2 2 2 1s i n 1 1 11 1 221 22 ! 1 2 ... 8 xx x xx xx                Comparing with (i), the terms agree up to the square term for small x. (verified) 1 mark apply approx 1 mark apply binomial 1 final expression and statement to conclude that its verified.