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 c