F06 - Numerical Methods - Assignment (Solutions)

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Numerical Methods (Assignment Solutions) Page 1 of 7 National Junior College 2016 – 2017 H2 Further Mathematics Topic F6: Numerical Methods (Assignment Solutions) Suggested Solutions 1 Let   2f x x and 2 1 1h n n   ,       f 1 2, 1 2f 1 , 1 2 2f 1 ,2 2f 1 , 1 2f 1 ,1 f 2 1. n n n n n n r n n n r n n n n n                              Using Trapezium Rule,           2 1 1 1 2 d 1 2 2 2 1 2 1 1 2 2 2 1 2 1 3 1 2 2 2 1 2 1 3 1 1 1 1 2 1 3 1 2 shown 2 1 22 3 22 2 22 2 n r xx n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n r                                                                        Since the curve 2y x from 1x to 2x concave upwards, applyi ng the trapezium rule will lead to an overestimation of the actual area as the sum of the areas of the n trapezia given by 1 1 3 1 22 n rn n r     , is more than the actual area, 2 1 2 dxx . x 1 2 www.KiasuExamPaper.com 361

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Numerical Methods (Assignment Solutions) Page 2 of 7   2 2 11 2 d 2ln 2ln 2 2ln1 2ln 2 x xx      Since the sum of the areas of the n trapezia, 1 1 3 1 22 n rn n r     , is more than the actual area, 1 1 3 1 2 2ln 22 n rn n r      . By considering n rectangles, 1 1 2 1 2 1 2 1 2total area of rectangles 1 2 3 2 2 2 2 1 2 3 12 n r n n n nn n n n n n n n n n n n n n n n r                                                                           Since the sum of the areas of the n rectangles, 1 12 n r n r  , is less than the actual area, 1 1 1 3 1 12 2ln 2 22 n n r rn n r n r          . x 2 1 11 n 11 n n  21 n  www.KiasuExamPaper.com 362

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Numerical Methods (Assignment Solutions) Page 3 of 7 2 From the graph of 2 2 2 xy x   and 1 e xy  , the graphs intersect exactly once. Hence, the equation 2 2 1 e2 xx x    has exactly one real root. Let   2 2f 1 e2 xxx x    , applying linear interpolation once,               1 1.9 f 1 1 f 1.9= f 1 f 1.9 1.0676 1.1 to 1 decimal place            Since  fy x has a vertical asymptote at 2x ,  f ' x changes rapidly in the interval  1.9, 1  , 1 is further away from  . Therefore, 1 is not a good approximation to  . 1 e xy  2 2 2 xy x   1y 1x 2y www.KiasuExamPaper.com 363

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Numerical Methods (Assignment Solutions) Page 4 of 7 3 (i) From the graph, for     2 2 1 1 x x k x      to have 2 negative roots, 1 1.k k    (ii) For 11 9 8k  , the graph of   2y x k   lies between 2 1y x  and 2 19 8y x  , it will intersect   2 1 1 xy x   at a point where x  and  1,3 . If 3k  ,     2 2 2 2 1 13 3 0 1 1 x xx x x x          . Let     2 2 1f 3 1 xx x x     ,     3 3f ' 2 1 xx x x    . By Newton-Raphson Method, 0 2x  ,      1 f 22 1.72477 1.725 to 3 decimal placesf ' 2x     . x = – 1 2 , where 1y x k k     2 1 1 xy x   www.KiasuExamPaper.com 364

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Numerical Methods (Assignment Solutions) Page 5 of 7              3 2 6 4 4 1 3 3 1f '' 2 1 2 10 2 1 52 1 1 x x xx x x x x x                   For  1,3x ,  f ' 0x  and  f '' 0x  . The curve strictly decreases and concave downwards for  1,3x . The tangent line at 0 2x  cuts the x-axis at 1x which is on the right of  . Therefore,  is smaller than 1x . 4 Let     4 3f 2,f ' 4x x x x   . By Newton-Raphson method,         1 2 1 1 4 1 1 3 1 1 1 3 1 1 3 1 f f ' 2 4 1 4 2 3 1 shown4 2                     Taking 1 1  ,   1 2 3 1 3 4 5 3 1 5 =4 2 4 1.1935 1.18923 1.18921 1.189 to 3 decimal placesx             x 31  1x 0x www.KiasuExamPaper.com 365

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Numerical Methods (Assignment Solutions) Page 6 of 7 1 4 4 1 4 2 0 2 2 1.189 x x      Taking 1 1  , 1 1 3 3 2 3 1 2 1 1 3 3 4 5 3 4 1 1 3 3 6 7 5 6 1 3 7 6 2 21.25992, 1.16653, 2 21.19686, 1.18667, 2 21.19006, 1.18892, 2 1.18930, x xx x x xx x x xx x x x                                                1 42 1.189  5 (i) Rate at which the pollutants enter the vessel = 100 qx litres/min. Rate at which the pollutants leave the vessel = 100 py litres/min. Let U be the volume of pollutant in the vessel at time t, then d d 100 U qx py t  . Also, d d V x w yt    and 100Up V  . Thus, d d 100 d d 100 100 d d d d d( ) d d d ( ) (shown)d d U qx py t pV qx py t V pp V qx pyt t pp x w y V qx py t p qx py px pw py p qx w x p t V t V                          www.KiasuExamPaper.com 366

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Numerical Methods (Assignment Solutions) Page 7 of 7 (ii) When x = y = w = 1000, and q = 40, d 1000 1000d V x w y V t Ct        . Using V = 10000 when t = 0, we get C = 10000. Thus, 1000 10000V t  .  40 1000 (1000 1000)d 40000 2000 40 2 d 1000 10000 10 pp p p t V t t        . Overestimate, because gradient is positive but decreasing. 6 (i) Exact solution is 21 1)( t ty   . So exact/true value of y(2) = 0.2. Using Euler’s method to approximate this solution over the interval 20  t for step size 2.0t , we obtain yapprox = 0.1858. The approximation is an under-estimate of the true value of y(2), and has an error e1 = 0.0142. Using Euler’s method to approximate this solution over the interval 20  t for step size 0.1t  , we obtain yapprox = 0.1933. The approximation is an under-estimate of the true value of y(2), and has an error e2 = 0.0067. Observe that 12.2 2 1 e e . (ii) Using improved Euler’s method with a step size of 1.0t , we obtain yapprox = 0.200695 which differs from the true value by 3e = 6.95  10−4. Observe that 5.9 3 2 e e . Thus, we can say that the improved Euler’s method is about 10 times more accurate than the Euler’s method. n tn pn, approx dp/dt 0 0.0 10.0000 2.0000 1 1.0 12.0000 1.4545 2 2.0 13.4545 1.0909 3 3.0 14.54545 www.KiasuExamPaper.com 367