Use Euler’s method to find an approximation for the value of c , using a step length of h = 0.1 . Give your answer to four decimal places.

You are told that if Euler’s method is used with h = 0.05 then \(c \simeq 2.7921\) , if it is used with h = 0.01 then \(c \simeq 2.8099\) and if it is used with h = 0.005 then \(c \simeq 2.8121\).

Plot on graph paper, with h on the horizontal axis and the approximation for c on the vertical axis, the four points (one of which you have calculated and three of which have been given). Use a scale of 1 cm = 0.01 on both axes. Take the horizontal axis from 0 to 0.12 and the vertical axis from 2.76 to 2.82.

Draw, by eye, the straight line that best fits these four points, using a ruler.

Use your graph to give the best possible estimate for c , giving your answer to three decimal places.

using \({x_0} = 1,{\text{ }}{y_0} = 1\)

\({x_n} = 1 + 0.1n,{\text{ }}{y_{n + 1}} = {y_n} + 0.1\sqrt {{x_n} + {y_n}} \)     (M1)(M1)(A1) 

Note: If they have not written down formulae but have \({x_1} = 1.1\) and \({y_1} = 1.14142…\) award M1M1A1.

gives by GDC \({x_{10}} = 2,{\text{ }}{y_{10}} = 2.770114792…\)     (M1)(A1)

so \(a \simeq 2.7701{\text{ (4dp)}}\)     A1     N6

Note: Do not penalize over-accuracy.

[6 marks]

points drawn on graph above     A1A1A1

Note: Award A1 for scales, A1 for 2 points correctly plotted, A1 for other 2 points correctly plotted (second and third A1 dependent on the first being correct).

[3 marks]

suitable line of best fit placed on graph     A1

[1 mark]

letting \({\text{h}} \to {\text{0}}\) we approach the y intercept on the graph so     (R1)

\(c \simeq 2.814{\text{ (3dp)}}\)     A1

Note: Accept 2.815.

[2 marks]

Part (a) was done well. We would recommend that candidates write down the equation they are using, in this case, \({y_{n + 1}} = {y_n} + 0.1\sqrt {{x_n} + {y_n}} \) , to ensure they get all the method marks. Beyond this the answer is all that is needed (or if a student wishes to show working, simply each of the values of \({{x_n}}\) and \({{y_n}}\)) . Many candidates wasted a lot of time by writing out values of each part of the function, perhaps indicating they did not how to do it more quickly using their calculators.

Part (b) Surprisingly when drawing the graph a lot of candidates had (0.01, 2.8099) closer to 2.80 than 2.81

Most realised that the best possible estimate was given by the y-intercept of the line they had drawn.

Part (a) was done well. We would recommend that candidates write down the equation they are using, in this case, \({y_{n + 1}} = {y_n} + 0.1\sqrt {{x_n} + {y_n}} \) , to ensure they get all the method marks. Beyond this the answer is all that is needed (or if a student wishes to show working, simply each of the values of \({{x_n}}\) and \({{y_n}}\)) . Many candidates wasted a lot of time by writing out values of each part of the function, perhaps indicating they did not how to do it more quickly using their calculators.

Part (b) Surprisingly when drawing the graph a lot of candidates had (0.01, 2.8099) closer to 2.80 than 2.81

Most realised that the best possible estimate was given by the y-intercept of the line they had drawn.

Part (a) was done well. We would recommend that candidates write down the equation they are using, in this case, \({y_{n + 1}} = {y_n} + 0.1\sqrt {{x_n} + {y_n}} \) , to ensure they get all the method marks. Beyond this the answer is all that is needed (or if a student wishes to show working, simply each of the values of \({{x_n}}\) and \({{y_n}}\)) . Many candidates wasted a lot of time by writing out values of each part of the function, perhaps indicating they did not how to do it more quickly using their calculators.

Part (b) Surprisingly when drawing the graph a lot of candidates had (0.01, 2.8099) closer to 2.80 than 2.81

Most realised that the best possible estimate was given by the y-intercept of the line they had drawn.

Part (a) was done well. We would recommend that candidates write down the equation they are using, in this case, \({y_{n + 1}} = {y_n} + 0.1\sqrt {{x_n} + {y_n}} \) , to ensure they get all the method marks. Beyond this the answer is all that is needed (or if a student wishes to show working, simply each of the values of \({{x_n}}\) and \({{y_n}}\)) . Many candidates wasted a lot of time by writing out values of each part of the function, perhaps indicating they did not how to do it more quickly using their calculators.

Part (b) Surprisingly when drawing the graph a lot of candidates had (0.01, 2.8099) closer to 2.80 than 2.81

Most realised that the best possible estimate was given by the y-intercept of the line they had drawn.