Date | May 2022 | Marks available | 3 | Reference code | 22M.2.AHL.TZ1.12 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 1 |
Command term | Show that | Question number | 12 | Adapted from | N/A |
Question
Consider the differential equation for and . It is given that when .
Use Euler’s method, with a step length of , to find an approximate value of when .
Use the substitution to show that .
By solving the differential equation, show that .
Find the actual value of when .
Using the graph of , suggest a reason why the approximation given by Euler’s method in part (a) is not a good estimate to the actual value of at .
Markscheme
attempt to use Euler’s method (M1)
, where
correct intermediate -values (A1)(A1)
Note: A1 for any two correct -values seen
A1
Note: For the final A1, the value must be the last value in a table or a list, or be given as a final answer, not just embedded in a table which has further lines.
[4 marks]
(A1)
replacing with and with M1
A1
(since )
AG
[3 marks]
attempt to separate variables and (M1)
(A1)
attempt to express in partial fraction form M1
A1
A1
Note: Condone absence of modulus signs throughout.
EITHER
attempt to find using M1
expressing both sides as a single logarithm (M1)
OR
expressing both sides as a single logarithm (M1)
attempt to find using M1
THEN
(since )
substitute (seen anywhere) M1
(since )
attempt to make the subject M1
A1
AG
[10 marks]
actual value at A1
[1 mark]
gradient changes rapidly (during the interval considered) OR
the curve has a vertical asymptote at R1
[1 mark]
Examiners report
Most candidates showed evidence of an attempt to use Euler's method in part a), although very few explicitly wrote down the formulae, they used in order to calculate successive y-value. In addition, many seemed to take a step-by-step approach rather than using the recursive capabilities of the graphical display calculator.
There were many good attempts at part b), but not all candidates recognised that this would help them to solve part c).
Part c) was done very well by many candidates, although there were a significant number who failed to recognise the need for partial fractions and could not make further progress. A common error was to integrate without a constant of integration, which meant that the initial condition could not be used. The reasoning given for the estimate being poor was often too vague and did not address the specific nature of the function given clearly enough.