P8.8 Interpreting differential equations

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question, which involved setting up and solving a differential equation using separation of variables, proved to discriminate across the cohort, with a wide range in quality of responses seen. The restart opportunities provided were well used, allowing many learners to access later parts of the question even if they had struggled in earlier parts. In part (c), it was pleasing that learners generally considered the behaviour of the curve as t tended to infinity. The use of appropriate notation was generally good and often lead to full marks for the justification. A large number of responses set their equation equal to 2 and attempted to solve it, interpreting the fact that there were no solutions as meaning that the tank would not become full. If a response gained the M1 mark, the A1ft mark was achieved with a high success rate, suggesting the practical problem that was being modelled was generally well understood by learners. Many learners who did not have a successful approach to part (b) were able to engage sufficiently with the problem and access the marks in part (c) using the given answer part (a). Those attempting the problem in this way did so by considering flow rates into and out of the tank either when the volume in the tank was 2 or by identifying that the flow rate was negative once the volume reached 1.5 (or some intermediate value).

Common mistakes

• as t → ∞ V → 1.5 (or dV/dt = 0 at V = 1.5) and 1.5 < 2, so it never becomes full
• setting V = 2 needs ln of a negative noted
• must conclude "no"

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

marks being lost in (b) and (c) due to having an impossible equation to solve. The vast majority of candidates remembered to include the constant of integration and use the given information to find it, likely supported by the question giving the required form of the answer as a reminder. Occasionally candidates missed out on the second method mark as they used 1.5 as the constant of integration rather than finding the constant of integration by correctly substituting H = 1.5 and t = 0, but the majority managed this with success, obtaining correct values for their constant and gaining the second method mark. A common reason for the loss of the final accuracy mark was failing to write out the final version of the model, generally omitting H = , or simply stating the values of A and B. Some candidates opted to differentiate the given answer and compare coefficients to find the values of A before going on to find B in the usual way. This verification strategy was not in line with the requirement of the question but candidates that used this method were allowed to score all but the final mark. Candidates who had a correct answer in part (a) generally performed well in part (b), with the majority able to rearrange the equation and use logarithms correctly to find the value of t. If these candidates were not awarded full marks, it was often because they were unable to convert their decimal into hours and minutes correctly, and therefore lost the final accuracy mark. Candidates who did not have a correct answer to part (a) Part (c) was generally less well attempted than parts (a) and (b) with some candidates failing to understand what the question was asking them to do. The most common error was for candidates to attempt to solve d H t = . A surprising proportion of candidates 0 d gave answers higher than the maximum height. For candidates with successful approaches, marks were frequently lost by forgetting units and therefore not satisfying the demand of the question and finding the height of the hole. Similarly, many decided to subtract an often correct 0.9 from the maximum height, which indicated a misunderstanding of the model.

Common mistakes

• 0.6e^(−0.2t) = 0.3 so t = 5 ln 2 = 3.47 h
• convert to 3 hours 28 minutes (conversion errors common)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

marks being lost in (b) and (c) due to having an impossible equation to solve. The vast majority of candidates remembered to include the constant of integration and use the given information to find it, likely supported by the question giving the required form of the answer as a reminder. Occasionally candidates missed out on the second method mark as they used 1.5 as the constant of integration rather than finding the constant of integration by correctly substituting H = 1.5 and t = 0, but the majority managed this with success, obtaining correct values for their constant and gaining the second method mark. A common reason for the loss of the final accuracy mark was failing to write out the final version of the model, generally omitting H = , or simply stating the values of A and B. Some candidates opted to differentiate the given answer and compare coefficients to find the values of A before going on to find B in the usual way. This verification strategy was not in line with the requirement of the question but candidates that used this method were allowed to score all but the final mark. Candidates who had a correct answer in part (a) generally performed well in part (b), Part (c) was generally less well attempted than parts (a) and (b) with some candidates failing to understand what the question was asking them to do. The most common error was for candidates to attempt to solve d H t = . A surprising proportion of candidates 0 d gave answers higher than the maximum height. For candidates with successful approaches, marks were frequently lost by forgetting units and therefore not satisfying the demand of the question and finding the height of the hole. Similarly, many decided to subtract an often correct 0.9 from the maximum height, which indicated a misunderstanding of the model.

Common mistakes

• as t → ∞, e^(−0.2t) → 0 so H → B = 0.9 m (units needed)
• 1.5 − 0.6 is a wrong method
• substituting a large t scores M0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question involved a balloon modelled as a sphere such that its radius was increasing at a rate that was inversely proportional to the square root of the radius. full marks in parts (c) and (d). this question to go on and successfully answer part (c). This question was generally very well attempted, and most candidates used correct processing to calculate the correct value. However, some failed to do so with the required accuracy or made an error in unit conversion. Many did not give the correct units, or any units, forfeiting the final 3 r = 118 mark. There was a significant minority who, having progressed to , failed to 2 3 118 and did not score use correct processing to solve the equation, incorrectly finding 2 the method mark as a result. A single mark was available in part (d), for suggesting a limitation to the model, and for

Common mistakes

• r = 118^(2/3) = 24.1 cm (nearest mm), units needed
• index errors e.g. √118³

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question involved a balloon modelled as a sphere such that its radius was increasing at a rate that was inversely proportional to the square root of the radius. full marks in parts (c) and (d). A single mark was available in part (d), for suggesting a limitation to the model, and for some candidates this was the only mark gained in the question. Very many candidates scored the mark for pointing out that the balloon would eventually “pop”, or that the assumption that the balloon was perfectly spherical was unrealistic. Some candidates offered a more sophisticated response involving taking the limit for large times, although this was only acceptable if there was some reference to context such as the radius or volume. The most common incorrect response involved commenting on the “rate of inflation of the balloon being constant” as unrealistic, but this aspect had already been addressed in the model.

Common mistakes

• balloon bursts / is not a perfect sphere / grows without limit
• comments on rate of inflation are already in the model and score 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question tested the candidates’ skill in various areas, i.e. finding partial fractions, integration, rules of logarithms, manipulating equations to eliminate logarithms using exponentials, indices, and commentating on the expected behaviour of a variable as another tends to infinity. This was a challenging question which in a number of instances, candidates struggled with after part (a). Those however, who were well prepared, were able to score full marks often with relatively concise solutions. In part (d), candidates sometimes identified that L = 25, even when little else of credit was presented throughout, but usually those who had not attempted part (c) also made no attempt at (d). The reasons given were often not rigorous enough to gain credit, even when 25 had been established as the required value of L. When correct explanations were seen they usually stated that L = 25 because 2.5 e 0 t - → as t → or that L = 25 because d 0. d V t =

Common mistakes

• L = 25
• reason: as t → ∞, e^(−2.5t) → 0 (or dV/dt = 0 when V = 25)
• reason must be rigorous

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Many candidates found this question challenging and there were a significant number of blank responses. attempt at part (c) and were generally able to gain the method mark. Relatively few lost the accuracy mark due to missing the units in their final answer. A number of candidates used an incorrect value for h, most commonly 1000, and so lost both marks.

Common mistakes

• full when h = 5 (not 1000): 5^(3/2) = 0.513t + 1.728 gives t = 18.4 minutes
• units needed

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (c) the solution of 30mins was only obtained by the strongest candidates although many did not attempt this part of the question or possibly failed to see it. Of those that did attempt this part, a common answer was to obtain ½ but not recognise the units of the question were in hours, some giving it incorrectly as 0.5 minutes As the equation for the volume was a given in this question most candidates were able to attempt this in spite of their attempt at part (b). Candidates gave the correct 6m3 although a substantial number thought that 3m3 was the answer. Although candidates were not required to provide units, many did so correctly and marks were rarely lost for stating incorrect units although a few instances of mm3 were seen.

Common mistakes

• (i) V = 0 at t = ½ hour = 30 minutes (not 0.5 or 0.5 minutes)
• (ii) V → 6 m³ as t → ∞ (3 is wrong)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question proved challenging for the majority of candidates, with many not answering it at all. Part (c) was attempted by many candidates, even if they hadn’t attempted (a) and (b).

Common mistakes

• as t → ∞, h → 4.8 m < 5 m so the tank never fills (or flow in = flow out at h = 4.8)
• no contradictory statements

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Students found aspects of this question difficult despite the fact that the differential equation to be solved in (b) was not too demanding. Only high scoring candidates tended to make any significant progress in this question. In part (c) many students were confused by the wording and did not realise that a limiting value for t was required. Many attempts merely stated that t could not be negative or that t >0. Those candidates who did proceed to a limiting value for t generally had the right idea and most of these obtained a value of t >5 and gained the A mark when their equation form part (b) was followed through.

Common mistakes

• the model is valid only while r ≥ 0: 64000 − 11200t ≥ 0 gives t ≤ 40/7 (5.71 s)
• "t > 0" or "t cannot be negative" scores nothing

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Q14 discriminated well between the higher ability candidates. Lower and medium ability candidates, however, found this question demanding with most of these candidates scoring no more than two marks. There were a few candidates who made no attempt at this question, either because the question was too demanding, or because their time had run out. Most higher ability candidates made good progress in Q14(a) and Q14(c) but they often struggled to make progress in Q14(b). 4 u h In Q14(a), most candidates who attempted this part differentiated to give a 1 d 1 u with a few finding a correct d h u , h 2(4 ) u correct and some 2 d d 2 h 1 d 1 u . h differentiating incorrectly to give At this stage, some candidates failed to 2 d 2 h make further creditable progress. Many candidates struggled to apply the substitution (4 ) d , k u completely to give an integral of the form with some making sign errors u u in the numerator of this integral. Those who progressed this far generally went on to A divide each term in the numerator by u to give an integral of the form d , B u u 8 with many of them obtaining a correct Most integrated correctly and 2 d .u u 4 u h applied to obtain a correct Some 8ln 4 2(4 ) . h h c candidates could not deal with the transition from 8 c to k to achieve the given and a large number of these candidates incorrectly stated 8ln 4 2 , h h k 2 8 d u k u Those few candidates who applied integration by parts on rarely 8. u progressed to give a correct solution to this part. In Q14(b), some candidates correctly set d h t 0 but only a few deduced the equation d h 4 0. h Some candidates used to state a correct allowable answer such as 16 h h 4 0 16, h 16. or A common mistake was to solve to give 0 16 h h 2. Material for Q14(c) was sometimes seen in Q14(b). Most candidates who attempted 0.25 t Q14(c) separated the variables correctly. Some candidates integrated incorrectly 1 5 0.25 0.75 1.25 , t t 4 t to give or while others integrated without reference to (4 ) h to find their constant of 0, 1 t h the result from Q14(a). Many candidates applied integration. Common errors at this stage included not applying a constant of integration to find their constant of integration. Most 0, 0 t h 1, 1 t h and using either or candidates who progressed this far applied a complete process of substituting their h constant of integration and into their integrated equation and solving this 12 equation to find their value for t. In many cases, sign errors, bracketing errors, manipulation errors or an incorrect method of solving a correct 1.25 221.2795202..., t t prevented candidates from obtaining the correct answer years. 75.2

Common mistakes

• tree grows while dh/dt > 0: 4 − √h = 0 gives h = 16, so 0 < h < 16
• common error h = 2

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Q14 discriminated well between the higher ability candidates. Lower and medium ability candidates, however, found this question demanding with most of these candidates scoring no more than two marks. There were a few candidates who made no attempt at this question, either because the question was too demanding, or because their time had run out. Most higher ability candidates made good progress in Q14(a) and Q14(c) but they often struggled to make progress in Q14(b). 4 u h In Q14(a), most candidates who attempted this part differentiated to give a 1 d 1 u with a few finding a correct d h u , h 2(4 ) u correct and some 2 d d 2 h 1 d 1 u . h differentiating incorrectly to give At this stage, some candidates failed to 2 d 2 h make further creditable progress. Many candidates struggled to apply the substitution (4 ) d , k u completely to give an integral of the form with some making sign errors u u in the numerator of this integral. Those who progressed this far generally went on to A divide each term in the numerator by u to give an integral of the form d , B u u 8 with many of them obtaining a correct Most integrated correctly and 2 d .u u 4 u h applied to obtain a correct Some 8ln 4 2(4 ) . h h c candidates could not deal with the transition from 8 c to k to achieve the given and a large number of these candidates incorrectly stated 8ln 4 2 , h h k 2 8 d u k u Those few candidates who applied integration by parts on rarely 8. u progressed to give a correct solution to this part. In Q14(b), some candidates correctly set d h t 0 but only a few deduced the equation d h 4 0. h Some candidates used to state a correct allowable answer such as 16 h h 4 0 16, h 16. or A common mistake was to solve to give 0 16 h h 2. Material for Q14(c) was sometimes seen in Q14(b). Most candidates who attempted 0.25 t Q14(c) separated the variables correctly. Some candidates integrated incorrectly 1 5 0.25 0.75 1.25 , t t 4 t to give or while others integrated without reference to (4 ) h to find their constant of 0, 1 t h the result from Q14(a). Many candidates applied integration. Common errors at this stage included not applying a constant of integration to find their constant of integration. Most 0, 0 t h 1, 1 t h and using either or candidates who progressed this far applied a complete process of substituting their h constant of integration and into their integrated equation and solving this 12 equation to find their value for t. In many cases, sign errors, bracketing errors, manipulation errors or an incorrect method of solving a correct 1.25 221.2795202..., t t prevented candidates from obtaining the correct answer years. 75.2

Common mistakes

• separate variables
• ∫t^0.25 dt = 0.8t^1.25
• use the result of (a)
• constant from t = 0, h = 1 (not h = 0)
• t^1.25 = 221.28 so t = 75.2 years

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was another modelling question, this time modelling the height of a roller coaster above Part (b) was found by almost all students, even those who could not attempt part (a). In this case the exact answer 5e 0.1 was accepted as well as 5.53 metres. Fewer students scored both t marks in (c), but many were aware that it would occur when sin(0.25 ) The most 1.

Common mistakes

• maximum when sin(0.25t) = 1: 5e^0.1 = 5.53 m (exact 5e^0.1 accepted)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was another modelling question, this time modelling the height of a roller coaster above t marks in (c), but many were aware that it would occur when sin(0.25 ) The most 1. or else use 450 rather than 5 thus , , common errors were to find the first value using 2 2 arriving at the incorrect answer of 1800 seconds.

Common mistakes

• second maximum when 0.25t = 5π/2 (radians), T = 10π ≈ 31.4 s
• using π/2 (first maximum) or 450 in degrees (giving 1800) is wrong

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question discriminated well between the average and more able students. There waere a significant minority of students, however, that made no creditable progress. In part (b), many students substituted 0 r in their equation from part (a) and rearranged to find a value for t, with a significant number obtaining the correct answer, 5 minutes 6 seconds. Some students lost the final mark by giving their final answer as either 5.10 minutes, 250 49 minutes or 306 seconds.

Common mistakes

• set r = 0: t = 250/49 minutes = 5 minutes 6 seconds
• 5.10 minutes, 250/49 or 306 seconds alone lose the mark

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question discriminated well between the average and more able students. There waere a significant minority of students, however, that made no creditable progress. Part (c) was generally well-answered with a variety of valid limitations given. Students commented on the shape of the mint while it was being sucked; on how the mint was sucked; or that the model predicted a negative radius for the mint after a time of 5 minutes 6 seconds after the mint was placed in the mouth. Some students who stated that ‘the radius of the mint may not decrease at a constant rate’ did not receive any credit because the radius of the mint in the original model was not decreasing at a constant rate.

Common mistakes

• e.g. mint may not stay spherical, bitten or swallowed, invalid after 5 min 6 s (negative radius)
• "radius does not decrease at a constant rate" is not credited