P7.6 Forming 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. Part (a) was generally well answered with most responses demonstrating good understanding of the problem. Responses generally deduced an expression for d d V t and went on to score full marks for multiplying through by 20. The few responses that did not score on this part often tried to incorporate a connected rates of change type approach and some deriving an expression for V = 0.45t – 0.3VT and attempted to differentiate it, usually incorrectly.

Common mistakes

• dV/dt = 0.45 − 0.3V (rate in minus rate out) then multiply by 20
• connected-rates approaches fail

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. Those that were unable to find a correct differential equation in part (a) were fairly limited in the number of marks that could be scored in part (b), but the given answer In part (a), 1 mark was available for writing down a correct differential equation to model the given situation. Very often this part was answered successfully. When the mark was not awarded it was usually because candidates had simply used d 1 r = d t r rather than d r k = . d t r differential equation from part (a). For those that offered an equation without a constant of proportionality, i.e. d 1 r = , d t r it was only possible to access two out of five marks in part (b). Generally, candidates

Common mistakes

• dr/dt = k/√r with a constant of proportionality (dr/dt = 1/√r scores 0)

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. limited in the number of marks that could be scored in part (b), but the given answer provided an excellent restart opportunity, and a significant proportion went on to score full marks in parts (c) and (d). In part (b) candidates were given some values for the model and were required to solve their differential equation and use the given values to show the given answer solved the differential equation from part (a). it was only possible to access two out of five marks in part (b). Generally, candidates seemed well practiced in separating the variables and integrating the resulting equation correctly. When attempted, the most common error was in the index of the radius, either by using an incorrect model or by separating variables incorrectly. The most successful responses found the constant of proportionality straight away, before integrating. Some of those who did not do this at the start struggled to know to go back to find it and therefore could not progress further by substituting for t and r in their integrated model as they had both k and c still to be found. Occasionally, candidates substituted 0.9 or another value for k, instead of substituting 0.9 for d r t . d The given answer to part (b) enabled many who were unsuccessful in the early parts of this question to go on and successfully answer part (c). This question was generally very

Common mistakes

• k = 3.6 from 0.9 = k/√16 (comparing coefficients with the given answer is invalid)
• separate √r dr = k dt
• ⅔r^(3/2) = 3.6t + c
• use t = 10, r = 16 for c

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

Many candidates found this question challenging and there were a significant number of blank responses. Candidates found part (a) particularly challenging, with many attaining either no marks or the first mark only. The concept of connected rates of change is one that candidates frequently struggle with. A good number found an expression for V and successfully differentiated to get d V h = 200 , although some confused the volume of water with the volume of the tank. Many d candidates failed to recognise that d V 1 and hence d V k = . Of those who had some d t d t h h concept of the rate being d V , many misunderstood the idea of inverse proportion and others d t missed the constant of proportionality. Even where candidates had correct expressions for both d and d V V , they often failed to use the chain rule correctly to find d h . Many candidates d h d t d t who did link their expressions correctly, achieving d 1 h k = d 200 t h , with k = or equivalent, stated that this could be expressed as and so gained all 200 h three marks. Others made no reference to 𝜆 and so lost the accuracy mark. A less common, but ( ) d 200 h d V = correct, approach used d d t t Approaches to part (b) of the question were very mixed. Many candidates failed to realise that,

Common mistakes

• V = 200h so dV/dh = 200
• dV/dt = k/√h (inverse proportion with a constant)
• chain rule dh/dt = dh/dV × dV/dt = k/(200√h), then λ = k/200 must be defined
• using λ for both constants loses the A mark

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. For those that did attempt a response, part (a) was reasonably well answered, with many candidates gaining at least one of the B marks but quite a few then failing to correctly apply the chain rule and complete the proof. Some weaker candidates recognised they had to consider "flow in - flow out” but didn’t identify this as d d V t , with d 0.48 0.1 d h h t = - sometimes seen. It was also not uncommon to see d d V t = 0.48 or d d V t = 'uncombined' in some responses. Some candidates even started with and “differentiated” to get the printed result. Responses to part (b) were mixed. Some candidates failed to separate the variables correctly or did not recognise the logarithmic form for the integration. It was not unusual for the “+c” to be missed completely, but if it was included candidates were able to gain the 2nd method mark despite poor integration. Some candidates changed to , making their following working much harder. There was some occasional use of Further Maths methods (e.g. integrating factors), potentially gaining full marks. Candidates who did not answer (a) rarely attempted (b), perhaps not realising (b) could be answered independently using the printed differential equation.

Common mistakes

• dV/dt = 0.48 − 0.1h (flow in − flow out)
• V = 24h so dV/dh = 24
• must show the chain rule and where 24 comes from

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

Examiner's comment

Candidates were expected to realise that ''when the rate of change is proportional to the y value, an exponential model should be used.'' The question seemed to take a great many candidates by surprise and many assumed that 'n was proportional to n' or that 'n was proportional to t', not that the rate of growth of 'n was proportional to n'. Those that did interpret the situation correctly could write down an equation very quickly whereas others attempted to solve a differential equation. Equations scoring one of two marks were common and included e , e t kt n A n = = and e kt n A b = +

Common mistakes

• rate of change proportional to n means an exponential model n = Ae^(kt)
• not n ∝ t
• the two constants must be different (n = e^(kt + c) fine, n = e^(kt) + A is not)

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 (a) many candidates failed to use the chain rule d d d d d d V V r t r t = x with an acceptable d d V t and d . d V r It was possible to proceed to the given answer without knowing the formula for a sphere, using 3 V r = λ but errors and blank solutions were common. Common errors were; • stating d d V k t = -, and then failing to provide a minimal explanation of how their k/4𝜋𝜋 was replaced with k. • quoting an incorrect formula for the volume of a sphere. Attempts included V = pi r3 , 2 pi r3 , 4 pi r3 and even 4 pi r2.

Common mistakes

• constant rate of decrease: dV/dt = −c
• chain rule dr/dt = dV/dt ÷ dV/dr with dV/dr = 4πr²
• explain replacing c/4π by k
• wrong sphere volume formula

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

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 (a), many students struggled to translate the description of the model into a mathematical statement. Some incorrect statements, such as 2 d 1 , d r t r 2 d d r r t or d 1 , d r t r did not contain a constant of proportionality; while other incorrect statements 2 d , d r kr t 2 d d r r t k or d d r k t r contained a constant of proportionality. Many students used a correct method to separate the variables and most proceeded to integrate their 2 d d r r k t to give an equation linking r and t which contained a constant of integration. At this stage, some students incorrectly integrated 2r to give 2 .r The majority of students applied the boundary conditions 0, 5 t r and 4, 3 t r (or 240, 3 t r ) to find both constants and proceeded to give an equation linking r with t. Some students who worked in minutes obtained correct answers of either 3 250 49 , 2 t r 3 49 125 2 r t or 3 250 2 . 49 r t A few students who worked in seconds obtained correct answers of either 3 15000 49 , 120 t r 3 49 125 120 r t or 3 15000 120 . 49 r t Many students lost the final mark in this part by not including sufficient detail in their definition of their variables. Some students defined r as the radius of the mint, in mm, but it was rare to see the time, t, defined from a correct starting point (i.e. from when the mint was placed in the mouth) and with the correct unit of time used.

Common mistakes

• dr/dt = −k/r² (constant of proportionality needed)
• separate to ∫r² dr = −∫k dt
• two conditions (t = 0, r = 5
• t = 4, r = 3) for k and c
• r³/3 = −49t/6 + 125/3
• define r (mm) and t (minutes from placing in the mouth)

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