P6.2 Gradient of eᵏˣ

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

Examiner's comment

Candidates’ performances in this exponential modelling question were varied with some able to obtain full or nearly full marks, whilst others made very little progress. Candidates who did In (d)(i) candidates were generally able to differentiate their equation of the model to obtain an expression of the correct form, earning the first mark. Many did not obtain the correct answer so were not awarded the second mark. A number of candidates incorrectly differentiated to e kt - which gained no credit in (d)(i). The first mark in (d)(i) was again at obtain expressions like aften awarded to candidates who made little progress elsewhere in this question. In (d)(ii)

Common mistakes

• dh/dt = Ake^(−kt) ≈ 0.698e^(−0.0223t)
• differentiating to −e^(−kt) or keeping the 31

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

Examiner's comment

Candidates’ performances in this exponential modelling question were varied with some able to obtain full or nearly full marks, whilst others made very little progress. Candidates who did aften awarded to candidates who made little progress elsewhere in this question. In (d)(ii) candidates knew they needed to set their d h t equal to 30cm, but often confused the units and d used d rather than d . Having set up an equation using d h h h t candidates were often 30 0.3 t = t = d d d able to use appropriate log rules to progress from a suitable equation to obtain a value for t. The final correct answer of 37.9 was obtained by some, but errors often appeared during the course of this long question, causing the final answer to be incorrect.

Common mistakes

• 30 cm per year = 0.3 m per year (not 30)
• set dh/dt = 0.3, not h = 0.3
• take ln: t ≈ 37.9

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

Examiner's comment

However, a small number retrieved this mark in part (b) by correctly stating the equation at the beginning of their working. There were only a few examples of not using enough significant figures or incorrectly rounding the value of k. Misreads were fairly common on this question (usually 1200 instead of 12000 or 15000 instead of 1500) which often meant that learners were unable to proceed to a value of k as they ended up with an equation with no real solutions. In part (b), most were able to identify the need for differentiation. The execution of the method was well-attempted, although some struggled to make the necessary links for the final part of the solution. The intention was that the question would be answered in terms of A and k but many used their own model from part (a) which could also score all the marks. Nearly

Common mistakes

• dV/dt = −kAe^(−kt) then substitute Ae^(−kt) = V − 1500 to get −k(V − 1500)
• dV/dt must be seen
• working backwards by separating variables rarely scored

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

Examiner's comment

This question assessed understanding of exponential modelling and differentiation. Performance across the question was generally very good, but common errors in dealing with exponential equations and differentiating exponential models persist. Part (c) was the least well answered part of the question with learners often not realising that differentiation was required to find the rate of increase. Finding two values of N using the gradient of a straight-line formula was seen. Another common error was to substitute t = 3 first and then try to differentiate which gained no credit. The final answer did not require units to be given but occasionally incorrect time periods were referred to, thus changing the rate, which scored A0.

Common mistakes

• rate of increase needs dN/dt = 375e^(−0.075t), not the gradient of a chord
• do not substitute t = 3 before differentiating
• 299 cars per month (wrong time unit loses the A mark)

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

Examiner's comment

This question was a modelling one relating to the price per gram of two different metals. It enabled candidates to demonstrate understanding of working with exponential functions and differentiating them. Part (b) was generally poorly answered, and a number of candidates made no attempt to answer this part. A common error was to set A B V V = . For those that found d d A V t and d d B V t some did not put them equal to each other, or they set one or both equal to 0 instead, but the most common mistake was to keep the negative and write 0.04 0.02 0.8e 4.8e t t - = - thus forfeiting the last three marks. The minority who did set d d A V t d d B V t = - were usually able to use a correct method to solve the equation. Some candidates did appreciate the issue of trying to take logs of a negative was not possible, but rather than go back to the original equation, they opted to either ignore it or scribble out the negative at some point of the solution which was not acceptable to score these final marks.

Common mistakes

• rate of increase of A equals rate of decrease of B: 0.8e^(0.04T) = 4.8e^(−0.02T) (drop the negative sign)
• e^(0.06T) = 6 so T = 29.9 months
• setting V_A = V_B or derivatives = 0
• logs of a negative

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

Examiner's comment

This question on exponential functions caused problems for the majority in part (b), as they failed to recognise that differentiation was required. However, generally parts (a) and (c) were answered well. In part (b), the majority of candidates failed to recognise that the 15.3 cm per year meant that they were required to differentiate to obtain a rate of change, and attempted instead to substitute t = 4 into the expression for h. This answer resembled 15.3 cm, so many incorrectly assumed that they had answered the question, even though the subsequent conversion from metres to cm made no sense. Some attempted to divide this value by 4, showing some appreciation for the units given in the question, but lacking an appreciation that the information given provided an instantaneous rate of change and not an average rate over the four years. Without an attempt at differentiation no marks could be scored in (b) but those candidates who did differentiate generally successfully answered this part of the question.

Common mistakes

• rate of growth means dh/dt = 0.34e^(−0.2t), not h at t = 4 (1.536 m) or h/4
• 0.153 m per year = 15.3 cm per year

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

Examiner's comment

Those scoring full marks in this question were in a significant minority of candidates. Part (b) had mixed outcomes for candidates. Interpretation of the given initial rate of cooling was problematic for many candidates. There was a general lack of appreciation that the rate of cooling was a continuous variable which represented d . d H t - Those candidates not recognising this and using a difference method (similar to those introduced at GCSE) between t = 0 and t = 1 were restricted to the initial B mark in part (a). The most common attempt at a response for these candidates was setting H(1) = 85 - 7.5 and working from there. Of those who recognised the need to first find d d H t and then substitute t = 0, many incorrectly used d d H t = 7.5 rather than the correct d d H t = -7.5 to express a negative increase in temperature. This led to the loss of the last mark. Another error was to forget to differentiate the + 30 to zero leaving a constant in their d d H t , only scoring at most the second method mark for substituting t = 0 and their A into their d d H t . Some differentiated incorrectly to obtain -55Bte-Bt. Several students missed out on the final mark despite correct work by not writing out the final equation. It is important that centres make candidates aware that when a question asks for the complete equation for a model that it is given as such and not just the values of any required constants.

Common mistakes

• rate of cooling is dH/dt = −55Be^(−Bt) = −7.5 at t = 0 (sign matters), B = 3/22 ≈ 0.136
• give the complete equation H = 55e^(−0.136t) + 30
• using H(1) = 85 − 7.5 (difference method) is wrong
• derivative of +30 is 0

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

Examiner's comment

This question was reasonably well attempted. Logs continue to be an area of challenge for many candidates but despite this, a significant proportion were able to earn a good number of marks here. Part (c) of this question was a good differentiator and was a challenge to many candidates. For a significant proportion of candidates, “rate” did not signal the need for rate of change 8 and hence differentiation. It was common to see candidates simply substituting in t = 2 into their equation for P to obtain P = 1.32. Of those candidates who did attempt to find d d P t many seemed to struggle with the chain rule. Some candidates obtained results of the form Ate-0.5t, whilst others retained the constant term 0.8 when differentiating. Others began by substituting in t = 2 and then proceeded from e-2 to e-1 or ‘differentiated’ e-1 to get e-2. It was common for candidates who were able to progress to a correct value to give the final answer of -0.258 rather than 0.258 but this was condoned.

Common mistakes

• rate means dP/dt = −0.7e^(−0.5t) (the constant k disappears
• not Ate^(−0.5t))
• substituting t = 2 into P is wrong
• 0.258 kg/cm² per minute

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

Examiner's comment

This was the third modelling question on the paper, this time based around an exponential model. This should have been expected given the Advance Information and candidates performed well on parts (a) and (b). Part (c) was more of a challenge with only higher achieving students scoring all 4 marks. Most candidates recognised the need to differentiate in part (b), and used appropriate notation. Most did this accurately before substituting in t =10 to obtain 18.1. A number of candidates failed to secure the final mark however, for not providing a minimal conclusion relating their answer to 18,000. Common reasons for not achieving full marks were • failing to use appropriate notation often using d d y x instead of d d N t • failing to recognise the need to use differentiation often subtracting the values of b N at t =9 and 10 or else t =10 and 11

Common mistakes

• dN/dt = 11e^(0.05t), at t = 10 gives 18.1
• conclude about 18 thousand per year
• subtracting N at t = 9 and 10 is not a rate
• use dN/dt notation, not dy/dx

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

Examiner's comment

This was the first significant modelling question on this paper. As with other series, candidates find aspects of modelling difficult. The barriers to success here were parts (b) and (c). In part (b) many students used an incorrect method to find the rate of increase in the number of bacteria. Some substituted t =8 into their expression for N, which just found the number of bacteria after 8 hours; others then went on to divide this number by 8 or even 1000 to try and find the rate of increase. Those who realised that the expression for N needed to be differentiated to find the rate of increase often scored at least the method mark.

Common mistakes

• rate of increase means dN/dt = 1000k e^(kt) at t = 8: 420 (2 s.f.)
• substituting t = 8 into N, or dividing N by 8, is wrong

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

Part (b) was the part candidates found most challenging, and many left it completely blank. Of those that attempted part (i), most knew to multiply by 0.25, but many struggled to construct a sufficiently thorough explanation to support their work to arrive at the given equation. The most common mistake was getting confused by the negative sign. Some candidates didn’t seem to have recognised that a decreasing rate of change would be negative, and quite often the negative sign was just ignored. Often the given equation was relied upon too heavily, with some appearing to work backwards – the mark scheme required some evidence that they had differentiated, and as a result, many scored no marks due to no reference to dV/dt (or at least dy/dx for the first M1). Occasionally candidates who had otherwise been successful here were not awarded the final accuracy mark due to a failure to replace t with T. It was encouraging to see that even when candidates were unable to make progress with (b)(i), they frequently had a go at part (ii), often with some success. Candidates’ use of logarithms was broadly good, with only a few forgetting to divide by 3925 before introducing logs. If attempted, most were able to get to T=8.24. There were several instances of this then being written as 8 years 2 months, instead of the correct 8 years 3 months, but the majority of those got to the decimal answer were able to convert it successfully.

Common mistakes

• dV/dt = −3925e^(−0.25t) must be seen and set equal to −500 (decreasing: negative rate)
• replace t by T
• divide by 3925 before taking ln
• T = 8.24 is 8 years 3 months, not 2 months