Paper 3 modelling: a logistic model Problem set B One question at a time

The number of registered members of a city's bike-share scheme, N thousand, is recorded at the end of each year. The time t is in years since the start of 2019, so t = 1 is the end of 2019. The number is modelled by \(N=\dfrac{64}{1+Ce^{-kt}}\), where C and k are positive constants. The table gives the data.

1. (a) In context, explain what the value 64 means in the model.

   • As t becomes large, \(e^{-kt}\to 0\), so \(N\to\dfrac{64}{1}=64\)
   • 64 thousand members is the maximum number of members the scheme will have (the carrying capacity). N gets closer to 64 as t increases, but never reaches it.

2. (b) Use the value of N at t = 2 to show that \(C=15e^{2k}\).

   • At t = 2, N = 4: \(4=\dfrac{64}{1+Ce^{-2k}}\)
   • \(1+Ce^{-2k}=64\div 4=16\)
   • \(Ce^{-2k}=15\)
   • \(C=15e^{2k}\)

3. (c) Use the value of N at t = 7 to find a second expression for C.

   • At t = 7, N = 40: \(40=\dfrac{64}{1+Ce^{-7k}}\)
   • \(1+Ce^{-7k}=64\div 40=1.6\)
   • \(Ce^{-7k}=0.6\)
   • \(C=0.6e^{7k}\)

In the model \(N=\dfrac{64}{1+Ce^{-kt}}\), the two data points at t = 2 and t = 7 give \(C=15e^{2k}\) and \(C=0.6e^{7k}\). Use a calculator to solve for k, and keep the unrounded value for later.

1. (a) Use your two expressions for C to find the value of k.

   • \(15e^{2k}=0.6e^{7k}\)
   • From the calculator: Solve \(15e^{2k}=0.6e^{7k}\) for k: k = 0.644

2. (b) Find the value of C.

   • \(C=15e^{2k}\) with the unrounded k
   • C = 54.4

The model is \(N=\dfrac{64}{1+Ce^{-kt}}\) with C = 54.4 and k = 0.644 (keep the unrounded values from your calculator). The model's predictions for t = 3 to 6 are compared with the data, to 1 decimal place. One prediction is missing.

1. (a) Calculate the value of a, the value the model gives at t = 5, correct to 1 decimal place.

   • \(N=\dfrac{64}{1+Ce^{-kt}}\) with t = 5, using the stored values of C and k
   • a = 20.2

2. (b) Using the value of a to 1 decimal place, find \(SS_{res}\) for the model's predictions.

   • Residuals (data − model) at t = 3, 4, 5, 6: 0.1, −0.3, 0.3, −0.4
   • Squares: 0.01, 0.09, 0.09, 0.16
   • \(SS_{res}\) = 0.01 + 0.09 + 0.09 + 0.16 = 0.35

The city will use the model if E < 0.35, where \(E=\sqrt{SS_{res}/n}\) and n is the number of predictions made. In the previous question, \(SS_{res}=0.35\) for the n = 4 predictions.

1. (a) By finding the value of E, show that the model will be used.

   • n = 4 predictions
   • \(E=\sqrt{0.35\div 4}\) = 0.296
   • 0.296 < 0.35, so the model will be used

The model is \(N=\dfrac{64}{1+Ce^{-kt}}\) with C = 54.4 and k = 0.644 (keep the unrounded values). N is in thousands of members. The city estimates that 22% of members ride on any one day, and that one bike is needed for every 5 daily riders.

1. (a) Use the model to predict the number of members at the end of 2034.

   • The end of 2034 is t = 16
   • N(16) = 63.9 thousand members
   • This is just below the carrying capacity 64

2. (b) Use the model to find an expression for the number of bikes, R, needed at time t.

   • 22% of the members ride each day: 0.22 × 1000N = 220N riders
   • One bike for 5 riders: R = 220N ÷ 5 = 44N
   • \(R=44\times\dfrac{64}{1+54.4e^{-0.644t}}\)
   • \(R=\dfrac{2816}{1+54.4e^{-0.644t}}\)

The number of bikes needed at time t is \(R=\dfrac{2816}{1+54.4e^{-0.644t}}\), using the model for the number of registered members.

1. (a) At the end of 2023 the scheme had 1100 bikes, and at the end of 2025 it had 1700. Find the average number of bikes added per year between the end of 2023 and the end of 2025.

   • t = 5 and t = 7: change = 1700 − 1100 = 600
   • Average = 600 ÷ 2 = 300 bikes per year

2. (b) Assume the number of bikes keeps increasing linearly at the rate you found in part (a). Find the value of t, where t ≥ 5, at which the number of bikes first becomes insufficient for the number of daily riders, and hence the year in which this happens.

   • Supply: 1100 at t = 5 and slope 300: \(S=300t - 400\)
   • Demand: \(R=\dfrac{2816}{1+54.4e^{-0.644t}}\)
   • From the calculator: graph R and S, then the first intersection with t ≥ 5: t = 6.56
   • t = 6.56 is 6 complete years after the start of 2019, so it is in 2025