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

The number of active users of a language-learning app, N thousand, is recorded at the end of each year. The time t is in years since the start of 2020, so t = 1 is the end of 2020. The number is modelled by \(N=\dfrac{1250}{1+Ce^{-kt}}\), where C and k are positive constants. The table gives the data.

1. (a) In context, explain what the value 1250 means in the model.

   • As t becomes large, \(e^{-kt}\to 0\), so \(N\to\dfrac{1250}{1}=1250\)
   • 1250 thousand users is the maximum number of active users the app will have (the carrying capacity). N gets closer to 1250 as t increases, but never reaches it.

2. (b) Use the value of N at t = 2 to show that \(C=24e^{2k}\).

   • At t = 2, N = 50: \(50=\dfrac{1250}{1+Ce^{-2k}}\)
   • \(1+Ce^{-2k}=1250\div 50=25\)
   • \(Ce^{-2k}=24\)
   • \(C=24e^{2k}\)

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

   • At t = 7, N = 250: \(250=\dfrac{1250}{1+Ce^{-7k}}\)
   • \(1+Ce^{-7k}=1250\div 250=5\)
   • \(Ce^{-7k}=4\)
   • \(C=4e^{7k}\)

In the model \(N=\dfrac{1250}{1+Ce^{-kt}}\), the two data points at t = 2 and t = 7 give \(C=24e^{2k}\) and \(C=4e^{7k}\). 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.

   • \(24e^{2k}=4e^{7k}\)
   • From the calculator: Solve \(24e^{2k}=4e^{7k}\) for k: k = 0.358

2. (b) Find the value of C.

   • \(C=24e^{2k}\) with the unrounded k
   • C = 49.1

The model is \(N=\dfrac{1250}{1+Ce^{-kt}}\) with C = 49.1 and k = 0.358 (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.

   • \(N=\dfrac{1250}{1+Ce^{-kt}}\) with t = 5, using the stored values of C and k
   • a = 136.0

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: 0.7, −0.3, 1, 0.1
   • Squares: 0.49, 0.09, 1, 0.01
   • \(SS_{res}\) = 0.49 + 0.09 + 1 + 0.01 = 1.59

The company will use the model if E < 0.8, where \(E=\sqrt{SS_{res}/n}\) and n is the number of predictions made. In the previous question, \(SS_{res}=1.59\) for the n = 4 predictions.

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

   • n = 4 predictions
   • \(E=\sqrt{1.59\div 4}\) = 0.630
   • 0.630 < 0.8, so the model will be used

The model is \(N=\dfrac{1250}{1+Ce^{-kt}}\) with C = 49.1 and k = 0.358 (keep the unrounded values). N is in thousands of users. The company estimates that 36% of its active users have a premium plan, and that one support agent is needed for every 600 premium users.

1. (a) Use the model to predict the number of active users at the end of 2037.

   • The end of 2037 is t = 18
   • N(18) = 1160 thousand users
   • This is just below the carrying capacity 1250

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

   • 36% of the users have a premium plan: 0.36 × 1000N = 360N premium users
   • One agent for 600 premium users: R = 360N ÷ 600 = 0.6N
   • \(R=0.6\times\dfrac{1250}{1+49.1e^{-0.358t}}\)
   • \(R=\dfrac{750}{1+49.1e^{-0.358t}}\)

The number of agents needed at time t is \(R=\dfrac{750}{1+49.1e^{-0.358t}}\), using the model for the number of active users.

1. (a) At the end of 2024 the company employed 120 support agents, and at the end of 2026 it employed 160. Find the average number of agents added per year between the end of 2024 and the end of 2026.

   • t = 5 and t = 7: change = 160 − 120 = 40
   • Average = 40 ÷ 2 = 20 agents per year

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

   • Supply: 120 at t = 5 and slope 20: \(S=20t + 20\)
   • Demand: \(R=\dfrac{750}{1+49.1e^{-0.358t}}\)
   • From the calculator: graph R and S, then the first intersection with t ≥ 5: t = 7.40
   • t = 7.40 is 7 complete years after the start of 2020, so it is in 2027