Mixed modelling Problem set A Worked solutions

Give each answer to 3 significant figures.

1. (a) A radioactive isotope has a half-life of 6 days. The mass is modelled by \(M = M_0e^{-kt}\), with t in days. Find k.

   • Half-life 6 days: \(e^{-6k} = \frac12\)
   • \(k = \frac{\ln 2}{6} = 0.116\)

2. (b) A population of yeast cells doubles every 40 minutes (t in minutes). The number is modelled by \(N = N_0e^{kt}\). Find k.

   • Doubling: \(e^{40k} = 2\)
   • \(k = \frac{\ln 2}{40} = 0.0173\)

3. (c) The size of a sample is \(S = 500 \times 2.5^{-t}\), with t in years. Find its half-life.

   • Half the start: \(2.5^{-t} = \frac12\)
   • Solve for t: \(t = \frac{\ln 2}{\ln 2.5} = 0.756\) years

A metal block cools in a workshop. Its temperature \(T\) °C, \(t\) minutes after it leaves a furnace, is \(T = B + Ae^{-kt}\). The block leaves the furnace at 140 °C. After 9 minutes it is 71 °C. The workshop is at 16 °C.

1. (a) Find the values of A and B. Say what each one means.

   • B is the surroundings: \(B = 16\)
   • At \(t=0\): \(140 = 16 + A\), so \(A = 124\)
   • A is the starting difference from the surroundings; B is the temperature the block approaches.

2. (b) Find the value of k.

   • \(71 = 16 + (124)e^{-9k}\)
   • Solve for k: \(k = 0.0903\)

3. (c) Find the time for the temperature to reach 30 °C.

   • \(30 = 16 + (124)e^{-kt}\)
   • Solve for t: \(t = 24.1\) minutes

The mass of a drug in a patient's blood is \(M = M_0e^{-0.18t}\) mg, where t is the time in hours after the dose. After 5 hours 12 mg remain.

1. (a) Find the initial amount \(M_0\).

   • \(12 = M_0e^{-0.18\times 5}\)
   • Solve for \(M_0\): \(M_0 = 29.5\) mg

2. (b) Find the time at which the amount reaches 2 mg.

   • \(2 = 29.5e^{-0.18t}\)
   • Solve for t: \(t = 15.0\) hours

The number of fish in a lake is modelled by \(N = 2400e^{-0.35t}\), where t is the time in years after a spill.

1. (a) Use the model to find the number of fish after 20 years.

   • \(N = 2400e^{-0.35\times 20} = 2.19\)

2. (b) State one assumption of the model.

   • Name something the model takes as fixed: for example no new arrivals, or the same conditions all the time.

3. (c) Explain why the model cannot be right for large values of t.

   • The model gives a fraction of one fish and never reaches 0, so it fails once the number is small.

The rating R of an earthquake is modelled by \(R = p\ln A + q\), where A is the amplitude of the ground movement in micrometres. When \(A = 20\), \(R = 4.1\). When \(A = 500\), \(R = 6.3\).

1. (a) Find the values of p and q.

   • \(4.1 = p\ln 20 + q\) and \(6.3 = p\ln 500 + q\)
   • Solve the pair: \(p = 0.683\), \(q = 2.05\)

2. (b) Find R when A = 8000.

   • \(R = 0.683\ln 8000 + 2.05 = 8.19\)

3. (c) Find the range of values of A for which \(5 \le R \le 6\).

   • \(A = e^{(R-q)/p}\), as R increases when A increases
   • \(R = 5\): \(A = 74.6\); \(R = 6\): \(A = 322\)
   • \(74.6 \le A \le 322\)

A paddling pool is filled. The volume \(V\) litres after \(t\) minutes is modelled by \(V = pt\) for \(0 \le t \le 6\) and \(V = qt^2\) for \(t > 6\). After 3 minutes there are 36 litres. The pool holds 700 litres.

1. (a) Find the value of p.

   • \(36 = p \times 3\)
   • \(p = 12.0\)

2. (b) The model is continuous at the switch. Find the value of q.

   • At \(t=6\): \(12.0 \times 6 = q \times 6^2\)
   • \(q = 2.00\)

3. (c) Find the time taken to fill the container.

   • \(2.00t^2 = 700\)
   • \(t = 18.7\) minutes (the positive root)

A rare plant has 6000 plants at \(t=0\), where t is in years. The number is halving every 4 years.

1. (a) Write down a model \(N = N_0e^{-kt}\), giving N0 and k.

   • \(N_0 = 6000\)
   • Half-life 4: \(e^{-4k} = \frac12\), so \(k = 0.173\)

2. (b) Use the model to find when the number first falls below 1.

   • \(6000e^{-0.173t} < 1\)
   • Solve \(6000e^{-kt} = 1\) for t: \(t = 50.2\) years

3. (c) Explain why the model cannot tell you exactly when the species dies out.

   • The curve never reaches 0 and N is not a whole number.
   • With few left, chance events decide the end.

A cup of coffee cools in a room at 20 °C. Its temperature is \(T = 20 + 70e^{-kt}\) °C, where t is in minutes. After 10 minutes, the difference between the coffee and the room has fallen to a fraction q of its starting value, where \(0 < q < 1\).

1. (a) Show that \(k = -\frac{1}{10}\ln q\).

   • After 10 minutes the difference is q times the start: \(70e^{-10k} = 70q\)
   • \(e^{-10k} = q\), so \(-10k = \ln q\)
   • \(k = -\frac{1}{10}\ln q\)

2. (b) Find an expression for the time to reach 35 °C in terms of q.

   • \(35 = 20 + 70e^{-kt}\), so \(e^{-kt} = \frac{15}{70}\)
   • \(-kt = \ln\frac{15}{70}\), and \(-k = \frac{\ln q}{10}\)
   • \(t = 10\ln\frac{15}{70} \div \ln q\)

3. (c) Find the time when \(q = 0.4\).

   • \(t = 10\ln\frac{15}{70} \div \ln 0.4 = 16.8\) minutes

The power used in a town is modelled by \(P\) megawatts, where t is the time in hours after midnight. The least power used in a day is 5 MW and the greatest is 9 MW. At midnight, \(P = 6\) and the power is increasing. The pattern repeats every day.

1. (a) Find a model \(P = a\sin(b(t-c)) + d\) with \(a > 0\) and \(0 \le c < 24\). Give c to 3 significant figures.

   • Amplitude \(a = \frac{9-5}{2} = 2.00\), middle line \(d = \frac{9+5}{2} = 7.00\)
   • Period 24: \(b = \frac{2\pi}{24} = \frac{\pi}{12}\)
   • \(P(0) = 6\): \(2.00\sin(-bc) + 7.00 = 6\), and P is increasing at \(t=0\)
   • \(c = 2.00\)

2. (b) A second model, Q, has the same period and the same middle value as P, half the amplitude, and its greatest value comes 3 hours later than P's. Write down Q in the same form.

   • Q has amplitude \(2.00\div 2 = 1.00\) and the same middle line \(d = 7.00\)
   • P is at its maximum at \(t = c + 6 = 8.00\); Q is 3 hours later
   • So the shift for Q is \(2.00 + 3 = 5.00\)
   • \(Q = 1.00\sin\left(\frac{\pi}{12}(t-5.00)\right) + 7.00\)

Use P and Q from the last question (power used by two towns, in MW). Use technology.

1. (a) Find the first time after \(t=0\) when \(P = Q\).

   • Graph \(y = P\) and \(y = Q\), or solve \(P = Q\) on the calculator
   • First intersection: \(t = 0.0883\)

2. (b) Find the maximum value of \(P + Q\) and the time it occurs, for \(0 \le t \le 24\).

   • Graph \(y = P + Q\) and use the maximum tool
   • Maximum \(= 16.8\) at \(t = 8.98\)

3. (c) Sketch \(P+Q\), labelling its maximum and minimum points.

   • Use the minimum tool as well
   • Minimum \(= 11.2\) at \(t = 21.0\)
   • The curve is a wave with one maximum and one minimum in 24 hours

4. (d) Write \(P + Q\) in the form \(a\sin(b(t-c)) + d\).

   • \(a = \frac{16.8-11.2}{2} = 2.80\), \(d = 14.0\), \(b = \frac{\pi}{12}\)
   • The maximum is at \(t = c + 6\), so \(c = 8.98 - 6 = 2.98\)

The population of a species in a reserve is modelled by \(P = \frac{5000}{1 + Ce^{-kt}}\), where t is the time in years after it is introduced.

1. (a) Given that \(P = 3000\) when \(t = 2\), show that \(C = \frac{2}{3}e^{2k}\).

   • \(3000 = \frac{5000}{1+Ce^{-2k}}\)
   • \(1 + Ce^{-2k} = 5/3\), so \(Ce^{-2k} = \frac{2}{3}\)
   • \(C = \frac{2}{3}e^{2k}\)

2. (b) The value of P when \(t = 6\) is 4500. Write down a second equation in C and k.

   • \(\frac{5000}{1+Ce^{-6k}} = 4500\)

3. (c) Use technology to solve the two equations for k and C.

   • Graph or solve the pair together on the calculator
   • \(k = 0.448\), \(C = 1.63\)

4. (d) Find the value of P when \(t = 0\).

   • \(P(0) = \frac{5000}{1+1.63} = 1900\)

5. (e) Find the least whole number of years for P to exceed 4800.

   • \(\frac{5000}{1+1.63e^{-0.448t}} = 4800\) gives \(t = 8.19\)
   • The least whole number of years is 9

The population of a town was 2000 at the start of 2020. Let t be the number of years after the start of 2020.

1. (a) A linear model assumes the population grows by 150 every year. Find the predicted number after 12 years.

   • \(2000 + 150 \times 12 = 3800\)

2. (b) Give one reason why a linear model is unlikely to be suitable for a long time.

   • The line keeps rising forever, but there is a limit (capacity) to the number.

3. (c) A logistic model \(P = \frac{5000}{1+Ce^{-kt}}\) is used instead. Find C.

   • At \(t=0\): \(\frac{5000}{1+C} = 2000\)
   • \(C = \frac{5000}{2000} - 1 = 1.50\)

4. (d) After 4 years the number is 3200. Find k, to 1 decimal place.

   • \(\frac{5000}{1+1.50e^{-4k}} = 3200\)
   • Solve for k: \(k = 0.245\), so \(k = 0.2\) to 1 d.p.

5. (e) Use your value of k to predict the number after 12 years. Sketch the graph of P against t.

   • \(P = \frac{5000}{1+1.50e^{-0.200\times 12}} = 4400\)
   • Sketch: an S-shaped curve starting at 2000 when \(t=0\), rising steeply and levelling off towards \(P = 5000\)

A colony of insects has 300 insects at \(t=0\), where t is in days. The number doubles every 4 days for \(0 \le t \le 12\), then halves every 5 days for \(t > 12\).

1. (a) Find the number of insects at \(t = 12\).

   • Doubling every 4 days: \(300\times 2^{12/4} = 2400\)

2. (b) Find the number of insects at \(t = 30\).

   • After \(t=12\) it halves every 5 days: \(2400\times\left(\tfrac12\right)^{18/5} = 198\)

3. (c) Find the time when the number of insects is back to 300.

   • \(2400\times\left(\tfrac12\right)^{(t-12)/5} = 300\)
   • \(\left(\tfrac12\right)^{(t-12)/5} = \frac{300}{2400}\)
   • Solve for t: \(t = 27.0\) days

4. (d) The doubling time is \(\alpha\) days and the halving time is \(\beta\) days. Find the condition on \(\alpha\) and \(\beta\) for the number of insects at \(t = 30\) to be greater than the starting number of insects.

   • At \(t=30\): \(N = 300\times 2^{12/\alpha}\times 2^{-18/\beta}\)
   • \(N > 300\) when \(2^{12/\alpha - 18/\beta} > 1\)
   • So \(\frac{12}{\alpha} > \frac{18}{\beta}\)