Mixed modelling Problem set B One question at a time

Give each answer to 3 significant figures.

1. (a) A medical tracer has a half-life of 14 days. The mass is modelled by \(M = M_0e^{-kt}\), with t in days. Find k.

   • Half-life 14 days: \(e^{-14k} = \frac12\)
   • \(k = \frac{\ln 2}{14} = 0.0495\)

2. (b) The population of a town doubles every 25 years (t in years). The number is modelled by \(N = N_0e^{kt}\). Find k.

   • Doubling: \(e^{25k} = 2\)
   • \(k = \frac{\ln 2}{25} = 0.0277\)

3. (c) The size of a sample is \(S = 900 \times 1.8^{-t}\), with t in years. Find its half-life.

   • Half the start: \(1.8^{-t} = \frac12\)
   • Solve for t: \(t = \frac{\ln 2}{\ln 1.8} = 1.18\) years

A bottle of milk is taken from a fridge at 4 °C and warms up in a kitchen. Its temperature \(T\) °C, \(t\) minutes later, is \(T = B + Ae^{-kt}\). After 25 minutes it is 15 °C. The kitchen is at 23 °C.

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

   • B is the surroundings: \(B = 23\)
   • At \(t=0\): \(4 = 23 + A\), so \(A = -19\)
   • A is the starting difference from the surroundings; B is the temperature the milk approaches.

2. (b) Find the value of k.

   • \(15 = 23 + (-19)e^{-25k}\)
   • Solve for k: \(k = 0.0346\)

3. (c) Find the time for the temperature to reach 20 °C.

   • \(20 = 23 + (-19)e^{-kt}\)
   • Solve for t: \(t = 53.3\) minutes

The mass of a radioactive waste sample is \(M = M_0e^{-0.045t}\) g, where t is the time in days since it was stored. After 14 days 80 g remain.

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

   • \(80 = M_0e^{-0.045\times 14}\)
   • Solve for \(M_0\): \(M_0 = 150\) g

2. (b) Find the time at which the amount reaches 25 g.

   • \(25 = 150e^{-0.045t}\)
   • Solve for t: \(t = 39.8\) days

The number of bacteria on a worktop after cleaning is modelled by \(N = 5000e^{-0.9t}\), where t is the time in hours.

1. (a) Use the model to find the number of bacteria after 8 hours.

   • \(N = 5000e^{-0.9\times 8} = 3.73\)

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 bacterium and never reaches 0, so it fails once the number is small.

A fitness score R is modelled by \(R = p\ln A + q\), where A is the number of hours of training in a month. When \(A = 2\), \(R = 35\). When \(A = 60\), \(R = 71\).

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

   • \(35 = p\ln 2 + q\) and \(71 = p\ln 60 + q\)
   • Solve the pair: \(p = 10.6\), \(q = 27.7\)

2. (b) Find R when A = 25.

   • \(R = 10.6\ln 25 + 27.7 = 61.7\)

3. (c) Find the range of values of A for which \(50 \le R \le 65\).

   • \(A = e^{(R-q)/p}\), as R increases when A increases
   • \(R = 50\): \(A = 8.25\); \(R = 65\): \(A = 34.0\)
   • \(8.25 \le A \le 34.0\)

A tank on a farm is filled by a pump. The volume \(V\) litres after \(t\) minutes is modelled by \(V = pt\) for \(0 \le t \le 5\) and \(V = qt^2\) for \(t > 5\). After 1.5 minutes there are 22.5 litres. The tank holds 900 litres.

1. (a) Find the value of p.

   • \(22.5 = p \times 1.5\)
   • \(p = 15.0\)

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

   • At \(t=5\): \(15.0 \times 5 = q \times 5^2\)
   • \(q = 3.00\)

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

   • \(3.00t^2 = 900\)
   • \(t = 17.3\) minutes (the positive root)

There are 900 frogs of a rare species in a nature reserve at \(t=0\), with t in years. The population halves every 6 years.

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

   • \(N_0 = 900\)
   • Half-life 6: \(e^{-6k} = \frac12\), so \(k = 0.116\)

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

   • \(900e^{-0.116t} < 1\)
   • Solve \(900e^{-kt} = 1\) for t: \(t = 58.9\) 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.

The temperature in a greenhouse is modelled by \(P\) °C, where t is the time in hours after midnight. The lowest temperature in a day is 2.5 °C and the highest is 6.5 °C. At midnight, \(P = 3\) and the temperature is rising. 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 \(a = \frac{6.5-2.5}{2} = 2.00\), middle line \(d = \frac{6.5+2.5}{2} = 4.50\)
   • Period 24: \(b = 360\div24 = 15\)
   • \(P(0) = 3.0\): \(2.00\sin(-bc) + 4.50 = 3.0\), and P is increasing at \(t=0\)
   • \(c = 3.24\)

2. (b) A second greenhouse is modelled by Q, with the same period and middle value as P, half the amplitude, and its greatest value 4 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 \(d = 4.50\)
   • P is at its maximum at \(t = c + 6 = 9.24\); Q is 4 hours later
   • So the shift for Q is \(3.24 + 4 = 7.24\)
   • \(Q = 1.00\sin\left(15(t-7.24)\right) + 4.50\)

Use P and Q from the last question (temperatures in two greenhouses, in °C). 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: \(t = 1.24\)

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 \(= 11.6\) at \(t = 10.5\)

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

   • Use the minimum tool as well
   • Minimum \(= 6.35\) at \(t = 22.5\)
   • The curve is a wave with one maximum and one minimum in 24 hours

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

   • \(a = \frac{11.6-6.35}{2} = 2.65\), \(d = 9.00\), \(b = 15\)
   • The maximum is at \(t = c + 6\), so \(c = 10.5 - 6 = 4.51\)

A school's online club had 800 members at the start of the year. Let t be the number of months after that.

1. (a) A linear model assumes the number of members grows by 90 every month. Find the predicted number after 8 months.

   • \(800 + 90 \times 8 = 1520\)

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 \(P = \frac{3000}{1+Ce^{-kt}}\) is used instead. Find C.

   • At \(t=0\): \(\frac{3000}{1+C} = 800\)
   • \(C = \frac{3000}{800} - 1 = 2.75\)

4. (d) After 5 months the number is 1900. Find k, to 1 decimal place.

   • \(\frac{3000}{1+2.75e^{-5k}} = 1900\)
   • Solve for k: \(k = 0.312\), so \(k = 0.3\) to 1 d.p.

5. (e) Use your value of k to predict the number after 10 months. Sketch the graph of P against t.

   • \(P = \frac{3000}{1+2.75e^{-0.300\times 10}} = 2640\)
   • Sketch: an S-shaped curve starting at 800 when \(t=0\), rising steeply and levelling off towards \(P = 3000\)