Exponential models, half-life and logarithmic models Problem set B One question at a time

The charge on a capacitor is \(Q=Q_0e^{-0.62t}\), with t in seconds. Find the time for the charge to halve.

• Half-life = −ln 2 ÷ k
• = −ln 2 ÷ (−0.62) = 1.12 seconds

Starting from \(P=P_0e^{kt}\), where k > 0, show that the doubling time is \(\dfrac{\ln2}{k}\).

• Set P = \(2P_0\): \(2P_0=P_0e^{kt}\)
• Divide by \(P_0\): \(2=e^{kt}\)
• Take ln: \(kt=\ln2\), so \(t=\ln2/k\)

A soil sample holds 120 units of salt where the background level is 45 units. The excess over the background halves every 5 days. Write the amount as \(S=d+Ae^{kt}\), with t in days.

• d = 45 and A = 120 − 45 = 75
• k = −ln 2 ÷ 5 = -0.139
• \(S=45+75e^{-0.139t}\)

A model \(y=a+b\ln x\) passes through (3, 2.5) and (30, 8). Find a and b.

• \(2.5=a+b\ln3\) and \(8=a+b\ln30\)
• Solve the pair on the calculator: a = −0.124, b = 2.39
• Check: \(-0.124+2.39\ln 30=8\)

A model is \(y=7.9-1.6\ln x\).

1. (a) Find y when x = 9.

   • \(y=7.9-1.6\ln 9=4.38\)

2. (b) Find x when y = 3.

   • \(7.9-1.6\ln x=3\)
   • Solve for x: x = 21.4

A tank of water is heated to 70°C in a cellar at 15°C. The temperature above the cellar halves every 45 minutes, where t is in minutes.

1. (a) Find a model \(T=d+Ae^{kt}\) for the temperature.

   • d = 15, A = 70 − 15 = 55, k = −ln 2 ÷ 45 = -0.0154
   • \(T=15+55e^{-0.0154t}\)

2. (b) Find when the temperature falls to 25°C.

   • \(15+55e^{-0.0154t}=25\)
   • Solve for t: t = 111 minutes

A sample of iodine-131 has a mass of 250 g at t = 0, where t is in days. Its mass halves every 8 days.

1. (a) Write a model \(M=Ae^{kt}\) for the mass.

   • Background 0, so d = 0 and A = 250
   • k = −ln 2 ÷ 8 = -0.0866
   • \(M=250e^{-0.0866t}\)

2. (b) Find the mass left after 20 days.

   • t = 20: \(M=250e^{-0.0866\times20}=44.2\)

The temperature T°C of a lake at depth d metres is modelled by \(T=a+b\ln d\). It is 18.5°C at 3 m and 9°C at 25 m.

1. (a) Find a and b.

   • \(18.5=a+b\ln 3\) and \(9=a+b\ln 25\)
   • Solve the pair: a = 23.4, b = −4.48

2. (b) Find the temperature at 10 m deep.

   • \(23.4-4.48\ln 10=13.1\)

3. (c) Find the depth at which the temperature is 12°C.

   • \(23.4-4.48\ln x=12\)
   • Solve for x: x = 12.8

The population of a town increases by 2.4% each year. It is modelled by \(P=P_0e^{kt}\), where t is in years.

1. (a) Find k.

   • \(e^k=1.024,\ k=\ln1.024=0.0237\)

2. (b) Find the time for the population to multiply by 2.5.

   • \(e^{0.0237t}=2.5\)
   • Solve for t: t = 38.6

A fossil shell has 28% of the carbon-14 of a living shell. The half-life of carbon-14 is 5730 years. The amount is \(M=M_0e^{kt}\), where t is in years.

1. (a) Find k.

   • \(k=-\ln2\div5730=-0.000121\)

2. (b) Find the age of the shell.

   • \(e^{-0.000121t}=0.28\)
   • Solve for t: t = 10500

The graph shows the depth D cm of water in a draining pool t hours after the plug is pulled. It is modelled by \(D=d+Ae^{kt}\).

1. (a) Write down d and A.

   • The curve levels off at 10, so d = 10. At x = 0 the value is 70, so A = 70 − 10 = 60

2. (b) Find k.

   • \(25=10+60e^{6k}\)
   • \(e^{6k}=15\div60\), so k = ln(0.25) ÷ 6 = -0.231

3. (c) Find D when t = 15.

   • \(10+60e^{-0.231\times15}=11.9\)

A model is \(y=a+b\ln x\).

1. (a) Show that trebling x adds \(b\ln3\) to y, whatever x is.

   • New y: \(a+b\ln(3x)=a+b\ln 3+b\ln x\)
   • Subtract the old y: \(b\ln 3\), which does not depend on x

2. (b) Find the increase in y when x trebles and b = 2.4.

   • \(2.4\ln 3=2.64\)