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AI HL · 12.7

Mathematical Models

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12.7.2

Exponential models, half-life and logarithmic models

12.7.2

Exponential models, half-life and logarithmic models

Target: I can show that the half-life of y = y₀e^(kt) is −ln 2 / k, and use it to find k or the half-life. (+5 more)

By the end of this section

Key points

Half-life

  • In \(y=y_0e^{kt}\), k > 0 is growth and k < 0 is decay. The half-life is the time for y to halve.
  • Half-life: \(t_{0.5}=-\dfrac{\ln 2}{k}\). It does not depend on the starting value \(y_0\).
  • To prove it, put \(y=\tfrac12y_0\), divide by \(y_0\), then take ln of both sides.
  • A half-life is a time, so it is positive. For decay k is negative, so \(-\ln2/k\) is positive.

A background level

  • Use \(y=d+Ae^{kt}\) when the quantity settles at a background level d, not at 0.
  • d is the background level. A is the initial value minus d. The curve has a horizontal asymptote at y = d.
  • The half-life applies to the part above the background, A e^(kt), not to y itself.
  • Find k from the half-life: \(k=-\ln2\div t_{0.5}\). Do not multiply.

Natural logarithm models

  • \(y=a+b\ln x\) has two parameters, so you need two points and two equations.
  • Put each point into the model. Solve the pair for a and b on the calculator.
  • The model passes through (1, a) because ln 1 = 0. It has the y-axis as a vertical asymptote and is defined for x > 0 only.
  • To find x for a given y, solve the equation on the calculator, or rearrange: \(x=e^{(y-a)/b}\).

Rates and ages

  • Grows by r% each unit of time: \(e^k=1+\dfrac{r}{100}\), so \(k=\ln\left(1+\dfrac{r}{100}\right)\). Decays by r%: \(k=\ln\left(1-\dfrac{r}{100}\right)\), which is negative.
  • The time to double: solve \(e^{kt}=2\), so \(t=\dfrac{\ln2}{k}\). The time to multiply by any factor f: \(t=\dfrac{\ln f}{k}\).
  • Carbon dating: k = −ln 2 ÷ half-life. If a fraction f of the carbon-14 is left, solve \(e^{kt}=f\): \(t=\dfrac{\ln f}{k}\).
  • Use the fraction that is left, not the fraction that was lost: 62% left means f = 0.62.

On the calculator

Solving the pair of equations for a and b

  • Write the two equations: put each point into \(y=a+b\ln x\).
  • Solve the pair for a and b on the calculator.
  • Keep full accuracy in a and b until the end. Round only the final answer.

Screenshot and video to come

Solving a model equation for t or x

  • Type the equation, such as \(40+160e^{kt}=55\), with k stored.
  • Solve for the variable.
  • Check the answer is sensible for the context.

Screenshot and video to come

Checking a model with a graph

  • Graph the model and the horizontal line at the value you want.
  • The intersection is the solution. For \(d+Ae^{kt}\) the curve levels off at y = d.

Screenshot and video to come

Explore: y = d + Aekt

Change A, k and d and watch the curve. The orange point marks the half-life (or the doubling time). Predict where it will be before you move a slider.



Initial value, A + d
Horizontal asymptote
Half-life

Choose an example or move a slider.

  1. Change only A. Does the half-life change? Look at the formula ln 2 ÷ |k| and say why.
  2. Press Cooling to room temperature. Why does the curve level off at 20 and not at 0? What is d for the tea?
  3. Make k twice as large, keeping its sign. What happens to the half-life? Check it with ln 2 ÷ |k|.
  4. Press Growth, then set d to 30. Which quantity now doubles at each doubling time: y, or the gap between y and d? How can you tell from the graph?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Where the half-life formula comes from

A medicine in the blood is modelled by \(C=80e^{-0.14t}\), where C is in mg/l and t is in hours. (a) Show that the time for C to halve does not depend on the 80. (b) Find the half-life.

Working
  1. Halve it

    \(\tfrac12y_0=y_0e^{kt}\)

    Write the initial value as \(y_0\) and set y to half of it. For this medicine \(y_0=80\).

  2. Divide by the start

    \(\tfrac12=e^{kt}\)

    The \(y_0\) cancels, so the start value does not matter. That is part (a).

  3. Take ln

    \(\ln\tfrac12=kt\), so \(t=\dfrac{-\ln2}{k}\)

    \(\ln\tfrac12=-\ln2\).

  4. Put in k

    \(t=\dfrac{-\ln2}{-0.14}=4.95\) hours

    Both signs are negative, so the half-life is positive.

Video walkthrough coming
Cooling towards the room temperature

A cup of tea starts at 85°C in a room at 20°C. The temperature above the room halves every 12 minutes. Model the temperature as \(T=d+Ae^{kt}\), t in minutes. (a) Find d, A and k. (b) Find when the tea reaches 40°C.

Working
  1. Background and start

    d = 20, A = 85 − 20 = 65

    The tea settles at room temperature, 20. A is the start minus the background.

  2. Find k

    \(k=-\ln2\div12=-0.0578\)

    The half-life belongs to the 65 above the room temperature.

  3. The model

    \(T=20+65e^{-0.0578t}\)

    Check: at t = 0, T = 85.

  4. Solve for t

    Solve for t: \(20+65e^{-0.0578t}=40\): t = 20.4

    The tea reaches 40°C after 20.4 minutes.

Video walkthrough coming
Fitting a logarithmic model

The number of fish species S in a lake of area x km² is modelled by \(S=a+b\ln x\). A 4 km² lake has 9 species. A 60 km² lake has 17 species. Find a and b.

Working
  1. Two equations

    \(9=a+b\ln4\) and \(17=a+b\ln60\)

    Put each point into the model.

  2. Solve

    Solve the pair for a and b: a = 4.90, b = 2.95

    Keep full accuracy for later use.

  3. The model

    \(S=4.90+2.95\ln x\)

    Check with the point (4, 9): 4.90 + 2.95 × ln 4 = 9.

Video walkthrough coming
Predicting from a fitted model

The typing speed w (words per minute) of a learner after t weeks of practice is modelled by \(w=11.4+9.2\ln t\). (a) Find w after 12 weeks. (b) Find when the speed reaches 40 words per minute.

Working
  1. Part (a)

    \(w=11.4+9.2\ln12=34.3\)

    Put t = 12 in. About 34.3 words per minute.

  2. Part (b)

    \(11.4+9.2\ln t=40\)

    Set w = 40.

  3. Solve

    Solve for t: t = 22.4

    Or rearrange: \(\ln t=3.11\), so \(t=e^{3.11}\). About 22 weeks.

Video walkthrough coming
From a percentage rate to k

A population grows by 3.5% each year. It is modelled by \(P=P_0e^{kt}\), where t is in years. (a) Find k. (b) Find the time for the population to double.

Working
  1. One year later

    \(P=1.035P_0\ \text{after 1 year}\)

    3.5% more: multiply by 1.035.

  2. Find k

    \(e^k=1.035,\ k=\ln1.035=0.0344\)

    Take the natural logarithm.

  3. Doubling

    \(e^{0.0344t}=2\)

    Solve for t: t = 20.1 years.

  4. Check

    \(20.1\times0.0344\approx0.693=\ln2\)

    The doubling time does not depend on the starting population.

Video walkthrough coming
Dating a bone

Carbon-14 has a half-life of 5730 years. A bone has 62% of the carbon-14 that a living bone has. The amount of carbon-14 is modelled by \(M=M_0e^{kt}\), where t is in years. (a) Find k. (b) Find the age of the bone.

Working
  1. Find k

    \(k=-\ln2\div5730=-0.000121\)

    Negative: the amount falls.

  2. The equation

    \(M_0e^{-0.000121t}=0.62M_0\)

    The fraction left is 0.62, so M₀ cancels.

  3. Solve

    Solve for t: t = 3950

    The bone is about 3950 years old.

Video walkthrough coming