Menu
Sign up Sign in

AI SL · 12.7

Mathematical Models

0 of 15 skills done

0
12.7.3

Mixed modelling practice

12.7.3

Mixed modelling practice

Target: I can find the constant k in a decay, growth or cooling model from a half-life, a doubling time, or a later value. (+5 more)

By the end of this section

Key points

Exponential models

  • Decay \(M = M_0e^{-kt}\), growth \(P = Ae^{kt}\): k is positive in both. The minus sign is in the model.
  • Half-life h: \(e^{-kh} = \frac12\), so \(k = \frac{\ln 2}{h}\). Doubling time d gives \(k = \frac{\ln 2}{d}\).
  • For \(S = S_0 b^{-t}\), the half-life is \(\frac{\ln 2}{\ln b}\). It does not depend on \(S_0\).
  • Newton cooling: \(T = B + Ae^{-kt}\). B is the surroundings, the value T settles to. A is the starting difference \(T(0) - B\), negative if the object starts colder than its surroundings.
  • Always subtract B before you divide: it is the difference from the surroundings that decays.

Other models

  • Logarithmic: \(R = p\ln A + q\). Two points give two equations in p and q. To find A from R, undo the ln: \(A = e^{(R-q)/p}\).
  • Piecewise: put each value into the piece it belongs to. Continuity means both pieces give the same value where they meet.
  • Sinusoidal: \(a\sin(b(t-c)) + d\) with \(a = \frac{\max - \min}{2}\), \(d = \frac{\max+\min}{2}\), \(b = \frac{360}{\text{period}}\). Use a starting condition to find c.
  • Logistic: \(P = \frac{L}{1+Ce^{-kt}}\). L is the limit. At \(t=0\), \(P = \frac{L}{1+C}\).

Assumptions and limits

  • An assumption is something the model takes as fixed: no new arrivals, constant surroundings, the same rate all the time.
  • A model breaks down where it gives impossible values: a fraction of an animal, a number below 0, or growth without limit.
  • Use unrounded values in later steps. Round only the final answer.

On the calculator

Solving for the time

  • Write the equation with the model on one side and the value on the other.
  • Use the solver on the equation and read the value of t.
  • Check that the answer is sensible: the graph should meet the line at that t.

Screenshot and video to come

Fitting the unknowns from two points

  • Write the two equations, one for each point.
  • Solve them together for the unknowns (p and q, or k and C).
  • Keep the unrounded values stored for the next part.

Screenshot and video to come

Intersections and maximum of a graph

  • Graph both models on the same axes in the window given by the domain.
  • Use the intersection tool for the first time two models are equal.
  • Graph the sum and use the maximum and minimum tools.

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?

Explore: when do two branches meet?

The orange branch contains the unknown a. Work out the value of a that makes the model continuous, type it in, then check the graph. A filled circle means the point belongs to that branch. An open circle means it does not.



Value of a needed

Move the slider or type a value.

  1. Before you touch the slider, put the boundary value of x into both branches and work out a. Then type it in. Does the graph agree?
  2. Look at the circles at the boundary. Which branch includes the boundary point, and how can you tell? What happens to the circles when the model is continuous?
  3. On the (x + 1)a model you cannot find a by dividing. What do you do instead, and why does it work?
  4. Try a value that is close to the right one but not equal. Does the model count as continuous? What does the gold bar show?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
The cooling constant

A cup of tea cools in a room at a constant temperature. Its temperature, \(T\) °C, \(t\) minutes after it is poured, is modelled by \(T = B + Ae^{-kt}\). The tea is poured at 85 °C and is 52 °C after 12 minutes. The room is at 20 °C. Find the value of k.

Working
  1. B and A

    \(B = 20\). At \(t=0\): \(85 = 20 + A\), so \(A = 65\)

    B is the room temperature, the value T settles to. A is the starting temperature difference.

  2. Use the later value

    \(52 = 20 + 65e^{-12k}\), so \(e^{-12k} = \frac{32}{65}\)

    Subtract 20 from both sides before you divide.

  3. Solve for k

    Solve for k: \(k = 0.0591\)

    Use the calculator's solver on \(52 = 20 + 65e^{-12k}\), or take ln of both sides by hand.

Video walkthrough coming
Time to cool

Use the model \(T = 20 + 65e^{-kt}\) with \(k = 0.0591\) (unrounded value stored) for the tea above. Find how long the tea takes to cool to 30 °C.

51015202530354020406080100t (minutes)T (°C)T
Working
  1. Write the equation

    \(30 = 20 + 65e^{-kt}\)

    Put T = 30 into the model.

  2. Solve for t

    Solve for t: \(t = 31.7\) minutes

    From the calculator. Check on the graph: the curve meets the line T = 30 at this t.

  3. Why it is above 20

    \(T \to 20\) as \(t \to \infty\)

    The tea never goes below the room temperature, so the equation only has an answer for T > 20.

Video walkthrough coming
Initial amount from a half-life

A radioactive substance has a half-life of 8 days. After 20 days, 50 g remain. The mass \(M\) g after \(t\) days is modelled by \(M = M_0e^{-kt}\). Find the initial mass \(M_0\).

Working
  1. The constant

    Half-life 8 days: \(\frac12 = e^{-8k}\), so \(k = \frac{\ln 2}{8} = 0.0866\)

    In one half-life the amount halves. Use this to find k.

  2. The equation

    \(50 = M_0e^{-20k}\)

    Put the later mass in.

  3. Solve

    Solve for \(M_0\): \(M_0 = 283\) g

    Keep k unrounded. The answer is larger than 50, as the substance has decayed.

Video walkthrough coming
Half-life with base 3

The mass of a sample is \(S = 600 \times 3^{-t}\) grams, where \(t\) is the time in hours. Find the half-life.

Working
  1. Half the start

    \(300 = 600 \times 3^{-t}\), so \(3^{-t} = \frac12\)

    The half-life is the time for S to fall to half of 600. The 600 cancels: it does not matter what the start is.

  2. Solve

    Solve for t: \(t = 0.631\) hours

    By hand: \(-t\ln 3 = -\ln 2\). The calculator gives the same.

Video walkthrough coming
Growth from a doubling time

A culture of bacteria doubles in size every 3.5 hours. The number of bacteria is modelled by \(P = Ae^{kt}\), where \(t\) is in hours. Find k.

Working
  1. Doubling

    \(Ae^{3.5k} = 2A\), so \(e^{3.5k} = 2\)

    After 3.5 hours the number is twice the number at the start, whatever A is.

  2. Solve

    \(k = \frac{\ln 2}{3.5} = 0.198\)

    k is positive for growth. A is the number at the start.

Video walkthrough coming
A logarithmic model

The strength R of a signal is modelled by \(R = p\ln A + q\), where A is the power. When \(A = 40\), \(R = 3.5\). When \(A = 900\), \(R = 6.1\). (a) Find p and q. (b) Find the value of A when \(R = 7\).

Working
  1. Two equations

    \(3.5 = p\ln 40 + q\) and \(6.1 = p\ln 900 + q\)

    Each point gives one equation in p and q.

  2. Solve

    Solve the pair: \(p = 0.835\), \(q = 0.420\)

    From the calculator. Or subtract: \(2.6 = p\ln 22.5\).

  3. Inverse

    \(7 = 0.835\ln A + 0.420\), so \(\ln A = 7.88\)

    To find A from R, undo the ln.

  4. Answer

    \(A = e^{7.88} = 2640\)

    For a range of R, find A for each end. A bigger R needs a bigger A, as p is positive.

Video walkthrough coming
Filling a tank, in two stages

Water runs into a tank. The volume \(V\) litres after \(t\) minutes is \(V = pt\) for \(0 \le t \le 4\), \(V = qt^2\) for \(t > 4\). After 2.5 minutes there are 30 litres. The model is continuous at \(t = 4\). The tank holds 500 litres. Find the time to fill the tank.

2468101214100200300400500600t (minutes)V (litres)V = ptV = qt²
Working
  1. Find p

    \(30 = p \times 2.5\), so \(p = 12.0\)

    2.5 is in the first stage, \(0 \le t \le 4\).

  2. Continuity

    \(12.0 \times 4 = q \times 4^2\), so \(q = 3.00\)

    The two pieces give the same V at \(t=4\): no jump.

  3. Full tank

    \(3.00t^2 = 500\), so \(t = 12.9\) minutes

    500 is larger than \(V(4)=48.0\), so we are in the second stage. Take the positive root.

Video walkthrough coming