Unit 6 · HL

Modelling closeout

Half-life · logarithmic models and scales · phase shift · piecewise models

SL students: enrichment only — these are not SL requirements.

Retrieval · 4 quick prompts

Bring the model back

1

For \(A(t)=A_0b^t\), which values of \(b\) produce growth or decay?

2

What operation reverses exponentiation?

3

What do \(A_0\) and \(b\) mean in context?

4

Predict the shape and long-term behaviour of exponential decay.

Retrieval answers

Exponential growth and decay

Growth: \(b>1\). Decay: \(0<b<1\). Logarithms reverse exponentiation. \(A_0\) is the initial amount; \(b\) is the multiplier per time unit. Exponential decay approaches 0 without reaching it in the model.

Predict first

A half-life is six hours

What fraction remains after 6, 12, 18 and 24 hours?

hours

Repeated multiplicative decay

Each interval keeps one half

\(1\;\to\;\frac12\;\to\;\frac14\;\to\;\frac18\;\to\;\frac1{16}\)

Meaning

Equal time intervals create equal ratios, not equal decreases.

Invariant

For exponential decay, half-life is constant and does not depend on the initial amount.

Two equivalent forms

Choose the form that fits the information

Multiplier form

\[A(t)=A_0b^t\]

For \(0<b<1\):

\[t_{1/2}=\frac{\ln(0.5)}{\ln(b)}\]

Continuous-rate form

\[A(t)=A_0e^{-kt}\]

For \(k>0\):

\[t_{1/2}=\frac{\ln 2}{k}\]

Predict before calculating

When will half remain?

\[A(t)=800(0.92)^t\]

Will the half-life be closer to 5, 10 or 20 time units? Explain your estimate.

Worked example

Make “half” an equation

Half of 800 is 400:

\[800(0.92)^t=400\]

\[(0.92)^t=0.5\]

\[t=\frac{\ln(0.5)}{\ln(0.92)}\approx 8.31\]

Interpretation: the modelled amount halves every 8.31 time units, regardless of the starting amount.

Unfamiliar attempt · predict first

Write the model in \(e^{-kt}\) form

A tracer begins at 450 mg and has a half-life of 11 days.

Find \(k\), write a model, then explain what the model predicts after 22 days.

\[k=\frac{\ln2}{11}\approx0.0630,\qquad A(t)=450e^{-0.0630t}\]

After 22 days, two half-lives have passed, so about \(450(\tfrac12)^2=112.5\) mg remains.

Predict the pattern

A slow-growth model

A skill score is modelled by

\[S(t)=12+5\ln(t+1),\qquad t\ge0\]

\(t\): weeks of practice · \(S\): score points

Before calculating

Will equal increases in \(t\) create equal increases in \(S\)?

Estimate \(S(0)\), \(S(4)\), \(S(20)\).

Calculate · then interpret

Growth continues, but slows

\(t=0\)

\(S(0)=12\)

\(t=4\)

\(S(4)=12+5\ln5\approx20.0\)

\(t=20\)

\(S(20)=12+5\ln21\approx27.2\)

The score keeps increasing, but the same additional practice produces smaller and smaller gains.

Solve and defend the domain

When does the model reach 25?

\[12+5\ln(t+1)=25\]

Undo each operation. Then state both mathematical and contextual domains.

\[\ln(t+1)=2.6\]

\[t=e^{2.6}-1\approx12.5\]

Mathematical domain: \(t>-1\). Contextual domain: \(t\ge0\), restricted further to the period for which practice data make the model credible.

Long term: logarithmic growth is unbounded but progressively slower; exponential growth accelerates and eventually dominates.

Do not confuse these

Model family vs transformation

Natural-logarithmic model

\[y=a+b\ln(x)\]

The logarithm is part of the model itself. The original \(y\) changes slowly as \(x\) grows.

Logarithmic transformation

\[y=ab^x\Rightarrow \ln y=\ln a+x\ln b\]

We transform another model to reveal a linear relationship and estimate its parameters.

Reading a logarithmic axis

Where does 50 belong?

Ordinary logarithmic axisbase 10 · positive values only

Horizontal axis: quantity (log scale), domain \(1\le x\le10000\)

Is 50 halfway between 10 and 100? What does equal spacing mean here?

Reading a logarithmic axis

Equal spacing means equal ratios

Spacing

\(1,10,100,1000,10000\) are equally spaced because each step multiplies by 10.

50 is not halfway

The midpoint between 10 and 100 on a log axis is \(\sqrt{10\cdot100}\approx31.6\).

Restriction

Zero and negative values cannot appear on an ordinary logarithmic axis.

Linearising an exponential model

How can a curve become a line?

\[y=ab^x\]

Take natural logs of both sides. Which parameter will control the gradient?

Semi-log linearisation

Plot \(\ln(y)\) against \(x\)

\[\ln(y)=\ln(a)+x\ln(b)\]

Straight-line form

\[Y=c+mx\]

\(Y=\ln y\), \(c=\ln a\), \(m=\ln b\)

Recover the model

\[a=e^{\text{intercept}},\qquad b=e^{\text{gradient}}\]

Predict before viewing the transform

A growing culture

Time, \(x\) (h)	0	1	2	3	4	5	6
Mass, \(y\) (mg)	24.0	37.2	57.7	89.4	138.5	214.7	332.8

Original representationlinear–linear

\(x\): time (h), linear, \(0\le x\le6\) · \(y\): mass (mg), linear

Transformed representationsemi-log linearisation

\(x\): time (h), linear · vertical: \(\ln(y/\mathrm{mg})\), \(3.0\le\ln y\le6.0\)

What does straightness suggest?

Evidence, not automatic validity

Mathematical evidence

The transformed points are approximately linear, with gradient \(0.438\) and intercept \(3.178\).

\[a=e^{3.178}\approx24.0,\qquad b=e^{0.438}\approx1.55\]

Contextual judgement

A possible model is \(y=24(1.55)^x\), but straightness alone does not prove unlimited growth, causal validity, or safe extrapolation.

Optional introduction · no independent mastery yet

Log-log linearisation

\[y=ax^n\quad\Rightarrow\quad \ln(y)=\ln(a)+n\ln(x)\]

Plot

\(\ln(y)\) against \(\ln(x)\)

Read

gradient \(=n\), intercept \(=\ln a\), so \(a=e^{\text{intercept}}\)

Checkpoint · 1 and 2

Calculate and interpret

1 · Half-life

\[P(t)=600e^{-0.18t}\]

Find the half-life and interpret it.

2 · Natural log

\[y=4+7\ln(x+1)\]

Interpret the initial value and describe the long-term behaviour for \(x\ge0\).

Checkpoint · 3 and 4

Represent and explain

3 · Log axis

On a base-10 log axis, which is farther right: 20 or 50? Explain why 50 is not halfway from 10 to 100.

4 · Semi-log evidence

A graph of \(\ln(y)\) against \(x\) is approximately straight. What does this suggest? Why is contextual evidence still needed?

Checkpoint answers

Check, correct, explain

1. \(t_{1/2}=\ln2/0.18\approx3.85\) time units; the modelled quantity halves every 3.85 time units.

2. \(y(0)=4\). The model grows without bound but at a decreasing rate.

3. 50 is farther right. Position depends on the logarithm; the log-midpoint is \(\sqrt{1000}\approx31.6\).

4. It suggests a possible exponential relationship. Check residuals, domain, mechanism and whether extrapolation is sensible.

Lesson 1 retrieval · 10 minutes

Four connections

1. Explain why half-life does not depend on \(A_0\).

2. Compare logarithmic growth with exponential growth.

3. In \(\ln y=c+mx\), recover \(a\) and \(b\) for \(y=ab^x\).

4. What can a straight transformed plot support — and what can it not prove?

Ratios drive half-life; logarithmic growth slows; \(a=e^c\), \(b=e^m\); transformed linearity is model evidence, not contextual proof.

Optional · predict first

A possible power model

A material test records radius \(r\) (mm) and supported mass \(M\) (mg).

\(r\) (mm)	1	2	4	8	16
\(M\) (mg)	3.20	9.05	25.60	72.41	204.80

Original datalinear–linear

\(r\): mm, linear, \(1\le r\le16\) · \(M\): mg, linear

Transformed datalog–log

horizontal \(\ln(r/\mathrm{mm})\) · vertical \(\ln(M/\mathrm{mg})\)

Recover the power model

Gradient gives the exponent

\[\ln M\approx1.163+1.500\ln r\]

Exponent

\[n=1.500\]

Coefficient

\[a=e^{1.163}\approx3.20\]

A possible model is \(M=3.20r^{1.50}\). The transformed line is mathematical evidence for a power model; it does not automatically validate the material mechanism, measurement range or extrapolation.

Retrieve what is known

Start without phase shift

\[f(t)=a\sin(bt)+d\]

Amplitude

\(|a|\)

Principal axis

\(y=d\)

Period

\(\frac{2\pi}{|b|}\) radians or \(\frac{360^\circ}{|b|}\) degrees

Add horizontal position

Phase shift is a meaningful event

\[f(t)=a\sin\bigl(b(t-c)\bigr)+d\]

At \(t=c\)

The sine input is 0, so \(f(c)=d\).

If \(a>0\) and \(b>0\)

The graph crosses the principal axis in the positive direction.

“Moves right” is incomplete: state what contextual feature occurs at the shifted time.

Predict before graphing

Daylight model

\[D(t)=3\sin\!\left(\frac{2\pi}{12}(t-3)\right)+12\]

\(t\): months after the start of a 24-month study · \(D\): daylight hours · \(0\le t\le24\)

Predict \(D(3)\) and the direction of crossing.

Identify amplitude, principal axis and period.

Graph and interpretation

At \(t=3\), the cycle is rising through its axis

Daylight modellinear axes

\(t\): months, \(0\le t\le24\) · \(D\): daylight hours, \(8\le D\le16\)

Amplitude \(=3\) h
Principal axis \(D=12\) h
Period \(=12\) months

\(D(3)=12\). The model crosses 12 daylight hours in the positive direction at month 3.

Common algebraic trap

The written angle is not automatically the shift

\[a\sin(bx-\varphi)+d=a\sin\!\left(b\left(x-\frac{\varphi}{b}\right)\right)+d\]

Do not say

“The phase shift is \(\varphi\)” automatically.

Instead

Factor \(b\). The phase shift is \(\varphi/b\), with units matching \(x\).

Predict the number of solutions

When is daylight 14 hours?

\[3\sin\!\left(\frac{2\pi}{12}(t-3)\right)+12=14,\qquad0\le t\le24\]

How many solutions should occur in two full periods? Estimate them from the graph before solving.

Technology: graph \(Y_1=D(t)\) and \(Y_2=14\), use intersections, then verify every result is inside \([0,24]\).

Solve within the contextual domain

Four crossings in 24 months

\[\sin\!\left(\frac{\pi}{6}(t-3)\right)=\frac23\]

One period gives two angles; the 24-month domain contains two periods.

\[t\approx4.39,\;7.61,\;16.39,\;19.61\]

The model predicts 14 daylight hours at approximately these four months of the study.

Predict the correct rule

A delivery charge changes after 4 km

\[C(d)=\begin{cases}8,&0<d\le4\\8+1.5(d-4),&d>4\end{cases}\]

\(d\): delivery distance (km) · \(C\): charge (RM)

Which rule applies at \(d=4\)? Why is the second rule written with \((d-4)\)?

Select · calculate · interpret

Evaluate the boundary deliberately

\(C(2)\)

Which interval?

\(C(4)\)

Which boundary symbol?

\(C(10)\)

How many km are charged extra?

\(C(2)=8\)

\(C(4)=8\)

\(C(10)=8+1.5(6)=17\)

Boundary and continuity

Do the rules meet at the switch?

From the first rule

\[C(4)=8\]

From the second expression

\[8+1.5(4-4)=8\]

The pieces meet at \((4,8)\), so this model is continuous at the switch. Continuity is not required for every piecewise model.

Model judgement

Why use more than one rule?

Different mechanisms

A flat base charge and a distance charge describe different regimes.

Contextual domain

Use \(0<d\le d_{\max}\), where \(d_{\max}\) is the service boundary.

Possible jumps

Tiered prices or entry fees may create discontinuities. That can be contextually correct.

Final check · 1 and 2

Decay and slow growth

1 · Half-life

\[M(t)=940(0.87)^t\]

Find and interpret the half-life.

2 · Natural logarithm

\[R(t)=9+4\ln(t+1)\]

Find \(R(8)\), then describe what the model says about future growth.

Final check · 3 and 4

Transform and position

3 · Semi-log evidence

\[\ln y=2.10+0.27x\]

Recover \(a\) and \(b\) for \(y=ab^x\). State one limitation.

4 · Phase shift

\[H(t)=2.5\sin\!\left(\frac{\pi}{6}(t-4)\right)+7\]

Interpret what happens at \(t=4\), and state the period.

Final check · 5 and 6

Piecewise evaluation and modelling decision

5 · Piecewise

\[F(d)=\begin{cases}6,&0<d\le3\\6+2(d-3),&d>3\end{cases}\]

Find \(F(8)\) and decide whether the model is continuous at \(d=3\).

6 · Justify a model

A transformed plot is nearly straight, but residuals grow at the largest values and the mechanism must eventually saturate.

Should you extrapolate far beyond the data? Justify with mathematical and contextual evidence.

Solutions · 1 to 3

Check the mathematics and the sentence

1. \(t_{1/2}=\ln(0.5)/\ln(0.87)\approx4.98\). The modelled mass halves every 4.98 time units.

2. \(R(8)=9+4\ln9\approx17.8\). Growth continues without bound but slows progressively.

3. \(a=e^{2.10}\approx8.17\), \(b=e^{0.27}\approx1.31\). The transform supports an exponential model only within a contextually defensible domain.

Solutions · 4 to 6

Interpret the structure

4. At \(t=4\), \(H=7\) and the model crosses the principal axis upward. Period \(=2\pi/(\pi/6)=12\) time units.

5. \(F(8)=6+2(5)=16\). Both pieces give 6 at the switch, so the model is continuous at \(d=3\).

6. Do not extrapolate far. Increasing residuals weaken the fit, while the saturation mechanism contradicts unlimited continuation of the transformed linear pattern.