Exponential Models and Thresholds | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Find the starting value at time zero

An exponential model describes repeated multiplication. In $N=N_0a^t$ , $N_0$ is the value at $t=0$ ; $a$ is the factor for each time unit. An 80% retained amount gives $a=0.8$ , a 20% decrease each unit.

In $M=M_0e^{-kt}$ with $k>0$ , the negative exponent gives decay. State what time unit $t$ uses before calculating.

Worked example

A mass is M = 80e⁻⁰·²ᵗ grams, with t in days. Find its half-life.

The half-life is the time needed to reach half the initial amount. Half of 80 is 40.

\begin{gathered}40=80e^{-0.2t},\\\ln(1/2)=-0.2t,\\t=\frac{\ln2}{0.2}\\t\approx3.47\text{ days}.\end{gathered}

The doubling time is the time needed to reach twice the initial amount. Set the model equal to that target and solve for time in the same way.

Your turnPractice

A population follows $P=300e^{0.1t}$ , with $t$ in hours. Find the initial population and the doubling time.

Show worked answer

Answer

\begin{gathered}P(0)=300\\t=\frac{\ln2}{0.1}\\t\approx6.93\text{ hours}\end{gathered}

Set $t=0$ for the initial value; set $P=600$ for doubling.

An integer time needs one final decision

Worked example

The amount after n complete days is 100(0.8)ⁿ. When is it first below 30?

\begin{gathered}(0.8)^n<0.3\\n>\frac{\ln0.3}{\ln0.8}\approx5.395\end{gathered}

The first whole number satisfying this is 6, not the nearest integer to 5.395. Check the adjacent days: day 5 gives 32.768; day 6 gives 26.2144.

Answer

\text{after 6 complete days}

Keep the exact boundary before choosing the integer. With a strict inequality, an integer exactly on the boundary is not allowed.

Your turnPractice

A quantity is $240(0.5)^n$ after $n$ complete hours. Find the first hour when it is below 30, and the first when it is at most 30.

Show worked answer

Answer

4\text{ hours};\quad3\text{ hours}

At hour 3 it equals 30, so it meets “at most” but not “below”. At hour 4 it equals 15.

Find the starting value at time zero

In $M=M_0e^{-kt}$ with $k>0$ , the negative exponent gives decay. State what time unit $t$ uses before calculating.

Worked example

A mass is M = 80e⁻⁰·²ᵗ grams, with t in days. Find its half-life.

The half-life is the time needed to reach half the initial amount. Half of 80 is 40.

\begin{gathered}40=80e^{-0.2t},\\\ln(1/2)=-0.2t,\\t=\frac{\ln2}{0.2}\\t\approx3.47\text{ days}.\end{gathered}

The doubling time is the time needed to reach twice the initial amount. Set the model equal to that target and solve for time in the same way.

Your turnPractice

A population follows $P=300e^{0.1t}$ , with $t$ in hours. Find the initial population and the doubling time.

Show worked answer

Answer

\begin{gathered}P(0)=300\\t=\frac{\ln2}{0.1}\\t\approx6.93\text{ hours}\end{gathered}

Set $t=0$ for the initial value; set $P=600$ for doubling.

An integer time needs one final decision

Worked example

The amount after n complete days is 100(0.8)ⁿ. When is it first below 30?

\begin{gathered}(0.8)^n<0.3\\n>\frac{\ln0.3}{\ln0.8}\approx5.395\end{gathered}

The first whole number satisfying this is 6, not the nearest integer to 5.395. Check the adjacent days: day 5 gives 32.768; day 6 gives 26.2144.

Answer

\text{after 6 complete days}

Keep the exact boundary before choosing the integer. With a strict inequality, an integer exactly on the boundary is not allowed.

Your turnPractice

A quantity is $240(0.5)^n$ after $n$ complete hours. Find the first hour when it is below 30, and the first when it is at most 30.

Show worked answer

Answer

4\text{ hours};\quad3\text{ hours}

At hour 3 it equals 30, so it meets “at most” but not “below”. At hour 4 it equals 15.