Exponential behavior (growth/decay) is defined by: equal changes in the input cause the output to be multiplied by a constant .
Once you've recognized exponential behavior then
you can always use
$\,P(t) = P_{\,0}{\text{e}}^{rt}\,$ as your model.
Here, $\,P(t)\,$ is the population at time $\,t\,,$ $\,P_{\,0}\,$ is the population at time zero, and $\,r\,$ is the
relative growth rate.
However, if an
(equal change,constant multiplier) = $\,(\Delta t,b)\,$ pair is readily available,
then the model $\,P(t) = P_{\,0}b^{t/\Delta t}\,$ is more natural and easier to use.
A variety of problems involving exponential functions were explored in
Solving Exponential Growth and Decay Problems.
This current section focuses on doubling time, half-life, and related problems.
Doubling time formula derived from
$\,P(t) = P_{\,0}{\text{e}}^{rt}\,$
Doubling time $\,d\,$ depends only on the relative growth rate, $\,r\,.$
Let $\,P(t) = P_{\,0}{\text{e}}^{rt}\,$ for $\,r > 0\,$ (so this is exponential growth).
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As above, doubling time does not depend on time!
It depends only on the (equal change,constant multiplier) = $\,(\Delta t,b)\,$ pair.
Let $\,P(t) = P_{\,0}b^{t/\Delta t}\,$ for $\,b > 1\,$ (so this is exponential growth).
Let $\,d\,$ denote the doubling time, so that $\,P(t+d) = 2P(t)\,$:
$\,\overbrace{\cancel{P_{\,0}}b^{(t+d)/\Delta t}}^{P(t+d)} = \overbrace{2\cancel{P_{\,0}}b^{t/\Delta t}}^{2P(t)}\,$
$\,\cancel{b^{t/\Delta t}} b^{d/\Delta t} = 2\cancel{b^{t/\Delta t}}\,$
(Note that $\,t\,$ disappears at this step!)
$b^{d/\Delta t} = 2$
$\displaystyle\ln b^{d/\Delta t} = \ln 2$
$\displaystyle\frac{d}{\Delta t}\ln b = \ln 2$
$\displaystyle d = \frac{\ln 2}{\ln b}\cdot \Delta t$
Half-life is the time it takes for a current amount to be cut in half.
Half-life formula derived from
$\,P(t) = P_{\,0}{\text{e}}^{rt}\,$
Again, half-life depends only on the relative growth rate, $\,r\,.$
Let $\,P(t) = P_{\,0}{\text{e}}^{rt}\,$ for $\,r < 0\,$ (so this is exponential decay). Note that since $\,r < 0\,$ for exponential decay, we have $\displaystyle \,h = -\frac{\ln 2}{r} > 0\,.$ |
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As above, half-life does not depend on time!
It depends only on the (equal change,constant multiplier) = $\,(\Delta t,b)\,$ pair.
Let $\,P(t) = P_{\,0}b^{t/\Delta t}\,$ for $\,0 < b < 1\,$ (so this is exponential decay).
Let $\,h\,$ denote the half-life, so that $\,P(t+h) = \frac 12P(t)\,$:
$\,\overbrace{\cancel{P_{\,0}}b^{(t+h)/\Delta t}}^{P(t+h)} = \overbrace{\frac 12\cancel{P_{\,0}}b^{t/\Delta t}}^{\frac 12P(t)}\,$
$\,\cancel{b^{t/\Delta t}} b^{h/\Delta t} = \frac 12\cancel{b^{t/\Delta t}}\,$
(Note that $\,t\,$ disappears at this step!)
$b^{h/\Delta t} = \frac 12 = 0.5$
$\displaystyle\ln b^{h/\Delta t} = \ln 0.5$
$\displaystyle\frac{h}{\Delta t}\ln b = \ln 0.5$
$\displaystyle h = \frac{\ln 0.5}{\ln b}\cdot \Delta t = \frac{\ln 2^{-1}}{\ln b}\cdot \Delta t = -\frac{\ln 2}{\ln b}\cdot \Delta t$
Note that since $\,0 < b < 1\,$ for exponential decay, we have $\,\ln b < 0\,,$ and thus $\displaystyle h = -\frac{\ln 2}{\ln b}\cdot \Delta t > 0\,.$
Let $\,g\,$ be the time it takes to go from $\,P(t)\,$ to $\,kP(t)\,$ for any $\,k > 0\,.$
The derivation is identical to those above, hence is greatly abbreviated here:
$$ \begin{gather} \cssId{s114}{P(t) = P_{\,0}{\text{e}}^{rt}}\cr\cr \cssId{s115}{P(t+g) = kP(t)}\cr\cr \cssId{s116}{P_{\,0}{\text{e}}^{r(t+g)} = kP_{\,0}{\text{e}}^{rt}}\cr\cr \cssId{s117}{{\text{e}}^{rg} = k}\cr\cr \cssId{s118}{g = \frac{\ln k}{r}} \end{gather} $$ | $$ \begin{gather} \cssId{s119}{P(t) = P_{\,0}b^{t/\Delta t}}\cr\cr \cssId{s120}{P(t+g) = kP(t)}\cr\cr \cssId{s121}{P_{\,0}b^{(t+g)/\Delta t} = kP_{\,0}b^{t/\Delta t}}\cr\cr \cssId{s122}{b^{g/\Delta t} = k}\cr\cr \cssId{s123}{\frac{g}{\Delta t}\ln b = \ln k}\cr\cr \cssId{s124}{g = \frac{\ln k}{\ln b}\cdot \Delta t} \end{gather} $$ |
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. |
PROBLEM TYPES:
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