Exponential models

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We investigate how the exponential functions model phenomena in the real world.

A differential equation is simply an equation with a derivative in it. Here is an example: $a\cdot f''(x) + b\cdot f'(x) + c\cdot f(x) = g(x).$

What is a differential equation?

An equation that you take the derivative of. An equation that relates the rate of a function to other values. It is a formula for the slope of a tangent line at a given point.

When a mathematician solves a differential equation, they are finding functions satisfying the equation.

Which of the following functions solve the differential equation $f''(x) = -f(x)?$

$e^x$ $\sin (x)$ $\cos (x)$

Exponential growth and decay

A function $f$ exhibits exponential growth if its growth rate is proportional to its value. As a differential equation, this means $f'(x) = k f(x)\qquad \text {for some constant of proportionality $k$.}$ We claim that this differential equation is solved by $f(x) = Ae^{kx},$ where $A$ and $k$ are constants. Check it out, if $f(x) = Ae^{kx}$ , then

$\begin{align*} f'(x) &= Ak e^{kx}\\ &= k\left(Ae^{kx} \right)\\ &= k f(x). \end{align*}$

A culture of yeast starts with $100$ cells. After $160$ minutes, there are $350$ cells. Assuming that the growth rate of the yeast is proportional to the number of yeast cells present, estimate when the culture will have $1000$ cells.

Since the growth rate of the yeast is proportional to the number of yeast cells present, we have the following differential equation $p'(t) = k p(t)$ where $p(t)$ is the population of the yeast culture at time $t$ with $t$ measured in minutes. We know that this differential equation is solved by the function $p(t) = A e^{k\cdot t}$ where $A$ and $k$ are yet to be determined constants. Since $100 = p(0) = Ae^{k\cdot 0}$ we see that $A = 100$ . So $p(t) = 100 e^{k\cdot t}.$ Now we must find $k$ . Since we know that $\answer [given]{350} = p(160) = 100e^{k\cdot 160}$ we need to solve for $k$ . Write $\begin{align*} 350 &= 100 e^{k\cdot 160}\\ 3.5 &= e^{k\cdot 160}\\ \ln(3.5) &= k\cdot 160\\ \ln(3.5)/160 &= k. \end{align*}$

Hence $p(t) = 100 e^{t\ln (3.5)/160} = 100 \cdot 3.5^{t/160}.$ To find out when the culture has 1000 cells, write

$\begin{align*} 1000 &= 100 \cdot 3.5^{t/160}\\ 10 &= 3.5^{t/160}\\ \ln(10) &= \frac{t\ln(3.5)}{160}\\ \frac{160\ln(10)}{\ln(3.5)} &= t. \end{align*}$

From this we find that after approximately $294$ minutes, there are around $1000$ yeast cells present.

One concept related to this sort of exponential models is that of doubling-time.

The doubling-time of an exponetial function $f(t) = c\cdot e^{k\cdot t}$ where $c\ne 0$ , $b>0$ , and $k>0$ is the value $T_2 =t$ such that $f(T_2+t) = 2 f(t)$ for all values of $t$ .

Find the doubling-time of the function $f(t) = c\cdot e^{k\cdot t}$ .

We want to find $T_2$ such that $f(T_2+t) = 2 f(t).$ Write with me: $\begin{align*} c\cdot e^{k\cdot (T_2+t)} &= 2 c\cdot e^{k\cdot t}\\ e^{k\cdot T_2+k\cdot t} &= 2 e^{k\cdot t}\\ k\cdot T_2+k\cdot t &= \ln (2) + k\cdot t\\ T_2 &= \frac{\ln (2)}{k} \end{align*}$

What is somewhat remarkable is that for the exponential function $f(t) = c\cdot e^{k\cdot t}$ where $c\ne 0$ , $b>0$ , and $k>0$ this doubling time only depends on $k$ !

It is worth seeing an example of exponential decay as well. Consider this: Living tissue contains two types of carbon, a stable isotope carbon-12 and a radioactive (unstable) isotope carbon-14. While an organism is alive, the ratio of one isotope of carbon to the other is always constant. When the organism dies, the ratio changes as the radioactive isotope decays. This is the basis of radiocarbon dating.

The half-life of carbon-14 (the time it takes for half of an amount of carbon-14 to decay) is about 5730 years. Moreover, the rate of decay of carbon-14 is proportional to the amount of carbon-14.

If we find a bone with $1/70$ th of the amount of carbon-14 we would expect to find in a living organism, approximately how old is the bone?

Since the rate of decay of carbon-14 is proportional to the amount of carbon-14 present, we can model this situation with the differential equation $f'(t) = k f(t).$ We know that this differential equation is solved by the function defined by $f(t) = A e^{k\cdot t}$ where $A$ and $k$ are yet to be determined constants. Since the half-life of carbon-14 is about $5730$ years we write $\frac {1}{2} = e^{k 5730}.$ Solving this equation for $k$ , gives $k = \frac {-\ln (2)}{5730}.$ Since we currently have $1/70$ th of the original amount of carbon-14 we write $\answer [given]{\frac {1}{70}} = 1\cdot e^{\frac {-\ln (2)t}{5730}}.$ Solving this equation for $t$ , we find $t \approx -35121$ . This means that the bone is approximately $35121$ years old.

In our example above we briefly defined half-life. Let’s do this justice.

The half-life of an exponetial function $f(t) = c\cdot e^{-k\cdot t}$ where $c\ne 0$ , $b>0$ , and $k>0$ is the value $T_{\frac {1}{2}} =t$ such that $f(T_{\frac {1}{2}}+t) = \frac {f(t)}{2}$ for all values of $t$ .

Find the half-life of the function $f(t) = c\cdot e^{-k\cdot t}$ .

We want to find $T_{\frac {1}{2}}$ such that $f(T_{\frac {1}{2}}+t) = \frac {f(t)}{2}.$ Write with me: $\begin{align*} c\cdot e^{-k\cdot (T_{\frac{1}{2}}+t)} &= \frac{1}{2} c\cdot e^{-k\cdot t}\\ e^{-k\cdot T_{\frac{1}{2}}-k\cdot t} &= \frac{1}{2} e^{-k\cdot t}\\ -k\cdot T_{\frac{1}{2}}-k\cdot t &= \ln \left(\frac{1}{2}\right) - k\cdot t\\ T_{\frac{1}{2}} &= \frac{\ln \left(\frac{1}{2}\right)}{-k}. \end{align*}$

Noting that $\ln (1/2) = -\ln (2)$ we see that $T_{\frac {1}{2}} = \frac {\ln (2)}{k}.$

Here again it is somewhat remarkable is that for the exponential function $f(t) = c\cdot e^{-k\cdot t}$ where $c\ne 0$ , $b>0$ , and $k>0$ the half-life only depends on $k$ !

While exponetial models are somewhat limited, they apply in many cases. Here are some examples of real-world phenomena that can be modeled by exponential growth:

The rate that a population grows is proportional to the size of the population.
The rate that a radioactive substance decays is proportional to the amount of the substance.
The rate that an investment grows is proportional to the amount of the investment.
The rate that a pathogen spreads is proportional to the number of people with the pathogen.
The rate that glucose is processed in one’s blood is proportional to the concentration of glucose in the blood.