### Exponential growth and decay: a differential equation

This little section is a tiny introduction to a very important subject
and bunch of ideas: *solving differential equations*. We'll just
look at the simplest possible example of this.

The general idea is that, instead of solving equations to find unknown
*numbers*, we might solve equations to find unknown *functions*. There are many possibilities for what this might mean, but
one is that we have an unknown function $y$ of $x$ and are given that
$y$ and its derivative $y'$ (with respect to $x$) satisfy a relation
$$y'=ky$$
where $k$ is some constant. Such a relation between an unknown
function and its derivative (or *derivatives*) is what is called a
**differential equation**. Many basic ‘physical principles’ can be
written in such terms, using ‘time’ $t$ as the independent variable.

Having been taking derivatives of exponential functions, a
person might remember that the function $f(t)=e^{kt}$ has exactly this
property:
$${d\over dt}e^{kt}=k\cdot e^{kt}$$
For that matter, any *constant multiple* of this function has the
same property:
$${d\over dt}(c\cdot e^{kt})=k\cdot c\cdot e^{kt}$$
And it turns out that these really are *all* the possible
solutions to this differential equation.

There is a certain buzz-phrase which is supposed to alert a
person to the occurrence of this little story: if a function $f$
has **exponential growth** or **exponential decay** then that is
taken to mean that $f$ can be written in the form
$$f(t)=c\cdot e^{kt}$$
If the constant $k$ is *positive* it has exponential *growth*
and if $k$ is *negative* then it has exponential *decay*.

Since we've described all the solutions to this equation,
what questions remain to ask about this kind of thing? Well, the usual
scenario is that some *story problem* will give you information in
a way that requires you to take some trouble in order to *determine the constants $c,k$*. And, in case you were wondering where
you get to take a derivative here, the answer is that you don't
really: all the ‘calculus work’ was done at the point where we granted
ourselves that all solutions to that differential equation are given
in the form $f(t)=ce^{kt}$.

First to look at some general ideas about determining the constants before getting embroiled in story problems: One simple observation is that $$c=f(0)$$ that is, that the constant $c$ is the value of the function at time $t=0$. This is true simply because $$f(0)=ce^{k \cdot 0}=ce^{0}=c\cdot 1=c$$ from properties of the exponential function.

More generally, suppose we know the values of the function at two different times: $$y_1=ce^{kt_1}$$ $$y_2=ce^{kt_2}$$ Even though we certainly do have ‘two equations and two unknowns’, these equations involve the unknown constants in a manner we may not be used to. But it's still not so hard to solve for $c,k$: dividing the first equation by the second and using properties of the exponential function, the $c$ on the right side cancels, and we get $${y_1\over y_2}=e^{k(t_1-t_2)}$$ Taking a logarithm (base $e$, of course) we get $$\ln y_1-\ln y_2=k(t_1-t_2)$$ Dividing by $t_1-t_2$, this is $$k={\ln y_1-\ln y_2\over t_1-t_2}$$ Substituting back in order to find $c$, we first have $$y_1=ce^{{\ln y_1-\ln y_2\over t_1-t_2}t_1}$$ Taking the logarithm, we have $$\ln y_1=\ln c+{\ln y_1-\ln y_2\over t_1-t_2}t_1$$ Rearranging, this is $$\ln c=\ln y_1-{\ln y_1-\ln y_2\over t_1-t_2}t_1= {t_1\ln y_2-t_2\ln y_1\over t_1-t_2}$$ Therefore, in summary, the two equations $$y_1=ce^{kt_1}$$ $$y_2=ce^{kt_2}$$ allow us to solve for $c,k$, giving $$k={\ln y_1-\ln y_2\over t_1-t_2}$$ $$c=e^{t_1\ln y_2-t_2\ln y_1\over t_1-t_2}$$

A person might manage to remember such formulas, or it might
be wiser to remember the way of *deriving* them.

#### Example 1

A herd of llamas has $1000$ llamas in it, and
the population is growing exponentially. At time $t=4$ it has $2000$
llamas. Write a formula for the number of llamas at *arbitrary*
time $t$.

**Solution:** Here there is no direct mention of differential equations, but use of
the buzz-phrase *‘growing exponentially’* must be taken as
indicator that we are talking about the situation $$f(t)=ce^{kt}$$
where here $f(t)$ is the number of llamas at time $t$ and $c,k$ are
constants to be determined from the information given in the
problem. And the use of language should probably be taken to mean that
at time $t=0$ there are $1000$ llamas, and at time $t=4$ there are
$2000$. Then, either repeating the method above or plugging into the
formula derived by the method, we find
$$c=\hbox{ value of $f$ at $t=0$ } = 1000$$
$$k={\ln f(t_1)-\ln f(t_2)\over t_1-t_2}={\ln 1000-\ln
2000\over 0-4}$$
$$={ \ln {1000\over 2000}}{-4}={\ln {1\over 2} \over -4 } = (\ln 2)/4$$
Therefore,
$$f(t)=1000\;e^{{\ln 2\over 4}t}=1000\cdot 2^{t/4}$$
This is the desired formula for the number of llamas at arbitrary time $t$.

#### Example 2

A colony of bacteria is growing exponentially. At time $t=0$ it has $10$ bacteria in it, and at time $t=4$ it has $2000$. At what time will it have $100,000$ bacteria?

**Solution:** Even though it is not explicitly demanded, we need to find the general
formula for the number $f(t)$ of bacteria at time $t$, set this
expression equal to $100,000$, and solve for $t$. Again, we can take a
*little* shortcut here since we know that $c=f(0)$ and we are
given that $f(0)=10$. (This is easier than using the bulkier more
general formula for finding $c$). And use the formula for $k$:
$$k={\ln f(t_1)-\ln f(t_2)\over t_1-t_2}=
{\ln 10 -\ln 2,000\over 0-4}={ \ln {10\over 2,000} \over -4 }=
{\ln 200\over 4} $$
Therefore, we have
$$f(t)=10\cdot e^{{\ln 200\over 4}\;t}=10\cdot 200^{t/4}$$
as the general formula. Now we try to solve
$$100,000=10\cdot e^{{\ln 200\over 4}\;t}$$
for $t$: divide both sides by the $10$ and take logarithms, to get
$$\ln 10,000={\ln 200\over 4}\;t$$
Thus,
$$t=4\,{\ln 10,000\over \ln 200}\approx 6.953407835.$$

#### Exercises

- A herd of llamas is growing exponentially. At time $t=0$
it has $1000$ llamas in it, and at time $t=4$ it has $2000$
llamas. Write a formula for the number of llamas at
*arbitrary*time $t$. - A herd of elephants is growing exponentially. At time $t=2$
it has $1000$ elephants in it, and at time $t=4$ it has $2000$
elephants. Write a formula for the number of elephants at
*arbitrary*time $t$. - A colony of bacteria is growing exponentially. At time $t=0$ it has $10$ bacteria in it, and at time $t=4$ it has $2000$. At what time will it have $100,000$ bacteria?
- A colony of bacteria is growing exponentially. At time $t=2$ it has $10$ bacteria in it, and at time $t=4$ it has $2000$. At what time will it have $100,000$ bacteria?

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##### Calculus Refresher

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