In certain situations, the rate at which a thing grows or decreases is proportional to the amount present.

When a substance undergoes radioactive decay, the release of decay particles precipitates additional decay as the particles collide with the atoms of the substance. The larger the mass, the larger the number of collisions per unit time.

Consider a collection of organisms growing without environmental constraints. The larger the population, the larger the number of individuals involved in reproduction --- and hence, the higher the reproductive rate.

Let P be the amount of whatever is being measured --- the mass of the radioactive substance or the number of organisms, for example. Let t be the time elapsed since the measurements were started. Then

If , then P *increases* with time (* exponential growth*); if , then P *decreases* with time (* exponential decay*). k is called the * growth constant* (or * decay
constant*).

I'll solve for P in terms of t using * separation of
variables*. First, formally move the P's to one side and the t's
to the other:

Integrate both sides and solve for P:

Let . This gives

Note that is the *initial amount*:
Setting , .

* Example.* A population of roaches grows
exponentially under Calvin's couch. There are 20 initially and 140
after 2 days. How many are there after 14 days?

If P is the number of roaches after t days, then

When , :

Hence,

When ,

* Remark.* We'll often have expressions of the
form " ",
and this can be inconvenient to write if "stuff" is
complicated. You can use the * exponential
function* notation to avoid this problem:

For instance, in the last example,

* Example.* A population of MU flu virus grows in
such a way that it triples every 5 hours. If there were 100
initially, when will there be 1000000?

Let N be the number of the little rascals at time t. Then

Since the amount triples in 5 hours, when , :

Hence,

Set :

The * half-life* of a radioactive substance is
the amount of time it takes for a given mass M to decay to . Note that in radioactive decay, this time
is *independent* of the amount M. That is, it takes the same
amount of time for 100 grams to decay to 50 grams as it takes for
1000000 grams to decay to 500000 grams.

* Example.* The half-life of radium is 1620
years. How long will it take 100 grams of radium to decay to 1 gram?

Let M be the amount of radium left after t years. Then

When , :

Hence,

Set :

* Example.* A population of bacteria grows
exponentially in such a way that there are 100 after 2 hours and 750
after 4 hours. How many were there initially?

Let P be the number of bacteria at time t. Then

There are 100 after 2 hours:

There are 750 after 4 hours:

I'll solve for k first. Divide the second equation by the first:

Plug this into the first equation:

There were 13 bacteria initially (if I round to the nearest bacterium).

* Newton's Law of Cooling.*

According to * Newton's law of cooling*, the rate
at which a body heats up or cools down is proportional to the
difference between its temperature and the temperature of its
environment.

If T is the temperature of the object and is the temperature of the environment, then

If , the object heats up (an oven). If , the object cools down (a refrigerator).

I'll solve the equation using separation of variables. Formally move the T's to one side and the t's to the other, then integrate:

Let :

Let be the initial temperature --- that is, the temperature when . Then

Thus,

* Example.* A bagel is placed in a room to cool. After 10 minutes, the bagel's
temperature is . When will its
temperature be ?

In this case, and , so

When , :

Hence,

Set :

* Example.* A pair of shoes is placed in a oven to bake. The temperature is after 10 minutes and after 20 minutes. What was the initial
temperature of the shoes?

I set in to obtain

When , :

When , :

Divide by and solve for k:

Plug this back into and solve for :

In the real world, things do not grow exponentially without limit. It's natural to try to find models which are more realistic.

* Logistic Growth.*

The * logistic growth* model is described by the
differential equation

P is the quantity undergoing growth --- for example, an animal
population --- and t is time. k is the growth constant, and the
constant b is called the * carrying capacity*;
the reason for the name will become evident shortly.

Notice that if P is less than b and is small compared to b, then . The equation becomes , which is exponential growth. Notice that the derivative is positive, so P increases with time.

If P is greater than b, then is negative, so the derivative is negative. This means that P decreases with time.

It's possible to solve the logistic equation using separation of variables, though it will require a little trick (called a partial fraction expansion).

Separate the variables:

Now

(You can verify that this is correct by adding the fractions on the right over a common denominator.) Therefore, I have

I can replace the 's with a on the right, then set . This yields

Now some routine algebra gives

Divide the top and bottom by , and set :

Suppose the initial population is . This means when :

Thus, the equation is

For example, consider the case where and . I've graphed the equation for P with , 1, 2, 3, 4, and 5:

shows initial exponential growth. As the population increases, growth levels off, approaching asymptotically.

and are already large enough that the population spends most of its time levelling off, rather than growing exponentially.

An initial population remains constant.

and yield populations that shrink, again approaching as .

You can see why b is called the * carrying
capacity*. Think of it as the maximum number of individuals that
the environment can support. If the initial population is smaller
than b, the population grows upward toward the carrying capacity. If
the initial population is larger than P, individuals die and the
population decreases toward the carrying capacity.

Copyright 2018 by Bruce Ikenaga