Barnes D.J., Chu D. Introduction to Modeling for Biosciences
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4.1 Differentiation 133
meaning. This new notation makes it much easier to refer to a point, but is still
somewhat cumbersome. If we want to measure distances very accurately, then using
place names may not be sufficient precise. Hence, it might be better to refer to
a position by a simpler label. Physicists often use x to refer to a specific place.
We could use different indices to distinguish between different place names; so for
example, x
1
could be Graz, x
5
could denote Presseggen, and so on.
This is not a complete solution, however, because we still need to worry about
mapping the label to the name for the place. Instead, it is much more expressive to
refer to the position in relation to the time when we were there. It is common to
use a notation like x(t) to indicate (in our case) the place where Kurt was at a time
t since starting the trip. If we adopt this notation then clearly x(t = 0),orsimply
x(0), is the place where Kurt was at time t =0 (a specific street in Graz). Similarly,
x(2 hours 20 min) is Hermagor, the destination. Since it took Kurt 7 minutes to get
there from Presseggen, x(2 hours 13 min) would stand for Presseggen.
This new notation is quite powerful and can make equations look much neater.
We can now write (4.1) in a concise form:
x(2 hrs 16 mins) −x(2 hrs 13 mins)
0.5hrs
=
3.3km
0.5hrs
=66 km/h (4.2)
Of course, with this new notation we can write the distance between any two points
along the route taken by Kurt as x(t
2
)−x(t
1
), commonly abbreviated as x(t), with
being pronounced “delta.” x(t) means the distance traveled over the period
of time t at some point in the journey. The delta notation is also used to indicate
the difference between two moments in time. In the above case t would be the
difference between times t
2
and t
1
. With this new notation, we can rewrite (4.2)in
an even more compact way.
x(t)
t
=
x(2hrs16min) −x(2hrs13min)
0.5hrs
=
3.3km
0.5hrs
=66 km/h (4.3)
So far, all that has changed from (4.1)to(4.3) is the notation; the meaning has
remained the same. All these equations say is how we calculate the speed given a
distance and the time taken to travel the distance.
Given the power of our new notation let us now return to the question of how
fast Kurt was traveling. By now, we know that our initial answer, namely that he
traveled at 94.3 km/h hardly reflects his actual speed at any particular point, but
rather is an average speed taken over the entire length of his trip (x(t) = 220 km
and t = 2 hr 20 min). Since there are varying road conditions and speed limits,
in reality the speed will vary very much across the journey. In order to get a better
estimate of the actual speed traveled at some of the time points, we had to reduce
the period of time over which we averaged; in other words we reduced our t.
Yet, we still do not know for sure how fast Kurt was traveling at a specific point
between Presseggen and Hermagor, although we know that this speed was closer
to about 65 km/h than to 93.4 km/h. As we continue to decrease t we get an
increasingly better idea of Kurt’s actual speed at a particular point along his trip.
Instead of considering the speed traveled between 2 hours 13 minutes and 2 hours
134 4 Differential Equations
16 minutes into the journey (corresponding to a t =3 min), we could look at the
speed traveled between 2 hours 13 minutes and 2 hours 13 minutes 20 seconds since
start of the journey (corresponding to t = 20 s). Maybe we could even consider
reducing our t further, perhaps even to milliseconds? Where is the limit to this
process?
Let us now come back to our original question from the beginning of the section:
“What is the speed Kurt traveled at?” By now it should be clear that our question
was not very precise. Either we should have asked about the average speed over a
given time period, or we should have asked about the speed at specific time—“At
what speed was Kurt traveling at time 2 hours and 13 minutes and 0 seconds into
the journey (and not an instant earlier or later)?”
There is an inherent paradox in this question. If we ask about Kurt’s speed then,
by definition, we are asking about the distance traveled. Yet asking about the speed
at a particular instant of time seems meaningless because in a particular instant of
time Kurt does not travel at all—he is frozen in time at a particular place. Therefore,
it appears, the question does not make sense.
In this case, the appearance is wrong. The question can be asked in a meaningful
way, and the discussion above has brought us half way towards being able to give
an answer. As we have seen, we can generate better estimates of the actual speed
traveled if we successively reduce the time over which we average, that is if we
reduce t. So, if we were to consider a time interval of 4 minutes around the time
we are interested then we would get a better estimate of the actual speed traveled
than if we considered an interval of 5 minutes. We would get an even better estimate
if we considered a 2 minute interval, or 1 minute, or even only one second around
the moment we are interested in. Eventually we will reach a t that gives an answer
that is sufficiently accurate for most practical purposes; if we keep reducing t
then we will eventually find a time interval over which Kurt’s speed is essentially
constant. At least as far as car journeys are concerned, it is rarely necessary to bother
with milliseconds.
In reality, hardly anyone will be interested in Kurt’s travels through Austria,
let alone his speed between Khünburg and Presseggen. However, the procedure
by which we determine his speed in a given moment, is of general importance in
many areas of science. Any quantity that changes can be analyzed following the
exact same procedure we used above. Instead of Kurt’s speed we could consider the
change of the size of a bacterial colony, or the change of temperature of a mountain
lake, or the change of the concentration of a protein over time. Reducing the interval
over which one averages the change of quantities to get the instantaneous change is
a very common procedure in physical sciences, engineering, mathematics and, of
course, biological modeling.
Important for the use of this idea is to develop a notation that allows us to effi-
ciently make precise statements. The process of making the time intervals smaller
and smaller is often written like:
lim
t→0
x(t) −x(t −t) = lim
t→0
x(t) (4.4)
4.1 Differentiation 135
Equation (4.4) looks very mathematical, but all it does is to formulate the idea of
measuring the distance between two points. The symbol, lim is read as “limit” and
encapsulates precisely the idea that we make t smaller and smaller until it nearly
reaches 0. As we do this, we consider the distance traveled between the time t and
just a small time before that, namely t −t.
Equation (4.4) only considers the distance traveled. As t gets smaller and
smaller, Kurt covers smaller and smaller distances. In the limiting case of t = 0,
the distance traveled goes to 0, which is what we expect: In an instant of time, one
is frozen in time and does not move at all. Hence, we get:
lim
t→0
x(t) =0
At this point now, the reader who is not familiar with basic calculus needs to take an
intellectual leap. While it is true that reducing t will also reduce the distance trav-
eled, the speed may not change. Just remember, above, that Kurt’s speed between
Presseggen and Hermagor was 60 km/h traveled over 7 minutes; the correspond-
ing t =7 min and x(t) =7 km. The much shorter distance between Presseggen
and Khünburg was covered at an average speed of 66 km/h but over a much shorter
t =3 min and a x(t) = 3.5 km. So, while both t and x(t) become smaller
and smaller, their ratio, the speed, does not necessarily decrease, but could converge
to a finite, non-vanishing value. So, maybe there is something wrong about the idea
of being frozen in time?
We now have all the necessary notation and concepts ready to ask the question
about Kurt’s speed at a particular point in time t. This is simply given by the distance
traveled divided by the time this has taken and this averaged over a vanishingly small
interval of time.
lim
t→0
x(t) −x(t −t)
t
= lim
t→0
x(t)
t
Physicists normally indicate the speed at a time t by the symbol v(t). And we can
now define this symbol using our new notation.
v(t)
.
= lim
t→0
x(t) −x(t −t)
t
= lim
t→0
x(t)
t
(4.5)
Here the symbol “
.
=” just means “is defined by.”
The problem of the current notation in the last equation is that it is somewhat
tedious to write. Hence, physicists have invented a more concise way to write this
equation which has the same meaning.
v(t)
.
=
dx(t)
dt
(4.6)
Despite looking very different, (4.5) and (4.6) have the same meaning. The function
v(t) measures the change that x(t) undergoes at a time t. In the case of a trip from
Graz to Hermagor, this is just the speed at a particular time t, but we can generalize
this to mean the change of any quantity. It does not even need to be the change
136 4 Differential Equations
as time passes, but it could be the change of a quantity as a different something
changes. Yet, we are getting ahead of ourselves.
For the moment, it is only important to remember, that
dx(t)
dt
is the instantaneous
rate of change of x(t) at time t Where the meaning is clear, it is quite common to
abbreviate x(t) as simply x.
4.1.1 A Mathematical Example
While Kurt’s road trip was a good example to illustrate the core idea of rates of
change, the power of (4.6) really only becomes apparent when we have a mathemat-
ical expression for the distance traveled. Let us start with the simplest case, namely
linear motion. In this case, we have x(t) given by:
x(t) =a ·t (4.7)
In this case, x(t) can be simply represented as a straight line. The parameter a could
be any number; its precise value will depend on the circumstances. Let us assume it
is simply 1 (see Fig. 4.1). We could think of x(t) as the distance traveled, just as in
the case of Kurt’s trip. Then the horizontal axis in Fig. 4.1 would represent time and
the vertical axis would be the point x at the given time t. However, it could really
represent anything else as well; the interpretations as “distance” and “point in time”
are perhaps helpful for the moment, but not necessary as the same kind of analysis
for the rate of change of any quantity can be performed no matter what x represents
and what it depends on.
Let us look more closely at the concrete example of the function x(t) given in
(4.7). This is essentially only an instruction of how to calculate the value of x de-
pending on t . If we set the value of t =3, then x(3) =3a = 3 when a has the value 1.
Similarly, we can calculate values at any other times. Now, we can go through the
same procedure as we did with Kurt’s travels. If we want to know how much x
changed from time, say, t =0tot =10, then we simply calculate,
x
t
=
x(10) −x(0)
10 −0
=
10a −0a
10
=1a =1
So, we calculated a rate of change between these two time points to be 1. Similarly,
we can now calculate the rate of change between t =6 and t =4, and, following the
same procedure, we will see that again the rate of change is a, which evaluates to 1
if we set a =1. In fact, the reader can easily convince herself that the rate of change
is always the same, at any point for any interval. This is not particularly surprising.
The graph of the linear motion is a simple straight line, which has a constant slope
throughout. The slope is just the rate of change, hence we would expect that the rate
of change is constant everywhere.
Let us now look at a non-trivial example and consider a new function of x(t)
given by,
x(t) =a ·t
3
(4.8)
4.1 Differentiation 137
Fig. 4.1 Function x(t) as in (4.7)
This curve is illustrated in Fig. 4.2. Unlike the case of (4.1) this one does not have
a constant slope throughout, which means that the rate of change varies; we would
now expect that the rate of change depends more strongly on the interval we average
over, very much as in the case of Kurt’s trip. To get a better handle on this, let us
focus on the point t =5. (This is exactly the same type of question as asking what
speed Kurt was traveling at exactly 2 hrs 13 mins after starting.)
At the time in question, the value of x(t) can be easily calculated to x(5) =
a ·5
2
=25a.Ifa =1 again then x(5) =25. This result is now the equivalent of the
position in the above example of Kurt’s trip. So, what is the rate of change (“speed”)
at this point? Following our now quite familiar procedure, let us start to hone in onto
the speed at t = 5. By looking at ranges centered on t = 5, and taking as the first
interval, t =0 and t =10 we get
x(10) −x(0)
t
=
1000a
10
=100a =100
So, the rate of change is 100. Repeating this process, but for a t = 4 centered
around t =5 we get,
x(7) −x(3)
t
=
343a −27a
4
=79 ·a =79
Similarly, we obtain for a t = 2 a value of 76, for t = 1 a value of 75.25, for
t =0.4avalueof75.04, and so on. Clearly, we get closer and closer to the rate of
change at x(5), but so far we have only calculated estimates.
138 4 Differential Equations
Fig. 4.2 Function x(t) as in (4.8)witha =1
4.1 Confirm the estimates of the slope given above.
Just as in the case of Kurt, if we want the exact value of change at the point x(5),
then we need to take the limit of a vanishingly small interval around that point, that
is we need to calculate lim
t→0
x(5)
t
, to obtain the expression for
dx(t)
dt
. As it turns
out, this expression is in fact very easy to obtain by simple differentiation.
1
˙x
.
=
dx(t)
dt
=
d(ax
3
)
dt
=3at
2
(4.9)
We arrived at this equation simply by applying the standard rules of differentiation
to the equation for the curve (4.8). Note also that we have introduced a new notation
here, namely ˙x. This is simply the rate of change of x; we will use this extensively
below.
If we now want to obtain the exact value of change of (4.8) at the point t =5, all
we need to do is to calculate it from (4.9)tobe
˙x(5) =3 ·a ·5
2
=75
1
The reader who is not familiar with the rules of differentiation is encouraged to consult the ap-
pendix for a refreshment of the rules.
4.1 Differentiation 139
when a =1. So, the exact rate of change, or slope (or speed) at this particular point
is 75, which is quite close to our last estimate with a t = 0.4 that suggested a
slightly higher value of 75.04.
In differentiation, we have a very effective tool to calculate the rate of change of
any function we like, and at any point. Not only in biology, but in science at large,
this is perhaps the most powerful and important mathematical technique. It is used
to formulate laws in physics, engineering, and crucially also in biology. Before we
are able to apply this powerful tool to concrete applications, we need to generate
a better understanding of what exactly it is doing and what we can do with it. It
would neither be possible nor useful to go through the entire theory of calculus
here, but there are a few key-points that are worth highlighting and that will also be
exceptionally relevant for what we will do later.
The first question we should clarify is what the “change” means. So, if at x(5)
the rate of change is 75, how should we interpret this number? In the case of Kurt’s
trip, the meaning was intuitively clear; the rate of change was a speed. In the more
abstract case of a simple curve it is not so evident how we should interpret the
number.
Of course, the exact meaning of the derivative of a variable will, to some extent,
depend on context, which can only be clarified with reference to the semantics of a
particular model. Yet, there is a general sense in which we can interpret derivatives
quite independent of the meaning of the variable. We will develop an understanding
of this by means of a small digression.
4.1.2 Digression
In pre-scientific times many people believed that the earth was flat. While today
this belief seems ridiculous, it is perhaps not that absurd if considered in the light
of every day empirical evidence. If we look around us, at least on a plain, then our
planet certainly looks very flat. Similarly, when we look out on the ocean, then we
cannot detect any curvature, but rather behold a level body of water. Even when we
travel long distances we see no obvious evidence contradicting the assumption that
we live on a flat surface (except for a few mountains and valleys of course). Even
from high up on the summits of mountains (or even aeroplanes), we look down on
a flat surface not a round body. So, really, pre-scientific man and woman must be
forgiven for their natural (and empirically quite justified) assumption that their home
planet is nothing but a big platter.
Perspectives only start to change when we take a large step backwards, in a way
that was impossible until relatively recently. When seen from space, the hypothesis
that the Earth is round seems much more natural given direct observation.
2
(Pre-
sumably, the reader has not had this first hand experience, but hopefully we can take
available footage on trust and believe that the Earth is actually round.)
2
Of course there are indirect ways to measure the curvature of the Earth. Based on triangulation
the ancient Greeks already knew that the Earth is (at least approximately) round.
140 4 Differential Equations
Fig. 4.3 Changing the point of view. As we zoom in to the point x(5),thecurveinFig.4.2 starts
to look more and more like a straight line, and we fail to detect the curvature that is so obvious to
us in the wider view. The dashed line is a straight line with slope 75. In the bottom panel the line
and t
3
are barely distinguishable
The failure of pre-scientific man and woman to recognize that they dwelled on a
curved object is due to an effect that is relevant in the context of differentiation as
well: Even very curved bodies look straight if the observer is small and close to the
curved surface. The same principle also applies in a two dimensions: if considered
from close enough, most kinds of curves appear as straight lines. If we imagine the
curve in Fig. 4.2 to be inhabited by some imaginary creatures, maybe of the size
of bacteria, then they would undoubtedly also assume that they live on a straight
line.
Mathematicians and physicists would summarize this observation in an efficient,
if somewhat unexciting way: Locally, a curve can be approximated by a straight
line. Figure 4.3 illustrates this using the example of the curve x(t) = t
3
.Aswe
zoom in on the point x(5) (although the same thing would work with any point)
the curve starts to look more and more like a straight line. This is exactly the same
effect that makes the Earth look flat to us. Yet, from a technical point of view, the
important question now is, what is the straight line that approximates the curve
locally?
We established above that straight lines are very simple things in that their rate
of change is constant and given by the parameter a in (4.7). We found this by direct
calculation of the slope. Another way to do this is to use the rules of differentiation
(see Appendix A.2). If a straight line is given by x(t) =at +b, then ˙x(t) =a.This
4.2 Integration 141
tells us that the slope is constant and does not depend on the time at all. (We know
that already, but it is worthwhile seeing this from a different point of view again.)
We have nearly arrived at an interpretation of the first derivative of a function.
All we need to do is to put together the two ideas, namely (i) that a curve locally
looks like a straight line and (ii) that the slope of a straight line is given by its
parameter a: Locally, any (differentiable) curve x(t) looks like a straight line with
parameter a =˙x(t). This gives us the interpretation of the derivative of a function:
The value of the derivative at a particular point on a curve is the slope of the straight
line (tangent) at that point.
From a mathematical point of view, there is of course much more to differentia-
tion and differentials than we are able to discuss here. The reader who really wants
to obtain a deep understanding of the theory is encouraged to consult a calculus text-
book. Certainly, in the long run it is necessary to develop a deeper understanding,
but it is fully possible to take the first steps into a new modeling adventure without
being versed in the sacred texts of mathematics, as long as one has developed some
intuition for the basic concepts.
4.2 Integration
Here is another simple exercise, similar to the one we have seen before, but subtly
different:
Kurt traveled from Graz to Hermagor. The trip took him exactly 2 hours and 20 minutes and
his average speed was 94.3 km/h. What distance did he travel?
This is clearly the inverse problem to our previous exercise on page 131, and we
know the answer: Kurt traveled 220 km. So, that is easy. Now we can ask a different
question.
Kurt traveled from Graz to Presseggen. The trip took him exactly 2 hours and 13 minutes
and his average speed was 96.1 km/h. He then continued from Presseggen to Hermagor.
The trip took him exactly 7 minutes and his average speed was 55.7 km/h. What is the total
distance he traveled?
Again, we know the answer (220 km); yet if we wanted to calculate the solution
we would have to figure out the distance from Graz to Presseggen, and then the
distance from Presseggen to Hermagor using the average speeds and travel times
we were given to get the total travel time. If we again denote the distance traveled
since Graz by x(t) then we can write this as:
x(2hrs20min) =v
1
t
1
+v
2
t
2
(4.10)
Here v stands for the average speed of a segment, and the subscript “1” stands for
the segment “Graz to Presssegen” and subscript “2” stands for segment “Presseggen
to Hermagor.” So, all we do is to calculate the distances traveled by multiplying the
average speeds with the times it took to complete the segments. No magic there.
142 4 Differential Equations
Imagine now that, instead of being given Kurt’s travel in 2 segments, we know
the average speeds and travel times in much finer grained form; say we know 1000
segments. Then, again, to obtain the total distance traveled, we calculate the dis-
tances of every individual segment and sum them up. Mathematically, we write this
as follows
x(2hrs20min) =
1000
i=0
v
i
t
i
(4.11)
The principle of solving this is the same as in the case of (4.10), even though the
actual practice of calculating the distances for 1000 records is much more tedious
than for a mere two. It is even more tedious, if, instead of 1000 entries, we are given
100000. Then our formula becomes,
x(2hrs20min) =
100000
i=0
v
i
t
i
(4.12)
and takes 100 times as long to compute. What we did here, at every step of refine-
ment was to split up Kurt’s trip into smaller and smaller segments, so we get the
required number of records. The question we want to ask now is, what will happen
when we are given Kurt’s trip split up into an infinite number of entries, each giving
his speed at a particular moment of time?
To see how we could deal with this situation, it will be helpful to remember what
v actually stands for. When we discussed the original problem (i.e., “What is Kurt’s
speed?”) we calculated v as follows:
v =
x(t)
t
Equation (4.12) looks almost trivial:
x(2hrs20min) =
100000
i=0
x
i
t
i
t
i
(4.13)
Or more generally, for any number N of intervals we can write this as:
x(2hrs20min) =
N
i=0
x
i
t
i
t
i
It is easy to see that it really just reduces to adding the length of the individual seg-
ments of the trip, given by x
i
(t) (though remember that we are not directly given
the length of the segments, only the speeds and times). As we now make the indi-
vidual segments shorter and shorter, we obtain more and more of them, or in other
words, our index i goes towards infinity, but the length and time of the individual
segments t
i
and x
i
(t) goes towards zero. We have already encountered this and