Preface: The definitions
of polynomial and rational functions are easy to understand.
They use only the arithmetic operations learned in
elementary school: addition, subtraction, multiplication, and
division. The definitions of the exponential, logarithmic, and
trigonometric functions are more advanced and sophisticated
requiring an understanding of powers, roots, and geometry for
similar triangles. What makes these latter functions important
today is the fact that they appear in many different modelling
contexts. One way in which these functions arise in models is
through the study of differential equations. The following
chart presents a summary of derivative results for the core
functions we have studied together with their related
characteristic differential equations with initial conditions.
Function
y =
f (x)
Derivative
y' = f
' (x )
Differential
equation
F(x, y,
y', y'') = 0
Initial
Conditions y(a)
= b,...
x^{n}
xy' -
ny = 0
y(0) =
0
sin(x)
cos(x)
y''
+ y = 0
y(0)
= 0 ; y'(0) = 1
cos(x)
-sin(x)
y''
+ y = 0
y(0)
= 1 ; y'(0) = 0
tan(x)
y(0)
= 0
exp(x) = e^{x}
y'-y
= 0
y(0)
= 1
ln(x)
... x>0
1/x
xy'-
1 = 0
y(1)
= 0
Looking at the differential
equations in this table notice that power functions have
characteristic differential equations that are more
complicated than those of the other functions on the table. In
this chapter we will explore further how differential
equations are connected to the functions that solve them. To
connect these functions to differential equations more
precisely, we will start the first sections of this chapter by
describing some modelling contexts. With each model we will
consider a differential equation defining a solution function
for that model. Studying these solution functions we
will recognize them as familiar functions from
properties and issues connected to the differential
equaitions.
VI.A. The Natural
Exponential Function and A Growth Model.
We begin our study models and differential
equations with exponential functions. First we'll examine a
model for populations that is based on a simple discrete
difference equation. Then we will turn our attention to a
comparable continuous calculus model. In our population model
we assume we have a population of biological objects [they may
be fruit flies, bacteria, rabbits, owls, fish, trees, and /or
even humans], and this population is changing with time so
that the measure of the population size, P, is a function of
time, t. Note that P(t) may be measured with integers only, as
with a population counter, or it may be measured with real
numbers, as with biomass for bacteria, tonnage for fish, or
aggregate height for trees.
n
P
0
*
1
**
2
* * * *
3
** ** ** **
Example VI.A.1. In a
laboratory experiment a single cell breaks into two distinct
cells every 8 hours. Starting with a single cell, this process
continues for 40 hours in the same fashion.
How many cells will be
present at the end of the time period?
If the process
continues without change for t = 8n hours, how many cells
will be present at time t?
Estimate how long it
will be before there are over 1000 cells.
Solution: Let P(n) be
the number of cells present after 8n hours. Then P(0) =
1 expresses the statement that the experiment starts with a
single cell and P(n+1) = 2 P(n) expresses the
statement that every eight hours the number of cells doubles.
Since 40 = 8^{ . }5, the first question asks us to
find the value of P(5). Following routine calculations we have
that P(1) = 2, P(2) = 4, P(3) = 8, P(4) = 16 and P(5)=32. So
if the growth continues as stated in the problem, 32 cells
will be present after 40 hours.
It is not hard to see from
the pattern in these calculations that
P(n) = 2^{ n}
when t = 8n.
This answers the second
question.
The last question can be
rephrased, "For what n is 2^{n }>1000?"
Here a quick check on the
powers of 2 shows that 2^{10 }= 1024, so n =
10 is the smallest n that will satisfy the inequality. So
P(10) = 2^{10 }>1000 shows we will need to wait
about t = 8 * 10 = 80 hours, a little more than 3 days, for
there to be 1000 cells.
Comment: In the example,
the change in populations between 8 hour periods is not
difficult to compute. If we let DP(n)
denote the change in population expressed by the formula
P(n+1)- P(n) then we can observe from the previous
computations that P(0) = 1, P(1) = 2, P(2) = 4 , and
The preceding example
illustrates a model in which the population changes occur
"discretely" in the sense that it is possible to isolate
particular instants in time when the changes occur. The
change in population,DP(n), from the n^{ th} period to the (n+1)^{
st} period is determined precisely to be the value of
the population at the end of the n^{ th} period.
The differential
equation model: We now turn our attention to a model for
populations in which the population measure changes as a
continuous function of time. Assume a biological population
for which at time t the population grows at a rate
proportional to the population at that time. The growth
rate of the population is expressed by the differential
equation
P'(t) = k P(t) or dP/dt
= kP,
with k being the
proportionality constant. When t = 0 the
population is P(0), the initial population.
Model units: We have
some flexibility in the choice of the population unit of
measurement in this model, so we will standardize our
population measurement unit so that P(0) = 1. You can
think of this standardization as measuring the size of the
population when our observation begins and comparing
subsequent populations in terms of that initial population. In
other words, the populations are measured using the initial
population as the unit for scaling.
The population growth rate,
P'(t), is expressed in terms of population units per unit
time. Since the unit of time is not yet determined for our
model, we can choose a measurement for time so that the
proportionality constant k is 1. Following this judicious
choice of units we can now study more thoroughly the simpler
model differential equation: P'(t) = P(t) .[See the comments at the end of this
section for a further exploration of the effects of
choosing units.]
In summary then, the
population
size in our model is characterized mathematically as a
particular solution to the differential equation P'(t) =
P(t) with the initial condition P(0) = 1.
Visualization: Using
the ideas of tangent fields from Chapter
IV.D, we begin to understand the differential equation
P'(t) = P(t) by sketching its tangent field.[See Figure 1.]By
focusing attention on the initial condition, P(0)=1, we obtain
a rough sketch of the graph of the solution function as an
integral curve for this field.
The integral curve sketch
confirms some information which the derivative gives us,
namely, that P is increasing when P(t)>0 and that the graph
of P is concave up [since P"(t) = P'(t) = P(t)] when
P(t)>0. This information allows us to conclude further that
the population in this model is increasing at ever increasing
rates.
Estimating P(1): One
immediate issue for this model is to learn the value of P(1).
This will tell us the size of growth from one unit of
population over one unit of time. We start to estimate P(1)
using Euler's method with n = 4, so Dt = 1/4 and dP(t) = P'(t) Dt = P(t) 1/4. Here are our calculations
in a table:
x
P
P
' = P
dP
= .25P
0
1
1
0.25
0.25
1.25
1.25
0.3125
0.5
1.5625
1.5625
0.390625
0.75
1.953125
1.953125
0.48828125
1
2.44140625
Using Euler's method to estimate f(1) with Sage.: Click
on the Evaluate button. dt = h=0.25.
You can change h to find a more accurate estimate.
You can work with this further yourself using a speadsheet.
It is also worth looking at these calculations with a little more
subtle technique, factoring as we go:
Okay. So our first estimate using Euler's
method with n = 4 is that `P(1) ~~ 2.4414`. But we can do better than that,
much better.
Note that since `P'(t) = P(t),
P(a + dt)~~ P(a) + P(a) dt = P(a)(1 + dt)` .
If we use `dt
= h = 1/n` for `n`, a positive
integer, then `P(1/n)~~ 1 + 1 . 1/n`, `P(2/n)~~
P(1/n) + P'(1/n) . 1/n = (1+1/n)^2` and continuing in this fashion we obtain
`P(1) = P(n/n)~~ (1+1/n)^ n.`
For `n` very large, `dt = h = 1/n` will
be close to 0 and we should obtain a much better approximation for
`P(1)` from Euler's method. In limit notation we write
`lim_{n to oo} (1+1/n)^n = lim_{h to 0}
(1+h)^{1/h} = P(1)`.
Using this estimation with `n =
1000`, we obtain `P(1)~~(1+.001)^{1000} ~~ 2.7169`
and with `n = 10,000`, we find
`P(1)~~(1+.0001)^10000 ~~ 2.7181`. With `n = 10^6,
P(1)~~ (1+10^{-6})^{1,000,000} ~~ 2.718280469`.
Using `(1+1/n)^n` to estimate f(1) with Sage:
Click on the Evaluate button.
You can change `n` to find a more accurate estimate. Caveat: Very large values for `n` (around `10^9`)will
give round off errors.
[In fact, the convexity of `P` helps explain
why in theory these are all underestimates. Can you see why?]
You may recognize from these
estimates that the exact value of `P(1)` is a number studied in
your pre-calculus work. It is a mathematical constant which has
been of interest for hundreds of years and, like the number
`pi`, has a special symbol
attached to it, `e`. [Actually the letter `e` was first used to represent this
constant extensively by Euler, whose textual treatment of both
pre-calculus and calculus topics were the standard for many
years.] Thus, by notational convention and our previous estimates
we have that `P(1) = e~~ 2.718` .
Properties of P: We turn now to explore
some of the important features of `P` as a function.
If it takes one unit of time for one unit of population to grow
to be e units of population, you might expect
that in one unit of time any population size will grow by
a factor of e.
Conjecture: For any `t, P(t+1) =
P(1)*P(t)`
Comment: We can make more sense of this
conjecture from our previous use of Euler's method:
Since `P(t + dt)~~ P(t) +
P(t) dt = P(t)(1+dt)` so `P(t
+ n *dt)~~P(t)(1+ dt)^n`.
With `dt=1/n` this shows that `P(t+1) = P(t+ n*
1/n)~~P(t)*(1+1/n)^n`.
When `n` is large we can see why `P(t+1) = P(t)*P(1)`.
Proof of Conjecture: We can prove that
the conjecture is true for all t in a more rigorous fashion by
considering the function F(t) = P(t+1)/P(t) .
Using the quotient and Chain Rules we find F
'(t) .
.
So [by Theorem IV.**,] F(t) is a constant
function, i.e., F(t) = K for all t.
But using t = 0, we see that F(0) = P(0+1)/P(0)
= P(1)/P(0) = e/1 = e.So, for all t, or .
EOP.
Comment: Note how this equation was
proven. First we examined a related function Fand showed that F
was a constant function because F '(t) = 0. Then we used the
particular value of F when t = 0 to find the constant, allowing
us then to prove the original equation was true.
From the conjectured and now
proven equation we see that
and in general we can show (using
mathematical induction) that for any natural number n,
P(n) = e^{ n}
.
It appears that P is
the exponential function with base e. ^{[1]}We
proceed now to justify that P satisfies the usual properties of
exponents using only its characterization from the differential
equation. [These results are also suggested by further
exploration of Euler's approximations for the values P(n) and
the definition of e. See Exercise 14.]
Theorem VI.A.1 Suppose
s and t are real numbers and P is the solution to the
differential equation P'(t) = P(t) with P(0) = 1. Then
But F(0)=[P(0)]^{ r}/P(r^{.}0)=1^{r}/1=1
so 1=[P(t))]^{ r}/P(rt)or P(rt) = [P(t)]^{ r}.
EOP.
Defining the
natural exponential function: We could give a
definition for the natural exponential function
characterizing it solely as the solution to the
differential equation y'=y where y(0)=1.
Defining a function by declaring it to be the
unique solution to a differential equation is
not as unusual as it may seem. Recall that in
the motion interpretation for the derivative a
differential equation can be interpreted as
giving a way to determine the velocity of a
moving object from the time and its position.
Knowing the velocity of the object at any time
and position would seem to enable us to
determine its position at any time just from
knowing its position at one time.
Comments:
1. P(t)=e^{ t}
: Applying the last part of the theorem to an rational number r,
we see that P(r) = P(1) ^{r} = e^{ r}. Since P
is a continuous function, this shows that P has the same values
as the exponential function with base e for any real number t,
so we can write P(t)=e^{ t} for any real number t.
Recognizing this identification of P with the exponential
function of base e, we express the key facts we have
demonstrated about P using the Liebniz notation for the
derivative and the exponential notation as follows:
2.Asymptotes: Note that
since e > 2 , e^{ 2} > 2^{ 2} , e^{ 3}
> 2^{ 3}and generally e^{ n} > 2^{
n}. So when n is very large, e^{ n} is very
large as well. This fact is expressed as a limit statement by
writing
.
Also, since e^{ -n}
= 1/e^{ n} , when n is very large , e^{ -n}
is the reciprocal of a large number. Thus for large
choices of n, e^{ -n }» 0. This is expressed as a limit statement by
writing
.
On the graph of P(t) =
e^{ t} this last fact is recognized by the horizontal
asymptote of the graph with the X - axis.
3. Looking at
integration: Since we can write the indefinite
integral form of this statement as
.
4. Existence:
The preceding discussion assumed that there was a solution to
the differential equation y' = y with y(0) = 1. The existence of
a solution is certainly suggested by the tangent field and the
computations using Euler's method. A rigorous demonstration of
the existence of this function independent of any assumptions
about exponential functions and their properties will be given
later in this chapter. For now, all the evidence we have
assembled suggesting the conclusion should suffice.
5.Other models: The
general issue of population models and differential equations is
studied in greater depth in more advanced courses in
differential equations and mathematical modeling. The
differential equation y' = ky with initial condition y(0)
= A is very important to the study of numerous other real
situations besides this simple population growth model. In exercise 8 below you are asked to check that
the general solution of this differential equation is y = A e^{
kt} .
6. Changing
units
to normalize the model: [This next comment is pretty tricky. Be sure to
read this with a pencil and paper at hand to make sure you
follow the changes in the units.]
As suggested initially, a
suitable choice of units shows the relation of the special case
when k = 1 and A = 1 to the general situation with y'(t)=ky(t)
and y(0) = A.
Here's how: Suppose the
units for y are called yugs and the time t is measured in
tocks.We'll let A yugs be our new unit for measuring the
population with the variable z, which we'll call zugs,
so A yugs is one zug. We'll let 1 / k tocks be our new time
units, we'll call them nocks, and we'll uses u to measure the
time in nocks, so k nocks is one tock. So y = A z and t =u.
Now what is the relation
between the measure of the population measure as y yugs at
time t tocks, y(t), and the measure of the same population
measure as z zugs at the same time but measured as u nocks,
z(u)? Well, Az(u) expresses the measure of the population in
yugs based on its measurement in mugs,while y() expresses the measurement of the population at
the same time measured in tocks also in the units of yugs. So
Az(u) = y(u). Now we suppose that y(0)=A, so Az(0)=y(0)=A
and z(0)=1. Furthermore since z(u) = 1/ A y(u) and we assume y'(t)= ky(t), we have
So if y satisfies the
original differential equation y '= ky then z
satisfies the differential equation z' = z with z(0) = 1.But our
work has shown then that z(u) = e^{ u}. Now translate
this back into a statement about y:
Az(u) = y(u) where u = t.
So y(t) = A z( kt ) = A e^{
kt}.
7. Other Applications of
y'=ky with y(0) = A
There are many other
situations where the differential equation beginning this
section arises in expressing the rate of change of a measured
quantity. One important application is to the measurement of
the amount of radioactive material present in a given mass of
material. Because of the nature of unstable radioactive
matter, the rate of change (decay) of the matter is assumed to
be proportional to the amount of matter present at the
specified time. This should make some sense from a physical
point of view. With less unstable material present there is
less opportunity for the interaction of unstable particles to
produce further radioactive decay.
Another situation
involving this basic differential equation concerns the change
in temperature of one object in a medium which remains at a
constant temperature. In this context we find a physical context
to which one of Newton's laws, called his law of cooling,
applies. The law says that the rate of change of the
temperature is proportional to the difference in temperatures
of the object and the medium.These situation will be
explored further in the exercises of section VI.D.
The exercises for
this section apply many of the concepts and techniques we have
developed already in the earlier chapters to a further
exploration of the exponential function and functions that use
the exponential function in their definition.
Exercises VI.A.:
1. Find the
derivatives of the functions as indicated. [Don't forget the
chain rule!]
a. f(x) = exp ( 5x
) . Find f '(0) , f '(1) and f '(t).
b. f(x) = exp ( -x
) . Find f '(0) , f '(1) and f'(t).
c. f(x) = exp ( x^{
2}) . Find f '(0) , f '(1) and f'(t).
d. f(x) = exp (-x^{
2}) . Find f '(0) , f '(1) and f'(t).
e. f(x) = exp
(sin(x)). Find f '(0) , f '(p) and f '(t).
2. Find the derivatives
of the functions as indicated. [Don't forget the chain rule!]
a. y = e^{ 3x}
. Find dy/dx when x = 0 and 1.
b. y = x e^{ x}.
Find dy/dx when x = 0 and 1.
c. y = e^{ x}
sin(x). Find dy/dx.
d. y = e^{ x}
+e^{ -x}. Find dy/dx when x = 0.
e. y = e^{ x}
-e^{ -x}. Find dy/dx when x = 0.
3. Find
the derivatives of the functions as indicated. [Don't forget the
chain rule!]
a. Find D_{ t}(sin(e^{
t})).
b. Find D_{ t}(exp(t^{
2} - 1)).
c. Find D_{ t}(
exp(t sin(t) ) ).
d. Find D_{ t}(e^{
t} / t^{ 2}).
e. Find D^{ t}(sec(e^{
2t} ) ).
4. Sketch a graph of y =
exp(-x^{ 2} / 2) showing all extrema and points of
inflection. Explain your work using first and second derivative
analysis.
5. Sketch a graph of y = x
e^{ -x} showing all extrema and points of inflection.
Explain your work using first and second derivative analysis.
6. Sketch a graph of y = x^{
2} e^{ -x} showing all extrema and points of
inflection. Explain your work using first and second derivative
analysis.
7. Sketch a graph of y =
sin(x) e^{ -x}for x e [0,4p],
showing all extrema and points of inflection. Explain your work
using first and second derivative analysis.
8. Show
that y = A e^{ kx} is a solution to the differential
equation y'=ky with y(0) = A.
9. Show that y = e^{ x }(sin(x)+cos(x))
is a solution to the differential equation y' = 2 e^{ x}cos(x)
and y'' - y' = -2e^{ x}sin(x).[Corrected
9-9-02 4:30 pm ]
10. Show
that y=e^{ x}(sin(x)+cos(x)) is a solution to the
differential equation y''-2y' +2y = 0.
11. Suppose
f(x) = k e^{ x} is the probability density function for
a random variable X on the interval [0,1]. Show that k =
1/(e-1).
12. Suppose
f(x) = k e^{ -x} is the probability density function for
a random variable X on the interval [0,1]. Find k .
13. Find
the following indefinite integrals:
a. ò e^{3x} dx
c. ò cos(x)e^{sin(x)}
dx
b.ò x exp(x^{2} )dx
d. ò e^{x}sin(e^{x})dx
14. Use Euler's method to explain why when
n >> 0, (1+1/n)^{ 2n }» e^{ 2}, (1+1/n)^{ 3n }»e^{ 3},
(1+1/n)^{ 4n }» e^{ 4}, and in general (1 + 1/n)^{ kn }» e^{ k}.
15. Find
the following definite integrals. You may express your answer
using the number e.
a. ò_{0}^{1} e^{3x} dx
b.
ò_{0}^{1}x
exp(x^{2} )dx
16. Find
the area of the region enclosed by the graph of Y= e^{ x }=
exp(X), the X axis, X= -1, and X= 1.
17. The
area of the region enclosed by the graph of , the X axis, X= -1,
and X= 1 is often of interest in the study of probability and
statistics. Estimate this area using Euler sums with a. n=4. b.
n=8. c. n = 100.
18. The
region in the plane bounded by the graph of , the X-axis, the Y
axis, and the line X=1 is rotated about the Y- axis to form a
region in space.Find the volume of this spatial region.
19. The
region in the plane bounded by the graph of , the X-axis, the
Y-axis, and the line X= t is rotated about the Y- axis to form a
region in space.Find the volume of this spatial region.Give an
estimate for the volume of the this region in space when t =
100.
[1] This confirms what we already
might have guessed from our previous knowledge from Chapter ***
about the derivative of this exponential function since the
function e^{x} does satisfy the differential
equation with the given initial condition. What is of interest
here is that we have arrived at this connection starting with a
model for population growth and a differential equation without
presuming what the solution would be.