MATH 192-01: Methods of Calculus (82792)
PL-211, MW 6 PM - 7:50 PM
SYLLABUS Fall 2014
John Sarli
JB-326
[email protected]
909-537-5374
O¢ ce Hours: MW 11:00AM-1:00PM, or by appt.
Text: Stewart/Clegg
Brief Applied Calculus
Prerequisites: MATH 110 or satisfactory score on the ELM exam
This an introductory course in the methods of calculus, with applications to the
social and life sciences as well as to business and economics. We will cover a wide
variety of topics, starting with a review of functions and working through the
di¤erential and integral calculus of one or more variables. In order to provide a
useful overview within ten weeks it is necessary to de-emphasize theory in favor of
technique that focuses on an intuitive exploration of quantitative skills requiring
calculus. Topics from every chapter of this text will be covered, but not every
topic in those chapters. The selection of topics may depend on expressed interest.
I will provide lecture notes as we proceed on the syllabus web page, and these
should be studied along with the textbook. Refer to the link MATH 192 on the
website: www.math.csusb.edu/faculty/sarli/
Grading will be based on two midterm exams, a cumulative …nal exam, and …ve
graded assignments, weighted as follows: First Midterm (15%), Second Midterm
(25%), Final Exam (40%), Graded Assignments (20%). To reinforce written
communication skills the Graded Assignment solutions should be clearly presented, either in a "bluebook" or sent to me as a PDF (do not scan in handwritten
work).
Experience shows that regular attendance and active participation signi…cantly
improve performance. I will take attendance until the class roster has been …nalized, however there is no attendance requirement for this class, except for the
following: Within the …rst two weeks you must complete the CSU/UC Mathematics Diagnostic Testing Project MR web test. Go to mdtp.ucsd.edu and
scroll down to MDTP Web Based Tests. Select the MR test. The items will
appear one at a time and the test will be scored instantly once you submit it.
Your response to each item will be provided to you and I can help you interpret
the results. Either print the test results or send them to me using the E-mail
option. I do not record the test results and they do not a¤ect your course
grade in any way.
A list of Suggested Exercises will be provided for each chapter. The exams
will be written at the level of these exercises though the format may not be
identical, mainly because the purpose of exams is to bring related ideas together
so that they can be used in relation to one another. I will list exercises that are
representative of a particular technique or concept, but you should attempt as
many similar exercises in the text as needed for understanding. In this way we
can avoid "practice exams" and other routines that use the time we need to cover
this material. It is your responsibility to bring questions to class that arise as you
work through exercises and notes.
After computing your total scores weighted according to the percentages above,
course grades will be assigned as follows:
A
91
A
86 90
B+
81 85
B
76 80
B
71 75
C+
66 70
C
61 65
C
51 60
D
45 50
F
< 45
Success in this course requires a balance of three activities:
1) Read the text and work the exercises regularly. Keep notes of your solutions.
If you have organized them e¢ ciently, bring them to the in-class exams.
2
2) Follow the lecture notes on my website and read the syllabus there. Bring
questions on these notes to class as they occur to you.
3) Participate in the class sessions as actively as you can. Lectures are more
useful to you if you use them to clarify ideas as we develop them.
Notes
1) After completing the MDTP MR web test I strongly suggest that you also
take the CR web test (same website). Some of the items on the CR test are not
required for success in this course but the results will give you a good idea of the
skill level that is expected.
2) Mid-term exam dates are subject to change. Due dates for the graded exercises
will be set as we progress.
3) Calculators: Own one and use it. Bring it to class and use it when you work
exercises. Exams will be written so that calculators are not required but their use
is not prohibited.
4) Please refer to the Academic Regulations and Policies section of your current
bulletin for information regarding add/drop procedures. Instances of academic
dishonesty will not be tolerated. Cheating on exams or plagiarism (presenting the
work of another as your own, or the use of another person’s ideas without giving
proper credit) will result in a failing grade and sanctions by the University. For
this class, all assignments are to be completed by the individual student unless
otherwise speci…ed.
5) If you are in need of an accommodation for a disability in order to participate
in this class, please let me know ASAP and also contact Services to
Students with Disabilities at UH-183, (909)537-5238.
Some important dates.
September 29: First day of class
October 1: Last day to add open classes over MyCoyote for Fall quarter
October 15: Fall CENSUS; last day to submit add/drop slips
October 15: First Exam
3
November 5: Second Exam
December 3: Last day of class
December 10: Final Exam (6-7:50PM)
Important ideas from Chapter 1: Functions and Models
Domain of the independent (input) variable
Range of the dependent (output) variable
Representations by graphs, tables, and mathematical expressions
Even and odd functions
Composition of functions
Transformation of functions by rigid motions and stretching
Rate of change
Polynomial, exponential, and logarithmic functions and models
4
Suggested Exercises for Chapter 1
The following exercises are not to be handed in. They represent skills required for
basic mastery.
1.1 (pages 13-17):
1; 7; 8; 23; 29; 33; 39; 43; 51; 55; 60
1.2 (pages 25-29):
3; 9; 11; 19; 25; 27; 33; 35; 55
1.3 (pages 37-41):
15; 25; 33; 39; 49; 53
1.4 (pages 49-53):
1; 9; 21; 29; 33; 37; 41; 49
1.5 (pages 60-63):
7; 9; 11; 13; 31 34; 35; 37
1.6 (pages 69-71):
15; 17; 19; 31; 45; 47
5
Suggested Exercises for Chapter 2
The following exercises are not to be handed in. They represent skills required for
basic mastery.
2.1 (pages 82-83):
5; 13; 15; 21
2.2 (pages 92-95):
3; 5; 13; 19; 37; 43; 49; 51
2.3 (pages 109-113):
3; 5; 7; 23; 27; 37; 50; 51
2.4 (pages 124-129):
1; 3; 10; 31; 37; 39; 46; 50; 57
First Graded Assignment
Due: October 13
To reinforce written communication skills the Graded Assignment solutions
should be clearly presented in a "bluebook" or provided in PDF format. Late
papers will not be graded.
First Graded Assignment. Do any one of the following:
Page
Page
Page
Page
Page
83: 18
95: 58
111: 52
127: 40
132: 44
6
Derivative of a Single-Variable Function
Average Rate of Change.
Suppose the interval [x1 ; x2 ] is in the domain of the function y = f (x). Let
y = f (x2 ) f (x2 ) and x = x2 x1 . The ratio
y
x
is called the di¤erence quotient for f on the interval. It measures the average
rate of change of the output y with respect to the input x. The idea comes from
computing the slope of a line, which is the graph of a linear function. The average
rate of change of a linear function does not depend on what interval we use. We
want to use this idea to get information about functions in general.
Example. Find the average rate of change for y = ln x on the interval
a) 12 ; 2
b) 1e ; e
In a),
ln 2 ln 12
y
=
x
2 12
4
=
ln 2
3
0:924 20
In b),
ln e ln 1e
y
=
x
e 1e
1 ( 1)
=
e 1e
2e
= 2
e
1
0:850 92
Here is a picture of the situation in this example:
7
y
1.6
1.4
1.2
1.0
0.8
0.6
0.4
0.2
0.0
-0.2
1
2
-0.4
3
4
x
-0.6
-0.8
-1.0
-1.2
-1.4
-1.6
y = ln x
The blue line, through the points 21 ; ln 12 and (2; ln 2), has slope
line, through the points 1e ; 1 and (e; 1), has slope 43 ln 2.
4
3
ln 2. The red
The concept of average rate of change was applied by Galileo to the more di¢ cult
problem of instantaneous rate of change. In modern terms, we can state this
problem by asking how a speedometer reads out speed in real time. This is where
calculus comes in. Galileo did not have calculus to work with, but his observations
led Newton and others to consider what happens when we have a function that
measures distance s from a …xed reference point as a function of time t. What they
discovered turned out to be fundamental tool for analyzing the behavior of any
function, regardless of what the independent and dependent variables represent.
Free Fall.
Galileo devised a model for the motion of a freely falling object: s, the distance
travelled, is proportional to the square of the elapsed time t. We can represent
this model with a quadratic function:
s(t) = 4:9t2
We have written the constant of proportionality in metric terms: 4:9 m= s2 . It is
equivalent to the constant that Galileo determined experimentally.
8
From this hypothesis he was able to infer a related law: v, the velocity of the
object, is proportional t. This is a linear model:
v(t) = 9:8t
How did he arrive at these conclusions? To …nd the function s he conducted careful
experiments to measure distance travelled over various periods of time. Measuring
v at a particular instant t, however, was very di¢ cult with the equipment available
in the early 1600’s. So he reasoned with di¤erence quotients as follows. Suppose
we want to know how fast the object is moving at a particular instant of time t0 .
We know how to compute the average speed of the object over an interval of time,
because that is just the average rate of change of the function s(t) = 4:9t2 . For
example, the average speed over the interval [3; 5] is
s (5) s (3)
s
=
t
5 3
4:9 (25 9)
=
2
= 39:2 m= s
Now consider the interval [t0 ; t0 + h]. This will be a long interval of time if h is
a large number and short interval of time if h is a small number. (Galileo was
interested in short intervals.) The di¤erence quotient is now
s (t0 + h) s (t0 )
s
=
t
(t0 + h) t0
4: 9
=
(h + t0 )2 t20
h
Remember, we are focusing on a given instant t0 but are thinking of h as a variable
that determines the length of the interval. Thus, this di¤erence quotient is itself a
function of h and its domain does not allow h = 0 (that’s why the instantaneous
rate of change problem is tricky). Let’s see what happens when we expand this
expression for st :
4: 9
(h + t0 )2
h
4: 9
(h + 2t0 ) h
h
= 4:9 (h + 2t0 )
t20
=
That is, for any non-zero value of h the average speed over the interval [t0 ; t0 + h]
is given by the expression 4:9 (h + 2t0 ). Galileo’s insight was to realize that he
9
could make the interval as short as he liked. As h gets smaller st gets closer to
4:9 (0 + 2t0 ) = 9: 8t0 , the speed at the instant t0 . He concluded that the speed of
an object in free fall is directly proportional to the time elapsed after the object
starts to fall:
v(t) = 9:8t
This process is an example of …nding a limit, which would become the central idea
in di¤erential calculus. (Galileo died in the year that Isaac Newton was born.)
Let’s look at the process geometrically. Suppose we want to …nd the instantaneous
velocity at t0 = 3 seconds. Let h = 1 and compute the average velocity on the
interval [2; 3] :
s (4) s (3)
s
=
t
4 3
= 34:3 m= s
The straight line (red) through the points (3; 44:1) and (4; 78:4) is a secant line
for the parabola that is the graph of s(t) = 4:9t2 . Compare this with the secant
line (blue) through (3; 44:1) and (5; 122:5), where we found st = 39:2 m= s. The
red line appears to be close to the tangent line (green) to the parabola at the
point (3; 44:1). This tangent line has slope v(3) = 29:4, which represents the
instantaneous velocity after 3 seconds.
10
s
180
160
140
120
100
80
60
40
20
0
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
s(t) = 4:9t2 near t0 = 3
Exercise. Compute the average velocity over the interval [2:5; 3]. With respect to
t0 = 3, what value of h does this interval correspond to? Sketch the secant line in
this case.
Limits of Functions.
In di¤erential calculus we often want to focus on the behavior of a function y =
f (x) in the vicinity of a particular value x0 . This value may or may not be in the
domain of f but we typically assume that points su¢ ciently close to x0 are in the
domain. Consider the example
f (x) =
ex
1
x
The domain of f is Rnf0g = ( 1; 0) [ (0; 1), so f (0) is nor de…ned. However,
if you graph this function on a calculator you will see:
11
6.0
t
y
30
20
10
-5
-4
-3
-2
-1
0
1
2
3
4
5
x
The calculator ’wants’to de…ne f (0) = 1 because f (0 + h) is very close to 1 for
1
1
jhj very small. For example, f 0 100
0:995 02 and f 0 + 100
1: 005. This
makes sense because if g(x) = ex then the average rate of change of g on [0; x] is
g(x)
x
ex 1
g(0)
=
= f (x)
0
x
and the slope of the tangent line to the graph y = ex at the point (0; 1) is 1:
12
y
8
7
6
5
4
3
2
1
-2
-1
0
1
2
x
Tangent line to y = ex at (0; 1)
How do we know where the output of a function is going based on inputs that are
near a particular value of interest? For most elementary functions, the output at
any value x0 in the domain is consistent with the outputs from nearby values. We
say:
The limit of f (x) as x approaches x0 is f (x0 ):
A function with this property is said to be continuous at x0 , because the graph will
x
appear unbroken at (x0 ; f (x0 )). Apparently, given f (x) = e x 1 we could simply
de…ne f (0) = 1 and then f would be continuous at any real number, including
x0 = 0. A complete study of limits is beyond the scope of this course, but we
need to be aware of the common ways that a function can fail to be continuous
at a particular value. These mainly fall into two categories, based on whether or
not the limit exists as we approach that value. In above example we write
lim f (x) = 1
x!0
13
so if we admit 0 into the domain and de…ne f (0) = 1 then f is continuous at 0.
Many applications involve functions with this type of behavior. But many involve
functions with discontinuities that cannot be removed. Here are some examples:
i) f unbounded at x0 : Examples include rational functions where x0 makes
the denominator, but not the numerator, 0. We expect a vertical asymptote at
x0 and there is no way to de…ne f (x0 ) so that limx!x0 f (x) exists.
ii) f (x) approaches di¤erent values as x approaches x0 from either side: Examples include step functions such as
1; x < 3
1; x 3
f (x) =
y
2
1
-5
-4
-3
-2
-1
1
2
3
4
-1
-2
Here, f (3) = 1 but we could de…ne f (3) = 1 or any other value for that
matter, yet the function would not be continuous at x0 = 3. When we write
limx!x0 f (x) = L we mean that the output gets close to the value L whether we
approach x0 from the left or the right.
14
5
x
Tangent Lines and Derivative Functions
The limit concept is fundamental to all aspects of calculus. We have used it
thus far to understand the limit of the di¤erence quotient of a function f at a
particular domain value x0 as x shrinks toward zero. If this limit exists we call
it the instantaneous rate of change of f at x0 and write it as
f 0 (x0 )
We read this notation as the derivative of f at x0 . We have seen that f 0 (x0 ) is
the slope of the tangent line to the graph y = f (x) at the point (x0 ; f (x0 )), and
so an equation for this tangent line is
y = f 0 (x0 ) (x
x0 ) + f (x0 )
Example. If f (x) = kx2 for some constant k then, as we saw in the
free-fall problem, f 0 (x0 ) = 2kx0 and so the tangent line at (x0 ; kx20 ) is
y = 2kx0 (x
= kx0 (2x
15
x0 ) + kx20
x0 )
y 120
100
80
60
40
20
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
-20
-40
Tangent line to the graph y = 4:9x2 at (3; 44:1)
Calculus is about processes. There are two important processes involved in understanding the derivative:
1) We began by constructing di¤erence quotients for a function
f over intervals where one of the endpoints is an input x0 that is
given and the other is x0 + h. We allow h to vary, so the di¤erence
quotient is a function of h. We might call this function m(h) because
it computes the slope of the secant line through the points (x0 ; f (x0 ))
and (x0 + h; f (x0 ) + h) on the graph.
2) We then try to …nd limh!0 m(h). If this limit exists we write
that number as f 0 (x0 ) and call it the derivative of f at x0 . If f 0 (x0 )
exists then so does the tangent line to the graph at (x0 ; f (x0 )), and
the slope of this tangent is f 0 (x0 ).
16
5.0
x
Now we can ask what happens if we decide to vary the input x0 . This produces
a new function f 0 (x), called the derivative function for f , but we usually call it
the derivative of f . For example, if f (x) = x2 then f 0 (x) = 2x. Let’s try another
example that is easy to compute.
Example. A function describes a quantity that varies in inverse
proportion to its input. Find the derivative of this function. Any
function of the type can be described algebraically as
k
x
for some constant k. It’s domain is Rnf0g. The di¤erence quotient
at an input x0 is a function of h
f (x) =
m(h) =
=
k
x0
k
x0 +h
h
k
x0 (h + x0 )
and so
k
x20
is the derivative at x0 . This derivative exists for any input in the
domain, so the derivative of f is
lim m(h) =
h!0
f 0 (x) =
k
x2
Notice that whenever a constant is a factor in an algebraic rule for a function
that it is also a factor in the derivative, because it multiplies each term in the
numerator of the di¤erence quotient. This will help us simplify calculations in
the future. Of even greater help is the fact that if we have a sum of functions
g(x) = f1 (x) + f2 (x) then
g 0 (x) = f10 (x) + f20 (x)
We say that …nding the derivative of a function is a linear process. This allows
us to focus on fundamental elementary functions from which we build general
functions.
17
Example. The derivative of f (x) = x2 +
1
x
is
1
x2
f 0 (x) = 2x
Therefore the tangent line to the graph at (x0 ; f (x0 )) is
y=
2x0
1
x20
(x
x0 ) + f (x0 )
y
14
12
10
8
6
4
2
-4
-3
-2
-1
1
2
3
-2
-4
-6
-8
-10
-12
-14
Tangent line is y =
55
x
9
29
3
at
3; 26
3
Shapes of Graphs
Since the derivative gives the slope of the tangent line it provides a lot of information about the behavior of a function. This goes both ways. If we only have a
18
4
x
graph of the function but no explicit rule for the output we can estimate its instantaneous rate of change at a point buy sketching the tangent line and estimating
its slope. On the other hand, if we have an expression for the output and are
able to derive an expression for the derivative then this can be used to determine
where the function is increasing or decreasing: if the derivative is positive on an
interval then the function must be increasing on that interval; if the derivative is
negative on an interval then the function must be decreasing on that interval; if
the derivative is zero at a particular value then the function is "leveling out" there
(its may be changing its growth behavior). The graph in our last example seems
to indicate that f is decreasing over the entire negative half of the domain. In
the positive half of the domain the decreasing behavior continues until the graph
bottoms out, after which f appears to be increasing. The derivative con…rms this
and even allows us to …nd the point where the graph changes its behavior. Let’s
…nd exactly where f is increasing:
f 0 (x) > 0
1
> 0
2x
x2
2x3 1 > 0
1
x3 >
2
Thus, f is increasing precisely when x > 2
1
3
0:793 7. At the point 2
1
3
p
; 32 3 2
1
= 0. We sometimes use critical
the tangent line will be horizontal because f 0 p
3
2
point to describe a place where the derivative is zero (or where the tangent line
cannot be de…ned).
Knowing where a function is increasing or decreasing is generally not enough to
get a good sketch of its graph because many functions share similar behavior. It
would help to know where the graph is concave up or concave down. The derivative
helps here as well. Since f 0 is itself a function we can try to …nd its derivative
f 00 . This second derivative now tells us where the …rst derivative f 0 is increasing
or decreasing. It is important to get a mental picture of what this means. Sketch
the graph of a typical concave up function, like y = x2 , and then sketch in some
tangent lines. What is happening to the slopes of these lines as x increases? The
slopes are increasing:
19
y
16
14
12
10
8
6
4
2
-4
-3
-2
-1
1
2
3
-2
Therefore, f 00 is positive. (Note that if f (x) = x2 then f 0 (x) = 2x and f 00 (x) = 2.)
Similarly, the slopes of the tangent lines to a concave down graph will decrease,
and so f 00 is negative. Recall that the free-fall model is a function for displacement
given by
s(t) = 4:9t2
We found that v(t) = s0 (t) = 9:8t, a linear function, and therefore s00 (t) = v 0 (t) =
9:8, a constant function. In kinematics, the derivative of velocity is called acceleration: s00 (t) = v 0 (t) = a(t). What Galileo discovered is that near the surface of
the Earth the acceleration due to gravity is constant: a(t) = 9:8 m= s2 .
Let’s summarize:
If f 0 (x) > 0 then f is increasing on that interval.
If f 0 (x) < 0 then f is decreasing on that interval.
If f 00 (x) > 0 then f is concave up on that interval.
20
4
x
If f 00 (x) < 0 then f is concave down on that interval.
Consider again the function f (x) = x2 + x1 . We found f 0 (x) = 2x x12 from which
we can directly compute f 00 (x) = 2+ x23 (use the di¤erence quotient for x12 ). Where
is the graph concave up?
f 00 (x) > 0
2
2+ 3 > 0
x
1
>
1
x3
This is certainly true for x > 0, as the graph indicates. If x < 0 then the inequality
becomes
x3 <
i.e., x <
1
1
and the graph does appear to be concave up for these values. Apparently the
concavity changes to down at the point ( 1; 0). A place where concavity changes
is called an in‡ection point.
Note. It is important to keep the logic straight: When a function
changes at x0 from increasing to decreasing (or decreasing to increasing) then f 0 (x0 ) = 0 if the derivative exists there. The converse can
fail, that is, f 0 (x0 ) = 0 does not necessarily imply that the growth
behavior is changing. Similarly, if there is an in‡ection point at x0
then f 00 (x0 ) = 0 if the second derivative exists there, but f 00 (x0 ) = 0
does not necessarily imply that concavity changes there.
Functions Described by Tables
Empirically, we do not always have a model for the behavior of a function. In
fact, sometimes we want to …t a model to given data. Suppose for example that
you have a chart of blood pressure readings on a surgery patient that have been
recorded every hour for a period of time. You are concerned not only about the
readings themselves but also how the readings are changing over time.
21
Time
12:00
1:00
2:00
3:00
4:00
5:00
Systolic
90
100
115
120
130
115
Diastolic
50
55
65
60
62
73
From this table can we get an estimate of the instantaneous rate of change in
the B/P at, for example, 3:00? We do not have enough data to estimate limits,
but we could make the assumption that these ratings …t some continuous model
and compute average rates of change. There are two questions here, one for the
systolic pressure S(t) and one for the diastolic D(t). The smallest we can take h
to be is 1 because the B/P was only taken every hour. For h = 1
130 120
S
=
= 10
t
1
For h =
1
115 120
S
=
=5
t
1
If we average these two average rates of change we obtain 7:5 per hour as an
estimate of the systolic rate of change at time 3:00. Note that the same procedure
applied to D(t) yields an estimate of 1:5 per hour, so the systolic seems to be
increasing but the diastolic decreasing at 3:00.
Exercise. Estimate the rates of change for S(t) and D(t) at time
2:00 by using h = 2 and h = 2.
Study Guide for First Exam
Domain and Range of a Function
Be able to identify these from either a function rule or its graph.
Average Rate of Change
Set up di¤erence quotient for f at x0 as a function of h
Interpretation with units in context
22
Instantaneous Rate of Change
Find limit of di¤erence quotient at x0 as h ! 0 for simple functions, e.g.,
free-fall function
Find tangent line to graph at (x0 ; f (x0 ))
Estimate instantaneous rate of change from values in a table
Limits and Continuity
Determine from a graph whether limit of f exists at x0 from both directions
Determine whether f is continuous at x0
Derivative Functions
Determine where f 0 (x) is positive, negative and relate to growth behavior of
f and its graph
Determine where f 00 (x) is positive, negative and relate to concavity of graph
of f . Practice exercise: The …gure below shows three graphs, y = f (x), y = f 0 (x),
y = f 00 (x). determine which is which.
y
1.0
0.8
0.6
0.4
0.2
-5
-4
-3
-2
-1
1
-0.2
-0.4
-0.6
-0.8
-1.0
23
2
3
4
5
x
A function and its …rst two derivatives.
24
Suggested Exercises for Chapter 3
The following exercises are not to be handed in. They represent skills required for
basic mastery.
3.1 (pages 143-145):
7; 9; 13; 19; 21; 29
45; 61; 63; 65
3.2 (pages 155-158):
5; 11; 13; 21; 27
3.3 (pages 164-166):
3; 5; 7; 15; 23; 27; 39; 41
3.4 (pages 175-177):
1; 5; 7; 13; 29; 37; 47; 53; 57
3.5 (pages 187-189):
1; 3; 9; 13; 17; 19; 31; 37; 41; 49
3.6 (pages 199-204):
1; 7; 17; 23; 25; 29; 31
Second Graded Assignment
Due: October 29
To reinforce written communication skills the Graded Assignment solutions
should be clearly presented in a "bluebook" or provided in PDF format. Late
papers will not be graded.
Second Graded Assignment. Do any one of the following:
Page
Page
Page
Page
Page
145:
165:
207:
207:
208:
62
42
52
60
64
25
Derivatives of Elementary Functions
We have directly computed the derivative of a few functions by taking the limit
of the di¤erence quotient as h ! 0. Newton’s general binomial theorem allows us
to …nd the derivative of any function of the form
f (x) = xa
where a is a constant. We get the same result as if a were a positive integer:
f 0 (x) = axa
1
Since the derivative operates linearly we can compute derivatives term by term
and multiplication of f by a constant multiplies f by the same constant.
Example. Find the derivative of
p
p
1
3
f (x) = 2x5
x+3 x+
x
Rewriting each term with exponents:
1
f (x) = 2x5
f 0 (x) = 10x4
1
x 3 + 3x 2
1 2
x 3+
3
+x 1
3 1
x 2
2
x
2
By an "elementary" function one usually means a function that occurs as a model
of some natural phenomenon. Such functions can exhibit very complicated behavior but many of them can be constructed from simpler functions such as logs
and exponentials, in addition to the functions that we build up from algebraic
processes as in the example above. Recall that the natural exponential function
is de…ned by
f (x) = ex
The di¤erence quotient at x is
y
f (x + h) f (x)
=
x
h
1 x
=
e
eh ex
h
eh 1
= ex
h
26
To …nd limh!0
y
x
we would need to determine
lim
eh
h
h!0
Note that
eh
eh
1
e0
f (0 + h) f (0)
h
h
h
so this limit would give us the slope of the tangent line to the graph at the point
(0; 1) :
1
=
=
y
8
7
6
5
4
3
2
1
-3
-2
-1
1
-1
2
x
This slope appears to equal 1 and experimenting with f (0+h)h f (0) for small values
of h seems to con…rm this. It can in fact be shown that the slope of this tangent
is exactly 1. We conclude that for f (x) = ex :
f 0 (x) = ex
that is, the natural exponential function is its own derivative.
Now consider the natural logarithmic function
f (x) = ln x
27
Its di¤erence quotient at x is
y
ln(x + h)
=
x
h
ln(x)
The laws of logarithms do not allow us to expand ln(x + h) easily, so we need to
…nd f 0 (x) another way. We remember that y = ex and y = ln x have graphs that
are re‡ections of each other in the line y = x, because the functions are inverses
of each other.
y
8
7
6
5
4
3
2
1
-3
-2
-1
1
2
3
4
-1
5
6
7
8
x
-2
-3
We just found that the tangent to the exponential graph at (a; ea ) has slope ea .
The re‡ection of this tangent in y = x has slope e a and is the tangent to y = ln x
at the point (ea ; a). We conclude that the slope of a tangent line to the graph of
y = ln x is the reciprocal of the input variable there. That is, if f (x) = ln x the
f 0 (x) =
1
x
The instantaneous rate of change of function f (x) = kex is directly
proportional to the output.
The instantaneous rate of change of function f (x) = k ln x is inversely proportional to the input.
28
Product, Quotient, and Chain Rules
Functions are constructed from simpler functions in two basic ways:
1) algebraic combinations such as sums, products and quotients;
2) function composition.
Suppose we have a function given by F (x) = f (x)g(x) where we already know
how to …nd f 0 (x) and g 0 (x). To …nd F 0 (x) set up the di¤erence quotient
y
F (x + h) F (x)
f (x + h)g(x + h) f (x)g(x)
=
=
x
h
h
f (x + h) f (x) [f (x + h)
g(x + h) g(x)
+ g(x)
+
= f (x)
h
h
Then limh!0 xy =
f (x)] [g(x + h)
h
f (x + h) f (x)
[f (x + h)
g(x + h) g(x)
+ g(x) lim
+ lim
h!0
h!0
h!0
h
h
0
0
0
= f (x)g (x) + g(x)f (x) + f (x) lim [g(x + h) g(x)]
f (x) lim
h!0
0
0
0
= f (x)g (x) + g(x)f (x) + f (x) 0
= f (x)g 0 (x) + g(x)f 0 (x)
This is the Leibniz (Product) Rule
(f g)0 (x) = f (x)g 0 (x) + g(x)f 0 (x)
Example. If y = x2 ex then
dy
= x2 ex + ex (2x)
dx
= x (x + 2) ex
If F (x) =
f (x)
g(x)
then f (x) = g(x)F (x)
f 0 (x) = g(x)F 0 (x) + F (x)g 0 (x)
f 0 (x) F (x)g 0 (x)
F 0 (x) =
g(x)
=
f (x) 0
g (x)
g(x)
f 0 (x)
g(x)
g(x)f (x) f (x)g 0 (x)
=
[g(x)]2
0
29
g(x)]
f (x)] [g(x + h)
h
g(x)]
This is the basic Quotient Rule
f
g
Example. If y =
0
(x) =
ex
ex x
g(x)f 0 (x) f (x)g 0 (x)
[g(x)]2
then
dy
(ex
=
dx
x) ex ex (ex
(ex x)2
1 x
= ex
(ex x)2
Determine the domain and range of f (x) =
tangent line to the graph horizontal?
1)
ex
.
ex x
Where is the
y
x
We can now …nd derivatives of a great many elementary functions constructed as
algebraic combinations of simple functions. However, we require a rule for …nding
derivatives of composite functions. For example, basic statistics uses many graphs
of functions such as
30
y
0.5
0.4
0.3
0.2
0.1
-5
-4
-3
-2
-1
0
1
2
3
4
5
x
1
(1+x2 )
p(x) =
and
y
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
-3
-2
-1
0
p(x) =
p1
1
e
2
3
x
x2
The derivatives of these functions compute the relevant statistical parameters,
such as standard deviation which is found from the in‡ection points of the graph.
31
We can …nd p0 (x) in the …rst example by using the quotient rule, but we do not
yet have a method to compute p0 (x) in the second example. However, we can
2
think of the function F (x) = e x as the composition of two simpler functions
F (x) = f g
g(x) =
x2
f (x) = ex
We already know that g 0 (x) = 2x and f 0 (x) = ex , so we expect that F 0 (x) can
be constructed from these two derivatives. We have
F (x + h) F (x)
f (g (x + h)) f (g (x))
=
h
h
What can we do with this di¤erence quotient? We can think of it in stages: Let
u = g(x) and y = F (x) = f (u). Then, since u = g (x + h) g (x), we can use
the ’rewriting strategy’:
f (g (x + h))
h
f (g(x) + [g (x + h) g (x)]) f (g(x)) g (x + h) g (x)
g (x + h) g (x)
h
f (u + u) f (u) g (x + h) g (x)
=
g (x + h) g (x)
h
f (u + u) f (u) g (x + h) g (x)
=
u
h
Since h = x, we can express this product of di¤erence quotients in a way that
is easier to remember:
y u
y
=
x
u x
Now
f (g (x))
u ! 0 as
=
x ! 0, so in the limit we get a product of derivatives
F 0 (x) = f 0 (u)g 0 (x)
= f 0 (g(x))g 0 (x)
In Leibniz notation
y
dy
dy du
!
=
x
dx
du dx
This is the Chain Rule for single-variable calculus.
32
Now we can …nd p0 (x) for p(x) =
p1
e
x2
. Since
p1
is a constant
1
p0 (x) = p f 0 (g(x))g 0 (x)
1
= p eg(x) g 0 (x)
1
= p e
x2
( 2x)
2
p xe
=
x2
Thus p0 (x) = 0 only when x = 0, which explains the single critical
point corresponding to the maximum p(0) = p1 . We can …nd p00 (x)
using the Chain Rule together with the Product Rule:
p00 (x) =
2
p
2
= p
x
2xe
2x2
1 e
x2
+e
x2
1
x2
Note that we expect two in‡ection points, and in fact p00 (x) = 0 for x = p12 .
Exercise. Find the coordinates on the graph of these two in‡ection points.
Find the equations of the tangent lines at these points.
33
Logarithmic Di¤erentiation
The Chain Rule provides a proof of Newton’s rule for di¤erentiating powers of the
input variable. Let f (x) = xa where a is any constant. Then
jf (x)j = jxa j = jxja
and so, if x 6= 0,
ln jf (x)j = a ln jxj
Using the Chain Rule:
1 0
a
f (x) =
f (x)
x
a
0
f (x) =
f (x) = axa
x
1
We have used logarithmic di¤erentiation here to compute a derivative implicitly
after taking the log in order to simplify the equation. This technique is often used
together with implicit di¤erentiation were we compute the derivative of a quantity
with respect to a given variable without solving for that quantity explicitly.
Example. Find the tangent line to the curve x3 + x + y 3 + y 2 = 0
at the point (0; 1).
In the xy-plane the slope of a tangent line is given by the value of
dy
.
However, we do not have y expressed explicitly as a function of x
dx
and, as is often the case, we cannot do so easily. But if we think of y
dy
varying as x varies then we can compute dx
using the Chain Rule:
3x2 + 1 + 3y 2
dy
dy
+ 2y
= 0
dx
dx
dy
3x2 + 1
=
dx
y (2 + 3y)
dy
j(0; 1) =
1
dx
The line through (0; 1) with slope
34
1 is
2
y
1
-2
-1
1
2
x
-1
-2
y=
x
1
Example. Let f (x) = xx on the domain (0; 1). Is the tangent line
to the graph ever horizontal? Are there any in‡ection points?
Here the domain variable appears in both the base and the exponent. We take the log on both sides of y = xx :
ln y = x ln x
1
1 dy
= x
+ (ln x) 1 = 1 + ln x
y dx
x
dy
= (1 + ln x) xx
dx
The derivative is zero only when x = 1e , so the tangent to the graph
at the point
1
;e
e
1
e
(0:367 88; 0:692 2) is horizontal. Further,
d2 y
1
= (1 + ln x)2 xx + xx
2
dx
x
1
=
+ (1 + ln x)2 xx
x
The second derivative is always positive so the graph is always
concave up.
35
y
4
3
2
1
0
0.0
0.5
1.0
1.5
2.0
x
Exponential Models
Many phenomena can be described mathematically with exponential models. Two
models that occur in a variety of contexts are:
1) Quantities that change at a constant percentage rate, that is, a quantity
that changes in proportion to its size;
2) Quantities that change logistically, that is, would grow at a constant percentage rate but are subject to a maximum size.
Constant percentage growth/decay.
If a quantity increases or decreases by k% each unit period of time. If the initial
amount, when t = 0, is P0 then after one unit of time the amount will be
P0 +
k
k
P0 = P0 1 +
100
100
36
Here k will be a positive number if the quantity is increasing and a negative
number if the quantity is decreasing. What happens after two units of time? We
take the amount after one unit of time and apply the percentage:
P0 1 +
k
100
+
k
k
P0 1 +
100
100
= P0 1 +
k
100
2
Similarly, after n units of time the amount is
P0
k
1+
100
n
We can turn this observation into a model by viewing t as a continuous variable:
P (t) = P0 1 +
k
100
t
is the growth (or decay) function that can now be analyzed using calculus.
Example. Suppose $1000 is invested at 5% annual interest. How
much is the investment worth after 64 months?
We are using the growth function
5
P (t) = 1000 1 +
100
= 1000 (1: 05)t
t
Since this is an annual interest rate we measure time in years, so
64 months is 64
= 16
years. Then
12
3
P
16
3
16
= 1000 (1: 05) 3
1297:2
The investment is worth $1297.20 after …ve and one-third years.
For any positive number a we can write
a = eln a
37
For example,
1:05 = eln(1:05)
e0:04879
By the laws of exponents we can write our growth function as
P (t) = 1000e(0:04879)t
By the Chain Rule we have
P 0 (t) = (48:79) e(0:04879)t
and therefore
P 0 (t)
= 0:04879
P (t)
Now 4:879% < 5%, so this model does not follow the rule that the ratio of
instantaneous rate of growth to amount present is equal to the annual percentage
rate. A model that does follow this rule is the continuous compounding model
P 0 (t) =
k
P (t)
100
Note that if our model had been
P (t) = 1000e(0:05)t
then in fact
P 0 (t)
= 0:05
P (t)
We can formalize this rule by setting P (0) = P0 and r =
k
.
100
Then
P (t) = P0 ert
is the continuous compounding model where t is measured in units corresponding
to r, typically years for an annual percentage rate k. This model applies to any
phenomenon where the ratio of derivative to value is constant. It is the exponential
growth/decay model
A(t) = A0 ect
The amount A is growing exponentially if c > 0 and decaying exponentially if
c < 0.
38
Newton’s Law of Cooling/Heating. Temperature di¤erence follows exponential growth or decay. Suppose a can of soda at 36 falls
into a hot tub at 104 and suppose that after 1 minute in the water the
soda has risen to 42 . How many minutes will it take to reach 50 ?
Newton’s model tells us
104
T (t) = (104
36) ect
and we are given that T (1) = 42. This allows us to solve for c :
104
T (1) = 68ec 1
62 = 68ec
c = ln 62
ln 68 = ln
31
34
We can write the temperature function either as
31
T (t) = 104
68e(ln 34 )t
T (t) = 104
68
or as
31
34
t
(In practice we would likely use the approximation c
Now we can solve
104
68ect = 50
27
54
=
ect =
68
34
27
ct = ln
34
ln 27 ln 34
t =
ln 31 ln 34
0:092373.)
2:4956
The soda will reach 50 after about 2 21 minutes in the water.
Logistic growth.
Quantities that would grow exponentially except for an imposed limitation often
follow a logistic model
M
P (t) =
1 + Aect
39
where M is the maximum population (carrying capacity) and
A=
M
P (0)
P (0)
The constant c < 0 so that as time goes on the denominator approaches 1 and P
approaches M . For a given carrying capacity and initial population P0 = P (0) the
constant c can be inferred from empirical data if the logistic model is postulated.
(Further data over time may con…rm the model or may refute it, requiring some
other model that better …ts the data.) The derivative
P 0 (t) =
AM c
ect
(1 + Aect )2
is always positive since c < 0, so the logistic function is always increasing asymptotically toward the value M . We therefore expect an in‡ection point, where the
second derivative will be zero:
00
P (t) =
(1 Aect ) ct
e
AM c
(1 + Aect )3
2
Note that P 00 (t) = 0 only when Aect = 1:
1 1
ln
c A
1
=
ln A
jcj
t =
At this time the population growth rate will be at its maximum and decrease
toward zero after that.
40
Suggested Exercises for Chapter 4
The following exercises are not to be handed in. They represent skills required for
basic mastery.
4.1 (pages 215-217):
1; 2; 5; 9; 15
4.2 (pages 225-228):
3; 5; 7; 21; 29; 37
4.3 (pages 236-238):
1; 17; 33; 37
4.4 (pages 247-249):
3; 5; 23; 31
4.5 (pages 254-255):
19
4.6 (pages 261-265):
3; 5; 9; 13; 17
Third Graded Assignment
Due: November 17
To reinforce written communication skills the Graded Assignment solutions
should be clearly presented in a "bluebook" or provided in PDF format. Late
papers will not be graded.
Third Graded Assignment. Do any one of the following:
Page
Page
Page
Page
Page
216:
217:
227:
237:
262:
24
32
58
42
16
41
Related Rates
An important application of the Chain Rule involves computing the rate of change
of a quantity that depends on the rate of change of other measurable quantities.
For example, in chemistry the ideal gas law relates pressure, volume and temperature by the equation
P V = kT
where k is a positive constant. Suppose the temperature is increasing steadily
at 5 per minute and the volume is decreasing exponentially. Find the rate of
change in pressure when the temperature is 30 and the volume is 20 liters. We
are thinking of P; V; T as functions of time so we di¤erentiate both sides with
respect to t :
P (t)V 0 (t) + V (t)P 0 (t) = kT 0 (t) = 5k
Dividing both sides by P (t)V (t) :
5k
5
V 0 (t) P 0 (t)
+
=
=
V (t)
P (t)
P (t)V (t)
T (t)
(Note that every term in the equation is now has the units 1/minute.) Since the
volume is decreasing exponentially we have
V 0 (t)
=
V (t)
c2
for some constant c, and so
P 0 (t)
5
=
+ c2
P (t)
T (t)
Finally, for T (t) = 30 and V (t) = 20 we have P (t) = 32 k atm (the unit of pressure
is 1 atmosphere) and so
3
5
k
+ c2
2
30
1
=
k 6c2 + 1
4
P 0 (t) =
42
so the pressure is increasing at this time at a rate of 41 k (6c2 + 1) atm= min: Note
that we did not need to know when this occurs nor did we need to solve for P
explicitly as a function of t. This is a fairly sophisticated example because it
was intended to model a realistic problem. To study the technique of computing
related rates we will mostly focus on situations where there are simple geometric
relations among the variables.
Example. A spherical weather balloon is being in‡ated at a constant rate of 10 cubic feet per second. How fast is the radius of the
balloon increasing when the volume has reached 2000 cubic feet? How
fast is the surface area of the balloon increasing?
The volume of a sphere is
4 3
r
3
where r is the radius. Di¤erentiating both sides of this equation
with respect to time t :
dV
dr
= 4 r2
dt
dt
dV
At any time t we have dt = 10 so
V =
dr
5
=
dt
2 r2
describes the rate of increase in the radius given its value. When
V = 2000 we have
4 3
2000 =
r
3
r
1
12 3
3 1500
r =
=5
7: 815 9 ft
Therefore
1
dr
5
1
12 3
=
2 =
dt
120
2 25 12 3
0:013 ft= sec 0:156 in= sec
43
The surface area of a sphere is
S = 4 r2
and so
dS
dr
=8 r
dt
dt
Therefore, at the same instant the rate of increase in surface area
is given by
1
dS
12 3 1
= 8 5
dt
120
2
2:559 ft = sec
12
1
3
=
2p
3
18
3
Note that it is a good idea to work with exact numbers and not approximate until
the end of the calculation. This helps minimize both arithmetic and rounding
errors. An approximation from a calculator is a good idea because it allows us to
see if our answer is reasonable.
Summary of Curve Sketching
The properties of derivatives establish some basic principles for sketching curves.
At this level we want to be able sketch the graphs of elementary functions y = f (x)
that are used frequently in applied models. Such sketches should include the
following:
Intercepts. The point (0; f (0)) where the graph crosses the yaxis, assuming 0 is in the domain, of course. Points where the graph
crosses the x-axis, assuming there are domain values x0 such that
f (x0 ) = 0.
Critical Points. These are points on the graph where either
f 0 (x) = 0 or where the derivative does not exist. If the derivative
does not exist at a point in the domain then there may be a discontinuity or a "cusp" so that a tangent line cannot be determined. Look
for these …rst since they are usually obvious. Then …nd the points
44
in the domain where f 0 (x) = 0 and investigate the behavior further:
Remember, if there is a maximum or minimum then f 0 (x) = 0;
if there is an in‡ection point and the second derivative also exists
there then f 00 (x) = 0.
Vertical Asymptotes. These occur at points x0 not in the domain of f where limx!x0 f (x) = 1.
Horizontal Asymptotes. These occur at limiting values of y as
x ! 1.
These features of a graph were highlighted by a few basic theorems about derivatives:
If f has a local maximum or minimum at x0 then (x0 ; f (x0 )) is a
critical point.
If f 0 (x) > 0 on an interval then f is increasing on that interval.
If f 0 (x) < 0 on an interval then f is decreasing on that interval.
If f 0 changes from positive to negative at x0 then f has a maximum
at x0 . If f 0 changes from negative to positive at x0 then f has a
minimum at x0 .
If f 00 (x) > 0 on an interval then the graph is concave up on that interval; thus, if f 00 is continuous and f 0 (x0 ) = 0 then f has a minimum
at x0 .
If f 00 (x) < 0 on an interval then the graph is concave down on
that interval; thus, if f 00 is continuous and f 0 (x0 ) = 0 then f has a
maximum at x0 .
Note. Points where f 0 and f 00 are zero or fail to exist identify candidates for the
behaviors that shape the graph of f . They do not guarantee that behavior. Two
simple examples will help you remember this.
45
Example. Let f (x) = x4 . Then f 0 (x) = 4x3 exists at all points
and is zero only when x = 0. This is a critical point that turns out
to be a minimum because f 0 changes from negative to positive at 0.
Note that f 00 (x) = 12x2 is always non-negative so the graph is always
concave up. We have f 00 (x) = 0 but the origin is not a point of
in‡ection.
y
6
5
4
3
2
1
-1.5 -1.0 -0.5 0.0 0.5 1.0 1.5
x
f (x) = x4
1
2
Example. Let f (x) = x 3 . Then f 0 (x) = 13 x 3 and f 00 (x) =
5
2
x 3 do not exist at the origin. What kind of critical point is this?
9
x = y3
46
y
2
1
-2
-1
1
-1
-2
1
f (x) = x 3
47
2
x
Study Guide for Second Exam
Derivatives of Basic Functions
Polynomials
Exponentials
Logarithms
Rules of Di¤erentiation
Product Rule
Quotient Rule
Chain Rule
Implicit and Logarithmic Di¤erentiation
Exponential and Logistic Models
Application of Chain Rule to Related Rates
48
Optimization
Di¤erential calculus is particularly useful for analyzing optimization problems.
These are situations where some quantity is to be maximized or minimized often,
but not always, on a restricted domain. At this stage we can only look at examples
where the quantity can be expressed as a function, implicitly or explicitly, of a
single variable, but even this case has many applications. As with related rate
problems, the key is usually to look for some relation among the variables that
arises from functional and/or geometric formulas.
Example. A work cubicle is to be constructed from straight partitions of a given height but which can be any length, however you must
use exactly 100 linear feet of partition material for the cubicle and the
resulting ‡oor plan must be a rectangle. What ration of length to width
provides the most area?
We choose to write L; W; A for length, width and area, respectively.
Then
A = LW
2L + 2W = 100
The second equation allows us to write one of the dimensions as a
function of the other, for example
W = 50
L
and so we can now write area as a function of length:
A (L) = L (50 L)
= 50L L2
Notice that the domain of A is restricted to 0
we would get a negative value for A.
49
L
50, otherwise
A 600
500
400
300
200
100
0
0
10
20
30
A (L) = L (50
L)
40
50
L
It appears that the maximum area occurs when L is the center of
this interval, but let’s use calculus to verify.
dA
= 50
dL
2L
so the only critical point is L = 25 ft and therefore A = 625 ft2 .
Consequently, a cubicle with a square ‡oor, L = W , plan provides the
most area.
This example shows more generally that a rectangle of a given perimeter has
maximum area when it is a square. Thus a fundamental geometric principle
seems to apply here. Also, whenever we have a restricted domain it is important
to check the value of the function at the endpoints of the interval because its
extreme values might be achieved at those points.
Example. A space plane is dropped from a jet travelling at 500
mph and accelerates so that its velocity over the …rst ten seconds is
v(t) = t3 3t2 + 500. Find the maximum and minimum velocities of
the space plane over this interval.
50
We …rst note that v(0) = 500 and v(10) = 1200. We will compare
these values for those at any critical points in the interval (0; 10). The
acceleration is the …rst derivative
v 0 (t) = 3t (t
2)
which is zero within (0; 10) only at t = 2, and
v(2) = 496
which is smaller than the initial velocity. We conclude that the
minimum velocity of 496 mph occurs after 2 seconds and the maximum
velocity of 1200 mph occurs after 10 seconds.
v
1200
1100
1000
900
800
700
600
500
400
0
1
2
3
4
v(t) = t3
5
6
7
8
9
10
t
3t2 + 500
Notice that it is di¢ cult to tell from the graph alone that the velocity drops
slightly before starting to increase.
Example. A dolphin, initially 100 meters from shore, can swim
parallel to the shore at 15 meters per second. Due to ocean currents it
51
can only swim at 10 meters per second if it swims at any other angle
to the shore. Suppose the dolphin wants to reach a beacon on the shore
400 meters away from the point on the shore closest to its starting
point in the ocean. Assuming it makes the swim in two straight line
segments and wants to reach the beacon in minimum time, how far
should the dolphin swim parallel to the shore before heading straight
toward the beacon?
The geometry of this problem is very simple (it is easier to describe
with math than with language).
y
100
50
0
0
50
100
150
200
250
300
350
400
x
It is natural to assign the variable x to the distance the dolphin
will swim parallel to the shore. The domain is 0 x 400 since the
dolphin will not swim beyond the beacon and then need to turn back.
The length of the send segment of the swim is then
q
1002 + (400 x)2
How long will it take the dolphin to swim each segment? For the
parallel segment we have
Tp =
x
seconds
15
52
and for the diagonal segment we have
q
1
Td =
1002 + (400 x)2 seconds
10
so the total time T = Tp + Td can now be expressed as a function
of x :
q
1
x
+
1002 + (400 x)2
T (x) =
15 10
First, note that
p
T (0) = 10 17 41:23 seconds
2
110
= 36 seconds
T (400) =
3
3
so it is better for the dolphin to swim the full 400 meters parallel
to the shore and then swim toward the beacon than to swim the single
diagonal segment of the rectangle. Now we check for critical points on
(0; 400) :
x 400
1
1
q
T 0 (x) =
+
10
15
(x 400)2 + 10 000
is zero when
q
x) = 2 (x
3 (400
400)2 + 10 000
Squaring both sides we obtain the quadratic equation
x2
800x + 152 000 = 0
which has the solutions
However, only 400
…nd
p
x = 400 40 5
p
40 5 is in the domain of the function T . We
p
10 p
40 5 =
5+8
34: 12 seconds
3
p
so in order to minimize time the dolphin should swim 400 40 5
310: 56 meters parallel to the shore and then head for the beacon. How
long did each segment of the swim take?
T 400
53
Example. For t in hours, suppose a drug concentration in mg= l
is given by
y
0.5
0.4
0.3
0.2
0.1
0.0
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
C(t) = e
2t
1.6
1.8
e
2.0
2.2
2.4
2.6
2.8
5t
Then C 0 (t) = 5e 5t 2e 2t = 0, when t = 13 ln 2 +
0:305 43, so the maximum concentration occurs after
60 (0:305 43)
3.0
x
1
3
ln 5
18: 3 min
Suppose we want the minimum value of C over the time period
[0:1; 0:8] :
C (0:1) = 0:212 2
C(0:8) = 0:183 58
So the minimum concentration between 6 and 48 minutes is approximately 0:184 mg= l.
54
Suggested Exercises for Chapter 5
The following exercises are not to be handed in. They represent skills required for
basic mastery.
5.1 (pages 293-296):
5; 9; 11; 15
5.2 (pages 306-308):
1 17 odd, 29; 41; 47; 57; 67; 71
5.3 (pages 314-316):
3; 7; 9; 17; 25; 35
5.4 (pages 321-323):
3; 5; 17; 19; 39; 51
5.5 (pages 327-328):
3; 11; 17; 29
Fourth Graded Assignment
Due: December 3
To reinforce written communication skills the Graded Assignment solutions
should be clearly presented in a "bluebook" or provided in PDF format. Late
papers will not be graded.
Fourth Graded Assignment. Do any one of the following:
Page
Page
Page
Page
Page
306:
315:
322:
328:
330:
20
26
52
28
4
55
Schedule
November 12
Area and the de…nite integral
FTC
November 17
Net change and average value
November 19
Substitution
Integration by parts
November 24
Areas between curves
Applications to biology
November 26
Di¤erential equations
Logistic Model
Probability
December 1
Functions of two variables
Partial derivatives and Extrema
December 3
Review
56
The De…nite Integral
The concept of the de…nite integral is motivated by the ancient problem of …nding
the area enclosed by curves that are not geometric …gures bounded by straight
lines, such as rectangles and triangles. The idea is to represent these curves
as graphs of functions and then apply limiting processes, a procedure originally
explored by Archimedes to …nd areas enclosed by conic curves. The rigorous proofs
of these ideas and their application to more general curves would have to wait for
the invention of calculus and its deeper development by Bernhard Riemann and
other mathematicians, but we can understand the concepts without these proofs.
Riemann’s approach is very simple. First consider the graph of a positive function
over a closed interval. For example, f (x) = 2x on [0; 3].
y
7
6
5
4
3
2
1
0
0
1
2
3
x
y = 2x on [0; 3]
The area under the graph y = 2x and above the interval [0; 3] is that of a triangle
and we know this area is 12 3 6 = 9. Now consider the function f (x) = x2 on the
same interval.
57
y
x
y = x2
Euclidean geometry does not provide a simple formula for this area, but we can
approximate it by dividing [0; 3] into a …nite number of sub-intervals (we call this
a partition) and selecting a point from each one. A convenient choice is the left
endpoint of each subinterval. We then evaluate f at each endpoint and multiply
by the length of the sub-interval: This will be the area of the rectangle whose base
is the sub-interval and whose height is the y-value on the graph. For example, if
we partition into six sub-intervals then each rectangle will have a base-length of
1
and be under the curve since this function is increasing on [0; 3].
2
58
y
x
The rectangle with base
3
;2
2
and height f
3
2
Now add up all six areas:
1
1
f (0) + f
2
2
1
2
1 3
1
1
1
+ f (1) + f ( ) + f (2) + f
2
2 2
2
2
5
2
=
55
8
Thus, we suspect that the actual area under the parabola is greater than 55
.
8
What happens if we choose the right endpoint of each sub-interval? Then we will
over-estimate the area under the parabola:
1
f
2
1
2
1
1 3
1
1
+ f (1) + f ( ) + f (2) + f
2
2 2
2
2
59
5
2
1
91
+ f (3) =
2
8
y
x
The rectangle with base
3
;2
2
and height f (2)
So the actual area under the parabola is less than 91
. A reasonable estimate of
8
the true area might be the average of these two approximations:
1
2
55 91
+
8
8
=
73
8
Thus far we have used no calculus, just basic geometry. Riemann’s idea was to
image the partition …ner and …ner by choosing an arbitrarily large number n of
sub-intervals of [a; b]. We set
b a
x=
n
He then proved that we can select the representative point xj in each sub-interval
any way we like and, as long as f is continuous on [a; b], the limit
lim [f (x1 ) x + f (x2 ) x +
n!1
will exist. We write this limit as
Z
+ f (xn ) x]
b
f (x)dx
a
and de…ne it to be the area under the graph y = f (x). As we will see,
Z 3
x2 dx = 9
0
60
so in this example the over-estimate was closer to the true area than the underestimate. This will not always be the case. As the number n increases both
approximations will get closer to 9.
This limit of so-called Riemann sums exists even if f changes sign on [a; b]. This
generalizes the area computation to the concept of net area: Areas below the
horizontal axis and above the graph carry a negative sign. For example,
Z 3
2xdx = 5
2
because the area of the triangle with base [ 2; 0] is 4.
y
6
4
2
-2
2
4
x
-2
-4
y = 2x on [ 2; 3]
An Application. If an object moves at a constant speed r then the
distance it travels over a period of time is the product of the speed and
the length of the period of time. If we graph the speed as a constant
function the graph will be a horizontal line at height r. Between time
t = a and t = b the distance travelled is
r (b
61
a)
which is the area of the rectangle with base [a; b] and height r. We
can apply this area concept to objects moving along a straight line
with velocity v given as a function of t. For example, if the velocity is
v(t) = 2t meters per second then then object starts out with velocity
v(0) = 0 and at time t = 3 has velocity v(3) = 6 meters per second.
The distance it has travelled over this interval of time is
Z 3
2tdt = 9 meters
0
which is the area of the triangular region under the graph of v. Notice
that the units of this de…nite integral are
( m= sec) sec = m
Many interpretations of the de…nite integral as net area of a graph over an interval
provide a wide variety of applications. First we explore velocity functions in
greater detail. When an object moves in a straight line we must remember that
the velocity function v(t) can change sign. This means that the object has changed
direction. The speed of the object at any time t is the absolute value of velocity
jv(t)j. The de…nite integral generalizes the formula d = rt for an object moving
at constant speed: The total distance travelled between t = a and t = b is
Z b
jv(t)j dt
a
Since v(t) is the …rst derivative of the position function s(t),
Z
b
v(t)dt
a
is the net change of position, which we call displacement.
Example. A freight train passes a station at time t = 0 and
travels along the track at a velocity v(t) = 1 t miles per hour. Where
is the train relative to the station after 90 minutes and what is the
total distance travelled?
62
The …rst question asks for the displacement so we …nd the de…nite
integral
Z 3
2
(1 t) dt
0
because 90 minutes is
v
3
2
hours.
1.0
0.8
0.6
0.4
0.2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
t
-0.2
-0.4
v(t) = 1
t
This integral is the net area obtained by subtracting the area of the
triangle below the interval 1; 23 from the area of the triangle above
the interval [0; 1] :
Z 3
2
1 1
3
(1 t) dt =
=
2 8
8
0
The train is 83 mile away from the station after 90 minutes. The total
distance travelled after 90 minutes is
Z 3
2
j1 tj dt
0
which is the sum of the two areas
5
1 1
+ =
2 8
8
63
The train travelled away from the station for the …rst hour, during
which it covered 21 mile. It stopped (v(1) = 0) and then started moving
back toward the station and travelled 18 mile in the last half-hour.
The Fundamental Theorem of Calculus
A close look at the previous examples reveals something that Newton realized
about Galileo’s experiments with motion. Since v(t) = s0 (t) we can infer the
displacement function for the train by asking the questions: What function s
would have a …rst derivative given by the rule s0 (t) = 1 t ? The answer is easy:
s(t) = t 21 t2 + c, where c is any constant (check that s0 (t) = 1 t). What the
inventors of calculus noticed was
#
"
2
1
3 1 3
3
+c
0
s (0) =
(0)2 + c
s
2
2 2 2
2
3
8
The de…nite integral of the velocity function over an interval is equal to the di¤erence of the displacement function at the endpoints of the interval. This observation
works for any continuous function f . If we want to compute
Z b
f (x)dx
=
a
then we think of f as the …rst derivative of some function F and just compute
F (b) F (a). We call F an anti-derivative of f and call this result the FTC
(Fundamental Theorem of Calculus). Any anti-derivative for f will work because
if F1 and F2 are antiderivatives for f then F1 F2 is just a constant, which will
go away when we take the di¤erence at the endpoints of the interval. Now we can
see why
Z
3
x2 dx = 9
0
An antiderivative for f (x) = x is F (x) = 31 x3 because F 0 (x) = x2 . Thus
2
F (3)
F (0) =
1 3
3
3
64
0=9
The FTC says that, in order to compute the de…nite integral of a function f over
a closed interval [a; b], we think of that function as the derivative of some other
function F . Then the Riemann sums look like
F 0 (x1 ) x + F 0 (x2 ) x +
+ F 0 (xn ) x
However, each
x) F (xk )
x!0
x
from which Riemann was able to prove, no matter how xk is chosen inside each
sub-interval (so x1 6= a and xn 6= b), that
F 0 (xk ) = lim
F (xk +
lim [F 0 (x1 ) x + F 0 (x2 ) x +
+ F 0 (xn ) x]
x!0
= [F (x1 ) F (a)] + [F (x2 )
= F (b) F (a)
F (x1 )] + [F (x3 )
F (x2 )] +
+ [F (b)
F (xn )]
Thus the de…nite integral of F 0 is the net change in F .
Example. With t 0 in hours suppose a battery loses charge at
the rate R(t) = e t . Find the amount of charge lost between 1 and 3
hours. We compute
Z
3
e t dt
1
by …rst …nding a function F (t) such that F 0 (t) = e t . All such
functions are of the form
F (t) =
e
t
+c
where c is any constant, so we choose F (t) =
F (3)
F (1) =
e
= e 1
3
e
e
3
e
t
and compute
1
0:318 09
We can interpret this result by assuming the battery starts with
100% charge. Between 1 and 3 hours it loses about 32% of its charge.
How much did it lose in the …rst hour?
Z 1
e t dt = F (1) F (0)
0
= 1
65
1
e
0:632 12
so in the …rst 3 hours the battery has lost over 95% of its charge.
66
Solving Initial-Value Problems
The general anti-derivative of a function f is denoted by
Z
f (x)dx
without any bounds on the integral sign. This symbol stands for all possible
functions F such that F 0 (x) = f (x). For example,
Z
1
x2 dx = x3 + c
3
where c is any constant. If F (x) = 13 x3 + c then F 0 (x) = x2 because the derivative
of a constant is zero. Any two speci…c anti-derivatives for f di¤er by a constant,
so the general anti-derivative is a family of functions whose graphs are all vertical
shifts of each other.
y
5
4
3
2
1
-2
-1
1
-1
2
x
-2
-3
-4
-5
The FTC says we can choose any speci…c anti-derivative to evaluate a de…nite
integral because
Z b
f (x)dx = F (b) F (a)
a
67
so the constant c does not matter for this purpose. However, it does matter if
initial conditions are present.
Rectilinear Motion. Suppose a dolphin swims in a straight line
subject to acceleration at time t in m= sec2 given by
a(t) = 8
t
Find the dolphin’s velocity at time t if its velocity at t = 0 is 5
meters per second. Since v 0 (t) = a(t) we …rst …nd
Z
1 2
(8 t) dt = 8t
t +c
2
Since v(0) = 5 we …nd
8 0
1 2
0 +c = 5
2
c = 5
so the velocity function is
v(t) = 8t
1 2
t +5
2
Now we can answer questions about the dolphin’s position:
1) What is the dolphin’s displacement over the interval 0 t
20
?
2) What is the total distance traveled by the dolphin over the interval 0 t 20 ?
3) What is the dolphin’s actual position after the …rst 20 seconds?
The …rst two questions can be answered by choosing any speci…c
anti-derivative for v, but the third cannot be answered without knowing the dolphin’s position at some speci…c time:
1) Since s0 (t) = v(t) the dolphin’s displacement is the de…nite
integral
Z 20
20
1 3
1 2
2
8t
t + 5 dt =
t + 4t + 5t
2
6
0
0
1100
=
366: 67 m
3
68
p
2) Since the dolphin’s velocity turned negative after 8 + 74 seconds we must …nd
Z 20
1 2
8t
t + 5 dt
2
0
Z 8+p74
Z 20
1 2
1 2
=
8t
t + 5 dt +
8t
+
t
5 dt
p
2
2
0
8+ 74
74 p
74 p
632
+
=
74 +
74 156
479: 05 m
3
3
3
v
40
30
20
10
0
2
4
6
8
10
12
14
-10
16
18
20
t
-20
-30
-40
v(t) = 8t
1 2
t
2
+5
3) Suppose the dolphin’s position after 10 seconds is known to be
30 meters from a reference beacon:
s(10) = 30
We have
s(t) =
=
Z
v(t)dt
1 3
t + 4t2 + 5t + c
6
69
for some constant c. To …nd c we use
1
(10)3 + 4 (10)2 + 5 (10) + c = 30
6
760
3
s(10) =
c =
The position function is
1 3
t + 4t2 + 5t
6
s(t) =
760
3
so after the …rst 20 seconds the dolphin’s position is
340
113: 33 m
3
from the reference beacon. Note that at the start of the time interval
s(20) =
760
253: 33 m
3
so the dolphin started out on the opposite side of the beacon from
where it ended up. To …nd when the dolphin was at the beacon we
would need to solve
s(0) =
s(t) =
1 3
t + 4t2 + 5t
6
760
=0
3
It is straightforward to show that the graph crosses the t-axis three
times, but only once in the interval [0; 20] :
70
s
200
100
-8
-6
-4
-2
2
4
6
8
10
12
14
16
18
20
22
24
t
-100
-200
s(t) =
1 3
t
6
+ 4t2 + 5t
760
3
at approximately t = 9:16 seconds. If the dolphin continues to
swim according to this law then it will pass the beacon again at about
t = 22:29 seconds but not again.
71
Methods for Finding Anti-Derivatives
There are two elementary methods for …nding anti-derivatives of functions. The
…rst (the method of substitution) uses the Chain Rule to reduce the integrand to
an easily recognized di¤erential, and the second (integration by parts) uses the
Product Rule to separate the integrand into simpler pieces.
Method of Substitution. Sometimes a factor in the integrand with di¤erential
dx is the composition of functions f (u (x)) where F 0 (u) = f (u) is easily deter. Since
mined and the remaining factor is du
dx
du =
du
dx
dx
the integrand can be re-written in terms of u. Then the anti-derivative can be
found and put back in terms of x.
Examples.
1)
Z
Here we notice that
2
xex dx
2
f (x) = ex = f (u (x))
where u(x) = x2 . Then
Now
R
du
dx
= 2x and so
du = 2xdx
1
xdx =
du
2Z
Z
1
2
xex dx =
eu du
2
eu du = eu + c and so
Z
1 2
2
xex dx = ex + c
2
The Chain Rule veri…es that this is in fact the general anti-derivative.
72
2) To …nd the anti-derivative of
f (t) =
t2
t
+1
we try u(t) = t2 + 1 so that du = 2tdt and then
Z
Z
t
1
1
dt =
du
2
t +1
2
u
1
=
ln juj + c
2
1
=
ln t2 + 1 + c
2
Again, the Chain Rule veri…es that F 0 (t) =
t
t2 +1
if F (t) = 12 ln (t2 + 1)+
c.
3) Suppose the concentration of a drug in the bloodstream, in mg
per liter, t hours after administration is given by
C(t) = te
t2
Find the average concentration during the …rst 45 minutes. We
use the following de…nition:
The average value of a function on [a; b] is
Z b
1
f (x)dx
jb aj a
First we compute
Z
0
3
4
te
1 t2
dt =
1
2
Z
1
1
=
e
2
7
16
1
e du =
2
u
7
e 16
Z
1
eu du
7
16
0:584 73
mg per liter. Note that we used the substitution u = 1 t2 so
that tdt = 12 du, and consequently the limits on the integral with
respect to u changed accordingly. Since we were only interested in
computing the de…nite integral it was not necessary to convert back
to the t variable.
73
C 1.2
1.0
0.8
0.6
0.4
0.2
0.0
0.0
0.1
0.2
0.3
0.4
C(t) = te1
t2
0.5
0.6
0.7
t
on 0; 43
y 1.4
1.2
1.0
0.8
0.6
0.4
0.2
0.0
0.5
0.6
0.7
C(u) = 21 eu on
0.8
0.9
1.0
x
7
;1
16
The area under each curve above over the corresponding intervals
is the same. However, we want the average concentration over the …rst
74
three-quarters of an hour. This is the t-interval so we must divide the
de…nite integral by 43 :
4 1
e
3 2
7
e 16
=
2
e
3
7
e 16
0:779 63
mg per liter.
Method of Parts.
The method of substitution was based on the Chain Rule for di¤erentiation. The
method of parts is based on the Leibniz (Product) Rule of di¤erentiation. Since
d
(f g) = f g 0 + gf 0
dx
we can write the corresponding di¤erential form as
d (f g) = f g 0 dx + gf 0 dx
If we were to …nd the de…nite integral of each side over [a; b] we would have the
equation
Z b
Z b
0
gf 0 dx
f g dx +
f (b)g(b) f (a)g(a) =
a
a
because f (x)g(x) is an anti-derivative for the integrand d(f g) on the left side.
The idea of the method of parts is to view the integral we are given to evaluate
as one of the terms on the right sideR with the presumption that the other term is
b
easier to evaluate. For example, if a gf 0 dx is known from earlier techniques but
Rb 0
the given integral can be viewed as a f g dx, then
Z b
Z b
0
gf 0 dx
f g dx = [f (b)g(b) f (a)g(a)]
a
Example. Compute
a
Z
3
xex dx
0
We want an anti-derivative for the function x 7! xex so we view
the di¤erential xex dx as f g 0 dx for some choice of f and g. If we choose
f (x) = x
g(x) = ex
75
then
f g 0 dx = xex dx
but
gf 0 dx = ex dx
which is easy to integrate. In fact, an anti-derivative for the original
function is
xex ex = (x 1) ex
It is easily checked that the derivative of this function is xex . Now
we can use the FTC to obtain
Z 3
xex dx = (x 1) ex j30 = 2e3 + 1
0
As with the method of substitution, we must be prepared to look carefully at the
given integrand to see if it can be separated into parts in this way. The general
pattern is often summarized by the anti-derivative equation
Z
Z
udv = uv
vdu
which reminds us to view the integrand as a product of a function u and a differential dv, the idea being to choose dv to be something easy to integrate with
respect to du. A variety of common examples helps us recognize a choice that
works, with the understanding that, even if the technique is combined with the
method of substitution, not all integrands can be decomposed in this way.
Example. Find the anti-derivative for f (t) = ln t. We want a
function F so that F 0 (t) = ln t. We will need the fact that f 0 (t) = 1t .
We can write
Z
Z
ln tdt =
udv
u = ln t
dv = dt
76
Then
Z
uv = (ln t) (t + c1 )
Z
1
vdu =
(t + c1 ) dt
t
= t + c1 ln t + c2
Putting this all together we obtain
Z
ln tdt = (ln t) (t + c1 )
= t ln t
(t + c1 ln t + c2 )
t+c
where we have simply used c as the undetermined constant. This
suggests a simplifying rule:
When …nding a general anti-derivative by this method just …nd any
anti-derivative at each stage (usually the simplest possible) and then
add the general constant at the very end. When computing a de…nite
integral no constant need be added since any anti-derivative works for
the FTC.
Example. Find the anti-derivative
Z
t2 et dt
Here we have more than one choice. For example, we could set
u = t
dv = tet dt
From the …rst example we have
1) et
v = (t
and, since du = dt,
Z
t2 et dt = t (t
1) e
= t (t
1) et
= t (t
= t2
1) et (t 1) et + et + c
2t + 2 et + c
t
77
Z
1) et dt
Z
Z
t
te dt + et dt
(t
Notice that we used the …rst example a second time. (See page
325 where the alternative choice u = t2 ; dv = et dt is used to …nd this
anti-derivative.)
78
Suggested Exercises for Chapter 6
The following exercises are not to be handed in. They represent skills required for
basic mastery.
6.1 (pages 338-340):
1; 3; 11; 17; 33; 37
6.3 (pages 353-354):
1; 7; 11; 13
6.4 (pages 362-363):
1; 3; 5; 15; 23; 27
6.5 (pages 368-369):
3; 5; 9; 11
Fifth Graded Assignment
Due: December 10
To reinforce written communication skills the Graded Assignment solutions
should be clearly presented in a "bluebook" or provided in PDF format. Late
papers will not be graded.
Fifth Graded Assignment. Do any one of the following:
Page
Page
Page
Page
354:
362:
376:
376:
12
28
6
10
79
Some Common Applications
Area RBetween Curves.
Rb
b
Since a f (x)dx is the net area under the graph y = f (x) over [a; b] and a g(x)dx
is the net area under the graph y = g(x) over [a; b] the net area under the graph
of f but over the graph of g is
Z b
[f (x) g(x)] dx
a
If we interchange the roles of f and g then the net area between the curves has
sign opposite this integral. The idea is a generalization of thepsimple geometric
picture where f (x)
g(x) on [a; b]. For example, if f (x) = x and g(x) = x2
then f (x) g(x) on [0; 1].
y
2.5
2.0
1.5
1.0
0.5
0.0
0.0
y=
0.5
1.0
1.5
x
p
x and y = x2
The area between these curves on [0; 1] is the positive number
Z 1
p
1
x x2 dx =
3
0
This concept has application to any problem where we want to interpret the
di¤erence between the anti-derivatives of two functions over an interval.
80
Example. Two dolphins pass
p a reference beacon at time t = 0.
One moves with velocity v1 (t) = t and the other with velocity v2 (t) =
t2 , meters per minute. What is the di¤erence in their positions after
1 minute?
Both velocity functions are non-negative on [0; 1] and v1
v2 on
this interval so we can interpret the answer as how much more distance the v1 dolphin has covered than the v2 dolphin. From the above
calculation we conclude that this di¤erence is 31 meter. Both dolphins
start out at velocity zero and after 1 minute each is moving at 1 meter
per second, but for all times in between the second dolphin is moving
more slowly than the …rst. What happens after 2 minutes?
Population Growth.
As we have seen, the exponential model of population growth is based on the
assumption
dP
= kP
dt
where k is a constant. in terms of di¤erentials
1
dP = kdt
P
so the anti-derivative relation is
ln P = kt + c
P (t) = ekt+c = ec ekt
The initial condition P (0) = P0 implies
P 0 = ec
which con…rms the exponential model is
P (t) = P0 ekt
If k > 0 we have exponential growth. If k < 0 we have exponential decay. These
models generally apply in situations where the growth or decay is not inhibited,
such as growth of a laboratory culture or decay of a radioactive isotope. Models for
populations subject to inhibiting factors can be derived from di¤erential equations
that include those considerations. One such model balances renewal and survival
rates.
81
Example. If an initial population P0 adds members at a rate R(t),
t in years, and the proportion of the population that remains after t
years is given by S(t) then the population after T years is
Z T
P0 S(T ) +
S(T t)R(t)dt
0
Suppose the initial population of rabbits on the campus is 3000
and they reproduce at the rate R(t) = 500e0:2t but the proportion
that survives after t years is S(t) = e 0:1t . Then the population after
T years is
Z
T
0:1T
3000e
e
+ 500
0:1(T t) 0:2t
e
dt
0
Since e
0:1(T t) 0:2t
e
Z
= e(0:3t
0:1T )
T
e(0:3t
0:1T )
and
10 (0:3t 0:1T ) T
e
j0
3
10 0:2T
e
e 0:1T
=
3
dt =
0
the population after T years is
P (T ) = 3000e
=
0:1T
1000
4e
3
5000 0:2T
e
3
+
0:1T
e
0:1T
+ 5e0:2T
For example, the population after 10 years is
P (10) =
1000
4e
3
1
+ 5e2
12806
The logistic model of population growth applies to a situation of exponential
growth inhibited by a maximum carrying capacity M . Our logistic growth model
was derived from the di¤erential equation
dP
dt
= kP
M
dP = kdt
P (M P )
82
1
P
M
Now
M
1
1
= +
P (M P )
P
M P
so an anti-derivative for the left side is
ln P
ln (M
P ) = ln
P
M
P
so we conclude
ln
P
M
= kt + c
P
P
M
= ec ekt
P
When t = 0 we have P = P0 and so
ekt
P = M e c kt
e e +1
c
but P = P0 when t = 0 so ec =
P0
M P0
=
1
A
in our previous discussion, thus
ekt
1 kt
e +1
A
M
=
1 + Ae kt
P (t) =
M
A
as before.
Poiseuille’s Law.
In 1840 this French physician discovered a law of laminar ‡ow in blood vessels
by assuming the velocity v of blood is greatest along the central axis of a tubular
vessel and decreases to 0 at the wall of the vessel. If the radius of the vessel is R
and its length is L then
P
v(r) =
R2 r 2
4 L
gives blood velocity as a function of distance r from the axis, where P is the
pressure di¤erence between the ends of the vessel and is the blood viscosity. An
83
important concept in measuring cardiac output is ‡ux, the rate of blood ‡ow as
volume per unit time. Flux for a vessel of radius R is the de…nite integral
Z R
2 rv(r)dr
F =
0
Z R
P
r
= 2
R2 r2 dr
4 L
0
Z R
P
=
r R2 r2 dr
2 L 0
P R4
=
8 L
This is Poiseuille’s Law: Flux is proportional to the fourth power of the radius
of the blood vessel. This law can be used to measure the narrowing of blood vessels
due to disease by injecting dye and using an empirical model for dye concentration.
For example, if 6 mg of dye are injected and the concentration model is
C(t) = 20te
mg per liter over 0
t
0:6t
10 seconds then the ‡ux through an artery is given by
F = R 10
0
6
20te
0:6t
which is approximately 6.60 liters per minute (see exercise 13 on page 354). If the
pressure di¤erence P at the ends of a length L of the artery is measured and the
blood viscosity is known then
6:60 =
R =
P R4
8 L
52: 8 L
P
1
4
2:025
r
4
L
P
This value can be converted to a fraction of the normal radius of the artery to
measure the degree of narrowing.
Probability Density Functions.
84
A measurement of a parameter related to a subject chosen at random is called a
continuous random variable X. The probability P that X lies between two values
a and b is given by a probability density function (pdf) f such that
P (a
X
b) =
Z
b
f (x)dx
a
In order for f to qualify as a pdf for some random variable it must have the
property that its de…nite integral over the entire domain of possible values of X
is equal to 1. That way the range of P is [0; 1], which means that the probability
ranges from 0% to 100%.
Example. Assuming credit scores X can range from 0 to 850
suppose that a pdf for the distribution of credit scores is of the form
f (x) = kx (8:50
x)
1
X. What is the value of k ? What is the probability
where x = 10
that the credit score of an individual chosen at random is between 600
and 750? What is the probability that the credit score is at least 750?
We must have
Z 8:5
kx (8:50 x) dx = 1
0
because any individual chosen at random must have 0
Since
Z
8:5
kx (8:5
x) dx = 102: 35k
0
we …nd
k=
1
102:35
and so the pdf is
1
x (8:50
102:35
x 2 [0; 8:5]
f (x) =
85
x)
X
850.
Then the probability that 600
X
750 is approximately 17%
because
Z 7:5
X
1
P 6
7:5 =
x (8:50 x) dx 0:17
10
102:35
6
y
0.16
0.14
0.12
0.10
0.08
0.06
0.04
0.02
0.00
0
1
P (600
X
2
3
4
5
6
7
750) is the area under the curve between the vertical lines
Now, P (X
750) =
Z 8:5
7:5
1
x (8:50
102:35
x) dx
0:038
or approximately 3.8% are expected to have credit scores above
750.
86
8
x
The mean
of a pdf over the domain of X is the de…nite integral of xf (x). Since
Z 8:5
1
x2 (8:50 x) dx 4:25
102:35 0
we would say that the mean credit score is approximately 425. Since many random
variables have in…nite domains it is conventional to consider the domain of a pdf
to be all real numbers by de…ning it to be identically 0 outside of its given domain.
We then have
Z 1
f (x)dx = 1
1
We then de…ne the median m of a pdf by the equation
Z 1
1
f (x)dx =
2
m
In the credit score example we have m 4:25 because the function is symmetrical
about the mean . The median is the value of X that divides the population in
half. Here is an example where the mean and median are not the same.
Example. Many random values, such as lifetimes of electronic
devices, have an exponential distribution, which is de…ned by a pdf of
the form
0; t < 0
f (t) =
ce ct ; t 0
for some positive constant c. To verify this is a pdf note that
Z 1
Z 1
f (t)dt = c
e ct dt
1
0
To evaluate this "improper" integral we take an anti-derivative
F (t) =
e
ct
and …nd
lim [F (b)
b!1
F (0)] = lim
b!1
87
e
cb
( 1) = 1
because e
cb
! 0 as b ! 1. The mean of this pdf is
Z 1
= c
te ct dt
0
=
lim
b!1
1
c
1
e
c
bc
(bc + 1) =
1
c
because bc+1
! 0 as b ! 1 (the exponential in the denominator
ebc
grows much more rapidly than the linear function in the numerator).
To compute the median we solve
Z 1
1
ce ct dt =
2
m
for m. The left side reduces to
0
e
cm
=e
=
1
2
cm
Solving
e
cm
cm =
ln 2
1
ln 2
m =
c
Thus, for an exponential distribution the median is less than the
mean because ln 2 0:693 15. Suppose, for example, that the lifetime
of LED bulbs is exponentially distributed with c = 0:25, where time
is measured in years. Then the mean lifetime is 4 years but half of the
bulbs are expected to fail before 4 ln 2 2: 8 years.
88
f
0.25
0.20
0.15
0.10
0.05
0.00
0
1
2
3
4
5
6
7
8
t
Median (vertical line) divides the area under the graph in half
89
Functions of Several Variables
Applications of calculus often require the analysis of a quantity that depends on
more that one input. We will look at functions of two inputs because functions of
more than two variables are analyzed in much the same way. The generic notation
y = f (x) now becomes
z = f (x; y)
We call x and y the independent (input) variables and z the dependent (output)
variable. It is important to respect the order of the input variables so we understand (x; y) to be an ordered pair. This allows us to view the input as a point
in the xy-plane, just as we interpreted f (x) as the output from a point on the
x-axis. The graph z = f (x; y) is therefore a surface in three-dimensional space
whose height above the domain plane is z.
Example. The graph of
f (x; y) = 2x
3y + 5
is a plane that crosses the z-axis at the point (0; 0; 5) :
12
10
8
6
z
4
-2
2
-2
-1
y
-1
0
0 0
-2
1
1
x
2
z = 2x
3y + 5
90
2
and the portion of this graph for the inputs (x; 0) is the line that
is its intersection with the plane containing the x and y axes.
12
10
8
6
z
4
2
-2
-1
y
0
0 0
-2
1
-2
-1
1
x
2
2
z = f (x; 0)
Traces, Level Curves, Sections
Most graphs of functions of two variables are more complicated than planes, so we
try to visualize them by slicing with planes. Computer tomography uses this technique in medicine to construct a three-dimensional image. Here is some common
terminology.
The intersection of a graph with a plane perpendicular to the domain plane is called a section (or vertical trace).
The intersection of a graph with a plane parallel to the domain
plane is called a trace (or horizontal section).
The curve in the domain plane corresponding to a constant value
z = c is called a level curve of value c (projection of trace at height c
into the domain plane).
Example. The graph of f (x; y) = x2
that contains the origin (0; 0; 0).
91
y 2 is a saddle-shaped surface
4
-2
2
-2
-1
z
1
2
y
0
0 0
-2
1x
-1
-4
2
z = x2
y2
The section of this graph by the vertical plane y = 1 is a parabola:
4
-2
2
-2
-1
-1
0
0 0
1 -2
2
1
2
-4
Section by the plane y = 1
The trace of this graph by the horizontal plane z = 1 is a hyperbola:
92
4
-2
2
-2
-1
z
1
2
y
0
0 0
-2
1x
-1
-4
2
z = x2
y2
This trace hyperbola becomes a level curve (sometimes called a
contour line) in the domain plane with equation f (x; y) = 1 :
y
5
4
3
2
1
-5
-4
-3
-2
-1
1
2
3
-1
-2
-3
-4
-5
Level curve x2
93
y2 = 1
4
5
x
An Application. According to Poiseuille’s Law of laminar ‡ow,
the velocity of blood ‡ow in an artery of radius R = 0:8 mm and
length L = 20 mm is a function of distance r from the central axis
and pressure di¤erence p in dynes per square millimeter. If the blood
viscosity is such that 4 L = 0:021 6 dyn sec = mm then the velocity
in mm= sec is
p
(0:8)2 r2
v (r; p) =
0:021 6
60
50
40
v
30
20
10
0
0.0 0.0
0.2
0.5
0.4
0.6
1.0
r
p
0.8
v (r; p) =
1.5
2.0
p
4 L
(0:8)2
r2 with 4 L = 0:021 6
The level curve in the domain for the velocity 20 mm= sec is
p=
0:432
(0:8)2 r2
94
p
4
3
2
1
0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
r
Level curve v (r; p) = 20 mm= sec
As r gets closer to R the pressure must increase without bound in
order to maintain the given velocity.
Partial Derivatives and Critical Points
For a function of several variables it is important to study the output when all but
one of the inputs is held constant. The function can then be analyzed relative to
the behavior of the remaining input variable. In particular, we can form the difference quotient for the remaining variable at a given point (x0 ; y0 ) in the domain.
There are two di¤erence quotients for a function of two independent variables.
f (x0 + h; y0 )
h
f (x0 ; y0 + h)
h
f (x0 ; y0 )
f (x0 ; y0 )
The limits, if they exist, as h ! 0 are called the partial derivatives of f at (x0 ; y0 ).
To compute the partial derivative with respect to x we treat y as a constant; to
compute the partial derivative with respect to y we treat x as a constant. The
95
notation is a variation on that used for derivatives of a single-variable function.
The Leibniz notation is
@f
(x0 ; y0 )
@x
@f
(x0 ; y0 )
@y
and the corresponding "prime" notation becomes
fx (x0 ; y0 )
fy (x0 ; y0 )
p
Example. For the Poiseuille function v (r; p) = 0:021
(0:8)2
6
the partial derivatives at the point (r0 ; p0 ) = (0:3; 2) are
r2
@v
(0:3; 2) =
92: 593prj(0:3;2) = 55: 556
@r
@v
(0:3; 2) = 29: 630 46: 296r2 j(0:3;2) = 25: 463
@p
How do we interpret these values? We have
v (0:3; 2) = 50: 926 mm= sec
at this domain point and this velocity is decreasing rapidly as r
increases; the slope of the vertical section curve p = 2 is 55: 556 at
this point. However, the velocity in increasing as p increases; the slope
of the vertical section curve r = 0:3 is 25: 463 at this point.
96
120
100
80
v
60
40
20
0
0 0.0
0.2
1
2
0.4
0.6
3
p
4
r
0.8
Behavior at the point (0:3; 2; 50:926)
Points in the domain where a partial derivative fails to exist or where both partial
derivatives are zero are called critical points. When both partial derivatives are
zero each section curve exhibits critical behavior. In the above example, this
happens at points (r; p) such that simultaneously
29: 630
92: 593pr = 0
46: 296r2 = 0
The only non-negative values that solve these equations are r = 0:8 and p = 0.
We have v = 0 for this radius and pressure. Here is an example with a critical
point that is not on the boundary of the domain.
Two-Variable Probability Density. The e¤ective lifetime t of
a type of battery varies with its voltage V . Suppose the lifetime and
voltage follow a distribution given by the probability density function
1
f (t; V ) = tV 3 e
6
97
t V
To …nd the critical points of the pdf we …rst compute
1 3
V (1 t) e V
6
1 2
fV (t; V ) =
V t (3 V ) e
6
ft (t; V ) =
t
V
t
Both partial derivatives are 0 if V = 0. Otherwise we must have
V = 3 and t = 1
Analyzing Critical Points.
To see what type of extreme behavior, if any, a smooth function exhibits at a
critical point we use the Second-Derivative Test:
Suppose fx (x0 ; y0 ) = fy (x0 ; y0 ) = 0 and let
2
fxy
(x0 ; y0 )
D (x0 ; y0 ) = fxx (x0 ; y0 ) fyy (x0 ; y0 )
If D (x0 ; y0 ) > 0 and fxx (x0 ; y0 ) > 0 then f has a local minimum
at (x0 ; y0 ).
If D (x0 ; y0 ) > 0 and fxx (x0 ; y0 ) ; 0 then f has a local maximum
at (x0 ; y0 ).
If D (x0 ; y0 ) < 0 the critical point (x0 ; y0 ) is called a saddle point.
If D (x0 ; y0 ) = 0 the test gives no information.
The second derivatives fxy and fyx are equal for smooth functions. In our pdf
example, the second partial derivatives are
1
@ 2 f (t; V )
= V3 e
2
@t
6
@ 2 f (t; V )
1
fV V (t; V ) =
= tV e
2
@V
6
2
@ f (t; V )
1
ftV (t; V ) =
= V2 e
@V @t
6
ftt (t; V ) =
98
V
t
V
t
V
t
(t
V2
(V
2)
6V + 6
3) (t
1)
These are all 0 when V = 0 but then so is f itself. A pdf cannot have negative
values so f has its smallest value at the points (t; 0). The only critical point of
interest is (1; 3) where we …nd
9
e
2
3
e
2
ftt (1; 3) =
fV V (1; 3) =
4
4
ftV (1; 3) = 0
9
e
2
D (1; 3) =
27
e
4
=
8
3
e
2
4
02
4
>0
Since D (1; 3) > 0 and ftt (1; 3) < 0 we conclude that f (1; 3) = 29 e
maximum value.
0.10
0.08
0.06
0.04
0.02
0.00
0 0
1
2
3
1
V
2
4
5
t
3
f (t; V ) = 16 tV 3 e
t V
for 0
t
3; 0
Extra Credit Assignment
99
V
5
4
is a local
Drug concentration in the blood is typically a function of both
time t and dosage q. The blood-brain barrier prohibits some drugs
from attaining high concentrations in brain tissue though the blood
may absorb high doses. Suppose, for initial doses greater than 10 mg,
the concentration in the brain of a certain drug administered at time
t = 0 is modeled by
2 (10
C(t; q) = q 3 tet
q)
with t in hours. This function has a single critical point for positive
values of t and q. Find the value of C at this critical point and interpret
your answer by applying the Second-Derivative Test.
100
Study Guide for Final Exam
The …nal exam is cumulative in the sense that topics we have studied require
mastery of earlier and more fundamental concepts. For example, on the …rst
exam you were asked to …nd information about the position of a dolphin from a
function for its velocity; on the second exam you analyzed related rates of two
dolphins. Similar examples will appear on the …nal exam but now we can …nd
information that requires the de…nite integral. Here is an outline of the progression
of concepts we have studied:
Domains and Ranges of functions of a single variable.
Average rate of change xy over an interval.
Derivatives:
Instantaneous rate of change at a given value of the input variable (limh!0 xy )
Slope of the tangent line at a point (x0 ; f (x0 )) on the graph
Equation of the tangent line
Product and Quotient Rules
Chain Rule
Implicit and logarithmic di¤erentiation
Applications to Related Rates and Optimization
Graphs:
Critical points (maxima, minima, in‡ections)
Intercepts and asymptotes
Integrals and Anti-Derivatives:
Riemann sums for areas
Evaluating de…nite integrals with the Fundamental Theorem of Calculus
Average value of a function over an interval
Interpretation of de…nite integrals, for example, computing displacement from
velocity, or change in population from a rate of growth
The above ideas are basic. Review your …rst two tests carefully
and bring them in with you in corrected form since similar items will
appear on the …nal exam. In addition, there will be a choice of items
on the …nal from the following topics:
101
Applications of the de…nite integral:
Area between curves
Finding displacement s(t) and velocity v(t) from acceleration a(t) and initial
conditions
Population growth
Blood ‡ow
Computing probabilities from density functions
Functions of two variables:
Sections and level curves
First and second partial derivatives
Critical points
Second-Derivative Test for extrema and saddle points
102