MATH 31, LECTURE 1
PROF. MICHAEL VANVALKENBURGH
• At the beginning of class, Carlos Pereyra spoke about the four PAL sections that
will be offered this semester for Math 31. Studies show that students who enroll in
the PAL program tend to have higher GPAs and higher pass rates.
• I gave out copies of the course syllabus, also available on the course webpage. I
went through it and answered questions.
How many people here did math over the summer? [A few people took a math class.
Others used math for construction, for computer programming, for “counting things,” . . . ]
This summer I drove from Grinnell, Iowa, to Sacramento, taking Interstate 80. For part of
the trip I listened to Feynman’s Lectures on Physics and had fun thinking about science.
Here is one math problem I encountered on the drive: for long stretches, the road was
straight and flat, so I could put on cruise control (and wished I had a self-driving car).
Question: If I am driving at 70 miles per hour, how far will I drive in 7 hours?
Answer: 70 mph × 7 hours = 490 miles. This is easy because “70 miles per hour” literally
means “if driving at this constant rate, after one hour you will have driven 70 miles.” The
number 490 may be interpreted as the area under the [horizontal] curve.
speed (in mph)
80
y = 70
60
40
20
0
0
1
2
3
4
5
time (in hours)
6
1
7
MATH 31, LECTURE 1
2
Next Question: If driving at 70 miles per hour, how far will I travel in one minute?
Answer: 70 mph × 1/60 hour = 7/6 mile, which may be interpreted as the area of the little
rectangle below. (I drew it wider than 1/60, so you can see it.)
speed (in mph)
80
y = 70
60
40
20
0
0
1
2
3
4
5
time (in hours)
6
7
More Realistic Question: What if my speed changes? Suppose at time t the speedometer says f (t) miles per hour. How far did I travel in 7 hours?
1
[In the figure I plotted the graph of f (t) = 70
arctan(7 − t) arctan(t).]
speed (in mph)
80
60
40
y = f (t)
20
0
0
1
2
3
4
5
time (in hours)
6
7
MATH 31, LECTURE 1
3
To get a first approximation, we look at the speedometer once every fifteen minutes and
record our speed. (We should keep our eyes on the road!)
speed (in mph)
80
60
40
y = f (t)
20
0
0
1
2
3
4
5
time (in hours)
6
7
Between hours 5 and 6 we have traveled approximately
f (5) ·
1
4
+ f ( 21
4 )·
1
4
+ f ( 22
4 )·
1
4
+ f ( 23
4 )·
1
4
miles.
This is the sum of the areas of the four rectangles above.
That approximation is bigger than the exact distance traveled. After all, between hours
5 and 6 the approximate speed is usually bigger than our actual speed. For a better
approximation, we check the speedometer more often; that is, we take more sample measurements. To be safe, we pick up a hitchhiker and make him stare at the speedometer
and take measurements.
In general, say we start at time a, we end at time b, and our speed at time t is f (t) miles
per hour. We take n equidistant sample measurements, so that they are b−a
n hours apart.
Common but possibly confusing notation: Write
∆t =
b−a
.
n
MATH 31, LECTURE 1
4
Here “∆t” is a single symbol. We then write our “times of measurement” as
t1 = a + ∆t
t2 = a + 2∆t
..
.
tj = a + j∆t
..
.
tn = a + n∆t = a + n( b−a
n ) = b.
Note: For no particular reason, I now decided to record a speed and use that as an
approximation for how fast we went in the previous interval, rather than for the next
interval.
From time a to time b we have traveled approximately
(†)
f (t1 )∆t + f (t2 )∆t + · · · + f (tn )∆t
miles.
We re-write (†) using the fancy “summation notation”
n
X
f (tj )∆t.
j=1
Fact: We get a better approximation by taking the sample points closer together—by
taking n to be bigger.
Fact: In the limit as n goes to infinity, we get the exact distance traveled:
n
X
lim
f (tj )∆t.
n→∞
j=1
We use more fancy notation and re-write this number as
Z b
f (t) dt.
a
This number is called “the integral of f from a to b.”
Moral: When traveling at speed f (t) mph at time t, from time a to time b we have traveled
Z b
L=
f (t) dt
miles.
a
Graphically, this number is the area under the graph of f between a and b.