MATH 221
FIRST SEMESTER
CALCULUS
Fall 2015
Typeset:August 20, 2015
1
2
MATH 221 1st Semester Calculus
Lecture notes version 1.3 (Fall 2015)
Copyright (c) 2012 Sigurd B. Angenent, Laurentiu Maxim, Evan Dummit, Joel Robbin.
Permission is granted to copy, di
Math 1210  Midterm Exam
Solution Manual
INSTRUCTIONS READ THIS NOW
OFFICIAL
USE ONLY
Each problem is worth 5 points. Partial credits will be given.
Give full credit if the student uses a different method (correctly) and
gets to the right answer.
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that the gas volume is expanding at a rate of 2inch3 per
minute, then what is the rate of change of the pressure?
(c) the slope of the tangent to the parabola at P ,
(d) the angle \OP Q where Q is the point (0, 3).
(b) The ideal gas law turns out to be on
Using derivatives to approximate numbers.
(a) Find the derivative of f (x) = x4/3 .
(b) Use (a) to estimate the number
127
4/ 3
(b) Show that for any pair of functions u and v one
has
(uv )0
u0
v0
=
+
uv
u
v
(u/v )0
u0
v0
=
u/v
u
v
4/ 3
125
2
approximatel
A depends on B depends on C depends on. . .
Someone is pumping water into a balloon. Assuming
that the balloon is spherical you can say how large it
is by specifying its radius R. For a growing balloon
this radius will change with time t.
The volume of th
13.4. Example where you really need the Chain Rule. We know what the derivative of sin x with
respect to x is, but none of the rules we have found so far tell us how to dierentiate f (x) = sin(2x).
The function f (x) = sin 2x is the composition of two sim
The direct approach goes like this:
0
f ( x) =
d1
=
1
4
=
1
4
x4
dx
1
4
x
x4 )
3/4 d(1
x4
1
1 /4
dx
3 /4
4x3
x3
=
1
x4
3/4
To nd the derivative using implicit dierentiation we must rst nd a nice implicit description of the
function. For instance, we could
For example, if y =
p
1
p
, then y = 1/(1 + u) where u = 1 + v and v = 9 + x2 so
2
1+ 9+x
dy
dy du dv
1
1
=
=
p 2x.
22v
dx
du dv dx
(1 + u)
so
dy
dx
=
x=4
dy
du
u=6
du
dv
v =25
dv
dx
11
8.
7 10
=
x=4
14. Exercises
p
149. Let y = 1 + x3 and nd dy/dx usin
Continuing this way one nds after n
(22)
u1 un
0
1 applications of the product rule that
= u0 u2 un + u1 u0 u3 un + + u1 u2 u3 u0 .
1
2
n
7.2. The Power rule . If all n factors in the previous paragraph are the same, so that the function f
is the nth powe
6.6. Dierentiating a constant multiple of a function . Note that the rule
(cu)0 = cu0
follows from the Constant Rule and the Product Rule.
by
(21)
6.7. Picture of the Product Rule. If u and v are quantities which depend on x, and if increasing x
x causes
9. Limits and Inequalities
This section has two theorems which let you compare limits of dierent functions. The properties in
these theorems are not formulas that allow you to compute limits like the properties (P1 ). . . (P6 ) from 5.
Instead, they allow
10.5. How to make functions discontinuous. Here is a discontinuous function:
(
x2 if x 6= 3,
f ( x) =
47 if x = 3.
In other words, we take a continuous function like g (x) = x2 , and change its value somewhere, e.g. at x = 3.
Then
lim f (x) = 9 6= 47 = f
Proving lim
x !0
B
sin
=1
The circular wedge OAC contains the
triangle OAC and is contained in the right
triangle OAB .
C
The area of triangle OAC is
sin
tan
1
2
sin .
The area of circular wedge OAC is
The area of right triangle OAB is
1
2
1
.
2
tan .
To see this try to compute the derivative at 0,
 x   0
 x
f 0 (0) = lim
= lim
= lim sign(x).
x!0 x
x!0 x
x!0
0
We know this limit does not exist (see 7.2)
If you look at the graph of f (x) = x then you see what is wrong: the graph has a corner at
There are no innitely small real numbers, and this makes Leibniz notation di cult to justify. In the
20th century mathematicians have managed to create a consistent theory of innitesimals which allows you
to compute with dx and dy as Leibniz and his conte
7.4. Trying to divide by zero using a limit. The expression 1/0 is not dened, but what about
lim
x!0
1
?
x
This limit also does not exist. Here are two reasons:
It is common wisdom that if you divide by a small number you get a large number, so as x & 0 t
y = f ( x)
L+"
How close must x be to a for f (x) to end up in this range?
L
L
"
a
y = f ( x)
L+"
L
L
"
For some x in this interval f (x) is not between L " and
L + ". Therefore the in this picture is too big for the
given ". You need a smaller .
a
a+
a
y
2.4. Show that limx!4 1/x = 1/4. Solution: We apply the denition with a = 4, L = 1/4 and
f (x) = 1/x. Thus, for any " > 0 we try to show that if x 4 is small enough then one has f (x) 1/4 < ".
1
4
We begin by estimating f (x)
 f ( x)
in terms of x
4.7. Example Limit of 1/x (again) . To prove that limx!1 1/x = 0 we apply the denition to
f (x) = 1/x, L = 0.
For given " > 0 we need to show that
1
x
(10)
L < " for all x > A
provided we choose the right A.
How do we choose A? A is not allowed to depend
p
p
p
6.3. Example Find limx!2 x . Of course, you would think that limx!2 x = 2 and you can
indeed prove this using & " (See problem 44.) But is there an easier way? There is nothing in the limit
properties which tells us how to deal with a square root, a
Something more complicated has happened. We did a calculation which is valid for all x 6= 1, and later
looked at what happens if x gets very close to 1. This is the concept of a limit and well study it in more
detail later in this section, but rst another
CHAPTER 2
Derivatives (1)
To work with derivatives you have to know what a limit is, but to motivate why we are going to study
limits lets rst look at the two classical problems that gave rise to the notion of a derivative: the tangent to
a curve, and the
CHAPTER 1
Numbers and Functions
The subject of this course is functions of one real variable so we begin by wondering what a real number
really is, and then, in the next section, what a function is.
1. What is a number?
1.1. Dierent kinds of numbers. The
2
Figure 2. To nd
p
1
0
1
p
22
2 on the real line you draw a square of sides 1 and drop the diagonal onto the real line.
Almost every equation involving variables x, y , etc. we write down in this course will be true for some
values of x but not for other
range of f
y = f ( x)
(x, f (x)
x
domain of f
Figure 3. The graph of a function f . The domain of f consists of all x values at which the function is
dened, and the range consists of all possible values f can have.
m
P1
y1
y1
y0
1
y0
P0
x1
n
x0
x0
x1
Figu
The graph of f
f ( c)
The graph of f
1
c
b
f ( b)
a
f ( a)
a
b
c
f ( a)
f ( b)
f (c)
Figure 7. The graph of a function and its inverse are mirror images of each other.
4.6. Examples. Consider the function f with f (x) = 2x + 3. Then the equation f (y ) =
3.8. Examples. The graph of f (x) = x3 x goes up and down, and, even though it intersects several
horizontal lines in more than one point, it intersects every vertical line in exactly one point.
The collection of points determined by the equation x2 + y 2