Mathematics 162 Class Notes
1
1
Introduction (September 5)
Goal: To remind ourselves as to what calculus is all about!
1.1
What is this course and who is it for?
• Second semester calculus.
• Advanced Placement.
• Honors.
1.2
What is calculus?
1.3
What is a derivative and what does it measure?
1.4
Expectations.
• What I expect of you.
• What you should expect from me.
Homework.
1. Do the following problems to turn in (Due Friday since we will be in the lab on Thursday).
The problem numbers in boxes are those for which special attention will be paid by the
grader. (See syllabus for details.)
pages 228–230
84, 85, 86 , 87, 99, 100 , 121, 122 , 125, 159, 160
2. Additionally, practice your differentiation skills by doing many of problems 1–80 on pages
228-229. (How many is many? As many as you need!) These do not have to be turned
in but you can be sure that your differentiation skills will be assessed on Wednesday.
Mathematics 162 Class Notes
2
2
Integration (September 6)
Goal: To review the definition and fundamental properties of the integral.
2.1
2.2
Some functions
(A)
(B)
(C)
(D)
(E)
(F)
(G)
(H)
(X)
The definition of the integral
• the area problem
• Riemann sums
• the definite integral
Mathematics 162 Class Notes
2.3
3
The integral and the antiderivative
Theorem 2.1 (Fundamental Theorem of Calculus). Suppose that F (x) is continuous on the
interval [a, b] and that F 0 (x) = f (x) for all x ∈ (a, b). Then
Z
b
f (x) dx = F (b) − F (a)
a
• The integral of rate of change is total change.
2.4
Z
Integration by Substitution
Z
2
2x cos(x ) dx
3x
e
Z
Z
dx
dx
2 + (x − 1)2
Z
dx
√
x 4x2 − 1
Homework.
1. Do the following problems (due date, September 11)
pages 382ff
1, 2 , 11, 12 , 15, 17, 35, 38 , 113, 121, 122
2. Practice integration. Do several of the problems 43–112 (not to turn in).
ln x
dx
x
Mathematics 162 Class Notes
3
4
Integration by Parts (September 8)
Goal: To develop the technique of integration by parts
3.1
Reversing the product rule
• Product rule
(f (x)g(x))0 = f 0 (x)g(x) + f (x)g 0 (x)
or
d(uv) = v du + u dv
• Integration by parts:
Z
Z
0
f (x)g (x) dx = f (x)g(x) −
3.2
Z
0
f (x)g(x) dx
Obvious Examples
Z
Z
x cos x dx
3.3
x ln x dx
Nonobvious Examples
Z
• dv = 1.
ln x dx
Z
• Parts twice.
ex cos x dx
Z
• Inverse functions (see p. 454)
3.4
•
cos−1 x dx
Reduction formulas
R
cosn x dx
Homework.
1.
pages 453ff
1, 3, 10 , 11, 25, 30 , 36a , 39, 44
or
Z
u dv = uv −
v du
Mathematics 162 Class Notes
4
Trigonometric Integrals (September 11)
Goal: To employ trigonometric identities craftily to find antiderivatives of trigonometric functions
4.1
Some Important Trig Identities
sin2 (x) + cos2 (x) = 1
tan2 (x) + 1 = sec2 (x)
1 − cos 2x
sin2 (x) =
2
1 + cos 2x
2
cos (x) =
2
1
sin mx sin nx =
[cos(m − n)x − cos(m + n)x]
2
1
sin mx cos nx =
[sin(m − n)x + sin(m + n)x]
2
1
cos mx cos nx =
[cos(m − n)x + cos(m + n)x]
2
Z
4.2
Z
4.3
Z
4.4
sinn (x) cosm (x) dx
tann (x) secm (x) dx
√
1 + cos ax dx
Z
4.5
Integrals like
sin nx sin mx dx
Homework.
1. Do problems in Section 7.2:
page 460: 4 ,7, 16 , 19, 21, 24, 34 , 40
5
Mathematics 162 Class Notes
5
6
Trigonometric Substitutions (September 12)
Goal: To employ crafty substitutions to integrate expressions involving sums and
differences of squares
5.1
Three Useful Substitutions
√
x2 + a2
x
a
x = a tan θ
dx = a sec2 θ dθ
√
x2 + a2 = a sec θ
π
π
− <θ<
2
2
5.2
x2
√
a2 − x2
x = a sin θ
√
dx = a cos θ dθ
a2
x2
−
= a cos θ
π
π
− ≤θ≤
2
2
dx
√
9 − x2
Z
(x2
√
dx
+ 1)3/2
x
dx
−3
x2
Homework.
1.
√
x
x2 − a2
x
Examples
Z
Z
a
Section 7.3, pages 463ff, 4 , 8 , 21 , 29, 37, 42
a
x = a sec θ
dx = a sec θ tan θ dθ
√
x2 − a2 = a| tan θ|
π
π
0 ≤ θ < or < θ ≤ π
2
2
Mathematics 162 Class Notes
6
7
Integration of Rational Functions (September 14)
Goal: To integrate, in principle (!), every rational function.
Definition 6.1. A quotient of two polynomials is called a rational function; i.e., a rational
f (x)
function is a function of the form
where f and g are polynomials.
g(x)
6.1
Case I: g(x) can be factored into distinct linear factors
Z
x
dx
(x − 1)(x − 2)
Key fact: There are numbers A and B such that
x
A
B
=
+
(x − 1)(x − 2)
x−1 x−2
Z
Then
x
dx = A ln |x − 1| + B ln |x − 2| + C.
(x − 1)(x − 2)
This works no matter how many distinct linear factors g has, provided the degree of f (x) is
less than the degree of g(x).
6.2
Case II: g(x) can be factored into linear factors, some multiple
Z
1
dx
(x − 1)(x − 2)2
Key fact: There are numbers A, B, and C such that
1
A
B
C
=
+
+
2
(x − 1)(x − 2)
x − 1 x − 2 (x − 2)2
In general, if (x − a)k is a factor of the denominator, there is a term of form
C
for each
(x − a)j
j ≤ k.
6.3
g(x) cannot be factored into linear terms
Not all polynomials can be factored into linear terms. Some might need quadratic terms.
Mathematics 162 Class Notes
8
Example 6.2.
x4 − 1 = (x2 + 1)(x2 − 1) = (x2 + 1)(x − 1)(x + 1)
Z
x3 + 2x − 1
dx
(x2 + 1)(x − 1)(x + 1)
Key fact: There are numbers A, B, C, and D such that
x3 + 2x − 1
A
B
Cx + D
=
+
+ 2
2
(x + 1)(x − 1)(x + 1)
x−1 x+1
x +1
Repeated quadratic factors just as repeated linear factors.
6.4
Degree of f (x) greater than or equal to that of g(x)
Proposition 6.3. If r(x) is a rational function, then r(x) = p(x) + f (x)/g(x) where p(x) is a
polynomial and the degree of f (x) is less than the degree of g(x).
Example 6.4 (Long Division).
x3 + 2x − 1
16x − 9
=x+4+ 2
x2 − 4x + 2
x − 4x + 2
Theorem 6.5 (Partial Fraction Decomposition Theorem). If r(x) is a rational function then
r(x) can be written as p(x) + f (x)/g(x) where p(x) is a polynomial, f (x) and g(x) are polynomials with the degree of f less than that of g, and f (x)/g(x) can be written as a sum of a
boatload of terms generated as follows:
1. If (x − r)k is one of the factors of g(x) we include terms of the form
A1
x−r
A2
(x − r)2
Ak
(x − r)k
...
2. and if (x + px + q)l is one of the quadratic factors of g(x) we include terms of the form
C1 x + D1
x + px + q
C2 x + D2
(x + px + q)2
Homework.
1. Do the following problems from Section 7.4:
pages 469 ff: 10 , 13 , 18, 21 , 31, 45, 49
···
Cl x + Dl
(x + px + q)l
Mathematics 162 Class Notes
7
9
0/0, ∞/∞, ∞ · 0, ∞ − ∞, 1∞ , 00 , ∞0
(September 15)
Goal: To evaluate limits that are indeterminate forms
7.1
0/0
Theorem 7.1 (L’Hôpital’s Rule). Suppose that lim f (x) = lim g(x) = 0. Then
x→a
x→a
f 0 (x)
f (x)
= lim 0
x→a g (x)
x→a g(x)
lim
if the limit on the right hand side of this equation exists.
cos2 x − 1
x→0
x2
sin x
x→0 x
lim
lim
L’Hôpital’s rule also works for one-sided limits and limits where x → ∞.
lim
x→0+
7.2
sin x
x2
∞/∞
Theorem 7.2 (L’Hôpital’s Rule). Suppose that lim f (x) = lim g(x) = ∞. Then
x→a
f (x)
f 0 (x)
= lim 0
x→a g(x)
x→a g (x)
lim
if the limit on the right hand side of this equation exists.
ln x
x→∞ x
lim
x2
x→−∞ e−x
lim
x→a
Mathematics 162 Class Notes
7.3
10
∞ · 0, ∞ − ∞
The key to indeterminate limits of these forms is usually to transform them to one of the
previous forms.
1
1
−x √
−
x
lim
lim e
x→∞
x→1+ ln x
x−1
7.4
1∞ , 00 , ∞0
Indeterminate forms of these sorts can often be computed by investigating the logarithm of the
limit and using the following fact that follows from the continuity of the exponential function.
Proposition 7.3. Suppose that lim ln f (x) = L. Then lim f (x) = eL .
x→a
x
lim (sin x)
x→0+
x→a
lim
x→∞
1
1+
x
x
Homework.
1. Read Section 4.6 and do the following problems. (due Tuesday, September 19)
pages 289ff
7, 15, 21, 24 , 28 , 42 , 47, 49, 57, 61, 63, 65
2. Extra Credit Integrals - due anytime
Z
Z
p
√
2
ln 1 + x dx
sin−1 x dx
Z
x
dx
1 + sin x
Mathematics 162 Class Notes
8
11
Improper Integrals (September 18)
Goal: To extend the definition of integral to cases where the domain of integration
or the range of the function on the domain is infinite
Z
∞
Z
b
Z
∞
f (x) dx
−∞
−∞
a
Example:
∞
f (x) dx
f (x) dx
8.1
Z
e−x dx
0
∞
Z
Z
f (x) dx = lim
b→∞ a
a
b
f (x) dx
If the right hand side exists and is finite, we say that the integral converges. If the right hand
side does not exist or is ∞, we say that the integral diverges.
Z +∞
1
Example:
dx
1
+
x2
−∞
Z
∞
Z
f (x) dx = lim
b→∞ c
−∞
Z
8.2
∞
Important special case:
1
Z
∞
Proposition 8.1. The integral
1
1. converges to
1
p−1
dx
xp
if p > 1 and
2. diverges (i.e., is equal to ∞) if p ≤ 1.
b
Z
f (x) dx + lim
a→−∞ a
dx
xp
c
f (x) dx
Mathematics 162 Class Notes
8.3
12
Infinite ranges (vertical asymptotes)
1
Z
Example:
0
dx
√
x
If f (x) is continuous on [a, b] except at the point x = a then
b
Z
Z
f (x) dx = lim
a
1
Z
Example:
0
8.4
c→a+
a
f (x) dx
c
dx
x
Important Property of the Real Numbers
Theorem 8.2. Suppose that g(x) is defined and increasing on some interval [a, ∞). Then
either
1. lim g(x) = ∞ or
x→∞
2. lim g(x) exists and is equal to some real number L.
x→∞
8.5
Testing for convergence when one cannot antidifferentiate
Theorem 8.3 (Comparison Test). Suppose that f (x) and g(x) are continuous functions
on
Z ∞
[a, ∞) and that there is a number c such that 0 ≤ f (x) ≤ g(x) for all x ≥ c. Then
f (x) dx
a
Z ∞
converges if
g(x) dx converges.
a
Very Important Example:
R∞
0
2
e−x dx converges.
Homework.
1. In Section 7.7 do the following problems (Due Thursday, September 21, but this assignment will not be collected)
pages 495 ff.
1, 3, 15, 24, 26, 42, 59, 65, 67, 69, p. 501 29
Mathematics 162 Class Notes
9
13
Geometric Series (September 19)
Goal: To introduce the concept of series by way of an important example.
9.1
The Geometric Series
Definition 9.1. A geometric series is an expression of the form
a + ar + ar2 + ar3 + ar4 + · · ·
for real numbers a and r.
Example 9.2. The following is a geometric series with a = 1 and r = 1/2
1+
9.2
1 1 1
1
+ + +
+ ···
2 4 8 16
The Geometric Series Summed
Consider the following sequence of numbers
1
=
1
1
2
=
3
2
1 1
+
2 4
=
7
4
1 1 1
+ +
2 4 8
=
15
8
1+
1+
1+
...
1
Notice that the nth number in this sequence is 2 − n . Therefore, no matter how many terms
2
of the series in Example 9.2 we add, we will never get a sum more than 2. On the other hand,
the sums are increasing and can be made as close to 2 as we like. That is
1
lim 2 − n = 2
n→∞
2
Thus it is reasonable to write
1+
In general
1 1 1
+ + + ··· = 2
2 4 8
Mathematics 162 Class Notes
14
Theorem 9.3. If |r| < 1 then
1. 1 + r + r2 + r3 + · · · =
2. for every a,
9.3
1
and
1−r
a + ar + ar2 + ar3 + · · · =
a
.
1−r
Infinite Series
Example 9.4. The following are examples of infinite series:
1+
1 1 1
1
+ + +
+ ···
2 4 8 16
(9.1)
1+
1 1 1 1 1
+ + + + + ···
2 3 4 5 6
(9.2)
1 + 1 + 1 + 1 + 1 + 1 + ···
(9.3)
1 + (−1) + 1 + (−1) + 1 + (−1) + · · ·
(9.4)
1+
1 1
1
+ +
+ ···
4 9 16
(9.5)
1−
1 1 1 1
+ − + + ···
2 3 4 5
(9.6)
A series is a sum of an infinite sequence of numbers. We will usually eschew the “dot-dot-dot
notation” and write series using the “sigma notation.”
The four series above are written
∞ n
∞
X
X
1
1
2
n
n=0
n=1
∞
X
n=1
1
∞
X
n=0
(−1)n
∞
X
1
n2
n=1
∞
X
(−1)n+1
n=1
1
n
In general, if for every n ≥ k, an is a number (given by some function of n)
∞
X
an = ak + ak+1 + ak+2 + · · ·
n=k
Definition 9.5. An infinite series is any expression of the form
∞
X
n=k
are called the terms of the series.
an . The numbers ak , ak+1 , ak+2 , . . .
Mathematics 162 Class Notes
9.4
15
Convergence and Divergence
An infinite series can do one of four things:
1. Converge to a finite number:
1+
1 1 1
+ + + ··· = 2
2 4 8
2. Diverge to ∞:
1 + 2 + 4 + 8 + 16 + · · · = ∞
3. Diverge to −∞:
1 − 3 + 2 − 4 + 3 − 5 + 4 − 6 + · · · = −∞
4. Diverge by bouncing around (“oscillation”):
1 − 1 + 1 − 1 + 1 − 1 + ···
Homework.
1. The following problems are due Tuesday, September 26, and will be discussed on Monday.
2. For each of the following series, write out the first 5 terms:
(a)
(b)
(c)
(d)
∞
X
n=1
∞
X
(−1)n+1 2n
ln n/n
n=1
∞
X
2n
n!
n=1
∞
X
n=0
(e)
∞
X
n=1
5
1
− n
n
2
3
1
(4n − 3)(4n + 1)
Mathematics 162 Class Notes
3. For each of the following series, write the series in Σ notation:
(a) ln
1
9
1
(c)
3
1
(d)
2
(b)
1
1
1
1
− ln + ln − ln + · · ·
2
3
4
5
1
1
1
−
+
−
+ ···
16 25 36
1 1 1
+ + + + ···
5 7 9
3 7 15
+ + +
+ ···
4 8 16
4. Compute the sums of the following geometric series (Hint: find a and r):
(a)
∞
X
e−2n
n=0
(b) .37 + .0037 + .000037 + .00000037 + · · ·
∞ X
1 n
√
(c)
2
n=3
(d) 1 −
1 1
1
+ −
+ ···
3 9 27
5. Do problems 41, 43, 49, 50 , 70 on page 522–3.
16
Mathematics 162 Class Notes
10
17
Infinite Series II (September 20)
Goal: To investigate the convergence and divergence of infinite series
10.1
Convergence of Infinite Series
• sequence of partial sums of a series
• a series converges
• a series diverges
10.2
Examples
• the geometric series
• “telescoping” series
∞
X
n=1
1
n(n + 1)
• “big” terms
Theorem 10.1 (The nth term test). If
does not exist or is not 0 then
∞
X
∞
X
an converges then lim an = 0. Therefore, if lim an
n=1
n→∞
n→∞
an diverges.
n=1
NB: If the terms do not get small, then the series diverges. However the series may still
diverge even if the terms do have limit 0.
Example 10.2. The following series is called the harmonic series:
∞
X
1 1 1
1
= 1 + + + + ···
n
2 3 4
n=1
The harmonic series diverges.
Mathematics 162 Class Notes
10.3
Combining Series
Theorem 10.3. If
∞
X
an = A and
n=1
1.
18
∞
X
∞
X
an = B, then
n=1
(an + bn ) = A + B
n=1
2.
∞
X
kan = kA for any constant k.
n=1
Homework.
1. Read Section 8.2 and do the following problems, due Wednesday, September 27.
pages 522 ff.
5, 15, 27, 33, 37, 62, 63 , 64 , 65 , 66 , 67
Mathematics 162 Class Notes
11
19
Integral Test (September 25)
Goal: To develop the integral test for convergence and divergence
11.1
Convergence Results So Far
1. Geometric Series:
∞
X
arn−1
n=1
(a) Converges if |r| < 1 to
a
1−r
(b) Diverges if |r| ≥ 1.
2. Harmonic Series:
∞
X
1
diverges.
n
n=1
3. Terms don’t go to zero: If lim an does not converge or does not equal 0 then
n→∞
n=1
diverges.
11.2
Series with Positive Terms
If each an > 0 then there are only two possibilities for
∞
X
an
n=1
1.
∞
X
n=1
2.
∞
X
an diverges to ∞; i.e., lim sn = ∞ where sn is the nth partial sum or
n→∞
an converges to a finite number.
n=1
11.3
Positive and Decreasing Terms
Suppose that a1 > a2 > a3 > · · · .
Example 11.1 (p-series). The series
∞
X
1
has this property for all p > 0.
np
n=1
∞
X
an
Mathematics 162 Class Notes
The key idea of this section is to write
20
∞
X
an as an area.
n=1
Theorem 11.2 (The Integral Test). If there is a continuous, positive, decreasing function f (x)
such that for every n ≥ N , f (n) = an , then
Z
∞
1. If
f (x) dx converges then so does
N
Z
an and
n=N
∞
f (x) dx diverges then so does
2. If
∞
X
N
∞
X
an .
n=N
We can also use the idea of integral to estimate the error made by sn in approximating the
infinite series for which it is a partial sum.
Homework.
1. Read Section 8.3.
2. Do the following problems.
pages 527 ff
1, 5, 6, 9 , 13, 17, 22 , 23, 27, 33 , 39 , 40
Mathematics 162 Class Notes
12
21
Ratio Test (September 26)
Goal: To show the convergence or divergence of a series by comparing it to a geometric series
12.1
Comparsion to a Convergent Series
Proposition 12.1. Suppose that 0 ≤ an ≤ bn for all n ≥ N . Then
1. if
∞
X
bn converges then
n=1
2. if
∞
X
an converges and
n=1
an diverges then
n=1
12.2
∞
X
∞
X
bn diverges.
n=1
Comparison to a Geometric Series – the Ratio Test
Note that a geometric series with positive terms,
∞
X
arn has the property that the ratio of
n=1
successive terms is r. We look at series that have that property in the limit:
Theorem 12.2 (The Ratio Test). Let
∞
X
an be a series with positive terms and suppose that
n=1
lim
n→∞
an+1
=ρ
an
Then
1. if ρ < 1, the series converges
2. if ρ > 1, the series diverges and
3. if ρ = 1, no conclusion is possible about the convergence or divergence of the series.
Homework.
1. Read about the ratio test in section 8.6 and do problems
pages 536 ff
1,3, 4 ,9,11,17,21, 22 ,27,29, 30 ,31, 44
Mathematics 162 Class Notes
13
22
Series with Positive and Negative Terms (September 27)
Goal: To investigate convergence of series with both positive and negative terms
13.1
Special Case - Alternating Series
An alternating series is one in which the terms alternate in sign from positive to negative.
Theorem 13.1 (Leibniz’s Theorem). Suppose that
∞
X
(−1)n+1 un is a series such that
n=1
1. for all n, un > 0 (so the series is alternating)
2. u1 ≥ u2 ≥ u3 ≥ · · ·
3. lim un = 0.
n→∞
Then the series converges.
13.2
Absolute and Conditional Convergence
Definition 13.2. A series
∞
X
an converges absolutely if
n
∞
X
|an | converges. A series that con-
n
verges but does not converge absolutely converges conditionally.
Theorem 13.3. If a series converges absolutely then it converges.
Homework.
1. Do the following problems in Section 8.6 – these problems will be discussed on Friday and
will not be collected.
pages 542 ff. 1,5,9,13,15,17,25,27
Mathematics 162 Class Notes
14
23
Power Series (September 29)
Goal: To define power series and introduce the notion of radius of convergence.
14.1
Power Series
A power series (about 0) is a series of the form
2
3
c0 + c1 x + c2 x + c3 x + · · · =
∞
X
cn xn
n=0
• The series converges for x = 0.
• It might not converge for any other x.
• The series defines a function f (x) =
∞
X
cn xn for all x such that the series does converge.
n=0
14.2
The Radius of Convergence
Given the power series
∞
X
cn xn , there is a number R ≥ 0 such that
n=0
• the series converges absolutely for all x such that 0 ≤ |x| < R and
• the series diverges for all x such that |x| > R.
• the series may converge or diverge for x = R and x = −R.
R is called the radius of convergence and the interval on which the series converges is called the
interval of convergence.
We can (almost) always find R by using the ratio test. To find the interval of convergence we
need to check the two endpoints x = R and x = −R separately.
Homework.
1. For the series in the following problems on page 552, determine only the radius of convergence: 1, 6 ,7,9,11, 12 ,19, 22 , 27 .
Mathematics 162 Class Notes
15
24
Power Series II (October 2)
Goal: To further investigate the properties of power series
15.1
Power Series About x = a
Definition 15.1. A power series about x = a is a series of form
∞
X
cn (x − a)n
n=0
A series about x = a has a radius of convergence and interval of convergence just as does a
series about x = 0 except that the interval is centered at x = a.
15.2
Differentiation and Integration of Power Series
For this section, suppose that f (x) is a function defined by the a power series
f (x) =
∞
X
cn xn
(15.7)
n=0
and that the power series has radius of convergence R > 0 (so that the domain of f contains
at least all x such that −R < x < R).
Theorem 15.2. If the function f (x) is defined by the power series in Equation 15.7 and the
series has radius of convergence R > 0. Then the series
∞
X
ncn xn−1
n=1
converges for all x such that −R < x < R and
0
f (x) =
∞
X
ncn xn−1
n=1
for all x such that −R < x < R.
Since the hypotheses of the theorem now apply to f 0 (x), we can continue to differentiate the
series to find derivatives of f of all orders, at least on the interval −R < x < R.
Mathematics 162 Class Notes
25
Theorem 15.3. Suppose f (x) is defined by Equation 15.7 and the radius of convergence of the
series is R > 0, we have that for all n,
cn =
f (n) (0)
n!
Theorem 15.4. Suppose that f (x) is defined by Equation 15.7 and the radius of convergence
of the series is R > 0. Then
Z x
∞
X
cn n+1
f (t) dt =
x
n+1
0
n=0
for all x such that −R < x < R.
Homework.
1. New Homework Policy: You do not have to turn in the problems that are not in
boxes. You will continue to turn in the “four problems” that are in boxes and you should
still pay attention to the way that you write these up. For the other problems, you should
still do them all and be able to participate in the discussion about them.
• To turn in: pages 552 ff
8 , 28 , 39 , 46
• Not to turn in: pages 552 ff 13, 15, 17, 21
Mathematics 162 Class Notes
16
26
Taylor Series (October 3)
Goal: To find series representations for important functions
16.1
The Taylor Series for f at x = a
Definition 16.1. If f is a function which has derivatives of all orders at x = a, the Taylor
Series for f at x = a is
∞
X
f (n) (a)
(x − a)n
n!
n=0
Important Notes:
1. The Taylor series has a radius of convergence R but it might be 0
2. Even if the radius of convergence R > 0, the function defined by Taylor series of f might
not equal f except at x = a
3. But, if R > 0, then for many “nice” functions f , the Taylor series for f equals f on its
whole interval of convergence
4. The polynomial that results from stopping the sum at the nth degree term is called the
Taylor polynomial of order n for f at x = a
5. If a = 0 the Taylor series is called the MacLaurin series
6. If f (x) equals any power series at all (on the interval of convergence) that power series
must be the Taylor series
16.2
Our Favorite Taylor Series
ex = 1 + x +
x2 x3
xn
+
+ ··· +
+ ···
2
3!
n!
sin x = x −
x3 x5
x2n+1
+
+ · · · + (−1)n
+ ···
3!
5!
(2n + 1)!
cos x = 1 −
x2 x4
x2n
+
+ · · · + (−1)n
+ ···
2!
4!
(2n)!
ln x = (x − 1) −
(x − 1)2 (x − 1)3
(x − 1)n
+
+ · · · + (−1)n+1
+ ···
2
3
n
Mathematics 162 Class Notes
16.3
27
The Substitution Method
Since
ex = 1 + x +
x2 x3
xn
+
+ ··· +
+ ···
2
3!
n!
for all x
we have
2
ex = 1 + x2 +
x2n
x4 x6
+
+ ··· +
+ ···
2
3!
n!
for all x
2
and so this last series is the MacLaurin series for ex .
Homework.
1. These are due Thursday, October 5, but will not be collected:
pages 558ff
3, 7, 9, 11, 13, 21, 27
Mathematics 162 Class Notes
17
28
Taylor Series II (October 3)
Goal: To determine if a function equals its power series
17.1
The Remainder Term
Suppose that f is a function that has n + 1 derivatives in an interval around x = a. We can
then write down the nth order Taylor polynomial for f about x = a.
Pn,a (x) = f (a) + f 0 (a)(x − a) +
f 00 (a)
f (3) (a)
f (n) (a)
(x − a)2 +
(x − a)3 + · · · +
(x − a)n
2!
3!
n!
Define the Remainder Term as follows:
Rn,a (x) = f (x) − Pn,a (x)
Theorem 17.1 (Lagrange). Given f , Pn,a and Rn,a as above, there is a number t ∈ (a, x) such
that
f (n+1) (t)
(x − a)n+1
Rn,a (x) =
(n + 1)!
Example 17.2. From Lagrange’s Theorem, we can show that sin x is equal to its power series
for all x. Note that for f (x) = sin(x), we have for every x
|Rn,0 (x)| =
|f (n+1) (t)| n+1
|x|
(n + 1)!
However, we know that no matter what t is
|f (n+1) (t)| ≤ 1
Therefore
Rn,0 (x) ≤
1
|x|n+1
(n + 1)!
But now fix x. it is easy to see that
lim Rn,0 (x) = 0
n→∞
This means that
lim Pn,0 (x) = sin x
n→∞
Finally, this means that the power series for sin x converges to sin x for every x.
Mathematics 162 Class Notes
29
Similar arguments work for all our favorite power series.
Example 17.3 (Weird Example). Let f be the function defined by
(
2
e−1/x x 6= 0
f (x) =
0
x=0
We can show that f (n) (0) = 0 for all n. Therefore, the power series for f is the zero power
series! Obviously, this power series doesn’t converge to f .
Homework.
1. There is no homework. This section is entirely optional.
Notes on Test.
1. The test covers sections 9–16 in the course notes.
2. The textbook sections covered are 8.2–8.8.
3. There will be questions that ask for explanations of concepts. For example, you might be
asked to state precisely what is meant by “a series coverges”
4. The tests for convergence and divergence of series that were emphasized include the nth term test, the integral test, the ratio test, and the alternating series (Leibniz’s) test.
Be able to say precisely what conditions are required to use each test and what those
conditions allow you to conclude.
5. Homework problems are an excellent guide to the sort of problems that will appear on
the test. There are additional sample problems that begin on page 573. For example, try
some of 19–68.
Mathematics 162 Class Notes
18
30
Multivariable Calculus (October 9)
Goal: To develop the basic concepts concerning functions of many variables
18.1
Definition and Examples
Definition 18.1. A function is a rule to assigns to each ordered n-tuple of real numbers
(x1 , . . . , xn ) in a certain set D a real number denoted f (x1 , . . . , xn ). D is called the domain of
the function.
We will often choose the names of the variables x1 , . . . , xn and the name of the function f to
reflect their meaning in the application at hand. We often use x, y if n = 2 and, naming the
output w (for example), we write w = f (x1 , . . . , xn ).
Example 18.2. The following are all examples of functions that come from a particular application.
1. v(r, h) = πr2 h
p
2. d(x, y, z) = x2 + y 2 + z 2
3. g(m1 , m2 , R) = Gm1 m2 /R2 (G is a constant)
4. s(t, e, f, s) = 6t + e + 3f + 2s
5. b(w, h) = 703w/h2
6. P (K, L) = AK 1/3 L2/3
7. w(P1 , P2 ) = P12 /(P12 + P22 )
8. A(V, M ) = .015 + .019M + .026V
18.2
Domains and Ranges
The domain of a function of n variables is the set of n-tuples on which the function is defined. If
the domain is not specified, it is assumed to be the largest set on which the function is defined
(the natural domain of the function). The range is the set of output values of the function.
The domain is a subset of n-dimensional space and so can be quite complicated. The range is
a subset of the real line.
Mathematics 162 Class Notes
18.3
31
Graphing Functions of Two Variables
There are two common ways to graph functions of two variables. The first is the analogue of
graphs of functions of single variables. To graph z = f (x, y) we need three coordinate axes,
two for the inputs and one for the output. And the graph of f is then a surface. The graph of
f (x, y) = y 2 − x2 is below.
4
2
0
-2
-4
-2
-2
-1
-1
0
0
1
1
22
The second common way of graphing functions of two variables is by contour plots. A contour
plot is a plot in domain space and so does not require three dimensions. To construct a contour
plot, we draw curves that go through points with the same value. Such curves are called level
curves of the function. Here is a contour plot of f (x, y) = y 2 − x2 .
2
1
-2
-1
1
-1
-2
2
Mathematics 162 Class Notes
32
Homework.
1. Read pages 702–704 and pages 614–615.
2. Do the following problems but not to turn in:
page 708 1–8 parts (a)-(c) and 13–18.
3. To turn in do problems 6 and 8 on page 558 and 22 and 26 on page 710. This
homework is due Thursday since Wednesday will be “special” day.
Mathematics 162 Class Notes
19
33
Limits (October 10)
Goal: To extend the definition of limit to functions of two and three variables
19.1
Some Geometry of the Domain Space
Definition 19.1 (Distance in <3 ). Suppose that (x1 , y1 , z1 ) and (x2 , y2 , z2 ) are points in <3 .
The distance between them is
p
(x1 − x2 )2 + (y1 − y2 )2 + (z1 − z2 )2
Definition 19.2.
1. A circle in <2 centered at (a, b) with radius R is the set of all points
(x, y) satisfying
(x − a)2 + (y − b)2 = R2
2. Given a circle C, the set of all points on or inside the circle is a closed disk. The set of
points inside but not on the circle is an open disk.
3. A sphere in <3 centered at (a, b, c) with radius R is the set of all points (x, y, z) satisfying
(x − a)2 + (y − b)2 + (z − c)2 = R2
4. The inside of a sphere is called a ball and can be closed or open according as to whether
it contains the sphere.
Definition 19.3. If R is a region of the plane and (x0 , y0 ) is a point in R,
1. (x0 , y0 ) is an interior point of R if there is a disk of positive radius centered at (x0 , y0 )
such that every point in the disk lies in R.
2. (x0 , y0 ) is a boundary point of R if every disk with positive radius centered at (x0 , y0 )
contains a point that is in R and a point that isn’t in R.
3. R is open if it consists entirely of interior points.
4. R is closed if it contains all of its boundary points.
5. R is bounded if it lies entirely inside a disk of finite radius.
Mathematics 162 Class Notes
19.2
34
Limits
Working Definition 19.4. The limit of f as (x, y) approaches (x0 , y0 ) is L, written
lim
f (x, y) = L
(x,y)→(x0 ,y0 )
if whenever (x, y) is close enough to (x0 , y0 ) then f (x, y) is close enough to f (x0 , y0 ).
Official Definition 19.5. The limit of f as (x, y) approaches (x0 , y0 ) is L if for every > 0
there is δ > 0 such that for all (x, y) in the domain of f ,
p
if 0 < (x − x0 )2 + (y − y0 )2 < δ then |f (x, y) − L| < The obvious limit rules apply. The obvious problems arise.
4xy 2
(x,y)→(0,0) x2 + y 2
lim
and
lim
4xy
+ y2
(x,y)→(0,0) x2
and
2x2 y
(x,y)→(0,0) x4 + y 2
lim
Homework.
1. Not to turn in: do problems 1,3,5,7,15,17 on page 711.
2. To turn in: pages 717 ff
40 , 64 ,
page 574
52 , 54
Mathematics 162 Class Notes
20
35
Partial Derivatives (September 12)
Goal: To define the partial derivatives of a function of many variables
20.1
Partial Derivatives
The idea is quite simple. Given a function of n-variables, treat all but one variable as constants.
Then we are back in single variable land.
Definition 20.1. The partial derivative of f (x, y) with respect to x at (x0 , y0 ) is
∂f f (x0 + h, y0 ) − f (x0 , y0 )
= lim
∂x (x0 ,y0 ) h→0
h
provided the limit exists.
Various notations:
∂f
(x0 , y0 )
∂x
fx (x0 , y0 )
∂z ∂x (x0 ,y0 )
Of course the partial derivative is a function of x and y in its own right. When we think of it
that way we write
∂f
or fx
∂x
We can also take partial derivatives with respect to y (or other variables if there are more than
2.
20.2
Meaning of Partial Derivative
• Algebraic – Rate of change in a given direction
• Geometric – Slope of line tangent to surface in a given direction
Homework.
1. Not to turn in: do problems 1–32 and 35–39 on pages 728-729.
2. To turn in: do 40 and 57 on page 729 and 22 and 32 on page 573.
Mathematics 162 Class Notes
21
36
Second Partial Derivatives (September 13)
Goal: To compute second partial derivatives and investigate properties of mixed
partials
21.1
Second Partial Derivatives
• “Pure” second partials:
∂2f
or fxx
∂x2
∂2f
2.
or fyy
∂y 2
1.
“Pure” Second partials measure concavity in the x or y direction.
• “Mixed” second partials:
1.
∂2f
or fxy
∂y∂x
2.
∂2f
or fyx
∂x∂y
It is more difficult to say exactly what mixed second partials measure.
Proposition 21.1. If f, fx , fy , fxy , fyx are defined on an disk containing (a, b) and are all
continuous at (a, b) then
fxy (a, b) = fyx (a, b)
We can also find higher order partial derivatives than second partials and we can include more
than two variables.
Example 21.2. Second partial derivatives often arise in models of physical systems. Suppose
that we observe a wave that travels along a line. Then the height of the wave w is a function of
where along the line that we look x and the time that we observe it t. An equation describing
w is
2
∂2w
2∂ w
=
c
∂t2
∂x2
An example of a solution of the wave equation is
w = sin(x + ct)
Mathematics 162 Class Notes
Homework.
1. Not to turn in: pages 729 ff. 47–50, 63–68
2. To turn in: page 729 ff. 61 , 74 , page 656 45 , page 574 36 .
37
Mathematics 162 Class Notes
22
38
Linear Approximations (October 16)
Goal: To find the best linear approximation to a function of several variables
NB: This topic is covered on pages 751–753 and 727–728 of the text with pages
245–247, 216, and AP 32–34 giving important background information.
22.1
What is a linear function?
• “Slope-intercept” form
y = b + mx
z = b + m0 x + m1 y
w = b + m0 x + m1 y + m2 z
• “Point-slope” form
y − y0 = m(x − x0 )
z − z0 = m0 (x − x0 ) + m1 (y − y0 )
w − w0 = m0 (x − x0 ) + m1 (y − y0 ) + m2 (z − z0 )
22.2
The best linear approximation
• y = f (x) and f is differentiable at x0 :
L(x) = f (x0 ) + f 0 (x0 )(x − x0 ) = f (x0 ) + f 0 (x0 )∆x
• z = f (x, y) and f is “differentiable” at (x0 , y0 ):
L(x, y) = f (x0 , y0 )+fx (x0 , y0 )(x−x0 )+fy (x0 , y0 )(y−y0 ) = f (x0 , y0 )+fx (x0 , y0 )∆x+fy (x0 , y0 )∆y
• w = f (x, y, z) and f is “differentiable” at (x0 , y0 , z0 ):
L(x, y, z) = f (x0 , y0 , z0 ) + fx (x0 , y0 , z0 )(x − x0 ) + fy (x0 , y0 , z0 )(y − y0 ) + fz (x0 , y0 , z0 )(z − z0 )
= f (x0 , y0 , z0 ) + fx (x0 , y0 , z0 )∆x + fy (x0 , y0 , z0 )∆y + fz (x0 , y0 , z0 )∆z
Mathematics 162 Class Notes
22.3
39
Using Linear Approximations to Approximate
• If y = f (x) and x0 is fixed, let ∆x = x − x0 and ∆y = f (y) − f (x0 ). Then
∆y ≈ f 0 (x0 )∆x
• If z = f (x, y) and (x0 , y0 ) is a fixed point, let ∆x = x − x0 , ∆y = y − y0 and ∆z =
f (x, y) − f (x0 , y0 ). Then
∆z ≈ fx (x0 , y0 )∆x + fy (x0 , y0 )∆y
• Example: Let f (x, y) = x2 y and consider the point x = 1, y = 1, z = 1. Then
∆z ≈ 2∆x + ∆y
• Example: The volume of a cylinder is given by V (r, h) = πr2 h. Since fr (r, h) = 2πrh
and fh (r, h) = πr2 , a change of height of ∆h inches and a change of radius of ∆r inches
produces a change in volume of approximately 2πrh∆r + πr2 ∆h cubic inches.
22.4
The differential notation
If y = f (x), we write dy = f 0 (x) dx. What does this mean?
• dx is an independent variable (think ∆x).
• dy is a dependent variable that is a function of both x and dx.
• The differential notation is another way of writing the linear approximation. It emphasizes
that we can produce the linearization at any point x.
If z = f (x, y), we write
dz = fx (x, y) dx + fy (x, y) dy
22.5
What is best about it?
Theorem 22.1. Suppose that y = f (x) and f is differentiable at x0 . Then
f (x) = f (x0 ) + f 0 (x0 )∆x + ∆x
for a function such that lim = 0.
∆x→0
Definition 22.2. Suppose that z = f (x, y) and both partials of f exist at (x0 , y0 ). Then we
say that f is differentiable at (x0 , y0 ) if there are functions 1 , 2 such that
f (x, y) = f (x0 , y0 ) + fx (x0 , y0 )∆x + fy (x0 , y0 )∆y + 1 ∆x + 2 ∆y
where lim 1 = 0 and lim 2 = 0 as (∆x, ∆y) goes to (0, 0).
Mathematics 162 Class Notes
22.6
40
When if f (x, y) differentiable?
Theorem 22.3 (Mean Value Theorem). Suppose that f (x) is continuous on [a, b] and differentiable on (a, b). Then there is a number c such that a < c < b and
f 0 (c) =
f (b) − f (a)
b−a
Theorem 22.4. Suppose that f, fx , fy are continuous in a disk around (x0 , y0 ). Then f is
differentiable at (x0 , y0 ).
Proof. The idea is to get from (x0 , y0 ) to (x0 + ∆x, y0 + ∆y) in two steps: first to (x0 + ∆x, y0 )
and then to (x0 + ∆x, y0 + ∆y). First note that
f (x0 +∆x, y0 +∆y)−f (x0 , y0 ) = f (x0 +∆x, y0 +∆y)−f (x0 +∆x, y0 )+f (x0 +∆x, y0 )−f (x0 , y0 )
The second difference on the right hand side is equal to fx (c, y0 )∆x for some c between x
and x + ∆x by the MVT applied to g(x) = f (x, y0 ). The first difference on the right hand
side is equal to fy (x0 + ∆x, d)∆y for some d between y0 and y0 + ∆y by the MVT applied to
h(y) = f (x0 + ∆x, y). Thus
f (x0 + ∆x, y0 + ∆y) − f (x0 , y0 ) = fx (c, y0 )∆x + fy (x0 + ∆x, d)∆x
Since fx is continuous at (x0 , y0 ) and c is between x and x + ∆x, we have that fx (c, y0 ) =
fx (x0 , y0 ) + 1 for 1 such that lim 1 = 0. Similarly, fy (x0 + ∆x, d) = fy (x0 , y0 ) + 2 for 2
∆x→0
such that
lim
(∆x,∆y)→(0,0)
2 = 0. The result follows.
Homework.
1. Read pages 751–753 and p727–728.
2. Not to turn in do problems 37,39,41 on page 754.
3. To turn in do problems 42 , 48 , 51 , 52 on page 754.
Mathematics 162 Class Notes
23
41
The Chain Rule (October 17)
Goal: To develop the various chain rules for functions of many variables
23.1
Review
If y is a function of x and x is a function of t
dy
dy dx
=
dt
dx dt
This is the chain rule for functions of a single variable.
Note here that we have
• An independent variable t
• A dependent variable y
• An intermediate variable x.
23.2
A Plethora of Chain Rules
Setting:
w = f (x, y), x = x(t), and y = y(t).
Chain Rule:
Setting:
w = f (x, y, z), x = x(t), y = y(t) and z = z(t).
Chain Rule:
Setting:
dw
∂w dx ∂w dy
=
+
dt
∂x dt
∂y dt
dw
∂w dx ∂w dy ∂w dz
=
+
+
dt
∂x dt
∂y dt
∂z dt
w = f (x, y), x = g(r, s), y = h(r, s)
Chain Rule:
∂w
∂f ∂x ∂f ∂y
=
+
∂r
∂x ∂r
∂y ∂r
Mathematics 162 Class Notes
23.3
42
A Different Approach to Implicit Differentiation
The equation F (x, y) = 0 defines y implicitly as a function of x. Let w = F (x, y) and differentiate w = 0 with respect to x:
Fx
dx
dy
+ Fy
=0
dx
dx
Thus
Fx
dy
=−
dx
Fy
Homework.
1. Read Section 12.4.
2. Not to turn in: pages 737 ff.:
1,3,6,7,9,11,15,25,26,40,47
3. Though the above problems are not to turn in due to the test Friday, pay particular
attention to 6,26,40,47.
Mathematics 162 Class Notes
24
43
Parametric Equations (October 18)
Goal: To “review” the essentials of parametric equations
24.1
Parametric Equations in the Plane (see pages 170-173)
Suppose that f (t), g(t) are continuous functions on some interval [a, b] Then the set of points
(x(t), y(t)) for a ≤ t ≤ b form a curve C traced out as t varies from a to b. The equations
x = f (t) y = f (t) a ≤ t ≤ b
are called the parametric equations of the curve C and t is called the parameter for the curve.
Application: Suppose that (x(t), y(t)) is the position of a particle at time t. Note that
•
dx
is the rate of change of the x coordinate as a function of time
dt
•
dy
is the rate of change of the y coordinate
dt
• If we think of y as a function of x, then
dy
dy/dt
=
dx
dx/dt
Many wonderful examples are on page 609 which you can investigate with MVT.
Similarly, we can describe a curve in 3-space by
x = f (t) y = g(t) z = h(t)
24.2
The Brachistochrone
The curve given by
x = a(t − sin t) y = a(cos t − 1)
0≤t≤π
is a brachistochrone curve. It is the curve which describes the shape of the “slide” we should
build so that a child reaches the bottom of the slide in the shortest possible time if the top and
bottom points of the slide are fixed.
Mathematics 162 Class Notes
24.3
44
Parametric Surfaces in Two or Three Dimensions
Similarly, we can define a surface parameterized by u and v by
x = f (u, v) y = g(u, v) z = h(u, v)
where u and v each range over some interval. Mathematica allows for the graphing of such
parametric surfaces.
Homework.
1. No homework.
Study Notes for Test III
1. The test covers the notes Sections 18–23 as well as Laboratory 2.
2. The textbook sections covered include
12.1, 12.2, 12.3, 12.4, 12.6 (only pages 751–753), 10.1, and 10.6
The latter two sections were more of background information for understanding the graphs
in 12.1.
3. There were several homework problems in Sections 18–23 on the material in the series
chapter. There will be three very similar problems on the test:
(a) I will ask you to write four nonzero terms of a Taylor series for a given function like
page 558 problems 6, 8.
(b) I will ask you to evaluate a certain series based on your knowledge of an important
Taylor series like problems 52 and 54 on page 574. You must know the Taylor series
1
for the functions
, ex , sin x and cos x. Any other series I would supply.
1−x
(c) I will ask you to do problem 32 on page 573! This should be a gift since I am telling
you the question. However, I will want a complete, correct, even gorgeous answer.
4. I will not ask for 3-dimensional graphs of functions of 2 variables. I might ask for sketches
of contour plots of functions of two variables. You might very well see a matching problem
related to graphing functions of two variables as in page 709 or page 656.
5. You had better know how to differentiate flawlessly!
6. Other than the above, let the kinds of homework problems that you did be your guide.
Mathematics 162 Class Notes
25
45
Optimization (October 23)
Goal: To find extrema of functions of two variables
25.1
Extrema of Functions of One Variable
1. (Extreme Value Theorem) If f is continuous on [c, d] then f has a maximum and minimum
value on [c, d].
2. A critical point of f is a point a in the interior of the domain of f such f 0 (a) = 0 or f 0 (a)
does not exist.
3. A function f has a local maximum (local minimum) at a if f (a) ≥ f (x) (f (a) ≤ f (x) for
all x in an open interval containing a.
4. If f is continuous on [c, d], the maximum on f on [c, d] occurs at a critical point or at an
endpoint of the interval.
25.2
Functions of Several Variables
Definition 25.1. The function f (x, y) has a local maximum at (a, b) if f (a, b) ≥ f (x, y) for all
(x, y) in a disk containing (a, b).
Theorem 25.2 (First Derivative Test). If f (x, y) has a local maximum (or local mimimum)
at an interior point (a, b) of its domain and if the fx (a, b) and fy (a, b) exist, then fx (x, y) =
fy (x, y) = 0.
Definition 25.3. A critical point of a function f (x, y) is a point (a, b) in the interior of the
domain of f such that either fx (a, b) = fy (a, b) = 0 or one or both of fx (a, b) and fy (a, b) do
not exist.
Important: A local maximum occurs at a critical point but not every critical point is a local
maximum or minimum.
Definition 25.4. A saddle point is a critical point at which the function has neither a local
maximum nor a local minimum.
Mathematics 162 Class Notes
25.3
46
The Second Derivative Test
Theorem 25.5. Suppose that f (x, y) and its first and second partial derivatives are continuous
on a disk centered at (a, b) and fx (a, b) = 0 = fy (a, b). Let
2
D(x, y) = fxx fyy − fxy
Then
1. f has a local maximum at (a, b) if D(a, b) > 0 and fxx (a, b) < 0
2. f has a local minimum at (a, b) if D(a, b) > 0 and fxx (a, b) > 0
3. f has a saddle point at (a, b) if D(a, b) < 0
4. if D(a, b) = 0 we can draw no conclusion as to whether f has a local maximum, minimum
or saddle point at (a, b).
Example 25.6. Find the local extreme values of f (x, y) = −x3 + 4xy − 2y 2 + 1.
y
3
2
1
-1
1
2
3
x
-1
Homework.
1. Not to turn in: do problems 1,3,7,11,17 on page 763.
2. To turn in: do problems 2, 18, 29, 44 on page 763.
Mathematics 162 Class Notes
26
47
Optimization Problems (October 26)
Goal: To solve optimization problems in two variables
26.1
Maximums and Minimums
Theorem 26.1 (Extreme Value Theorem). If f (x, y) is a function that is continuous on a
closed and bounded region R, then f has a both a maximum and minimum on R.
Just as in the single variable case, if f is continuous on a closed and bounded region R, its
maximum either occurs at a critical point or a boundary point of R.
26.2
Contstrained Optimization of a Function of Three Variables
Example 26.2. US Mail postage rates for packages depend on the size of the package. The
“large package” rate applies to packages that have length plus girth no more than 108 inches.
(The length of a package is its longest dimension and the girth is the perimeter of a cross section
perpendicular to the length.) What are the dimensions of the package of largest volume that
can be mailed for those rates?
Maximize
subject to
lwh
l + 2w + 2h ≤ 108
The steps to solve such a problem are:
1. Use the constraints to eliminate variables in the function to be optimized
2. Find the local maximums and minimums on the interior of the appropriate domain
3. Determine the value of the function on the boundary of the domain to determine whether
the optimum occurs there
26.3
Maximizing Profit
The cost to manufacture a product is often expressed as fixed cost and variable cost. So if x
units are produced, the cost of producing those units is often something like
C(x) = A + Bx
The revenue received from selling x units is proportional to the price, but the number sold
usually depends on the price. For example, we might have a relationship between price p and
number sold x such as
D
p(x) =
1+x
Mathematics 162 Class Notes
48
This function is called the demand function. With this price function, the revenue is R(x) =
Dx/(1 + x). Finally, the profit function is P (x) = R(x) − C(x). The goal is to choose x to
maximize P (x). This is a calculus I problem.
Example 26.3. A pharmaceutical company produces a product that is sold in two different
markets, foreign and domestic. Suppose that its price functions in the two markets are given
by
p1 = 25 − .2x
p2 = 10 − .05y
where x and y are the quantities sold in each market. If the cost function for producing these
units is
C(x, y) = 15 + .2(x + y)
find the values of x and y that maximize the profit.
Homework.
1. To turn in, do the following four problems:
(a) Page 764, 47.
(b) Find three positive numbers such that the sum is 32 and the quantity xy 2 z is maximum.
(c) A bicycle company makes two different models of bicycles. Suppose that x is the
quantity of Madone bikes that this company sells in a year and that y is the quantity
of Pilot bicycles that they sell. The company estimates that the demand curve for
the Madona bike satisfies
p1 = 3000 − .1x − .1y
where p1 is the price of the Madona bike and the demand curve for the the Pilot
bicycle is
p2 = 4000 − .1x − .2y
where p2 is the price of the Pilot bicycle. Find the quantity of Madone bicycles that
they will sell if the prices are chosen to maximize the revenue.
(d) A water line is to be built from point P to point S and must pass through regions
where construction costs differ. Find x and y so that the total cost will be minimum
if the cost per mile in dollars is 3k from P to Q and 2k from Q to R and k from R
to S. Distances below are in miles.
2
Ps
Q
6Q
Q
Q Q
? x
QP
sP
PP
1 6
y
?
Ps
R
10
s
-S
Mathematics 162 Class Notes
27
49
Checking the Boundaries (October 27)
Goal: To find the extreme values of a function on a closed, bounded region
To find the absolute maximum (minimum) of a function f (x, y) defined on a closed and bounded
region R
1. Find the critical points of f (the points (a, b) on the interior of the region R such that
f 0 (a, b) = 0 or f 0 (a, b) does not exist
2. Find the points on the boundary curves of R where f has a maximum when restricted to
the boundary
3. For all these points, evaluate f and choose the point (a, b) where f (a, b) is greatest
We concentrate on step 2. The strategy is (usually) to examine each “piece” of the boundary
and think of f as a function of a single variable on that piece.
Example 27.1. Find the extreme values of 3x2 − x + 2y 2 − 2y + 1 on the triangular region
defined by
x≥0 y ≥0 x+y ≤1
Homework.
1. Not to turn in: do problems 31, 33 on page 763.
2. To turn in: do problems 32 , 35 , 38 on page 763.
Mathematics 162 Class Notes
28
50
Vectors October 27 continued
Goal: To define vectors and develop the elementary properties thereof
28.1
Vectors in 2 and 3 dimensions
Definition 28.1. A vector is a directed line segment. A directed line segment from the initial
−−→
point A to the terminal point B is denoted AB. A vector has a length and a direction.
• Names: bold-faced letters v, letters with arrows ~v
• Component form: put initial point of the vector v at the origin. Then the terminal point
is at a point (v1 , v2 , v3 ) in space. We write v = hv1 , v2 , v3 i. This is the component form of
v. The numbers v1 , v2 , v3 are the components of v.
p
• If v = hv1 , v2 , v3 i then the length of v is v12 + v22 + v32 . Length is sometimes called
magnitude and it is written |v|.
28.2
Vector Operations
Vector Addition: If v = hv1 , v2 , v3 i and w = hw1 , w2 , w3 i then
v + w = hv1 + w1 , v2 + w2 , v3 + w3 i
Scalar Multiplication: If v = hv1 , v2 , v3 i and k is a real number (a scalar), then
kv = hkv1 , kv2 , kv3 i
Properties of Vector Operations:
Unit Vectors: A unit vector is a vector with length 1.
The component unit vectors are i = h1, 0, 0i j = h0, 1, 0i, and k = h0, 0, 1i
If v = hv1 , v2 , v3 i, then u = v1 i + v2 j + v3 k.
Homework.
1. Not to turn in on page 626 ff: 1–45, every fourth problem (1,5,etc.).
2. To turn in: page 627 46 .
Mathematics 162 Class Notes
29
51
The Dot Product (October 30)
Goal: To define the dot product and investigate its properties
29.1
The Dot Product
• If u = hu1 , u2 , u3 i and v = hu1 , u2 , u3 i are vectors then
u · v = u1 v1 + u2 v2 + u3 v3
• Note that u · v is a number not a vector.
• Properties
1. u · v = v · u
2. cu · v = cu · v = u · cv
3. u · (v + w) = u · v + u · w
4. 0 · v = 0
5. v · v = |v|
Theorem 29.1. If u and v are nonzero vectors and the angle between them is θ then
cos θ =
u·v
|u||v|
Definition 29.2. Two vectors u and v are orthogonal or perpendicular if u · v = 0.
29.2
Projections
Definition 29.3. Given vectors u and v the projection of u onto v is given by
u·v
projv u =
v
|v|2
The component of u in the direction of v is
|u| cos θ =
u·v
|v|
Note that u can be written as the sum of two orthogonal vectors, projv u and u − projv u.
Application:
Work equals Force times Distance.
Mathematics 162 Class Notes
52
Definition 29.4. If F is a force vector and D is a displacement vector, the work done by the
force F in displacing an object along D is
W =F·D
Homework.
1. On pages 634 ff (not to turn in): 1, 5, 11, 15, 41
2. On pages 634 ff to turn in: 6 , 17 , 20 , 28
Mathematics 162 Class Notes
30
53
The Gradient (October 31)
Goal: To define the gradient and use it to gain information about a function of two
variables
30.1
The gradient
Definition 30.1. If f (x, y) is a function of two variables, then the gradient of f at (x0 , y0 ) is
the vector
∂f −
→ ∂f −
→
∇f =
i +
j
∂x
∂y
There is a natural extension of this to three variables.
30.2
The directional derivative
Definition 30.2. Let u = hu1 , u2 i be a unit vector. The directional derivative of f at P0 =
→
(x0 , y0 ) in the direction of −
u is
df
f (x0 + su1 , y0 + su2 ) − f (x0 , y0 )
= lim
ds u,P0 s→0
s
if the limit exists.
Theorem 30.3. If f is differentiable on a disk containing P0 = (x0 , y0 ) then
df
= ∇f (x0 , y0 ) · u
ds u,P0
Therefore at any given point
1. f increases most rapidly in the direction of ∇f
2. f decreases most rapidly in the direction of −∇f
3. the rate of change of f is zero in any direction orthogonal to ∇f if ∇f 6= 0.
Homework.
1. Not to turn in do problems 9,13,15 on page 746.
2. To turn in do problems 10 , 14 , 28 , 30 on page 746.
Mathematics 162 Class Notes
31
54
The Gradient Continued (November 2)
Goal: To investigate further the properties of the gradient
31.1
The gradient and rates of change
Since
df
ds
= ∇f (x0 , y0 ) · u
u,P0
we have that if θ is the angle between ∇f and u,
df
= |∇f (x0 , y0 )| cos θ
ds u,P0
So as we let θ vary
1. f increases most rapidly in the direction of ∇f
2. f decreases most rapidly in the direction of −∇f
3. the rate of change of f is zero in any direction orthogonal to ∇f (if ∇f 6= 0)
4. so ∇f at the point (x0 , y0 ) is orthogonal to the level curve passing through the point
(x0 , y0 ).
31.2
Differentiation Rules for Gradients
1. ∇(kf ) = k∇f (k a constant)
2. ∇(f + g) = ∇f + ∇g
3. ∇(f g) = f ∇g + g∇f
g∇f − f ∇g
f
4. ∇( ) =
g
g2
Homework.
1. to turn in do problems 19 , 21 , 31 and 32 on page 747.
Mathematics 162 Class Notes
55
20.
15.
2
12.
11.
10.
1
-2
-1
1
-1
11.
10.
8.
-2
5.
2. 1.
0
-3
2
3
Mathematics 162 Class Notes
32
56
Lagrange Multipliers (November 3)
Goal: To find the maximum and minimum values of a function f subject to a
constraint
32.1
The Method of Lagrange Multipliers
Problem:
One method:
Maximize f (x, y, z) subject to g(x, y, z) = 0
Solve the constraint equation for one variable and substitute back into f
The Method of Lagrange Multipliers
The maximum and minimum of f (x, y, z) subject to the constraint g(x, y, z) = 0 occurs at a
point (x0 , y0 , z0 ) such that there is a λ 6= 0 with
∇f (x0 , y0 , z0 ) = λ∇g(x0 , y0 , z0 )
In other words, at a point such that the gradients of f and g are parallel.
Example 32.1. A company’s production (in some convenient units) is given by
P (x, y) = x1/3 y 2/3
where x and y are units of capital and labor respectively. Suppose that the company’s resources
are limited so that x + 5y/4 = 1000. Find the values of x and y that maximize P .
Solution: find x, y, λ such that x + 5y/4 = 1000 and
1 −2/3 2/3
x
y
= 1λ
3
2 1/3 −1/3
x y
= (5/4)λ
3
Example 32.2. Find the point on the ellipsoid
y2 z2
+
=1
4
9
that maximizes the product f (x, y, z) = xyz.
x2 +
y2 z2
+
= 1 and
4
9
yz = λ2x
Solution: Find x, y, z, λ such that x2 +
xz = λy/2
xy = λ2z/9
Mathematics 162 Class Notes
57
Homework.
1. To turn in do problems 5, 9, 19, 29 on page 773.
1000
800
600.
600
500.
400.
400
300.
200
200.
200
400
600
800
1000
Mathematics 162 Class Notes
33
58
Cross Product (November 6)
Goal: To define the cross product of two vectors and to use it to solve problems
33.1
The Cross Product
u × v = (|u||v| sin θ)n
where
1. u and v are nonzero vectors
2. n is a normal vector perpendicular to both u and v and in the direction determined by
the “right-hand rule.”
3. θ is the angle between u and v.
4. If u = 0 or v = 0 then we define u × v = 0
5. If u and v are parallel, we define u × v = 0.
The cross product can be computed using the following symbolic determinant
i
j k u × v = u1 u2 u3 v1 v2 v3 33.2
Properties
1. i × j = k
j×k=i
k×i=j
2. u × v = −v × u
3. (ru) × (sv) = (rs)(u × v)
4. The cross product is not associative!
33.3
Applications
1. |u × v| is the area of a parallelogram determined by u and v
2. |(u × v) · w| is the volume of the parallelopided determined by u, v and w.
Mathematics 162 Class Notes
Homework.
1. Not to turn in do problems 1,5,15,21 on page 641
2. To turn in do problems 7 , 23 , 27 , and 43 on pages 641–642.
59
Mathematics 162 Class Notes
34
60
Lines and Planes (November 7)
Goal: To solve problems concerning lines and planes using vectors
34.1
Vector Equations of a Line
To specify a line we need
1. a point P0 = (x0 , yo , z0 ) and
2. a direction, say specified by a vector v
Vector Equation of the Line
r(t) = r0 + tv
−∞<t<∞
Parametric Equations for a Line
x = x0 + tv1
y = y0 + tv2
z = z0 + tv3
Application 1: Distance from a Point to a Line
S a point, P a point on a line, v a direction vector of the line
−→
|P S × v|
d=
|v|
34.2
Vector Equation of a Plane
To specify a plane we need
1. a point P0 = (x0 , y0 , z0 ) on the plane and
2. a vector n = Ai + Bj + Ck that is normal to the plane
Equation for the Plane
A(x − x0 ) + B(y − y0 ) + C(z − z0 ) = 0
We could generate this equation from other data. For instance three points determine a plane.
Mathematics 162 Class Notes
61
Application 2: Distrance from a Point to a Plane
If P is a point on the plane that has normal n, then the distance to that plane from the point
S is
−→ v P S ·
|v| Homework.
1. not to turn in, on page 650 do 3,21,35,47,61,67
2. to turn in, on page 650 do 24 , 26 , 36 , 42
Mathematics 162 Class Notes
35
62
Lines and Planes (November 8)
Goal: To solve some problems about lines and planes
1. Find the distance between the point (1, 2, 3) and the plane
2(x − 1) + y + 3(z − 2) = 0
2. Find the equation of a line that contains the point (1, 2, 3) but does not intersect the
plane
x + 2y + 3z = 6
Mathematics 162 Class Notes
63
3. Find an equation for the line of intersection of the planes
2x − y − z = 4
3x + 2y + 4z = 12
4. Find a plane through the origin that meets the plane
2x + 3y + z = 12
in a right angle.
Homework.
1. There is no further homework assignment. Recall that there are assignments due both
Monday and Tuesday as well as a test due Friday.
Mathematics 162 Class Notes
36
64
The Tangent Plane (November 13)
Goal: To view the linear approximation to a function of two variables geometrically
– namely as the tangent plane.
36.1
Tangent Plane to the Surface f (x, y, z) = c
1. Fix a point (x0 , y0 , z0 ) on the surface of f (x, y, z) = c.
2. The gradient, ∇f is normal to the surface at (x0 , y0 , zo ).
3. The plane containing (x0 , y0 , z0 ) with normal ∇f is the tangent plane to the surface at
(x0 , y0 , z0 ).
4. Equation of tangent plane is
fx (x0 , y0 , z0 )(x − x0 ) + fy (x0 , y0 , z0 )(y − y0 ) + fz (x0 , y0 , z0 )(z − z0 ) = 0
5. Example:
x2 + y 2 − z 2 = 1
36.2
Special Case – Surface determined by z = f (x, y)
1. The graph has equation f (x, y) − z = 0
2. So gradient of g(x, y, z) = 0 is
fx (x0 , y0 )i + fy (x0 , y0 )j − k
3. Tangent Plane is
fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 ) − (z − f (x0 , y0 )) = 0
or
z − f (x0 , y0 ) = fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 )
4. This is exactly the equation of the best linear approximation to f at the point (x0 , y0 ).
Homework.
1. Not to turn in do problems 1a, 3a, 9a, 19 on page 754.
2. To turn in do problems 8a, 10, 20, 22 on page 754.
Mathematics 162 Class Notes
37
65
Double Integrals
Goal: To integrate functions of two variables over rectangles
37.1
Definition
ZZ
1. We define
f (x, y) dA where
R
(a) f (x, y) is a function of two variables defined on the
(b) rectangle R defined by a ≤ x ≤ b, c ≤ y ≤ d
(c) and dA represents that the measure on the domain is area
ZZ
2. To compute
f (x, y) dA define Riemann sums as follows.
R
(a) Divide the rectangle into n subrectangles.
(b) Pick a point (xk , yk ) in the k th subrectangle.
(c) For each subrectangle compute f (xk , yk )∆Ak where ∆Ak is the area of the subrectangle
(d) The Riemann sum determined by this choice of subrectangles and points is
n
X
f (xk , yk ) ∆Ak
k=1
3.
Definition 37.1. The double integral of f (x, y) over R is defined by
ZZ
f (x, y)dA = lim
n→∞
R
n
X
f (xk , yk )∆Ak
k=1
where the limit is taken over all possible ways of partitioning R into subrectangles such
that lim ∆Ak = 0 and of choosing the points (xk , yk ) in each subrectangle.
n→∞
4. The double integral computes (among other things) the volume underneath the surface
determined by z = f (x, y) and above the rectangle R.
Mathematics 162 Class Notes
37.2
66
Iterated Integrals
For nice functions f (x, y), Fubini’s Theorem says that we can compute double integrals by
iterated integrals.
Z bZ
ZZ
d
dZ b
f (x, y) dx dy
f (x, y) dy dx =
f (x, y) dA =
a
R
Z
c
c
a
“Proof:”
Rb
For every y such that c ≤ y ≤ d we have A(y) = a f (x, y) dx is the cross-sectional area
underneath the surface z = f (x, y) between x = a and x = b for that fixed y. And then the
volume under the surface is the integral of cross-sectional area. That is
ZZ
Z
Z
dZ b
f (x, y) dx dy
A(y) dy =
f (x, y) dA =
R
d
c
c
a
Homework.
1. Not to turn in do problems 1, 5, 17 on pages 789-790.
2. To turn in do problems 4, 18, 20, 28 on pages 789-790.
Mathematics 162 Class Notes
38
67
Double Integration - Interesting Regions (November 16)
Goal: To define the double integral of a function of two variables over more general
regions.
38.1
RR
f (x, y) dA for General Regions
R
1. IfRRR is a closed region in the plane (not necessarily a rectangle) we can still define
f (x, y) dA by Riemann sums. An added twist is that the rectangles do not exactly fill
R
all of R.
2. We have a Fubini theorem for certain kinds of regions R.
Case I. R has left and right boundaries x = a, x = b, and bottom and top boundaries
y = g1 (x), y = g2 (x). Then
Z bZ
ZZ
g2 (x)
f (x, y) dA =
f (x, y) dy dx
a
R
g1 (x)
Example 38.1. Suppose that f (x, y) = x2 + y 2 and the region R is the region between
the two curves y = 3 − x2 and y = x + 1.
y
3
2.5
2
1.5
0.2
Then a = 0 and b = 1. So
ZZ
2
2
0.4
Z
0.6
1 Z 2−x2
x + y dA =
R
0
x+1
0.8
1
x
x2 + y 2 dy dx
Mathematics 162 Class Notes
68
Case II. R has bottom and top boundaries y = c, y = d, and left and right boundaries
x = h1 (y), x = h2 (y). Then
ZZ
d Z h2 (y)
Z
f (x, y) dA =
f (x, y) dx dy
c
R
h1 (y)
Example 38.2. The region below has bottom and top y = −1 and y = 1. The left boundary
is x = 0 and the right boundary is x = 1 − y 2 .
1
0.5
-1
-0.5
0.5
1
-0.5
-1
Then
ZZ
Z
1
Z
f (x, y) dA =
R
1−y 2
f (x, y) dx dy
−1
0
Homework.
1. Do problems 15, 17, 21, 23 on page 797. Note that you do not have to integrate, just
change the limits of integration!
2. To turn in do problems 2, 26, 40, 42 on page 797.
Mathematics 162 Class Notes
39
69
Applications of Double Integration (November 17)
Goal: To use double integrals to solve problems
39.1
Area
If R is a closed region of the plane, it is obvious that
RR
1 dA is equal to the area of R. (Think
R
of the Riemann sums for this integral.)
39.2
Average Value of a Function
The average value of a function f on a region R is defined by
ZZ
1
f (x, y) dA
area of R
R
39.3
Function as Density
Suppose that f (x, y) gives the density (of something) of a region R at each point (x, y) in the
region. Then the total of that something in the region is
ZZ
f (x, y) dA
R
Homework.
1. To turn in do problems 6, 16, 18 and 19 on pages 801-2.
2. Not to turn in do problems 1, 3 on page 801.
Mathematics 162 Class Notes
40
70
Integration in Polar Coordinates (November 27)
Goal: To change to polar coordinates to compute some double integrals
40.1
Preliminaries
1. Pep talk.
2. Christian perspectives on mathematics?
40.2
Recall u-Substitution
If x = g(u) then
Z
b
Z
g −1 (b)
f (x) dx =
f (g(u))g 0 (u) du
g −1 (a)
a
1. Change the function
2. Change the limits of integration
3. Stick in the du.
40.3
Polar Integrals
Motivation. There are two sources of difficulty in evaluating double integrals:
1. The region might result in “messy” limits of integration,
2. The function might be complicated.
Changing to polar coordinates sometimes results in simpler limits and/or a simpler function.
ZZ
2
2
Example 40.1. Evaluate
ex +y dA where R is the circle of radius 1 centered at the origin.
R
2
2
2
In this case, the function f (x, y) = ex +y is er in polar coordinates. And the region R is
defined by the inequalities 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π. So we might expect to evaluate our
integral by a double integral of the form
Z 1 Z 2π
2
er ? dθ dr
0
0
where ? plays the same role that du did in u substitution. Indeed ? = r.
Mathematics 162 Class Notes
71
Theorem 40.2. Suppose that R is a region in the plane defined by the polar inequalities
α≤θ≤β
Then
Z
ZZ
g1 (θ) ≤ r ≤ g2 (θ)
β
Z
g2 (θ)
f (r cos(θ), r sin(θ))r dr dθ
f (x, y) dx dy =
α
R
g1 (θ)
Thus if R is the circle of radius 1,
Z
ZZ
x2 +y 2
e
dx dy =
2π
Z
0
R
Remember to evaluate an integral
RR
1
2
er r dr dθ = π(e − 1)
0
f (x, y) dA in polar coordinates
R
1. Change the limits
2. Change the function
3. Stick in an extra r.
Example 40.3. Find the volume of the part of the paraboloid z = 9 − x2 + y 2 that lies above
the disk x2 + y 2 ≤ 4.
We need to evaluate
ZZ
9 − x2 − y 2 dA
R
where R is the disk. Since R is given by the inequalities 0 ≤ θ ≤ 2π and 0 ≤ r ≤ 2 we have
ZZ
Z 2π Z 2
9 − x2 − y 2 dA =
(9 − r2 )r dr dθ
0
R
0
Example 40.4. Find the area within the curve given by r2 = sin 2θ. The region has two pieces.
0.75
0.5
0.25
-0.75-0.5-0.25
-0.25
-0.5
-0.75
0.25 0.5 0.75
Mathematics 162 Class Notes
72
We will evaluate the piece in the first quadrant: this is defined by
√
0 ≤ θ ≤ π/2
0 ≤ r ≤ sin 2θ
We have
Z
ZZ
π/2 Z
1 dA =
R
√
sin 2θ
r dr dθ
0
0
Homework.
1. To turn in (on Wednesday) do problems 2, 8, 18, 24 on page 806.
Test 5 - Thursday
1. Covers the following sections in the text (with approximate percentages of points on test)
(a) (30%) Vectors: 10.3, 10.4, 10.5 (notes sections 29,33,34,35)
(b) (20%) Polar Coordinates: 9.1, 9.2 (no notes - pinch hitter)
(c) (50%) Double Integrals: 13.1, 13.2, 13.3 (notes sections 37, 38, 39)
2. You will need a bluebook.
3. For this test, calculators are not allowed.
Mathematics 162 Class Notes
41
73
Triple Integrals
Goal: To integrate functions of three variables over bounded three-dimensional regions
41.1
Definition
ZZZ
1. We define
F (x, y, z) dV where
D
(a) F (x, y, z) is a function of three variables defined on the
(b) closed, bounded region D in three-dimensional space
(c) and dV represents that the measure on the domain is volume
ZZZ
2. To compute
F (x, y, z) dA define Riemann sums as follows.
D
(a) Divide the region into n “boxes”.
(b) Pick a point (xk , yk , zk ) in the k th box.
(c) For each box compute F (xk , yk , zk )∆Vk where ∆Vk is the volume of the box
(d) The Riemann sum determined by this choice of boxes and points is
n
X
F (xk , yk , zk ) ∆Vk
k=1
3.
Definition 41.1. The triple integral of f (x, y, z) over D is defined by
ZZZ
F (x, y, z)dA = lim
n→∞
D
n
X
F (xk , yk , zk )∆Vk
k=1
where the limit is taken over all possible ways of coveing D with disjoint boxes such that
lim ∆Vk = 0 and of choosing the points (xk , yk , xk ) in each box.
n→∞
4. The triple integral computes (among other things) the mass of the solid D if the density
at each point is F (x, y, z).
Mathematics 162 Class Notes
41.2
74
Iterated Integrals
For nice functions F (x, y, z), Fubini’s Theorem says that we can compute triple integrals by
iterated integrals just as in the two-dimensional case. We need to write the boundaries of the
region as follows. We need to write the “top” and “bottom” of D as two surfaces z = f2 (x, y)
and z = f1 (x, y) where the possible values for x and y are in some region R of the plane. We
then need to write the region R as we did for double integrals. Namely, the y values in R range
from g1 (x) to g2 (x) for a ≤ x ≤ b.
Z bZ
ZZ
g2 (x) Z f2 (x,y)
f (x, y, z) dV =
D
f (x, y, z) dz dy dx
a
g1 (x)
f1 (x,y)
Homework.
1. This homework is due on Monday, December 4, and will be discussed on Friday, December
1.
2. Do problems 4, 9, 22, 42 on pages 815 ff.
Mathematics 162 Class Notes
42
75
The Center of Mass
Goal: To develop an important physical application of triple integrals
42.1
Mass
If δ(x, y, z) measures mass per unit volume of a solid region D at each point (x, y, z), the mass
of D is
ZZZ
δ(x, y, z) dV
M=
D
42.2
Center of Mass
• The center of mass of a solid object is a theoretical concept. In many physics applications,
we can treat the solid object as a point mass with all the mass concentrated at the center
of mass. We denote the center of mass by (x̄, ȳ, z̄).
• We first define the first moments about the coordinate planes
ZZZ
ZZZ
ZZZ
Myz =
xδ(x, y, z) dV
Mxz =
yδ(x, y, z) dV
Mxy =
zδ(x, y, z) dV
D
D
D
• Intuitively, the moments measure the total distance to the coordinate planes weighted by
mass.
• The coordinates of the center of mass are then
x̄ =
Myz
M
ȳ =
Mxz
M
z̄ =
Mxy
M
• If D has constant density, the center of mass is called the centroid of D.
Homework.
1. Not to turn in, do problems 24, 25, 30, 33 on pages 823–824.
2. As one take-home problem for the final exam, do problem 34 on page 824. This problem
is due at the final exam on December 15 at 9 AM. No collaboration is allowed for this
problem.
Mathematics 162 Class Notes
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76
Cylindrical and Spherical Coordinates
Goal: To compute triple integrals by changing coordinate systems
43.1
Cylindrical Coordinates
Definition 43.1. The cylindrical coordinates of a point (x, y, z) in space are (r, θ, z) where
(r, θ) are the polar coordinates of (x, y).
Equations relating cylindrical and Cartesian coordinates
x = r cos θ
y = r sin θ
r2 = x2 + y 2
z=z
tan θ = y/x
To compute a triple integral over a region D in space, we have
ZZZ
ZZZ
f (x, y, z) dV =
D
43.2
f (r cos θ, r sin θ, z)r dz dr dθ
D
Spherical Coordinates
Definition 43.2. The spherical coordinates of a point P = (x, y, z) in space are (ρ, φ, θ) where
1. ρ is the distance of the point to the origin
−−→
2. φ is the angle that OP makes with the positive z-axis (0 ≤ φ ≤ π)
3. θ is the angle determined by (x, y) in polar coordinates
Equations relating spherical, Cartesian, and cylindrical coordinates
r = ρ sin φ
x = r cos θ = ρ sin φ cos θ
z = ρ cos φ
y = r sin θ = ρ sin φ sin θ
p
p
ρ = x2 + y 2 + z 2 = r2 + z 2
To compute a triple integral over a region D in space, we have
Mathematics 162 Class Notes
77
ZZZ
ZZZ
f (x, y, z) dV =
D
f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ)ρ2 sin φ dρ dφ dθ
D
Homework.
1. Not to turn in do problems 15, 17, 19, 31, 49, 51, 71 on pages 834ff.
2. As a take-home problem for the final exam, do problem 80 on page 837. This is due at
the final exam and no collaboration is allowed.