<
Chapter
5
V
ecto
rs
in
Calculus
This
chapter
intro
duces
vecto
rs
and
their
applications
to
calculus.
W
e
will
use
them
to
compute
direc-
tional
derivatives,
to
differentiate
comp
ositions
of
functions,
and
to
find
minimum
and
maximum
values
of
a
function.
Contents
5.1
V
ecto
rs
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306
5.2
The
Dot
Product
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321
5.3
No
rmal
Equations
of
Planes
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330
5.4
The
Gradient
V
ecto
r
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342
5.5
The
Chain
Rule
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359
5.6
Maximum
and
Minimum
Values
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375
5.7
Lagrange
Multipliers
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393
Section
5.1
V
ecto
rs
Goals:
1
Distinguish
vecto
rs
from
scalars
(real
numb
ers)
and
p
oints.
2
Add
and
subtract
vectors,
multiply
by
scala
rs.
3
Exp
ress
real
w
o
rld
vecto
rs
in
terms
of
their
comp
onents.
Calculus
is
the
study
of
change.
W
e
defined
the
partial
derivative
to
be
instantaneous
rate
of
change
of
a
multi-variable
function
when
one
va
riable
changed
but
the
other
sta
y
ed
constant.
If
w
e
w
ant
to
describ
e
a
more
complicated
change,
we
will
need
new
notations
and
vo
cabula
ry
to
describ
e
them.
We
will
need
vecto
rs.
Question
5.1.1
What
is
a
Vecto
r?
A
vector
is
a
w
a
y
of
describing
a
change
in
p
osition
in
n
-space.
T
o
k
eep
things
simple,
w
e’ll
sta
rt
with
vecto
rs
in
the
plane.
We
need
tw
o
pieces
of
information
to
identify
a
vector.
Definition
A
vecto
r
in
2
-space
consists
of
a
magnitude
(length)
and
a
direction.
Two
vecto
rs
with
the
same
magnitude
and
the
same
direction
are
equal
.
Example
Here
a
re
four
vectors
in
2
-space
(the
plane)
represented
by
arro
ws.
Tw
o
of
these
vectors
are
equal.
Here
a
re
some
vec
to
rs
3
miles
south
306
The
fo
rce
that
a
magnetic
field
applies
to
a
charged
particle
The
velocity
of
an
airplane
Here
a
re
some
non-vectors
17
The
mass
of
an
automobile
3:15
PM
A
tlanta,
GA
Question
5.1.2
Ho
w
Do
W
e
Denote
Vecto
rs?
When
defining
a
new
type
of
object,
we
need
to
agree
on
a
notation.
This
allows
us
to
communicate
clea
rly
which
vecto
r
w
e
are
referring
to.
One
wa
y
of
denoting
a
vector
is
by
its
endp
oints.
Endp
oint
Notation
The
vecto
r
v
from
p
oint
A
to
point
B
can
be
represented
by
the
notation
−
−
→
AB
.
A
is
the
initial
point
and
B
is
the
terminal
p
oint
.
Ho
w
does
this
notation
interact
with
the
idea
of
equal
vecto
rs?
Theo
rem
−
−
→
AB
=
−
−
→
C
D
if
and
only
if
AB
DC
is
a
parallelogram
(p
erhaps
a
squished
one).
The
plane
has
a
co
ordinate
system.
We
can
take
advantage
of
this
to
produce
a
more
quantitative
notation
fo
r
vecto
rs.
307
Question
5.1.2
Ho
w
Do
W
e
Denote
Vecto
rs?
Co
o
rdinate
Notation
W
e
can
represent
a
vector
in
the
Cartesian
plane
b
y
the
x
and
y
components
of
its
displacement.
If
A
=
(2
,
3)
and
B
=
(5
,
1)
,
then
−
−
→
AB
increases
x
by
5
−
2
=
3
and
y
by
1
−
3
=
−
2
.
W
e
can
represent
−
−
→
AB
=
⟨
3
,
−
2
⟩
Click to Load Applet
Figure:
The
x
and
y
comp
onents
of
a
vector
W
e
can
use
co
ordinate
notation
to
quickly
test
whether
tw
o
vectors
are
equal.
Theo
rem
v
=
u
if
and
only
if
their
co
o
rdinate
rep
resentations
match
in
each
comp
onent.
W
e
can
also
measure
slop
e
using
the
co
ordinate
notation.
Fo
r
the
vector
v
=
⟨
a,
b
⟩
:
b
rep
resents
the
displacement
in
the
y
-direction
(rise).
a
rep
resents
the
displacement
in
the
x
-direction
(run).
The
slope
of
v
is
rise
run
=
b
a
.
V
ecto
rs
a
re
not
points,
but
their
coordinate
notations
lo
ok
a
wfully
similar.
W
e
can
connect
them
mo
re
fo
rmally
.
Every
p
oint
in
a
Cartesian
coordinate
system
has
a
p
osition
vecto
r
,
which
gives
the
displacement
of
that
p
oint
from
the
o
rigin.
The
comp
onents
of
the
vector
are
the
coordinates
of
the
p
oint.
308
Click to Load Applet
Figure:
There
is
only
one
p
oint
equal
to
(
−
5
,
1)
,
but
there
are
many
vectors
equal
to
⟨−
5
,
1
⟩
.
Question
5.1.3
What
Arithmetic
Can
We
P
erfo
rm
with
Vecto
rs?
Unlik
e
lo
cations
(points),
displacements
(vectors)
can
b
e
added
and
multiplied.
This
a
rithmetic
allo
ws
unlo
cks
a
variet
y
of
computations
and
measurements,
sp
ecifically
it
will
allow
us
to
do
calculus.
Since
we
have
multiple
wa
ys
of
representing
vectors,
w
e
will
want
to
understand
ho
w
to
p
erfo
rm
these
op
erations
with
each
of
those
representations.
309
Question
5.1.3
What
Arithmetic
Can
We
P
erfo
rm
with
Vecto
rs?
V
ecto
r
Sums
The
sum
of
tw
o
vectors
v
+
u
is
calculated
by
p
ositioning
v
and
u
head
to
tail.
The
sum
is
the
vector
from
the
initial
p
oint
of
one
to
the
terminal
point
of
the
other.
In
co
ordinate
notation,
we
just
add
each
comp
onent
numerically
.
Click to Load Applet
⟨
1
,
3
⟩
+
⟨
3
,
−
1
⟩
⟨
4
,
2
⟩
Scala
r
Multiples
Given
a
numb
er
(called
a
scala
r)
λ
and
a
vector
v
we
can
produce
the
scala
r
multiple
λ
v
,
which
is
the
vecto
r
in
the
same
direction
as
v
but
λ
times
as
long.
If
λ
is
negative
then
λ
v
extends
in
the
opp
osite
di-
rection.
Either
wa
y
,
we
say
λ
v
is
parallel
to
v
.
Click to Load Applet
In
coordinates
scala
r
multiplication
is
distributed
to
each
comp
onent.
F
o
r
example:
2
.
5
⟨
6
,
4
⟩
=
⟨
15
,
10
⟩
310
Example
5.1.4
P
erfo
rming
V
ecto
r
Arithmetic
Given
diagrams
of
tw
o
vecto
rs
u
and
v
,
how
would
we
calculate
1
2
u
+
v
?
What
if
w
e
are
instead
given
the
comp
onents
u
=
⟨
a,
b
⟩
and
v
=
⟨
c,
d
⟩
?
Solution
After
drawing
a
random
u
and
a
random
v
,
w
e
draw
1
2
u
in
the
same
direction
as
u
but
is
half
as
long.
W
e
place
it
head
to
tail
with
v
,
and
1
2
u
+
v
completes
the
triangle.
In
coordinates
the
computation
is
as
follows.
1
2
u
+
v
=
1
2
⟨
a,
b
⟩
+
⟨
c,
d
⟩
=
1
2
a,
1
2
b
+
⟨
c,
d
⟩
=
1
2
a
+
c,
1
2
b
+
d
311
Question
5.1.5
What
Is
Standa
rd
Basis
Notation?
V
ecto
r
arithmetic
gives
us
another
notation
that
takes
advantage
of
our
algeb
raic
intuition.
W
e
can
rep
resent
any
vector
in
the
plane
as
a
sum
of
scala
r
multiples
of
the
follo
wing
standa
rd
basis
vectors
.
Standa
rd
Basis
V
ecto
rs
The
emphstanda
rd
basis
vectors
in
R
2
a
re
i
=
⟨
1
,
0
⟩
j
=
⟨
0
,
1
⟩
F
o
r
example,
the
vecto
r
⟨
3
,
−
5
⟩
can
b
e
written
as
3
i
−
5
j
.
Y
ou
can
check
y
ourself
that
the
sum
on
the
right
gives
the
correct
vector.
Question
5.1.6
Ho
w
Do
W
e
Measure
the
Length
of
a
Vecto
r?
A
vecto
r
consists
of
tw
o
pieces
of
info
rmation:
magnitude
and
direction.
Ho
w
do
we
measure
these?
Length
is
the
distance
b
et
ween
the
endpoints.
We
already
have
a
metho
d
for
measuring
distance
in
the
plane.
Definition
The
length
or
magnitude
of
a
vecto
r
is
calculated
using
the
distance
fo
rmula
and
notated
|
v
|
.
If
v
=
a
i
+
b
j
,
then
|
v
|
=
p
a
2
+
b
2
312
Example
5.1.7
The
Length
of
a
Vecto
r
If
v
=
⟨
3
,
−
5
⟩
calculate
|
v
|
Solution
|
v
|
=
p
3
2
+
(
−
5)
2
=
√
34
Definition
A
unit
vecto
r
is
a
vector
of
length
1
.
Given
a
vector
v
the
scalar
multiple
1
|
v
|
v
is
a
unit
vector
in
the
same
direction
as
v
.
Question
5.1.8
Ho
w
Do
W
e
Measure
the
Direction
of
a
Vecto
r?
Direction
cannot
b
e
described
as
clea
rly
as
length.
Ho
w
do
we
even
measure
it?
A
partial
answer
is
to
measure
the
difference
in
direction
b
etw
een
tw
o
vectors.
Angles
are
a
go
o
d
wa
y
of
comparing
directions.
In
general,
tw
o
vectors
will
not
intersect
to
form
an
angle,
so
w
e
use
the
following
definition:
Definition
The
angle
b
etw
een
tw
o
vectors
is
the
angle
they
make
when
they
a
re
placed
so
their
initial
p
oints
are
the
same.
If
they
make
a
right
angle,
w
e
call
them
orthogonal
.
If
they
mak
e
an
angle
of
0
or
π
,
they
a
re
pa
rallel.
313
Question
5.1.9
Ho
w
Do
W
e
Denote
Vecto
rs
in
Higher
Dimensions?
Higher
dimensional
vecto
rs
represent
displacements
in
higher
dimensional
spaces.
W
e
can
call
a
vecto
r
in
n
-space
an
n
-vector.
W
e
can
still
denote
and
n
-vector
b
y
its
endp
oints.
We
can
also
denote
it
in
coordinate
notation,
but
we
need
more
comp
onents.
Example
If
A
=
(2
,
4
,
1)
and
B
=
(5
,
−
1
,
3)
then
−
−
→
AB
=
⟨
3
,
−
5
,
2
⟩
.
In
three
space,
we
add
another
standard
basis
vector
k
.
Standa
rd
basis
fo
r
3-vectors
i
=
⟨
1
,
0
,
0
⟩
j
=
⟨
0
,
1
,
0
⟩
k
=
⟨
0
,
0
,
1
⟩
Example
⟨
3
,
−
5
,
2
⟩
=
3
i
−
5
j
+
2
k
Higher
dimensions
still
have
a
standa
rd
basis,
but
at
this
p
oint
the
naming
conventions
a
re
less
standa
rd.
{
e
1
,
e
2
,
e
3
,
.
.
.
,
e
n
}
is
common
fo
r
n
-vectors.
Length
of
a
Vecto
r
The
length
of
an
n
-vector
derives
from
the
distance
formula
in
n
-space.
|⟨
a
1
,
a
2
,
a
3
,
.
.
.
,
a
n
⟩|
=
q
a
2
1
+
a
2
2
+
a
2
3
+
·
·
·
+
a
2
n
W
e
might
b
e
concerned
that
direction
b
ecomes
an
even
more
difficult
concept
to
wo
rk
with
as
the
dimension
increases.
Ho
w
ever,
angles
a
re
a
valid
a
w
a
y
of
compa
ring
directions
any
dimension
(though
they
ma
y
be
more
difficult
to
compute).
314
Angles
Bet
w
een
V
ecto
rs
Any
tw
o
vectors
with
the
same
initial
point
lie
in
a
plane.
Their
angle
is
a
t
w
o-dimensional
measurement.
Ho
w
ever
there
is
no
go
o
d
w
a
y
to
measure
clo
ckwise
in
3
or
more
dimensions.
The
angle
betw
een
t
w
o
vecto
rs
is
never
negative,
nor
more
than
π
.
Click to Load Applet
Figure:
Tw
o
3
-vectors
with
a
common
initial
p
oint,
the
plane
that
contains
them,
and
the
angle
b
et
w
een
them
Section
5.1
Exercises
Summa
ry
Questions
Q1
Ho
w
is
a
vector
simila
r
to
a
p
oint?
T
o
a
numb
er?
Q2
Ho
w
is
a
vector
different
from
a
p
oint?
From
a
numb
er?
Q3
Ho
w
can
y
ou
t
ell
if
tw
o
vecto
rs
point
in
the
same
direction?
Opp
osite
directions?
Q4
If
u
and
v
are
p
osition
vectors
of
the
p
oints
P
and
Q
,
how
are
u
and
v
related
to
−
−
→
P
Q
?
315
Section
5.1
Exercises
5.1.1
Q5
Which
of
the
following
a
re
vectors?
i.
The
reading
on
a
sp
eedometer.
ii.
The
intersection
of
t
w
o
lines.
iii.
Five
miles
to
w
a
rd
A
tlanta.
iv.
The
length
of
a
string.
v.
The
velo
cit
y
of
a
projectile.
Q6
Which
of
the
following
a
re
vectors?
i.
The
displacement
of
a
key
on
a
keyboard,
when
pressed.
ii.
The
sp
eed
of
light.
iii.
The
center
of
the
earth.
iv.
The
fo
rce
applied
b
y
a
ro
cket
engine.
v.
The
mass
of
five
hipp
op
otamuses.
Q7
If
−
−
→
AB
=
−
→
AC
,
what
does
that
tell
us
ab
out
the
p
oints
B
and
C
?
Explain.
Q8
If
−
−
→
AB
=
−
−
→
B
A
,
what
do
es
that
tell
us
ab
out
the
p
oints
A
and
B
?
Explain.
5.1.2
Q9
If
A
=
(8
,
7
,
11)
and
B
=
(2
,
3
,
15)
write
the
vector
−
−
→
AB
a
in
terms
of
its
comp
onents
b
in
standa
rd
basis
notation
Q10
If
P
=
(
−
2
,
3
,
5)
and
Q
=
(
−
2
,
0
,
−
4)
write
the
vector
−
−
→
P
Q
a
in
terms
of
its
comp
onents
b
in
standa
rd
basis
notation
Q11
What
is
the
slop
e
of
the
vector
−
4
i
+
10
j
?
Q12
Give
three
different
vectors
of
slop
e
3
7
.
316
Q13
Supp
ose
t
w
o
different
vectors
have
the
equal
slop
es.
How
are
they
related?
Q14
Given
a
number
m
,
give
tw
o
different
vectors
with
slop
e
m
.
5.1.3
Q15
Let
u
b
e
a
vec
to
r.
Ho
w
are
the
magnitude
and
direction
of
u
and
2
u
related?
Q16
Ho
w
is
the
direction
and
magnitude
of
u
related
to
the
direction
and
magnitude
of
−
u
?
Q17
Given
diagrams
of
tw
o
vecto
rs
u
and
v
,
how
would
we
draw
u
−
v
?
What
it
its
significance?
Q18
If
u
is
a
vecto
r
and
2
u
=
u
,
what
do
es
that
tell
us
ab
out
u
?
Explain.
Q19
If
u
=
−
−
→
AB
,
v
=
−
→
AC
,
and
1
2
u
+
1
2
v
=
−
−
→
AD
,
where
is
D
?
Q20
If
u
=
−
−
→
AB
,
v
=
−
→
AC
,
and
1
5
u
+
4
5
v
=
−
−
→
AD
,
where
is
D
?
5.1.4
Q21
Let
u
=
4
i
+
3
j
and
v
=
5
i
−
2
j
.
Com
pute
u
+
v
.
Q22
Let
w
=
⟨
5
,
−
1
⟩
and
v
=
⟨
12
,
10
⟩
.
Compute
w
−
v
.
Q23
F
o
r
Lindsey
to
get
from
her
house
to
Sam’s
house,
she
travels
5
mi
north
and
3
mi
w
est.
T
o
get
to
Russel’s
house,
she
travels
2
mi
due
south.
What
displacement
w
ould
get
her
from
Sam’s
house
to
Russel’s
house?
Q24
One
can
get
from
Atlanta
to
Decatur
b
y
travelling
8
k
m
east
and
2
k
m
no
rth.
T
o
get
from
Decatur
to
Covington,
one
can
travel
43
k
m
east
and
20
k
m
south.
Describ
e
how
to
get
directly
from
A
tlanta
to
Covington.
Q25
Using
the
diagram
below,
describ
e
each
vecto
r
in
terms
of
u
and
v
using
vector
addition
and
scala
r
multiplication.
Use
the
fact
that
AC
D
B
and
AC
B
E
are
parallelograms.
317
Section
5.1
Exercises
a
−
−
→
E
B
b
−
−
→
C
G
c
−
−
→
B
C
d
−
→
AF
e
−
−
→
GB
Q26
Using
the
diagram
below,
describ
e
each
vecto
r
in
terms
of
u
and
v
using
vector
addition
and
scala
r
multiplication.
Use
the
fact
that
AC
B
D
is
a
parallelogram,
and
the
mark
ed
segments
a
re
congruent.
a
−
−
→
B
D
b
−
→
E
A
c
−
−
→
D
C
d
−
−
→
B
G
e
−
→
AG
f
−
−
→
C
F
5.1.5
Q27
W
rite
⟨
5
,
2
⟩
in
standard
basis
notation.
Q28
F
o
r
any
numbers
a
and
b
,
use
the
definition
of
i
and
j
to
show
that
a
i
+
b
j
=
⟨
a,
b
⟩
.
318
5.1.6
Q29
Compute
the
length
of
u
=
⟨−
5
,
12
⟩
.
Q30
Given
a
nonzero
vector
u
,
many
vecto
rs
of
length
5
are
parallel
to
u
?
Explain.
Q31
Find
a
unit
vector
in
the
direction
of
3
i
−
j
.
Q32
Find
a
unit
vector
in
the
direction
of
⟨
12
,
−
16
⟩
.
5.1.7
Q33
If
u
and
v
a
re
vectors
in
R
2
whose
comp
onents
are
all
positive,
what
is
the
largest
p
ossible
angle
b
et
w
een
u
and
v
?
Q34
Explain
the
difference
b
etw
een
the
terms
“p
erp
endicular”
and
“orthogonal.”
Q35
Supp
ose
tw
o
vecto
rs
do
not
have
the
same
inital
p
oint,
but
when
we
rep
resent
them
b
y
arro
ws,
the
a
rro
ws
happ
en
to
cross.
Is
the
angle
made
in
the
crossing
equal
to
the
angle
betw
een
the
vecto
rs
(as
w
e
defined
it)?
Q36
Describ
e
all
the
vectors
that
make
an
angle
of
π
4
with
v
=
−
j
.
5.1.8
Q37
If
u
=
⟨
2
,
0
,
3
⟩
and
v
=
⟨
5
,
6
,
0
⟩
,
compute
3
u
−
4
v
.
Q38
If
a
=
10
i
−
25
k
and
b
=
8
i
−
4
j
+
10
k
,
compute
3
5
a
+
1
2
b
.
Q39
Compute
the
magnitude
of
v
=
2
i
−
7
j
+
6
k
.
Q40
Compute
t
w
o
unit
v
ecto
rs
pa
rallel
to
v
=
⟨
4
,
−
4
,
2
⟩
.
319
Section
5.1
Exercises
Q41
a
Ho
w
many
different
(nonequal)
unit
vectors
are
orthogonal
to
a
given
vector
in
R
2
?
Ho
w
a
re
they
related
to
each
other?
b
Ho
w
many
different
(nonequal)
unit
vectors
are
orthogonal
to
a
given
vector
in
R
3
?
Ho
w
a
re
they
related
to
each
other?
Q42
Let
u
and
v
be
non-parallel
vecto
rs
in
R
3
.
How
many
unit
vecto
rs
in
R
3
a
re
orthogonal
to
b
oth
u
and
v
?
Synthesis
and
Extension
Q43
Is
the
vecto
r
v
=
2
i
+
3
j
+
8
k
parallel
to
the
plane
p
whose
slop
e-intercept
equation
is
z
=
x
+
2
y
−
7
?
Q44
F
o
r
a
t
w
o-va
riable
function
f
(
x,
y
)
,
f
x
(
x
0
,
y
0
)
is
the
slop
e
of
the
line
tangent
to
z
=
f
(
x,
y
)
at
(
x
0
,
y
0
,
f
(
x
0
,
y
0
))
in
the
x
-direction.
Write
a
vector
v
that
is
parallel
to
this
line.
Q45
If
u
=
−
−
→
AB
and
v
=
−
→
AC
,
show
that
for
any
scala
r
t
,
t
u
+
(1
−
t
)
v
=
AD
where
D
is
a
p
oint
on
the
line
through
B
and
C
.
Q46
If
u
,
v
and
w
are
p
osition
vectors
of
the
three
vertices
A
,
B
and
C
of
a
triangle,
then
1
3
(
u
+
v
+
w
)
is
the
p
osition
vector
of
K
,
the
center
of
mass
of
the
triangle.
V
erify
this
by
sho
wing
that
K
lies
on
the
line
b
etw
een
A
and
the
midp
oint
of
the
side
B
C
.
Q47
Supp
ose
we
become
interested
in
studying
vecto
rs
of
infinite
dimension
(y
es
this
is
something
mathematicians
actually
do).
a
Explain
what
trouble
we
might
run
computing
the
length
of
the
vector
⟨
1
,
1
,
1
,
1
,
1
,
.
.
.
⟩
.
b
What
w
ould
the
length
of
the
vector
⟨
1
,
1
2
,
1
4
,
1
8
,
1
16
,
.
.
.
⟩
b
e?
320
Section
5.2
The
Dot
Pro
duct
Goals:
1
Calculate
the
dot
product
of
tw
o
vectors.
2
Determine
the
geometric
relationship
b
etw
een
tw
o
vecto
rs
based
on
their
dot
product.
3
Calculate
vecto
r
and
scalar
p
rojections
of
one
vector
onto
another.
The
a
rithmetic
of
vecto
rs
app
ears
to
have
ro
om
for
expansion.
While
w
e
can
add
and
subtract
vecto
rs,
w
e
only
defined
ho
w
to
multiply
them
b
y
scalars,
not
b
y
other
vectors.
There
are
in
fact
p
ro
ducts
of
tw
o
vecto
rs.
The
simplest
and
most
useful
is
the
dot
product.
The
dot
product
takes
t
w
o
n
-vecto
rs
and
outputs
a
single
numb
er.
Despite
this
appa
rent
loss
of
info
rmation,
the
dot
product
is
the
key
to
ol
in
computing
the
angle
b
etw
een
vectors,
the
wo
rk
done
b
y
a
force,
or
the
illumination
in
a
digital
scene.
Question
5.2.1
What
Is
the
Dot
Pro
duct?
Definition
The
dot
p
ro
duct
of
t
w
o
vectors
is
a
numb
er.
F
o
r
t
w
o
dimensional
vecto
rs
v
=
⟨
v
1
,
v
2
⟩
and
u
=
⟨
u
1
,
u
2
⟩
w
e
define
v
·
u
=
v
1
u
1
+
v
2
u
2
F
o
r
three
dimensional
ve
cto
rs
v
=
⟨
v
1
,
v
2
,
v
3
⟩
and
u
=
⟨
u
1
,
u
2
,
u
3
⟩
w
e
define
v
·
u
=
v
1
u
1
+
v
2
u
2
+
v
3
u
3
This
pattern
can
b
e
extended
to
any
dimension.
Example
5.2.2
Computing
a
Dot
Pro
duct
a
Calculate
⟨
2
,
3
,
−
1
⟩
·
⟨
4
,
1
,
5
⟩
b
Calculate
(
−
2
i
+
4
k
)
·
(
i
+
2
j
−
k
)
321
Example
5.2.2
Computing
a
Dot
Pro
duct
Solution
a
⟨
2
,
3
,
−
1
⟩
·
⟨
4
,
1
,
5
⟩
=
(2)(4)
+
(3)(1)
+
(
−
1)(5)
=
6
b
(
−
2
i
+
4
k
)
·
(
i
+
2
j
−
k
)
=
(
−
2)(1)
+
(0)(2)
+
(4)(
−
1)
=
−
6
Question
5.2.3
What
Are
the
Algebraic
Properties
of
the
Dot
Pro
duct?
Theo
rem
The
follo
wing
algeb
raic
properties
hold
for
any
vectors
u,
v
and
w
and
scalars
m
and
n
.
Commutative
u
·
v
=
v
·
u
Distributive
u
·
(
v
+
w
)
=
u
·
v
+
u
·
w
Asso
ciative
m
u
·
n
v
=
mn
(
u
·
v
)
Question
5.2.4
What
Is
the
Geometric
Significance
of
the
Dot
Pro
duct?
u
·
v
enco
des
key
information
about
the
magnitude
and
direction
of
u
and
v
.
This
geometric
relationship
can
b
e
derived
from
the
algebraic
properties
w
e’ve
established.
We
begin
with
the
idea
that
u
·
u
=
|
u
|
2
.
This
doesn’t
tell
us
the
value
of
every
dot
product,
but
we
can
extend
the
reasoning
to
any
pair
of
parallel
vecto
rs.
322
Theo
rem
If
u
and
v
are
parallel
then
u
·
v
=
(
|
u
||
v
|
if
u
and
v
have
the
same
direction
−|
u
||
v
|
if
u
and
v
have
opp
osite
directions
Since
u
and
v
are
parallel,
w
e
can
write
v
=
m
u
for
some
scala
r
m
.
v
is
m
times
as
long
as
u
.
Both
lengths
a
re
positive,
so
this
means
if
m
>
0
then
|
v
|
=
m
|
u
|
,
but
if
m
<
0
,
then
|
v
|
=
−
m
|
u
|
u
·
v
=
u
·
(
m
u
)
=
m
u
·
u
=
m
|
u
|
2
=
|
u
|
m
|
u
|
=
(
|
u
||
v
|
if
u
and
v
have
the
same
direction
−|
u
||
v
|
if
u
and
v
have
opp
osite
directions
W
e
can
establish
the
dot
product
in
another
sp
ecial
case:
when
the
vecto
rs
are
o
rthogonal.
Theo
rem
If
u
and
v
are
orthogonal
then
u
·
v
=
0
.
In
this
case,
we
place
u
and
v
head
to
tail
and
draw
u
+
v
.
Since
u
and
v
make
a
right
angle,
these
three
vecto
rs
mak
e
a
right
triangle.
The
Pythagorean
theorem
applies
to
the
lengths
of
the
vecto
rs.
Figure:
Orthogonal
vectors
and
their
sum
making
a
right
triangle
|
u
+
v
|
2
=
|
u
|
2
+
|
v
|
2
(
Pythago
rean
theo
rem
)
(
u
+
v
)
·
(
u
+
v
)
=
u
·
u
+
v
·
v
u
·
u
+
u
·
v
+
v
·
u
+
v
·
v
=
u
·
u
+
v
·
v
(
distributive
p
roperty
)
u
·
v
+
v
·
u
=
0
2
u
·
v
=
0
(
commutative
p
rop
ert
y
)
u
·
v
=
0
323
Question
5.2.4
What
Is
the
Geometric
Significance
of
the
Dot
Pro
duct?
Tw
o
vecto
rs
need
not
b
e
parallel
o
r
o
rthogonal,
but
given
vecto
rs
u
and
v
w
e
can
alw
a
ys
write
v
=
v
proj
+
v
orth
.
W
e
cho
ose
v
proj
to
be
parallel
to
u
and
v
orth
to
be
orthogonal
to
u
.
Click to Load Applet
The
p
roperties
of
the
dot
product
tell
us
that
u
·
v
=
u
·
(
v
proj
+
v
orth
)
=
±
|
u
||
v
proj
|
+
0
Definition
The
numb
er
u
·
v
|
u
|
is
called
the
scala
r
projec-
tion
of
v
onto
u
.
The
scala
r
p
rojection
is
equal
to
the
length
of
v
proj
if
v
proj
is
in
the
same
direction
as
u
.
Otherwise,
it
is
the
negative
of
the
length.
Theo
rem
Let
u
and
v
have
the
same
initial
point
and
meet
at
angle
θ
.
The
following
fo
rmula
holds
in
any
dimension:
u
·
v
=
|
u
||
v
|
cos
θ
Click to Load Applet
Recall
that
cos
θ
is
p
ositive
when
θ
<
π
/
2
negative
when
θ
>
π
/
2
zero
when
θ
=
π
/
2
.
So
the
sign
of
u
·
v
tells
us
whether
θ
is
acute,
obtuse
o
r
right.
Example
5.2.5
Using
the
Cosine
Fo
rmula
What
is
the
angle
b
etw
een
⟨
1
,
0
,
1
⟩
and
⟨
1
,
1
,
0
⟩
?
324
Solution
W
e’ll
apply
the
cosine
formula,
compute
all
of
the
comp
onents
b
esides
θ
and
solve.
⟨
1
,
0
,
1
⟩
·
⟨
1
,
1
,
0
⟩
=
|
⟨
1
,
0
,
1
⟩
||
⟨
1
,
1
,
0
⟩
|
cos
θ
(1)(1)
+
(0)(1)
+
(1)(0)
=
p
1
2
+
0
2
+
1
2
p
1
2
+
1
2
+
0
2
cos
θ
1
=
√
2
√
2
cos
θ
1
2
=
cos
θ
cos
−
1
1
2
=
θ
π
3
=
θ
W
e
can
verify
this
by
noting
that
these
vectors
are
diagonals
in
a
unit
cub
e.
We
could
connect
them
with
a
third
diagonal
to
make
an
equilateral
triangle.
W
e
ma
y
recall
that
an
equilateral
triangle
has
angles
of
π
3
.
Click to Load Applet
Figure:
Tw
o
vectors
in
a
unit
cub
e
Application
5.2.6
W
o
rk
In
physics,
we
say
a
force
w
o
rks
on
an
object
if
it
moves
the
object
in
the
direction
of
the
fo
rce.
Given
a
fo
rce
F
and
a
displacement
s
,
the
fo
rmula
fo
r
w
o
rk
is:
W
=
F
s
325
Application
5.2.6
W
o
rk
In
higher
dimensions,
displacement
and
force
are
vectors.
If
the
force
and
the
displacement
a
re
not
in
the
same
direction,
then
only
F
proj
contributes
to
w
o
rk.
W
=
F
proj
·
s
=
F
·
s
Click to Load Applet
Section
5.2
Exercises
Summa
ry
Questions
Q1
What
algeb
raic
p
roperties
do
es
a
dot
product
share
with
real
numb
er
multiplication?
Q2
What
is
the
significance
of
the
dot
product
of
tw
o
parallel
vectors?
Q3
Ho
w
is
the
angle
b
etw
een
tw
o
vecto
rs
related
to
their
dot
product?
Q4
What
is
a
scalar
p
rojection,
and
how
do
you
compute
it?
326
5.2.1
Q5
What
do
v
·
i
and
v
·
j
measure
ab
out
v
?
Q6
Elaine
computes
u
·
v
and
gets
⟨
15
,
4
⟩
.
How
can
you
tell
that
Elaine
got
the
wrong
answ
er
without
even
kno
wing
what
u
and
v
are?
5.2.2
Q7
Compute
the
follo
wing
dot
products.
a
⟨
4
,
5
⟩
·
⟨−
1
,
−
2
⟩
b
(5
i
+
6
j
)
·
(
i
−
2
j
)
c
⟨
2
,
4
,
−
10
⟩
·
⟨
0
,
−
1
,
−
2
⟩
Q8
Compute
the
follo
wing
dot
products.
a
⟨
4
,
5
⟩
·
⟨−
1
,
−
2
⟩
b
(5
i
+
6
j
)
·
(
i
−
2
j
)
c
(2
i
−
3
k
)
·
(7
j
−
k
)
5.2.3
Q9
Let
u
=
⟨
2
,
3
⟩
,
v
=
⟨
4
,
−
1
⟩
and
w
=
⟨−
5
,
2
⟩
.
a
Compute
u
·
u
and
u
·
v
and
u
·
w
.
b
Compute
v
·
u
.
Ho
w
do
es
it
compare
to
u
·
v
?
327
Section
5.2
Exercises
c
Ho
w
is
u
·
u
related
to
|
u
|
?
d
Compute
3
u
and
3
v
then
take
their
dot
product.
How
is
it
related
to
u
·
v
?
e
Compute
v
+
w
then
compute
u
·
(
v
+
w
)
.
Ho
w
is
it
related
to
u
·
v
and
u
·
w
?
f
Why
do
y
ou
think
we
call
this
op
eration
a
“dot
product”
and
not
a
“dot
sum?”
g
If
you
wanted
to
prove
that
relationships
your
noticed
in
b
-
e
w
o
rk
for
all
possible
vecto
rs,
ho
w
w
ould
y
ou
do
that?
Q10
Expand
the
pa
rentheses
2
u
·
(3
v
−
w
)
.
Q11
Expand
the
pa
rentheses
(
a
−
3
b
)
·
(5
c
+
2
d
)
.
Q12
F
acto
r
a
·
a
+
6
a
·
b
+
9
b
·
b
.
5.2.4
Q13
Supp
ose
w
e
kno
w
that
u
and
v
are
pa
rallel,
that
|
v
|
=
4
and
that
u
·
v
=
−
28
.
a
What
is
the
length
of
u
?
b
What
can
y
ou
say
about
the
directions
of
u
and
v
?
Q14
If
|
u
|
=
12
,
|
v
|
=
9
,
and
u
·
v
=
0
,
what
is
the
magnitude
of
the
vector
w
=
u
+
v
?
Q15
If
|
u
|
=
5
and
u
·
v
=
15
,
what
a
re
the
possible
values
of
|
v
|
?
Q16
If
|
u
|
=
6
and
|
v
|
=
10
what
are
the
greatest
and
least
p
ossible
values
of
u
·
v
?
Q17
Let
v
=
7
i
−
2
j
+
k
,
what
unit
vecto
r
u
produces
the
largest
p
ossible
dot
product
u
·
v
?
Q18
Argue
that
u
·
v
cannot
b
e
any
larger
than
|
u
||
v
|
.
328
5.2.5
Q19
Compute
the
angle
b
etw
een
⟨
6
,
1
,
4
⟩
and
⟨
7
,
0
,
2
⟩
.
Q20
Compute
the
angle
b
etw
een
⟨
0
,
3
,
−
5
⟩
and
⟨
3
,
−
4
,
3
⟩
.
Q21
Let
A
b
e
the
vertex
of
a
cub
e.
Let
B
the
a
vertex
closest
to
A
and
C
b
e
the
vertex
farthest
from
A
.
Compute
the
angle
b
etw
een
−
−
→
AB
and
−
→
AC
.
Q22
Let
A
b
e
the
vertex
of
a
cub
e,
and
B
and
C
be
any
tw
o
other
points
on
the
cub
e.
Use
a
dot
p
ro
duct
to
explain
why
the
angle
b
et
w
een
−
−
→
AB
and
−
→
AC
cannot
b
e
larger
than
π
2
.
(Hint,
put
A
at
(0
,
0
,
0)
.)
Synthesis
and
Extension
Q23
Ho
w
could
you
use
the
dot
product
to
determine
whether
tw
o
vecto
rs
a
re
parallel?
How
do
es
this
compa
re
with
the
metho
ds
we
already
have?
Q24
Use
dot
p
roducts
to
find
at
least
one
vector
that
is
orthogonal
to
b
oth
⟨
5
,
−
1
,
2
⟩
and
⟨
4
,
4
,
1
⟩
Q25
“Think
of
a
vector
v
”
sa
ys
Raphael,
“tell
me
its
dot
product
with
the
vector
of
my
choice,
and
I’ll
tell
y
ou
what
your
vector
w
as.”
a
Is
there
any
mathematical
wa
y
to
make
such
a
trick
wo
rk?
Explain.
b
Ho
w
many
dot
products
would
you
need
to
ask
for
to
uniquely
identify
an
unknown
vector?
What
dot
p
roducts
would
you
ask
for?
329
Section
5.3
No
rmal
Equations
of
Planes
Goals:
1
Give
equations
of
planes
in
b
oth
vector
and
normal
forms.
2
Use
no
rmal
vecto
rs
to
measure
the
distance
to
a
plane.
Question
5.3.1
What
is
a
Normal
V
ecto
r
to
a
Plane?
In
algeb
ra,
you
learned
t
he
no
rmal
equation
of
a
line:
e.g.
2
x
+
3
y
−
12
=
0
.
Why
is
it
called
this?
Figure:
A
line
and
one
of
its
normal
vectors
The
vecto
r
⟨
2
,
3
⟩
is
a
normal
vector
to
the
line,
meaning
it
is
orthogonal
to
any
vector
contained
in
the
line.
W
e
can
extend
this
definition
to
planes
in
3
-space.
A
no
rmal
vecto
r
to
a
plane
is
o
rthogonal
to
every
vecto
r
in
the
plane.
Theo
rem
In
three-dimensional
space,
every
plane
has
normal
vectors.
They
are
all
parallel
to
each
other.
330
Click to Load Applet
Figure:
A
plane
,
its
normal
vector
n
,
and
a
vector
−
−
→
P
Q
in
the
plane
This
gives
us
an
avenue
to
test
whether
a
p
oint
Q
lies
on
the
plane
or
not.
If
−
−
→
P
Q
is
o
rthogonal
to
n
,
then
Q
lies
on
the
plane.
If
−
−
→
P
Q
and
n
make
a
different
angle,
then
Q
is
not
on
the
plane.
W
e’d
like
to
rewrite
this
relationship
terms
of
the
coordinates
of
Q
.
If
r
0
is
the
p
osition
vector
of
P
and
r
is
the
p
osition
vector
of
Q
,
then
−
−
→
P
Q
=
r
−
r
0
.
The
dot
product
gives
us
a
simple
test
to
see
whether
this
vecto
r
is
orthogonal
to
n
.
Theo
rem
If
r
0
=
⟨
x
0
,
y
0
,
z
0
⟩
describ
es
an
kno
wn
p
oint
on
a
plane,
and
n
=
⟨
a,
b,
c
⟩
is
a
normal
vector.
Then
the
no
rmal
equation
of
the
plane
is
(
r
−
r
0
)
·
n
=
0
o
r
a
(
x
−
x
0
)
+
b
(
y
−
y
0
)
+
c
(
z
−
z
0
)
=
0
Notice
that
since
x
0
,
y
0
and
z
0
a
re
constants,
w
e
can
distribute
and
collect
them
into
a
single
term:
d
.
ax
+
by
+
cz
−
ax
0
−
by
0
−
cz
0
=
0
ax
+
by
+
cz
+
d
=
0
This
reasoning
wo
rks
in
any
dimension
to
define
a
set
of
p
oints
whose
displacement
from
a
known
p
oint
is
o
rthogonal
to
some
normal
vector.
331
Question
5.3.1
What
is
a
Normal
V
ecto
r
to
a
Plane?
Example
a
(
x
−
x
0
)
+
b
(
y
−
y
0
)
=
0
defines
a
line.
a
(
x
−
x
0
)
+
b
(
y
−
y
0
)
+
c
(
z
−
z
0
)
=
0
defines
a
plane.
a
1
(
x
1
−
c
1
)
+
a
2
(
x
2
−
c
2
)
+
·
·
·
+
a
n
(
x
n
−
c
n
)
=
0
defines
a
hyp
erplane
.
Example
5.3.2
Computing
a
No
rmal
Vecto
r
Find
the
no
rmal
equation
of
the
plane
with
intercepts
(4
,
0
,
0)
,
(0
,
3
,
0)
and
(0
,
0
,
8)
.
Compute
a
no
rmal
vecto
r.
Solution
The
normal
equation
of
a
plane
has
the
fo
rm
ax
+
by
+
cz
+
d
=
0
.
Each
of
these
p
oints
must
satisfy
this
equation.
W
e
will
plug
them
in
and
see
what
they
tell
me
ab
out
the
co
efficients.
a
(4)
+
b
(0)
+
c
(0)
+
d
=
0
4
a
+
d
=
0
d
=
−
4
a
a
(0)
+
b
(3)
+
c
(0)
+
d
=
0
3
b
+
d
=
0
d
=
−
3
b
a
(0)
+
b
(0)
+
c
(8)
+
d
=
0
8
c
+
d
=
0
d
=
−
8
c
There
a
re
infinitely
many
solutions
to
this
system
of
equations.
This
mak
es
sense,
b
ecause
there
are
infinitely
many
normal
vectors
to
a
plane.
Different
choices
of
d
give
n
’s
that
are
scalar
multiples
of
each
other.
A
convenient
choice
for
d
is
−
24
,
but
any
nonzero
value
will
w
o
rk.
d
=
−
24
gives
6
x
+
8
y
+
3
z
−
24
=
0
The
no
rmal
vecto
r
is
⟨
6
,
8
,
3
⟩
.
332
Synthesis
5.3.3
Using
the
No
rmal
Vecto
r
to
Compute
Distance
Consider
the
line
2
x
+
3
y
−
12
=
0
.
This
is
the
line
with
normal
vector
n
=
⟨
2
,
3
⟩
and
known
p
oint
P
=
(3
,
2)
.
Example
Let
P
1
=
(7
,
2)
and
P
2
=
(4
,
0)
.
1
Dra
w
the
vecto
rs
−
−
→
P
P
1
and
−
−
→
P
P
2
.
2
If
y
ou
didn’t
have
a
picture,
how
could
you
use
the
values
of
n
·
−
−
→
P
P
1
and
n
·
−
−
→
P
P
2
to
determine
which
side
of
the
line
P
1
and
P
2
lie
on?
Solution
Since
n
is
a
normal
vector,
its
angle
with
any
vector
in
the
line
is
π
2
.
The
vecto
rs
on
the
same
side
of
the
line
as
n
make
an
acute
angle
with
n
.
The
vecto
rs
on
the
far
side
make
an
obtuse
angle.
Thus
when
n
·
−
−
→
P
P
i
<
0
,
P
i
lies
on
the
far
side
of
the
line
from
n
.
When
n
·
−
−
→
P
P
i
>
0
,
P
i
lies
on
the
same
side
as
n
.
W
e
can
get
mo
re
detailed
information
than
just
the
sign
of
the
dot
product.
We
can
actually
compute
a
distance.
333
Synthesis
5.3.3
Using
the
No
rmal
Vecto
r
to
Compute
Distance
Theo
rem
Given
a
line,
plane,
or
hyp
erplane
with
normal
equation
L
(
x
1
,
.
.
.
,
x
k
)
=
0
and
co
rresp
onding
normal
vecto
r
n
,
the
signed
distance
from
the
hyp
erplane
to
the
p
oint
Q
=
(
q
1
,
.
.
.
,
q
k
)
is
L
(
q
1
,
.
.
.
,
q
k
)
n
.
Let
P
b
e
a
known
point
on
the
hyp
erplane.
The
scala
r
p
rojection
of
−
−
→
P
Q
onto
n
is
equal
to
the
signed
distance
from
the
hyp
erplane
to
Q
.
Click to Load Applet
Figure:
The
scalar
projection
of
−
−
→
P
Q
onto
the
normal
vector
of
a
line
Distance
=
−
−
→
P
Q
·
n
|
n
|
(
fo
rmula
fo
r
scala
r
projection
)
=
L
(
q
1
,
.
.
.
,
q
k
)
|
n
|
(
no
rmal
equation
of
the
plane
)
This
formula
is
esp
ecially
pow
erful
b
ecause
we
do
not
need
to
know
a
p
oint
on
the
hyp
erplane.
The
equations
a
(
x
−
x
0
)
+
b
(
y
−
y
0
)
+
c
(
z
−
z
0
)
=
0
ax
+
by
+
cz
+
d
=
0
a
re
equivalent,
and
co
rrespond
to
the
same
no
rmal
vecto
r.
We
can
use
whichever
one
w
e
happ
en
to
have
in
our
signed
distance
formula.
334
Example
5.3.4
The
Distance
from
a
Plane
Compute
the
geometric
distance
from
the
origin
to
the
plane
6
x
+
8
y
+
3
z
−
24
=
0
.
Solution
n
=
⟨
6
,
8
,
3
⟩
.
The
signed
distance
from
the
plane
to
the
o
rigin
is
L
(0
,
0
,
0)
|
n
|
=
(6)(0)
+
(8)(0)
+
(3)(0)
−
24
√
36
+
64
+
9
=
−
24
√
109
Geometric
distance
cannot
b
e
negative,
so
it
is
24
√
109
.
Application
5.3.5
Supp
o
rt
V
ecto
r
Machines
One
type
of
machine
learning
involves
training
a
computer
to
distinguish
b
et
w
een
tw
o
states.
Fo
r
example,
a
computer
might
b
e
trained
to
distinguish
b
etw
een
a
cancerous
tumor
and
a
b
enign
one.
T
o
do
this
the
computer
is
given
a
large
set
of
cases.
Each
case
is
measured
by
numerical
data,
such
as:
The
size
of
the
tumor
The
location
of
the
tumor
The
age
of
the
patient
Results
of
bloo
d
tests
The
b
rightness
of
each
pixel
in
a
CT
scan
or
MRI
Each
data
t
ype
is
a
dimension,
and
each
case
is
a
p
oint
in
a
(probably
very
high)
dimensional
space.
The
computer
would
like
a
simple
test
to
divide
these
cases
into
cancerous
and
b
enign.
The
test
will
b
e
which
side
of
a
hyp
erplane
they
lie
on.
It
is
unlik
ely
that
any
such
hyperplane
exists
initially
,
so
the
computer
attempts
a
sequence
of
transformations
of
the
data
until
they
are
separated
b
y
a
hyp
erplane
with
some
degree
of
reliability
.
335
Application
5.3.5
Supp
o
rt
V
ecto
r
Machines
Click to Load Applet
Section
5.3
Exercises
Summa
ry
Questions
Q1
What
info
rmation
do
you
need
in
order
to
write
the
normal
equation
of
a
plane?
Q2
Ho
w
a
re
the
normal
vecto
rs
of
a
plane
related
to
each
other?
Q3
What
is
the
significance
of
the
co
efficients
in
the
normal
equation
of
a
plane?
Q4
Ho
w
do
w
e
compute
the
signed
distance
from
a
p
oint
to
a
plane?
336
5.3.1
Q5
Is
v
=
⟨
8
,
−
3
,
−
10
⟩
parallel
to
the
plane
6
x
+
6
y
+
3
z
+
11
=
0
?
Explain.
Q6
Is
v
=
9
i
−
15
j
+
6
k
no
rmal
to
the
plane
−
6
x
+
10
y
−
4
z
+
23
=
0
?
Explain.
Q7
Name
a
no
rmal
vector
to
the
following
planes:
i.
3
x
−
8
y
+
10
z
−
4
=
0
ii.
z
−
2
=
4(
x
+
7)
−
5(
y
+
1)
Q8
Supp
ose
that
n
is
a
normal
vector
to
6
x
−
3
y
+
2
z
−
4
=
0
,
that
happ
ens
to
also
b
e
a
unit
vector.
Give
all
possible
values
of
n
.
Q9
W
rite
a
no
rmal
equation
of
a
plane
parallel
to
7
x
−
11
y
+
8
z
+
15
=
0
that
passes
through
the
o
rigin.
Q10
W
rite
a
normal
equation
of
a
plane
pa
rallel
to
10
x
−
11
y
+
z
+
20
=
0
that
passes
through
(2
,
3
,
5)
.
Q11
Given
that
the
plane
ax
+
by
+
cz
+
d
=
0
passes
through
the
origin,
what
can
you
say
ab
out
a
,
b
,
c
,
and
d
?
Q12
Given
that
plane
ax
+
by
+
cz
+
d
=
0
contains
the
x
-axis,
what
can
you
say
about
a
,
b
,
c
,
and
d
?
Q13
Are
the
planes
4
x
+
6
y
+
8
z
+
15
=
0
and
10
x
+
15
y
+
20
z
−
7
=
0
parallel?
Explain
ho
w
you
kno
w.
Q14
Supp
ose
we
know
the
planes
12
x
+
18
y
+
6
z
−
15
=
0
and
ax
+
by
+
4
z
+
d
=
0
are
pa
rallel.
What
can
y
ou
say
about
the
values
of
a
,
b
and
d
?
Q15
The
equations
3
x
−
y
+
4
z
+
10
=
0
and
−
6
x
+
2
y
−
8
z
+
k
=
0
describ
e
the
same
plane.
What
is
the
value
of
k
?
Q16
Consider
the
plane
with
normal
equation
7
x
+
y
−
2
z
=
5
.
a
Give
t
w
o
other
no
rmal
equations
of
this
plane.
b
What
are
the
no
rmal
vectors
corresponding
to
the
orginal
equation
and
your
t
w
o
equations
in
a
?
337
Section
5.3
Exercises
c
Ho
w
a
re
these
vecto
rs
in
b
related
to
each
other?
5.3.2
Q17
Give
a
no
rmal
equation
of
the
plane
with
intercepts
(10
,
0
,
0)
,
(0
,
−
5
,
0)
and
(0
,
0
,
2)
.
Q18
Give
a
no
rmal
equation
of
the
plane
with
intercepts
(
−
18
,
0
,
0)
,
(0
,
9
,
0)
and
(0
,
0
,
−
4)
.
Q19
Give
a
no
rmal
equation
of
the
plane
through
(4
,
3
,
0)
,
(5
,
1
,
1)
and
(
−
2
,
5
,
2)
.
Q20
Give
a
no
rmal
equation
of
the
plane
through
(1
,
1
,
1)
,
(8
,
1
,
4)
and
(0
,
0
,
4)
.
5.3.3
Q21
Katie
is
computing
the
distance
from
the
p
oint
(6
,
3)
to
the
line
2
x
+
3
y
−
12
=
0
.
She
notices
that
(6
,
0)
is
the
x
-intercept
of
the
line.
Since
(6
,
3)
is
3
units
aw
a
y
from
(6
,
0)
she
concludes
the
distance
from
the
p
oint
to
the
line
is
3
.
What
do
y
ou
think
of
Katie’s
reasoning?
Q22
Consider
the
line
L
with
normal
equation
2
x
+
3
y
−
12
=
0
and
the
p
oint
Q
=
(6
,
3)
.
a
What
is
the
slop
e
of
L
?
b
What
w
ould
be
the
slop
e
of
a
line
p
erp
endicular
to
L
?
c
W
rite
an
equation
(in
any
fo
rm
you’d
like)
of
a
line
K
that
passes
through
Q
and
is
p
erp
en-
dicula
r
to
L
.
d
Compute
the
intersection
p
oint
of
P
of
L
and
K
.
e
What
is
the
distance
from
P
to
Q
?
f
Check
that
your
answer
to
e
matches
the
distance
formula
we
derived.
Which
metho
d
do
y
ou
lik
e
better?
338
5.3.4
Q23
Ho
w
fa
r
is
(5
,
2
,
1)
from
3
x
+
2
y
−
5
z
+
10
=
0
?
Q24
Ho
w
fa
r
is
(0
,
0
,
1)
from
3
x
+
12
y
−
4
z
+
20
=
0
?
Q25
Are
(6
,
7
,
1)
and
(5
,
−
3
,
−
4)
on
the
same
or
different
sides
of
3
x
−
10
y
+
9
z
+
46
=
0
?
Q26
The
p
oint
(
x,
4
,
5)
lies
on
the
same
side
of
the
plane
2
x
+
y
−
2
z
+
10
=
0
as
the
origin
do
es.
What
does
that
tell
you
ab
out
the
value
of
x
?
5.3.5
Q27
W
e
have
six
images
of
dogs
and
cats.
We
measure
four
things
about
each,
and
have
collected
the
data
b
elow.
We
would
like
to
use
the
hyp
erplane
2
x
1
+
5
x
2
−
4
x
3
+
10
x
4
+
k
=
0
to
sepa
rate
the
images
of
dogs
from
the
images
of
cats.
T
yp
e
Measurements
Cat
(5
,
1
,
3
,
6)
Dog
(7
,
3
,
7
,
2)
Dog
(7
,
2
,
6
,
4)
Dog
(9
,
1
,
8
,
5)
Cat
(6
,
4
,
5
,
5)
Cat
(9
,
2
,
7
,
6)
a
What
values
of
k
w
ould
cause
the
hyp
erplane
to
correctly
separate
the
dog
images
from
the
cat
images?
b
If
you
intended
to
use
the
hyp
erplane
to
guess
whether
a
future
image
was
a
dog
or
cat,
what
k
w
ould
y
ou
choose?
Why?
Q28
Supp
ose
we
have
a
hyp
erplane
that
we
would
lik
e
to
separate
tw
o
sets
of
p
oints,
but
it
do
esn’t
quite
wo
rk.
We
measure
the
error
of
this
separation
by
taking
the
sum
of
the
geometric
distances
from
the
hyp
erplane
of
each
point
that
is
on
the
wrong
side
of
the
hyp
erplane.
Suppose
we
w
ere
hoping
that
the
line
2
x
+
3
y
−
12
=
0
w
ould
separate
the
points
of
t
yp
e
T
from
the
p
oints
of
t
yp
e
S.
339
Section
5.3
Exercises
T
yp
e
Co
o
rdinates
T
(6
,
2)
T
(2
,
1)
T
(5
,
3)
T
(4
,
4)
S
(1
,
5)
S
(1
,
1)
S
(4
,
0)
S
(4
,
2)
a
Create
a
diagram
of
these
p
oints
(lab
elled
or
colored
by
type)
and
the
line.
b
W
e
did
not
sp
ecify
which
side
of
the
line
should
b
e
T
and
which
should
be
S
.
Use
y
our
diagram
to
decide
which
choice
of
sides
will
give
less
error.
c
Compute
the
erro
r
in
this
metho
d
of
separation.
d
Supp
ose
we
were
trying
to
find
a
better
line
of
the
form
ax
+
by
+
c
=
0
.
When
a
=
2
,
b
=
3
and
c
=
−
12
,
would
increasing
a
increase
or
decrease
the
error?
Justify
your
answer
with
a
derivative.
Synthesis
and
Extension
Q29
W
rite
the
equation
of
a
plane
that
contains
all
the
p
oints
equidistant
from
A
=
(1
,
−
2
,
7)
and
B
=
(7
,
0
,
5)
Q30
Tw
o
planes
a
re
p
erp
endicular
if
their
normal
vectors
a
re
orthogonal.
a
Are
4
x
−
7
y
+
z
−
3
=
0
and
5
x
+
y
+
13
z
+
25
=
0
p
erp
endicular?
b
If
tw
o
planes
are
p
erp
endicula
r,
is
every
vector
in
the
first
plane
orthogonal
to
every
vector
in
the
second
plane?
Q31
W
rite
the
normal
equation
of
a
plane
that
contains
the
x
and
z
axes.
Where
have
we
seen
this
plane
before?
340
Q32
What
trouble
do
you
run
into
if
you
try
to
write
the
equation
of
the
plane
through
(6
,
0
,
0)
,
(0
,
8
,
0)
and
(3
,
4
,
0)
?
Explain
geometrically
why
this
mak
es
sense.
341
Section
5.4
The
Gradient
V
ecto
r
Goals:
1
Calculate
the
gradient
vecto
r
of
a
function.
2
Relate
the
gradient
vector
to
the
shap
e
of
a
graph
and
its
level
curves.
3
Compute
directional
derivatives
.
Armed
with
ideas
ab
out
vectors,
w
e
have
the
vo
cabulary
to
discuss
mo
re
complex
changes
in
the
va
riables
of
a
function.
Rather
than
having
one
va
riable
change
and
the
other
stay
constant,
we
can
indicate
a
change
in
b
oth
va
riables
with
a
vecto
r.
When
exploring
these
computations,
we
will
construct
one
of
the
most
imp
ortant
to
ols
for
multivariable
calculus.
Question
5.4.1
Ho
w
Do
W
e
Compute
Rates
of
Change
in
Another
Direction?
The
partial
derivatives
of
f
(
x,
y
)
give
the
instantaneous
rate
of
change
in
the
x
and
y
directions.
This
is
realized
geometrically
as
the
slop
e
of
the
tangent
line.
What
if
we
want
to
travel
in
a
different
direction?
Click to Load Applet
Figure:
The
tangent
line
to
z
=
f
(
x,
y
)
in
the
x
direction
Definition
Let
f
(
x,
y
)
be
a
function
and
u
b
e
a
unit
vecto
r
in
R
2
.
The
directional
derivative
,
denoted
D
u
f
,
is
the
instantaneous
rate
of
change
of
f
as
we
move
in
the
u
direction.
This
is
also
the
slop
e
of
the
tangent
line
to
y
=
f
(
x,
y
)
in
the
direction
of
u
.
342
Click to Load Applet
Figure:
The
tangent
line
to
f
(
x,
y
)
in
the
direction
of
u
Recall
that
w
e
compute
D
x
f
b
y
comparing
the
values
of
f
at
(
x,
y
)
to
the
value
at
(
x
+
h,
y
)
,
a
displacement
of
h
in
the
x
-direction.
D
x
f
(
x,
y
)
=
lim
h
→
0
f
(
x
+
h,
y
)
−
f
(
x,
y
)
h
T
o
compute
D
u
f
for
u
=
a
i
+
b
j
,
we
compare
the
value
of
f
at
(
x,
y
)
to
the
value
at
(
x
+
ta,
y
+
tb
)
,
a
displacement
of
t
in
the
u
-direction.
Limit
F
o
rmula
D
u
f
(
x,
y
)
=
lim
t
→
0
f
(
x
+
ta,
y
+
tb
)
−
f
(
x,
y
)
t
Questions:
1
What
direction
p
roduces
the
greatest
directional
derivative?
The
smallest?
2
Ho
w
a
re
these
directions
related
to
the
geometry
(sp
ecifically
the
level
curves)
of
the
graph?
3
Ho
w
these
directions
related
to
the
partial
derivatives?
W
e
can
explo
re
these
questions
with
an
applet
in
the
Other
Cross
Sections
activit
y
.
343
Question
5.4.1
Ho
w
Do
W
e
Compute
Rates
of
Change
in
Another
Direction?
Click to Load Applet
Figure:
A
cross
section
of
z
=
f
(
x,
y
)
and
a
tangent
line
in
the
direction
of
u
Question
5.4.2
What
Is
the
Gradient
Vecto
r?
The
relationship
b
et
w
een
the
direction
of
maximum
increase
and
the
partial
derivatives
suggest
that
w
e
could
treat
the
partial
derivatives
like
comp
onents
of
a
vector.
Definition
The
gradient
vecto
r
of
f
at
(
x,
y
)
is
∇
f
(
x,
y
)
=
⟨
f
x
(
x,
y
)
,
f
y
(
x,
y
)
⟩
Rema
rks:
1
The
gradient
vecto
r
is
a
function
of
(
x,
y
)
.
Different
p
oints
have
different
gradients.
2
u
max
,
which
maximizes
D
u
f
,
p
oints
in
the
same
direction
as
∇
f
.
3
u
0
,
which
is
tangent
to
the
level
curves,
is
orthogonal
to
∇
f
.
344
Rema
rk
Students
often
wonder:
what
is
the
geometric
intuition
b
ehind
the
gradient
vecto
r
and
its
properties?
The
answer
is
often
disapp
ointing,
but
imp
ortant.
The
gradient
vecto
r
does
not
have
a
geometric
motivation.
W
e
a
rtificially
created
the
gradient
vecto
r
b
ecause
it
has
convenient
algebraic
properties.
If
that
were
the
end
of
the
story
,
w
e
w
ouldn’t
b
other
learning
ab
out
it.
How
ever,
the
gradient
turns
out
to
be
so
useful
that
we
will
study
it
intensely
,
despite
its
uncompelling
origins.
Question
5.4.3
Ho
w
Do
W
e
Compute
a
Directional
Derivative?
There
a
re
several
wa
ys
to
derive
a
formula
for
the
directional
derivative.
One
approach
is
to
apply
algeb
ra
and
limit
la
ws
to
the
limit
definition.
A
more
geometric
metho
d
is
to
exploit
our
previous
wo
rk
with
the
tangent
plane.
The
directional
derivative
is
the
slop
e
of
a
tangent
line.
The
tangent
lines
live
in
the
tangent
plane.
We
can
compute
their
slop
e
by
rise
over
run.
Let
u
b
e
a
unit
vector
from
(
x
0
,
y
0
)
to
(
x
1
,
y
1
)
.
Let
the
asso
ciated
z
values
in
the
tangent
plane
b
e
z
0
and
z
1
resp
ectively
.
D
u
f
(
x
0
,
y
0
)
=
rise
run
=
z
1
−
z
0
|
u
|
=
f
x
(
x
0
,
y
0
)(
x
1
−
x
0
)
+
f
y
(
x
0
,
y
0
)(
y
1
−
y
0
)
=
∇
f
(
x
0
,
y
0
)
·
u.
Click to Load Applet
F
unctions
of
Mo
re
Va
riables
W
e
can
also
define
directional
derivatives
of
higher
variable
functions
with
analogous
results.
f
(
x
1
,
.
.
.
,
x
n
)
is
a
differentiable
function.
u
is
a
unit
vecto
r
in
R
n
.
D
u
f
denotes
the
directional
derivative
in
the
direction
of
u
.
∇
f
=
⟨
f
x
1
,
.
.
.
,
f
x
n
⟩
is
an
n
-dimensional
vector
function
on
R
n
.
D
u
f
=
∇
f
·
u
345
Synthesis
5.4.4
Directional
Derivative
and
the
Cosine
Fo
rmula
No
w
that
we
have
a
formula
for
directional
derivatives,
we
can
verify
our
observations
from
earlier.
Supp
ose
f
(
x,
y
)
is
a
differentiable
function
and
we
can
cho
ose
any
unit
vector
u
.
a
W
rite
D
u
f
(
x,
y
)
in
terms
of
the
length
of
a
vector
and
an
angle.
b
In
what
direction
u
will
f
increase
fastest?
c
What
will
be
the
value
of
D
u
f
(
x,
y
)
in
that
direction?
d
In
what
direction
u
will
D
u
f
(
x,
y
)
=
0
?
Solution
a
Since
the
directional
derivative
is
a
dot
product,
we
can
apply
our
fo
rmula
that
relates
the
dot
p
ro
duct
to
the
lengths
of
the
vectors
and
the
angle
b
etw
een
them.
D
u
f
(
x,
y
)
=
∇
f
(
x,
y
)
·
u
dot
p
roduct
formula
=
|∇
f
(
x,
y
)
||
u
|
cos
θ
cosine
fo
rmula
=
|∇
f
(
x,
y
)
|
cos
θ
u
is
a
unit
vecto
r
b
Given
a
pa
rticula
r
(
x,
y
)
,
|∇
f
(
x,
y
)
|
cos
θ
is
la
rgest
when
θ
=
0
This
means
that
D
u
f
(
x,
y
)
is
maximized
when
u
is
in
the
direction
of
∇
f
(
x,
y
)
.
The
formula
for
a
unit
vector
in
the
direction
of
the
gradient
is
u
=
1
|∇
f
(
x,
y
)
|
∇
f
(
x,
y
)
c
In
this
direction,
cos
θ
=
1
so
D
u
f
(
x,
y
)
=
|∇
f
(
x,
y
)
|
.
d
W
e
can
solve
for
θ
D
u
f
(
x,
y
)
=
0
|∇
f
(
x,
y
)
|
cos
θ
=
0
b
y
part
(a)
cos
θ
=
0
as
long
as
∇
f
(
x,
y
)
=
0
θ
=
π
2
W
e
conclude
that
u
must
b
e
orthogonal
to
∇
f
(
x,
y
)
.
346
Click to Load Applet
Figure:
The
angle
b
etw
een
the
gradient
of
f
and
a
unit
vector
Main
Ideas
The
cosine
fo
rmula
for
the
dot
product
lets
us
relate
the
directional
derivative
to
an
angle.
f
increases
fastest
in
the
direction
of
∇
f
(
x,
y
)
.
D
u
f
(
x,
y
)
=
0
when
∇
f
(
x,
y
)
and
u
are
orthogonal.
Example
5.4.5
A
Directional
Derivative
Let
f
(
x,
y
)
=
p
9
−
x
2
−
y
2
and
let
u
=
⟨
0
.
6
,
−
0
.
8
⟩
.
a
What
a
re
the
level
curves
of
f
?
b
What
direction
does
∇
f
(1
,
2)
p
oint?
c
Without
calculating,
is
D
u
f
(1
,
2)
p
ositive
o
r
negative?
d
Calculate
∇
f
(1
,
2)
and
D
u
f
(1
,
2)
.
347
Example
5.4.5
A
Directional
Derivative
Solution
a
The
level
curves
have
the
equations
p
9
−
x
2
−
y
2
=
c
.
These
solve
to
x
2
+
y
2
=
9
−
c
2
.
As
c
increases
from
0
to
3
these
are
circles
starting
at
radius
3
and
shrinking
to
the
origin.
Fo
r
c
outside
this
range,
the
level
curve
has
no
p
oints.
b
∇
f
points
in
the
direction
of
increase
and
normal
to
the
level
curves.
Since
higher
level
curves
a
re
smaller
circles,
closer
to
the
origin,
∇
f
(1
,
2)
p
oints
tow
a
rd
the
origin.
c
D
u
f
(1
,
2)
=
∇
f
(1
,
2)
·
u
.
Since
u
app
ea
rs
to
make
an
acute
angle
with
∇
f
(1
,
2)
,
w
e
exp
ect
this
dot
p
roduct
to
b
e
p
ositive.
d
First
w
e
need
to
compute
∇
f
(1
,
2)
.
∇
f
(
x,
y
)
=
⟨
f
x
(
x,
y
)
,
f
y
(
x,
y
)
⟩
=
*
1
2
p
9
−
x
2
−
y
2
(
−
2
x
)
,
1
2
p
9
−
x
2
−
y
2
(
−
2
y
)
+
(
chain
rule
)
∇
f
(1
,
2)
=
1
2
√
9
−
1
2
−
2
2
(
−
2)(1)
,
1
2
√
9
−
1
2
−
2
2
(
−
2)(2)
=
−
1
2
,
−
1
No
w
w
e
use
the
dot
product
formula
to
compute
D
u
f
(1
,
2)
.
D
u
f
(1
,
2)
=
∇
f
(1
,
2)
·
u
=
−
1
2
,
−
1
·
⟨
0
.
6
,
−
0
.
8
⟩
348
=
−
0
.
3
+
0
.
8
=
0
.
5
This
confirms
our
intuition
that
D
u
f
(1
,
2)
is
positive.
Example
5.4.6
Dra
wing
the
Gradient
Let
h
(
x,
y
)
give
the
altitude
at
longitude
x
and
latitude
y
.
Assuming
h
is
differentiable,
draw
the
direction
of
∇
h
(
x,
y
)
at
each
of
the
p
oints
lab
eled
b
elo
w.
Which
gradient
is
the
longest?
A
B
C
Figure:
A
top
ographical
map
Solution
The
gradient
vector
at
each
p
oint
is
normal
to
the
level
curves,
p
ointing
uphill.
The
hill
is
steep
est
at
B
,
b
ecause
the
level
curves
a
re
closer
together.
This
tells
us
that
the
partial
derivatives
a
re
larger.
Thus
∇
h
(
B
)
is
longer
than
∇
h
(
A
)
and
∇
h
(
C
)
.
A
B
C
349
Application
5.4.7
Edge
Detection
Rep
resenting
an
image
by
defining
a
b
rightness
(or
color)
function
on
the
pixels
is
simple
enough,
but
can
a
computer
be
taught
to
make
sense
of
what
it
sees?
Image
recognition
is
an
exciting
field
that
p
romises
to
automate
and
improve
tasks
from
medical
diagnosis
to
driving
a
vehicle.
The
problem
is
daunting.
What
algorithm
can
p
ossibly
take
a
set
of
pixels
and
lo
cate
a
tumor
o
r
a
p
edestrian?
The
first
step
is
to
identify
the
objects
in
the
image.
The
first
step
of
object
identification
is
edge
detection,
determining
where
one
object
ends
and
another
b
egins.
W
e
can
do
this
b
y
appro
ximating
the
partial
derivatives
at
each
pixel.
We
compare
each
pixel
to
nea
rb
y
pixels
and
compute
rise
over
run
(ho
w
these
a
re
chosen
and
averaged
can
significantly
affect
the
accuracy
of
the
algo
rithm).
The
length
of
the
gradient
of
a
b
rightness
function
detects
the
edges
in
a
picture,
where
the
b
rightness
is
changing
quickly
.
∂
B
∂
x
(336
,
785)
≈
185
−
187
1
∂
B
∂
y
(336
,
785)
≈
179
−
187
1
∇
B
(336
,
785)
≈
(
−
2
,
−
8)
∂
B
∂
x
(340
,
784)
≈
97
−
139
1
∂
B
∂
y
(340
,
784)
≈
72
−
139
1
∇
B
(340
,
784)
≈
(
−
42
,
−
67)
∇
B
∇
B
Figure:
A
long
gradient
vector
indicates
a
swift
change
in
brightness.
Its
direction
suggests
the
shap
e
of
the
edges.
Notice
that
the
gradient
is
long
nea
r
the
edge
of
the
iris
in
Mona
Lisa’s
eye.
It
is
much
shorter
at
a
p
oint
in
the
white
of
her
eye.
Moreover,
the
gradient
at
the
edge
of
the
iris
is
appro
ximately
normal
to
the
edge
of
her
iris,
b
ecause
gradients
are
normal
to
level
curves
.
This
info
rmation
can
b
e
used
by
an
algo
rithm
to
detect
not
only
the
lo
cation
of
the
edges,
but
also
their
direction.
Application
5.4.8
T
angent
Planes
to
a
Level
Surface
Use
a
gradient
vector
to
find
the
equation
of
the
tangent
plane
to
the
graph
x
2
+
y
2
+
z
2
=
14
at
the
point
(2
,
1
,
−
3)
.
There
a
re
t
w
o
solutions
wo
rth
compa
ring
here.
350
Solution
1
W
e
can
write
z
as
a
function
of
x
and
y
and
apply
the
tangent
plane
formula.
x
2
+
y
2
+
z
2
=
14
z
2
=
14
−
x
2
−
y
2
z
=
−
p
14
−
x
2
−
y
2
(
z
=
−
3
is
on
the
negative
branch
of
the
function
)
f
x
(
x,
y
)
=
−
1
2
p
14
−
x
2
−
y
2
(
−
2
x
)
f
x
(2
,
1)
=
2
3
f
y
(
x,
y
)
=
−
1
2
p
14
−
x
2
−
y
2
(
−
2
y
)
f
y
(2
,
1)
=
1
3
Equation:
z
+
3
=
2
3
(
x
−
2)
+
1
3
(
y
−
1)
Solution
2
Define
F
(
x,
y
,
z
)
=
x
2
+
y
2
+
z
2
.
The
graph
x
2
+
y
2
+
z
2
=
14
is
a
level
surface
of
F
.
∇
F
(2
,
1
,
−
3)
is
no
rmal
to
the
level
surface,
meaning
it
is
also
a
normal
vector
for
the
tangent
plane.
∇
F
(
x,
y
,
z
)
=
⟨
2
x,
2
y
,
2
z
⟩
∇
F
(2
,
1
,
−
3)
=
⟨
4
,
2
,
−
6
⟩
W
e
no
w
have
a
normal
vector
n
=
∇
F
(2
,
1
,
−
3)
.
Our
known
p
oint
is
(
x
0
,
y
0
,
z
0
)
=
(2
,
1
,
−
3)
.
The
no
rmal
equation
of
the
plane
is
4(
x
−
2)
+
2(
y
−
1)
−
6(
z
+
3)
=
0
.
Solution
2
requires
more
conceptual
reasoning,
but
is
computationally
much
easier.
In
fact,
in
some
cases
we
cannot
use
Solution
1
at
all
b
ecause
we
do
not
kno
w
how
to
solve
for
z
.
Once
we
are
comfo
rtable
with
the
concepts
involved,
the
second
metho
d
is
generally
sup
erior
for
graphs
of
implicit
equations.
351
Application
5.4.8
T
angent
Planes
to
a
Level
Surface
Main
Idea
The
graph
of
an
implicit
equation
can
b
e
written
as
a
level
set
of
a
function.
The
gradient
of
that
function
is
a
normal
vecto
r
to
the
level
set
and
also
to
its
tangent
line/plane/hyp
erplane.
Click to Load Applet
Figure:
The
level
surface
x
2
+
y
2
+
z
2
=
14
,
its
tangent
plane
and
∇
F
.
Section
5.4
Exercises
Summa
ry
Questions
Q1
What
does
the
direction
of
the
gradient
vector
tell
you?
Q2
What
does
the
directional
derivative
mean
geometrically?
Q3
Ho
w
do
y
ou
com
pute
a
directional
derivative?
Q4
Ho
w
is
the
gradient
vector
related
to
a
level
set?
352
5.4.1
Q5
Supp
ose
that
f
(3
,
7)
=
12
and
f
(7
,
4)
=
10
.
a
What
is
the
distance
from
(3
,
7)
to
(7
,
4)
?
b
App
ro
ximate
the
rate
of
change
of
f
at
(3
,
7)
travelling
tow
a
rd
(7
,
4)
Q6
Supp
ose
g
(0
,
2)
=
15
and
g
(4
,
1)
=
17
.
a
What
is
the
distance
from
(0
,
2)
to
(4
,
1)
?
b
App
ro
ximate
the
rate
of
change
of
g
at
(0
,
2)
travelling
tow
a
rd
(4
,
1)
.
c
If
y
ou
wanted
to
express
the
previous
rate
of
change
as
an
app
ro
ximation
of
D
u
g
(0
,
2)
,
what
w
ould
the
unit
vector
u
be?
5.4.2
Q7
If
f
(
x,
y
)
=
x
2
sin(
xe
y
)
,
what
is
∇
f
(
x,
y
)
?
Q8
If
g
(
x,
y
)
=
p
6
x
2
+
5
y
4
,
what
is
∇
g
(
x,
y
)
?
Q9
If
∇
f
(
x
0
,
y
0
)
is
orthogonal
to
∇
g
(
x
0
,
y
0
)
,
what
can
w
e
say
ab
out
the
level
curves
of
f
and
g
?
Be
specific.
Q10
Ha
rriet
sa
ys
“The
gradient
vector
of
f
is
tangent
to
the
graph
of
z
=
f
(
x,
y
)
.”
“No,”
sa
ys
Ma
rcus,
“it
is
normal
to
the
graph
of
z
=
f
(
x,
y
)
.”
Who
is
correct?
353
Section
5.4
Exercises
5.4.3
Q11
Consider
our
computation
of
the
directional
derivative
as
a
dot
product.
a
Where
did
w
e
use
the
fact
that
u
is
a
unit
vecto
r?
b
If
u
w
ere
not
a
unit
vector,
then
∇
f
·
u
w
ould
no
longer
represent
rise
over
run.
What
would
it
rep
resent
instead?
Q12
Supp
ose
the
linea
rization
of
f
(
x,
y
)
at
(
−
3
,
9)
has
the
equation
L
(
x,
y
)
=
4
+
2(
x
+
3)
−
1
3
(
y
−
9)
.
What
is
the
slop
e
of
L
from
(
−
3
,
9)
to
(5
,
3)
?
5.4.4
Q13
Given
a
function
f
(
x,
y
)
and
a
p
oint
(
x,
y
)
,
in
what
direction
u
is
f
decreasing
fastest?
Compute
an
exp
ression
fo
r
u
.
Q14
If
D
u
f
(
x,
y
)
<
0
,
what
can
you
say
ab
out
the
directions
of
∇
f
(
x,
y
)
and
u
?
Q15
If
f
x
(3
,
5)
=
f
y
(3
,
5)
in
what
direction(s)
from
(3
,
5)
could
f
increase
most
quickly?
Q16
Explain
why
it
makes
sense
that
if
D
u
f
(
a,
b,
c
)
=
0
,
then
u
is
tangent
to
the
level
surface
of
f
through
(
a,
b,
c
)
.
Q17
If
f
(
x,
y
,
z
)
=
3
xy
+
z
2
,
find
the
unit
vector
u
that
maximizes
D
u
f
(2
,
1
,
−
4)
.
What
is
the
value
of
D
u
f
(2
,
1
,
−
4)
for
this
u
?
Q18
Let
f
(
x,
y
)
=
2
x
2
y
−
10
x
−
y
2
.
a
What
unit
vecto
r
u
maximizes
the
quantity
D
u
f
(
−
1
,
3)
?
b
Compute
D
u
f
(
−
1
,
3)
fo
r
the
u
you
found
in
part
a
.
354
5.4.5
Q19
If
u
=
2
3
,
−
1
3
,
−
2
3
and
f
(
x,
y
,
z
)
=
xe
y
z
,
compute
D
u
f
(3
,
0
,
4)
.
Q20
If
u
=
3
7
,
6
7
,
−
2
7
and
f
(
x,
y
,
z
)
=
xy
+
y
z
+
z
x
,
compute
D
u
f
(7
,
−
7
,
14)
.
Q21
If
u
is
a
unit
vector
in
the
direction
of
⟨
2
,
3
⟩
and
f
(
x,
y
)
=
x
2
+
3
xy
+
2
,
calculate
D
u
f
(
−
1
,
4)
.
Q22
Compute
the
directional
derivative
of
g
(
x,
y
)
=
e
x
2
−
y
at
(3
,
7)
in
the
direction
of
⟨−
12
,
5
⟩
.
5.4.6
Q23
In
this
diagram,
we
have
several
level
sets
of
f
(
x,
y
)
.
a
Which
w
a
y
does
∇
f
(
−
4
,
1
.
25)
p
oint?
b
Ma
rk
all
the
p
oints
(
x,
y
)
that
satisfy
f
(
x,
y
)
=
30
∇
f
(
x,
y
)
p
oints
in
the
p
ositive
y
-direction
Q24
Some
level
curves
of
f
are
drawn
b
elow.
Indicate
the
direction
of
the
gradient
of
f
at
each
lab
elled
point.
355
Section
5.4
Exercises
5.4.7
Q25
If
∇
B
(
x
0
,
y
0
)
=
⟨
13
,
−
17
⟩
,
would
you
exp
ect
the
pixels
ab
ove
(
x
0
,
y
0
)
to
b
e
brighter
or
dimmer
than
(
x
0
,
y
0
)
?
Explain.
Q26
The
b
rightness
function
on
the
Mona
Lisa
image
ranges
from
0
to
255
.
If
we
use
adjacent
p
oints
to
apporixmate
the
gradient
as
in
the
example,
what
is
the
longest
gradient
vecto
r
we
could
theo
retically
p
roduce
?
5.4.8
Q27
Calculate
a
no
rmal
equation
of
a
tangent
line
to
x
3
+
8
y
3
−
12
xy
=
0
at
(3
,
1
.
5)
.
Q28
Let
P
be
a
point
on
the
circle
x
2
+
y
2
=
r
2
.
Sho
w
that
the
position
vector
of
P
is
no
rmal
to
the
circle
at
P
.
Q29
Pro
duce
an
equation
of
the
tangent
plane
to
z
3
−
xz
2
−
y
x
2
=
24
at
(4
,
−
2
,
2)
.
Q30
Give
an
equation
of
the
tangent
plane
to
the
graph
z
2
x
+
2
y
z
−
x
2
y
2
=
59
at
(3
,
2
,
5)
.
356
Synthesis
and
Extension
Q31
Supp
ose
f
(
x,
y
)
is
a
differentiable
function,
and
we
kno
w
that
for
u
=
⟨−
0
.
6
,
0
.
8
⟩
,
D
u
f
(5
,
−
1)
=
4
and
fo
r
v
=
⟨
0
,
−
1
⟩
w
e
kno
w
that
D
v
f
(5
,
−
1)
=
−
2
.
What
is
∇
f
(5
,
−
1)
?
Q32
Supp
ose
the
point
P
=
(
x
0
,
y
0
,
z
0
)
lies
on
the
graph
z
=
f
(
x,
y
)
.
a
Give
the
fo
rmula
for
tangent
plane
to
this
graph
at
P
.
b
z
=
f
(
x,
y
)
is
a
level
surface
of
F
(
x,
y
,
z
)
=
f
(
x,
y
)
−
z
.
Use
the
gradient
of
F
to
write
the
equation
of
the
tangent
plane
to
F
(
x,
y
,
z
)
=
0
at
P
.
c
Are
these
equations
equivalent?
Justify
your
answer
with
algebra.
Q33
Ho
w
could
y
ou
use
the
gradient
of
f
to
rewrite
the
fo
rmula
for
the
linea
rization
L
(
x,
y
)
of
f
(
x,
y
)
at
(
x
0
,
y
0
)
?
Q34
Supp
ose
f
(
x,
y
)
is
a
differentiable
function
and
∇
f
(
a,
b
)
is
not
the
zero
vector.
Ho
w
many
unit
vecto
rs
u
exist
such
that
D
u
f
(
a,
b
)
=
0
.
Ho
w
are
they
related
geometrically?
Q35
Supp
ose
f
(
x,
y
,
z
)
is
a
differentiable
function
and
∇
f
(
a,
b,
c
)
is
not
the
zero
vector.
How
many
unit
vecto
rs
u
exist
such
that
D
u
f
(
a,
b,
c
)
=
0
.
Ho
w
are
they
related
geometrically?
Q36
Supp
ose
that
f
(
x,
y
,
z
)
is
a
differentiable
function,
and
f
(3
,
5
,
−
2)
=
13
.
Suppose
further
that
the
vectors
⟨
3
,
1
,
0
⟩
and
⟨
0
,
2
,
5
⟩
b
oth
lie
in
the
tangent
plane
to
the
surface
f
(
x,
y
,
z
)
=
13
at
(3
,
5
,
−
2)
.
If
the
maximum
value
of
D
u
f
(3
,
5
,
−
2)
is
20
,
find
all
p
ossible
values
of
∇
f
(3
,
5
,
−
2)
.
Q37
Consider
the
function
h
(
x,
y
)
=
x
2
+
2
x
+
4
y
3
/
2
a
Compute
all
possible
unit
vectors
u
such
that
D
u
h
(2
,
3)
=
6
b
What
angle
do
these
vectors
u
make
with
the
tangent
line
to
the
level
curve
h
(
x,
y
)
=
8
+
12
√
3
at
(2
,
3)
.
Q38
Let
f
(
x,
y
)
=
x
4
y
+
3
x
−
y
3
.
a
Give
an
equation
of
the
level
curve
of
f
through
the
point
(
−
1
,
2)
.
b
Give
an
equation
of
the
tangent
line
to
the
level
curve
of
f
at
(
−
1
,
2)
.
W
rite
y
our
equation
in
no
rmal
fo
rm.
357
Section
5.4
Exercises
c
Give
an
exp
ression
for
the
linearization
of
f
at
(
−
1
,
2)
.
358
Section
5.5
The
Chain
Rule
Goals:
1
Use
the
chain
rule
to
compute
derivatives
of
comp
ositions
of
functions.
2
P
erfo
rm
implicit
differentiation
using
the
chain
rule.
Motivational
Example
Supp
ose
Jinteki
Corporation
makes
widgets
which
is
sells
for
$100
each.
It
commands
a
small
enough
p
o
rtion
of
the
ma
rk
et
that
its
p
roduction
level
do
es
not
affect
the
demand
(p
rice)
for
its
p
roducts.
If
W
is
the
numb
er
of
widgets
produced
and
C
is
their
operating
cost,
Jinteki’s
profit
is
mo
deled
by
P
=
100
W
−
C
The
pa
rtial
derivative
∂
P
∂
W
=
100
do
es
not
correctly
calculate
the
effect
of
increasing
p
ro
duction
on
p
rofit.
Ho
w
can
we
calculate
this
correctly?
Question
5.5.1
Ho
w
Can
W
e
Visualize
a
Comp
osition
with
a
Multivariable
Function?
Y
ou
ma
y
recall
parametric
equations
from
high
school
algebra.
A
parametric
equation
actually
consists
of
tw
o
o
r
mo
re
equations.
Each
expresses
a
va
riable
in
our
co
o
rdinate
system
in
terms
of
a
pa
rameter
t
.
W
e
can
visualize
a
parametric
equation
as
particle
traveling
through
space.
The
va
riable
t
rep
resents
time.
x
(
t
)
and
y
(
t
)
rep
resent
the
coordinates
of
the
p
osition
at
time
t
.
The
vecto
r
⟨
x
′
(
t
)
,
y
′
(
t
)
⟩
rep
resents
velo
cit
y
.
It
p
oints
in
the
direction
of
travel.
Click to Load Applet
Figure:
A
particle
whose
p
osition
is
defined
by
x
(
t
)
and
y
(
t
)
,
the
path
it
follows
and
its
velo
city
vector
359
Question
5.5.1
Ho
w
Can
W
e
Visualize
a
Comp
osition
with
a
Multivariable
Function?
Given
a
function
f
(
x,
y
)
where
x
=
x
(
t
)
and
y
=
y
(
t
)
,
we
can
ask
how
f
changes
as
t
changes.
W
e
can
visualize
this
change
by
drawing
the
graph
z
=
f
(
x,
y
)
over
the
path
given
by
the
parametric
equations
x
(
t
)
and
y
(
t
)
.
Click to Load Applet
Figure:
The
comp
osition
f
(
x
(
t
)
,
y
(
t
))
,
represented
by
the
height
of
z
=
f
(
x,
y
)
over
the
path
(
x
(
t
)
,
y
(
t
))
Question
5.5.2
Ho
w
Do
W
e
Compute
the
Derivative
of
a
Comp
osition
of
Functions?
Theo
rem
[The
Chain
Rule]
Consider
a
differentiable
function
f
(
x,
y
)
.
If
we
define
x
=
x
(
t
)
and
y
=
y
(
t
)
,
b
oth
differential
functions,
w
e
have
d
f
dt
=
∂
f
∂
x
dx
dt
+
∂
f
∂
y
dy
dt
o
r
d
f
dt
=
∇
f
(
x,
y
)
·
⟨
x
′
(
t
)
,
y
′
(
t
)
⟩
360
Rema
rks
f
(
x
(
t
)
,
y
(
t
))
is
a
function
(only)
of
t
.
Because
of
this,
d
f
dt
is
an
ordina
ry
derivative,
not
a
pa
rtial
derivative.
d
f
dt
is
not
the
slop
e
of
the
comp
osition
graph.
slop
e
=
rise
in
z
run
in
xy
-plane
d
f
dt
=
rise
in
z
change
in
t
The
chain
rule
is
easy
to
rememb
er
b
ecause
of
its
similarit
y
to
the
differential:
dz
=
∂
z
∂
x
dx
+
∂
z
∂
y
dy
.
The
p
roof
is
more
complicated
than
just
sticking
a
dt
under
each
term.
Example
5.5.3
Using
the
Chain
Rule
If
P
=
R
−
C
and
we
have
R
=
100
w
and
C
=
3000
+
70
w
−
0
.
1
w
2
,
calculate
dP
dw
.
Solution
The
chain
rule
says
dP
dw
=
∂
P
∂
R
dR
dw
+
∂
P
∂
C
dC
dw
W
e
compute
the
required
partial
derivatives:
∂
P
∂
R
=
1
∂
P
∂
C
=
−
1
dR
dw
=
100
dC
dw
=
70
−
0
.
2
w
W
e
plug
these
into
the
formula
to
get
dP
dw
=
(1)(100)
+
(
−
1)(70
−
0
.
2
w
)
=
30
+
0
.
2
w
361
Example
5.5.3
Using
the
Chain
Rule
Rema
rk
Notice
we
don’t
need
the
chain
rule
when
w
e
have
expressions
for
each
function.
We
can
write
the
comp
osition
ourselves
and
tak
e
an
o
rdina
ry
derivative.
In
this
example
we
could
just
differentiate
P
=
100
w
−
(3000
+
70
w
−
0
.
1
w
2
)
.
Question
5.5.4
What
If
W
e
Have
More
Va
riables?
The
chain
rule
wo
rks
just
as
well
if
x
and
y
a
re
functions
of
more
than
one
variable.
In
this
case
it
computes
pa
rtial
derivatives.
Theo
rem
If
f
(
x,
y
)
,
x
(
s,
t
)
and
y
(
s,
t
)
,
are
all
differentiable,
then
∂
f
∂
s
=
∂
z
∂
x
∂
x
∂
s
+
∂
z
∂
y
∂
y
∂
s
o
r
∂
f
∂
s
=
∇
f
(
x,
y
)
·
∂
x
∂
s
,
∂
y
∂
s
W
e
can
also
mo
dify
it
for
functions
of
more
than
tw
o
variables.
Theo
rem
Given
f
(
x,
y
,
z
)
,
x
(
t
)
,
y
(
t
)
and
z
(
t
)
,
all
differentiable,
we
have
d
f
dt
=
∂
f
∂
x
dx
dt
+
∂
f
∂
y
dy
dt
+
∂
f
∂
z
dz
dt
o
r
d
f
dt
=
∇
f
(
x,
y
,
z
)
·
⟨
x
′
(
t
)
,
y
′
(
t
)
,
z
′
(
t
)
⟩
362
Example
5.5.5
A
Composition
with
More
Va
riables
Recall
that
for
an
ideal
gas
P
(
n,
T
,
V
)
=
nRT
V
.
R
is
a
constant.
n
is
the
number
of
molecules
of
gas.
T
is
the
temp
erature
in
Celsius.
V
is
the
volume
in
meters.
Supp
ose
w
e
want
to
understand
the
rate
at
which
the
pressure
changes
as
an
air-tight
glass
container
of
gas
is
heated.
a
Apply
the
chain
rule
to
get
an
expression
for
dP
dT
.
b
What
is
dn
dT
?
c
What
is
dT
dT
?
d
Supp
ose
that
dV
dT
=
(5
.
9
×
10
−
6
)
V
.
Calculate
and
simplify
the
expression
you
got
for
dP
dT
.
Solution
a
dP
dT
=
∂
P
∂
T
dT
dT
+
∂
P
∂
n
dn
dT
+
∂
P
∂
V
dV
dT
b
The
container
is
sealed
so
no
molecules
are
getting
in
or
out.
dn
dT
=
0
.
c
If
w
e
write
T
as
a
function
of
T
,
we
get
T
=
T
.
dT
dT
=
1
.
d
W
e’ll
compute
the
partial
derivatives
and
then
plug
them
into
our
chain
rule
expression.
∂
P
∂
T
=
nR
V
∂
P
∂
V
=
−
nRT
V
2
dP
dT
=
nR
V
(1)
+
0
−
nRT
V
2
(5
.
9)(10
−
6
)
V
=
nR
(1
−
0
.
0000059
T
)
V
363
Example
5.5.6
A
Composition
with
Limited
Information
Supp
ose
g
(
p,
q
,
r
)
=
r
e
p
2
q
.
Given
that
p,
q
,
r
are
all
differentiable
functions
of
x
with
the
values
in
the
follo
wing
table,
compute
dg
dx
when
x
=
2
.
x
0
1
2
3
p
(
x
)
3
1
5
10
p
′
(
x
)
−
3
2
3
4
q
(
x
)
6
2
−
2
3
q
′
(
x
)
−
1
−
5
2
3
r
(
x
)
10
11
7
3
r
′
(
x
)
1
0
−
1
−
3
Solution
The
chain
rule
says
dg
dx
=
∂
g
∂
p
dp
dx
+
∂
g
∂
q
dq
dx
+
∂
g
∂
r
dr
dx
W
e
require
the
partial
derivatives
of
g
∂
g
∂
p
=
2
pq
r
e
p
2
q
∂
g
∂
q
=
p
2
r
e
p
2
q
∂
g
∂
r
=
e
p
2
q
No
w
w
e
plug
in
the
partial
derivatives,
along
with
the
derivatives
of
p
,
q
and
r
from
the
table.
dg
dx
=
2
pq
r
e
p
2
q
(3)
+
p
2
r
e
p
2
q
(2)
+
e
p
2
q
(
−
1)
This
is
correct,
but
not
sufficiently
simplified.
We
have
left
p
’s,
q
’s
and
r
’s
in
the
exp
ression,
but
the
table
tells
us
what
value
these
have
when
x
=
2
.
We
can
make
these
subsitutions:
dg
dx
=
2(5)(
−
2)(7)
e
(5)
2
(
−
2)
(3)
+
(5)
2
(7)
e
(5)
2
(
−
2)
(2)
+
e
(5)
2
(
−
2)
(
−
1)
=
−
420
e
−
50
+
350
e
−
50
−
e
−
50
=
−
71
e
−
50
364
Application
5.5.7
Implicit
Differentiation
Recall
that
an
implicit
equation
on
n
va
riables
is
a
level
curve
of
a
n
-va
riable
function.
Consider
the
graph
x
3
+
y
2
−
4
xy
=
0
.
How
can
we
use
this
to
calculate
dy
dx
at
the
point
(3
,
3)
?
Solution
First,
note
that
(3
,
3)
do
es
lie
on
the
graph.
When
we
plug
x
=
3
and
y
=
3
into
our
equation,
we
get
27
+
9
−
36
=
0
,
which
is
true.
Now
supp
ose
that
for
every
x
near
3
,
we
can
define
y
(
x
)
to
b
e
the
y
co
o
rdinate
on
the
graph
x
3
+
y
2
−
4
xy
=
0
.
Define
F
(
x,
y
)
=
x
3
+
y
2
−
4
xy
.
The
p
oints
(
x,
y
(
x
))
lie
on
the
graph
F
(
x,
y
)
=
0
.
We
can
use
this
equation
to
obtain
an
expression
for
dy
dx
.
When
we
differentiate
F
(
x,
y
(
x
))
,
b
oth
comp
onents
change
as
x
changes,
so
w
e
cannot
use
a
partial
derivative.
We
need
the
chain
rule.
F
(
x,
y
(
x
))
=
0
d
dx
F
(
x,
y
(
x
))
=
d
dx
0
differentiate
both
sides
∂
F
∂
x
dx
dx
+
∂
F
∂
y
dy
dx
=
0
apply
chain
rule
∂
F
∂
x
+
∂
F
∂
y
dy
dx
=
0
dx
dx
=
1
∂
F
∂
y
dy
dx
=
−
∂
F
∂
x
solve
fo
r
dy
dx
dy
dx
=
−
∂
F
∂
x
∂
F
∂
y
W
e
compute
the
partial
derivatives
at
(3
,
3)
,
then
plug
them
into
the
fo
rmula
w
e
derived.
F
x
(
x,
y
)
=
3
x
2
−
4
y
F
x
(3
,
3)
=
15
F
y
(
x,
y
)
=
2
y
−
4
x
F
y
(3
,
3)
=
−
6
dy
dx
=
−
15
−
6
=
5
2
Figure:
The
graph
of
F
(
x,
y
)
=
x
3
+
y
2
−
4
xy
=
0
,
its
tangent
line
at
(3
,
3)
,
and
the
gradient
of
F
365
Application
5.5.7
Implicit
Differentiation
Main
Ideas
dy
dx
is
the
slope
of
the
tangent
line
to
F
(
x,
y
)
=
c
.
The
chain
rule
allows
us
to
derive
dy
dx
=
−
F
x
F
y
−
F
x
F
y
is
the
negative
reciprocal
of
F
y
F
x
,
which
is
the
slop
e
of
∇
F
.
In
o
rder
to
solve
fo
r
dy
dx
w
e
had
to
assume
that
y
w
as
a
differentiable
function
of
x
.
Ho
w
do
w
e
kno
w
that’s
even
true?
There
is
an
advanced
and
p
o
w
erful
theo
rem
that
tells
us
when
we
can
write
one
va
riable
in
an
implicit
equation
as
a
function
of
the
others.
Here
is
the
tw
o-va
riable
version.
Theo
rem
[The
Implicit
Function
Theo
rem]
Supp
ose
w
e
have
a
p
oint
(
x
0
,
y
0
)
on
the
graph
of
F
(
x,
y
)
=
c
.
Supp
ose
that
1
The
pa
rtial
derivatives
of
F
exist
and
are
continuous
at
(
x
0
,
y
0
)
2
F
y
(
x
0
,
y
0
)
=
0
Then
there
is
a
function
y
=
f
(
x
)
that
agrees
with
the
graph
of
F
(
x,
y
)
=
c
in
some
neighb
o
rho
o
d
a
round
(
x
0
,
y
0
)
.
F
urthermo
re
1
f
is
continuous
2
f
is
differentiable
3
f
′
(
x
0
)
=
−
F
x
(
x
0
,
y
0
)
F
y
(
x
0
,
y
0
)
In
the
case
of
our
example,
the
pa
rtial
derivatives
in
question
a
re
p
olynomials.
As
long
as
F
y
(
x
0
,
y
0
)
=
0
,
w
e
a
re
gua
ranteed
that
our
graph
has
a
tangent
line
at
(
x
0
,
y
0
)
,
and
its
slop
e
is
−
F
x
(
x
0
,
y
0
)
F
y
(
x
0
,
y
0
)
.
Application
5.5.8
Indirect
Profit
F
unctions
Supp
ose
a
firm
cho
oses
how
much
quantity
q
to
p
ro
duce,
but
their
profit
Π(
q
,
α
)
dep
ends
on
some
pa
rameter
α
outside
their
control
(maybe
a
tax
o
r
a
measure
of
regulatory
burden).
The
firm,
once
it
knows
the
value
of
α
,
will
cho
ose
the
q
that
maximizes
p
rofit.
How
will
their
p
rofit
change
as
α
changes?
366
Solution
The
change
in
the
firms
profit
is
d
Π
dα
.
Since
q
is
also
a
function
of
α
we
will
need
the
chain
rule.
d
Π
dα
=
∂
Π
∂
q
dq
dα
+
∂
Π
∂
α
dα
dα
W
e
can
substitute
dα
dα
=
1
.
W
e
can
also
a
rgue
that
∂
Π
∂
q
=
0
.
Why?
Because
q
is
the
choice
that
maximizes
p
rofit,
and
maximums
o
ccur
at
critical
p
oints.
If
∂
Π
∂
q
>
0
then
the
firm
could
increase
q
to
increase
p
rofit
(without
changing
α
,
which
it
has
no
control
over).
Simila
rly
,
If
∂
Π
∂
q
<
0
then
reducing
p
ro
duction
w
ould
increase
profit.
P
erfo
rming
these
substitutions
give
s:
d
Π
dα
=
∂
Π
∂
α
This
suggests
that
in
this
case,
the
total
derivative
is
equal
to
the
pa
rtial
derivative.
W
e
can
verify
this
equality
graphically
as
well.
Pick
a
particula
r
α
0
and
let
q
0
=
q
(
α
0
)
.
Notice:
The
graph
π
(
q
0
,
α
)
is
never
ab
ove
π
(
q
(
α
)
,
α
)
for
any
α
,
since
q
(
α
)
is
the
optimal
choice
of
q
.
The
graphs
π
(
q
0
,
α
)
and
π
(
q
(
α
)
,
α
)
meet
at
α
0
,
since
q
0
=
q
(
α
0
)
.
If
tw
o
graphs
meet
but
one
stays
b
elow
the
other,
they
are
tangent.
They
have
the
same
tangent
line
and
thus
the
same
derivative.
Click to Load Applet
Figure:
Tw
o
graphs
of
z
=
Π(
q
,
α
)
,
one
where
q
changes
to
be
the
optimal
choice
for
each
α
and
one
where
q
is
fixed
at
q
0
,
the
optimal
choice
for
α
0
367
Application
5.5.8
Indirect
Profit
F
unctions
Rema
rk
If
we
had
an
exp
ression
for
q
(
α
)
and
an
expression
fo
r
Π
,
w
e
could
substitute
and
use
o
rdina
ry
differen-
tiation.
Since
we
did
not,
w
e
needed
the
chain
rule.
Even
with
such
an
expression,
to
find
d
Π
dα
directly
w
e
w
ould
need
to
1
Solve
fo
r
q
as
a
function
of
α
2
Substitute
q
(
α
)
into
Π(
q
,
α
)
3
Differentiate
the
result
T
aking
a
pa
rtial
derivative
is
less
w
o
rk.
Our
result
(which
economists
call
the
envelop
e
theo
rem
)
is
b
oth
a
useful
abstraction
and
a
computational
shortcut.
Section
5.5
Exercises
Summa
ry
Questions
Q1
Ho
w
can
w
e
visualize
f
(
x,
y
)
,
when
x
and
y
a
re
functions
of
t
?
Q2
Explain
why
d
f
dt
cannot
be
interpreted
as
a
slop
e
of
f
over
the
xy
-plane.
Q3
What
is
the
difference
b
etw
een
dz
dx
and
∂
z
∂
x
?
Ho
w
is
the
first
one
computed?
Q4
Ho
w
do
y
ou
use
the
chain
rule
to
differentiate
implicit
functions?
5.5.1
Q5
Plug
in
a
few
different
t
values
and
plot
the
corresponding
p
oints
of
x
(
t
)
=
3
+
5
t
y
(
t
)
=
−
2
+
4
t
What
is
the
resulting
curve?
What
is
the
significance
of
the
t
coefficients?
368
Q6
Consider
the
curve
defined
by
x
(
t
)
=
t
y
(
t
)
=
e
t
a
Plot
a
few
p
oints
on
the
curve
by
plugging
in
different
values
of
t
.
b
In
general,
what
curve
do
es
x
(
t
)
=
t
y
(
t
)
=
f
(
t
)
seem
to
p
roduce?
Q7
A
pa
rticle
is
travelling
according
to
the
parametric
equations
x
(
t
)
=
2
cos
t
y
(
t
)
=
3
sin
t
What
is
the
sp
eed
(magnitude
of
velo
city)
at
t
=
π
3
?
Q8
Pro
duce
a
tangent
vector
to
the
curve
defined
by
x
(
t
)
=
t
3
y
(
t
)
=
t
2
at
the
point
(
−
27
,
9)
.
Q9
Is
the
graph
of
x
(
t
)
=
t
2
y
(
t
)
=
sin(
t
)
the
graph
of
a
function?
How
can
you
tell
without
graphing
it?
Q10
Ho
w
are
the
graphs
of
the
follo
wing
tw
o
pa
rametric
equations
related?
Can
you
generalize
your
answ
er
to
simila
r
pairs
of
parametric
equations?
x
(
t
)
=
cos
t
x
(
t
)
=
cos(
t
3
)
y
(
t
)
=
ln
t
y
(
t
)
=
ln(
t
3
)
369
Section
5.5
Exercises
5.5.2
Q11
Let
f
(
x,
y
)
b
e
a
funtion.
Under
what
conditions
is
d
f
dt
equal
to
the
directional
derivative
of
f
in
the
direction
of
the
tangent
vector
⟨
x
′
(
t
)
,
y
′
(
t
)
⟩
?
Q12
Liam
sa
ys
“If
f
is
a
function
of
x
and
y
and
x
and
y
a
re
increasing,
then
f
is
increasing.”
We
all
kno
w
Liam
is
incorrect.
Ho
w
could
we
use
the
chain
rule
to
refute
him?
5.5.3
Q13
The
angula
r
sp
eed
of
an
object
is
given
by
ω
=
v
r
where
r
is
the
distance
from
the
center
of
rotation
and
v
is
the
linear
sp
eed.
Supp
ose
an
object
is
orbiting
ea
rth
at
a
radius
of
8400000
m
and
a
sp
eed
of
6900
m/s
.
If
the
radius
is
increasing
at
a
rate
of
100
m/s
and
the
linear
sp
eed
is
decreasing
b
y
60
m/s
2
,
ho
w
quickly
is
the
angular
sp
eed
changing?
Q14
Let
x
=
t
2
and
y
=
sin
t
.
Let
f
(
x,
y
)
=
xy
.
a
Compute
d
f
dt
using
the
multiva
riable
chain
rule.
b
Compute
d
f
dt
b
y
substituting
and
using
single-variable
differentiation.
c
What
earlier
rule
of
differentiation
can
we
recover
by
applying
the
chain
rule
to
f
(
x,
y
)
=
xy
?
5.5.4
Q15
Supp
ose
h
(
x
1
,
x
2
,
x
3
,
x
4
)
is
a
four-variable
function
and
each
x
i
(
x,
t
)
is
a
function
of
parameters
s
and
t
.
Ho
w
would
the
multivariable
chain
rule
compute
∂
h
∂
t
?
Q16
Supp
ose
k
(
x
)
is
a
function
and
x
(
r
,
s,
t
)
is
a
function
of
pa
ramters
r
,
s
,
and
t
.
How
do
es
the
multiva
riable
chain
rule
say
w
e
should
compute
∂
k
∂
r
?
370
5.5.5
Q17
Agula
r
momemtum
is
given
by
L
=
rmv
where
r
is
the
radius
of
roatation,
m
is
the
mass
of
the
object,
and
v
is
its
linear
sp
eed.
At
a
certain
time
t
0
,
r
is
42
million
meters
and
increasing
at
80
,
000
meters
p
er
second,
m
is
6000
k
g
and
not
changing,
and
v
is
3100
m/s
and
increasing
at
20
m/s
2
.
Ho
w
quickly
is
angular
momentum
increasing?
Q18
Let
f
(
x,
y
)
=
x
2
−
y
2
.
If
x
(
r,
θ
)
=
r
cos
θ
and
y
(
r,
θ
)
=
r
sin
θ
,
compute
∂
f
∂
θ
at
(
r
,
θ
)
=
4
,
π
6
.
5.5.6
Q19
Supp
ose
x
(
t
)
and
y
(
t
)
are
differentiable
functions
of
t
such
that
x
(2)
=
3
x
′
(2)
=
2
y
(2)
=
−
5
y
′
(2)
=
10
If
f
(
x,
y
)
=
y
e
(
x
2
y
)
,
sho
w
ho
w
to
compute
d
f
dt
at
t
=
2
.
Q20
Supp
ose
that
x
and
y
a
re
functions
of
t
such
that
when
t
=
2
:
x
=
3
y
=
1
dx
dt
=
5
dy
dt
=
2
If
g
(
x,
y
)
=
3
xy
2
−
x
2
+
2
y
,
compute
dg
dt
t
=2
.
5.5.7
Q21
Compute
dy
dx
at
(4
,
2)
,
if
x
and
y
satisfy
y
3
−
xy
+
x
2
−
4
=
0
Q22
Compute
dy
dx
at
(3
,
0)
,
if
x
and
y
satisfy
xe
xy
=
3
Q23
What
is
the
slop
e
of
the
tangent
line
to
x
−
y
2
=
9
at
(18
,
−
3)
?
Q24
Compute
the
slope
of
the
tangent
line
to
x
3
=
y
2
at
(4
,
−
8)
.
371
Section
5.5
Exercises
Q25
Angula
r
momentum
is
given
by
L
=
rmv
.
One
law
of
physics
states
that
angula
r
momentum
of
an
object
is
conversed
(unchanged)
unless
the
a
force
(b
esides
gravity)
acts
to
sp
eed
up
or
slow
do
wn
the
object.
Use
the
chain
rule
to
derive
an
exp
ression
for
dv
dr
,
the
amount
of
linear
sp
eed
an
object
gains
or
loses
p
er
unit
that
its
radius
of
rotation
increases.
What
do
you
notice
about
the
role
of
mass
in
your
answer?
Q26
Another
p
rinciple
in
physics
is
the
conservation
of
energy
.
Kenetic
energy
is
given
by
E
=
1
2
mv
2
,
where
m
is
the
mass
and
v
is
the
linear
sp
eed
of
the
object.
Supp
ose
that
we
have
a
rock
drifiting
through
space.
Supp
ose
it
impacts
stationary
ro
cks
and
the
combined
mass
sticks
together
(without
releasing
any
energy
as
heat,
light
o
r
sound).
Thus
the
mass
of
the
total
travelling
object
increases,
while
the
total
energy
sta
ys
the
same.
Derive
an
exp
ression
for
ho
w
sp
eed
changes
per
unit
of
increase
in
mass.
5.5.8
Q27
Supp
ose
that
x
is
a
function
of
t
and
that
when
t
=
9
,
we
have
x
=
7
and
dx
dt
=
−
3
.
Define
f
(
x,
t
)
=
√
x
+
t
.
a
Compute
the
pa
rtial
derivate
∂
f
∂
t
(7
,
9)
.
b
Compute
the
total
derivative
d
f
dt
(7
,
9)
.
c
In
a
few
sentences,
explain
what
these
tw
o
quantities
compute
and
why
they
are
different
from
each
other.
Q28
A
firm
with
a
monop
oly
produces
gets
to
set
the
price
of
its
products
and
decide
ho
w
much
to
p
ro
duce.
There
is
a
demand
function
p
such
that
if
the
firm
produces
q
units,
it
must
set
its
p
rice
at
p
(
q
)
to
get
consumer
to
buy
all
of
its
production.
Each
unit
costs
c
to
p
roduce.
The
p
rofit
function
of
the
firm
is
π
(
q
,
c
)
=
p
(
q
)
q
−
cq
W
e
can
assume
that
once
the
firm
has
wo
rked
out
what
c
is,
it
chooses
the
q
to
maximize
profit.
Ho
w
much
will
the
firm’s
actual
profit
change
p
er
unit
of
increase
in
c
?
372
Synthesis
and
Extension
Q29
Find
the
slop
e
of
the
tangent
line
to
x
2
+
2
x
−
y
2
=
8
at
(5
,
−
3)
using
each
of
the
follo
wing
t
w
o
metho
ds.
a
Using
a
gradient
vector
to
write
the
normal
equation
of
the
line
and
solving
for
the
slop
e.
b
Using
implicit
differentiation.
Q30
Supp
ose
the
position
of
a
particle
at
time
t
is
given
by
x
(
t
)
=
t
2
y
(
t
)
=
3
−
t
z
(
t
)
=
√
t
A
t
t
=
4
,
how
quickly
is
pa
rticle
travelling
a
w
a
y
from
the
plane
x
+
2
y
−
2
z
=
10
?
Q31
Here
is
a
diagram
of
the
level
curves
of
h
(
x,
y
)
for
certain
values
of
c
.
a
Is
h
y
(2
,
1)
p
ositive
or
negative?
Explain
in
a
sentence
o
r
t
w
o.
b
Add
a
vecto
r
to
the
diagram
that
indicates
the
direction
of
greatest
increase
of
h
at
(
−
2
,
0)
.
c
Supp
ose
x
=
4
−
5
t
and
y
=
3
t
2
.
Determine,
with
the
aid
of
a
relevant
calculation,
whether
dh
dt
is
positive
or
negative
at
t
=
1
.
Q32
Let
f
(
x,
y
)
=
x
5
+
20
xy
+
5
y
2
.
a
Give
an
equation
of
the
level
curve
of
f
through
the
point
(1
,
−
1)
.
373
Section
5.5
Exercises
b
Give
an
equation
of
the
tangent
plane
to
z
=
f
(
x,
y
)
at
the
point
(1
,
−
1
,
−
14)
.
c
Use
the
differential
of
f
to
estimate
ho
w
much
the
z
value
of
z
=
f
(
x,
y
)
would
change
from
(1
,
−
1
,
−
14)
,
if
x
increased
b
y
3
and
y
decreased
by
1
.
If
y
ou
don’t
rememb
er
differential
notation,
y
ou
ma
y
use
another
notation
for
partial
credit.
374
Section
5.6
Maximum
and
Minimum
V
alues
Goals:
1
Find
critical
p
oints
of
a
function.
2
T
est
critical
p
oints
to
find
lo
cal
maximums
and
minimums.
3
Use
the
Extreme
Value
Theo
rem
to
find
the
global
maximum
and
global
minimum
of
a
function
over
a
closed
set.
F
unctions
can
b
e
used
to
model
a
va
riet
y
of
real-wo
rld
quantities.
A
company’s
profit,
a
disease’s
infection
rate,
o
r
the
impact
of
a
government
p
rogram.
In
these
cases,
the
most
pressing
question
is:
what
choice
of
indep
endent
va
riables
will
maximize
or
minimize
the
value
of
the
function?
Answering
this
question
w
as
one
of
the
headline
applications
of
single-variable
calculus.
In
this
section
w
e
will
generalize
those
methods
to
functions
of
multiple
variables.
Question
5.6.1
What
Are
Local
Extremes?
The
lo
cal
extremes
of
a
function
are
the
lo
cal
minimums
and
maximums.
Definition
Given
an
n
-va
riable
function
f
(
x
1
,
x
2
,
.
.
.
,
x
n
)
w
e
sa
y
that
a
p
oint
P
in
n
-space
is
1
a
lo
cal
maximum
if
f
(
P
)
≥
f
(
Q
)
fo
r
all
Q
in
some
neighb
orhoo
d
around
P
.
2
a
lo
cal
minimum
if
f
(
P
)
≤
f
(
Q
)
fo
r
all
Q
in
some
neighb
orhoo
d
around
P
.
Question
5.6.2
Where
Do
Local
Extremes
Lie?
A
t
a
lo
cal
maximum
(or
minimum)
D
u
f
cannot
be
p
ositive
(or
negative)
in
any
direction.
Thus
at
a
lo
cal
extreme,
∇
f
(
P
)
=
0
,
the
zero
vector.
In
other
wo
rds,
all
the
pa
rtial
derivatives
of
f
are
0
at
P
.
In
the
case
of
a
tw
o-va
riable
function,
we
can
visualize
this
condition.
If
f
x
(
P
)
=
0
,
then
we
could
travel
in
the
x
direction
to
increase
o
r
decrease
f
.
If
f
x
(
P
)
=
0
,
then
we
could
travel
in
the
y
direction
to
increase
o
r
decrease
f
.
Thus
at
a
local
maximum
o
r
lo
cal
minimum,
the
tangent
plane
must
b
e
375
Question
5.6.2
Where
Do
Local
Extremes
Lie?
ho
rizontal.
Click to Load Applet
Figure:
T
angent
lines
must
have
slop
e
0
at
a
local
max.
This
a
rgument
w
o
rks
anywhere
that
∇
f
exists.
That
motivates
the
following
definition:
Definition
W
e
sa
y
P
is
a
critical
p
oint
of
f
if
either
1
∇
f
(
P
)
=
0
o
r
2
∇
f
(
P
)
do
es
not
exist
(b
ecause
one
of
the
pa
rtial
derivatives
does
not
exist).
Theo
rem
The
local
maximums
and
minimums
of
a
function
can
only
o
ccur
at
critical
points.
Example
5.6.3
Finding
Critical
P
oints
The
function
z
=
2
x
2
+
4
x
+
y
2
−
6
y
+
13
has
a
minimum
value.
Find
it.
376
Solution
W
e
kno
w
the
minimum
value
exists,
so
it
must
lie
at
a
critical
p
oint.
W
e
compute
∇
f
(
x,
y
)
=
⟨
4
x
+
4
,
2
y
−
6
⟩
One
type
of
critical
p
oint
is
where
this
is
undefined,
but
no
value
of
(
x,
y
)
mak
es
these
exp
ressions
undefined.
The
other
t
yp
e
of
critical
p
oint
o
ccurs
when
these
comp
onents
are
0
.
We
can
solve
that
system
of
equations.
4
x
+
4
=
0
2
y
−
6
=
0
x
=
−
1
y
=
3
The
only
p
oint
that
satisfies
this
requirement
is
(
−
1
,
3)
.
Since
there
is
only
one
critical
p
oint,
and
the
p
romised
minimum
lies
at
a
critical
p
oint,
(
−
1
,
3)
must
b
e
that
point.
The
minimum
value
is
z
=
(2)(
−
1)
2
+
(4)(
−
1)
+
3
2
−
(6)(3)
+
13
=
2
¿
Question
5.6.4
Ho
w
Do
W
e
Identify
Two-V
a
riable
Local
Maximums
and
Minimums?
Once
we
have
found
a
critical
p
oint,
how
do
we
kno
w
whether
it
is
a
lo
cal
minimum,
a
lo
cal
maximum
o
r
neither?
Consider
a
function
f
(
x,
y
)
and
a
critical
point
P
.
There
a
re
tw
o
p
ossibilities
for
∇
f
(
P
)
.
In
the
case
that
∇
f
(
P
)
do
es
not
exist,
calculus
can
b
e
no
further
use
to
us.
If
∇
f
(
P
)
=
⟨
0
,
0
⟩
,
there
are
a
few
different
shap
es
the
graph
could
take.
Since
we
a
re
wo
rking
with
tw
o-va
riables,
we
can
visualize
these
shapes.
A
critical
point
could
b
e
a
lo
cal
maximum.
In
this
case
f
curves
downw
a
rd
in
every
direction.
Click to Load Applet
Figure:
A
lo
cal
maximum
at
(0
,
0)
377
Question
5.6.4
Ho
w
Do
W
e
Identify
Two-V
a
riable
Local
Maximums
and
Minimums?
A
critical
point
could
b
e
a
lo
cal
minimum.
In
this
case
f
curves
upw
a
rd
in
every
direction.
Click to Load Applet
Figure:
A
lo
cal
minimum
at
(0
,
0)
A
critical
point
could
b
e
neither.
f
curves
up
w
a
rd
in
some
directions
but
downw
a
rd
in
others.
This
configuration
is
called
a
saddle
p
oint
.
Click to Load Applet
Figure:
A
saddle
p
oint
at
(0
,
0)
Curvature
is
measured
by
the
second
derivatives.
This
matches
our
exp
erience
with
single-variable
critical
p
oints,
where
the
second
derivative
test
classifies
critical
p
oints
as
lo
cal
maximums
or
local
minimums.
W
e
have
a
similar
test
fo
r
tw
o-va
riable
functions,
though
the
computation
is
more
involved.
378
Theo
rem
[The
Second
Derivatives
T
est]
Supp
ose
f
is
differentiable
at
(
P
)
and
f
x
(
P
)
=
f
y
(
P
)
=
0
.
Then
w
e
can
compute
D
=
f
xx
(
P
)
f
y
y
(
P
)
−
[
f
xy
(
P
)]
2
1
If
D
>
0
and
f
xx
(
P
)
>
0
then
P
is
a
lo
cal
minimum.
2
If
D
>
0
and
f
xx
(
P
)
<
0
then
P
is
a
lo
cal
maximum.
3
If
D
<
0
then
P
is
a
saddle
p
oint.
Unfo
rtunately
,
if
D
=
0
,
this
test
gives
no
information.
Definition
The
quantity
D
in
the
second
derivatives
test
is
actually
the
determinant
of
a
matrix
called
the
Hessian
of
f
.
f
xx
(
P
)
f
y
y
(
P
)
−
[
f
xy
(
P
)]
2
=
det
f
xx
(
P
)
f
xy
(
P
)
f
y
x
(
P
)
f
y
y
(
P
)
|
{z
}
H
f
(
P
)
H
f
follo
ws
a
logical
pattern
and
can
b
e
a
useful
mnemonic
fo
r
the
second
derivatives
test.
Example
5.6.5
Classifying
a
Critical
Point
Let
f
(
x,
y
)
=
cos(2
x
+
y
)
+
xy
a
V
erify
that
∇
f
(0
,
0)
=
⟨
0
,
0
⟩
.
b
Is
(0
,
0)
a
lo
cal
minimum,
a
lo
cal
maximum,
o
r
neither?
379
Example
5.6.5
Classifying
a
Critical
Point
Solution
a
f
x
(
x,
y
)
=
−
sin(2
x
+
y
)(2)
+
y
(
chain
rule
)
f
x
(0
,
0)
=
−
sin((2)(0)
+
0)(2)
+
0
=
0
f
y
(
x,
y
)
=
−
sin(2
x
+
y
)(1)
+
x
(
chain
rule
)
f
y
(0
,
0)
=
−
sin((2)(0)
+
0)(1)
+
0
=
0
∇
f
(0
,
0)
=
⟨
0
,
0
⟩
b
F
o
r
the
second
derivatives
test,
we
need
to
compute
f
xx
,
f
xy
and
f
y
y
at
(0
,
0)
.
f
xx
(
x,
y
)
=
−
2
cos(2
x
+
y
)(2)
(
chain
rule
)
f
xx
(0
,
0)
=
−
2
cos((2)(0)
+
(0))(2)
=
−
4
f
xy
(
x,
y
)
=
−
2
cos(2
x
+
y
)(1)
+
1
(
chain
rule
)
f
xy
(0
,
0)
=
−
2
cos((2)(0)
+
(0))(1)
+
1
=
−
1
f
y
y
(
x,
y
)
=
−
cos(2
x
+
y
)(1)
(
chain
rule
)
f
y
y
(0
,
0)
=
−
cos((2)(0)
+
(0))(1)
=
−
1
D
=
f
xx
(0
,
0)
f
y
y
(0
,
0)
−
[
f
xy
(0
,
0)]
2
=
(
−
4)(
−
1)
−
(
−
1)
2
=
3
Since
D
>
0
and
f
xx
<
0
,
(0
,
0)
is
a
lo
cal
maximum
of
f
.
Click to Load Applet
Figure:
The
graph
z
=
cos(2
x
+
y
)
+
xy
with
a
lo
cal
maximum
at
(0
,
0)
380
Rema
rk
Why
do
es
the
final
determination
b
etw
een
maximum
and
minimum
rely
on
f
xx
(
P
)
instead
of
f
y
y
(
P
)
?
Actually
it
do
esn’t
matter
which
we
test.
In
order
fo
r
D
to
b
e
p
ositive,
f
xx
(
P
)
and
f
y
y
(
P
)
must
have
the
same
sign.
Question
5.6.6
Ho
w
Do
W
e
Find
Global
Extremes?
The
second
derivatives
test
can
categorize
lo
cal
extremes,
but
what
ab
out
a
global
extreme?
Definition
Given
an
n
-va
riable
function
f
(
x
1
,
x
2
,
.
.
.
,
x
n
)
w
e
sa
y
that
a
p
oint
P
in
n
-space
is
1
a
lo
cal
maximum
if
f
(
P
)
≥
f
(
Q
)
fo
r
all
Q
in
the
domain
of
f
.
2
a
lo
cal
minimum
if
f
(
P
)
≤
f
(
Q
)
fo
r
all
Q
in
the
domain
of
f
.
In
a
real-wo
rld
application,
we
are
much
more
interested
in
finding
global
extremes
than
local
ones.
Many
abstract
functions
do
not
even
have
global
extremes.
y
=
e
x
has
no
global
maximum.
It
increases
without
b
ound.
y
=
1
x
2
has
no
global
minimum.
It
approaches
0
but
never
reaches
it.
The
following
theo
rem
gua
rantees
that
ce
rtain
functions
will
have
global
extremes
for
us
to
try
to
find.
Theo
rem
[The
Extreme
Value
Theo
rem]
A
continuous
function
f
on
a
closed
and
bounded
domain
D
has
a
global
maximum
and
a
global
minimum
somewhere
in
D
.
Tw
o
of
the
wo
rds
in
this
theorem
have
not
b
een
defined
yet.
Here
are
their
definitions.
Definition
Let
D
b
e
a
subset
of
n
-space.
D
is
closed
if
it
contains
all
of
the
p
oints
on
its
b
ounda
ry
.
D
is
b
ounded
if
there
is
some
upp
er
limit
to
how
far
its
p
oints
get
from
the
o
rigin
(or
any
other
fixed
point).
If
there
are
p
oints
of
D
arbitra
rily
far
from
the
origin,
then
D
is
unb
ounded
.
381
Question
5.6.6
Ho
w
Do
W
e
Find
Global
Extremes?
F
o
r
one-variable
functions.
The
EVT
requires
that
the
domain
b
e
a
union
of
finite,
closed
intervals
(and
ma
ybe
finitely
many
isolated
p
oints).
Figure:
A
union
of
finite,
closed
intervals
In
2
-space,
w
e
can
get
a
better
sense
of
what
these
requirements
mean.
The
boundary
of
D
is
the
set
of
p
oints
from
which
y
ou
can
find
p
oints
in
D
and
p
oints
outside
D
arbitra
rily
close
by
.
The
b
ounda
ry
of
a
disc
is
a
circle.
If
the
disc
includes
the
circle,
it
is
closed.
If
it
do
es
not
include
the
circle,
it
is
not
closed.
Figure:
x
2
+
y
2
≤
9
is
closed.
Figure:
x
2
+
y
2
<
9
is
not
closed.
Containing
part
of
the
b
oundary
is
not
enough.
Any
missing
p
oint
means
that
D
is
not
closed.
Even
removing
an
isolated
p
oint
from
the
interior
of
D
is
a
problem.
That
p
oint
is
arbitra
rily
close
to
points
in
D
.
It
is
also
arbitra
rily
close
to
a
p
oint
outside
D
,
itself.
Thus
it
is
a
b
ounda
ry
p
oint
not
contained
in
D
,
and
D
is
not
closed.
382
Figure:
−
2
≤
x
≤
2
and
−
3
<
y
<
3
is
not
closed.
Figure:
−
2
≤
x
≤
2
and
−
3
≤
y
≤
3
and
(
x,
y
)
=
(1
,
2)
is
not
closed.
Bounded
regions
are
easier
to
understand.
If
we
can
enclose
the
region
in
a
sufficiently
large
circle,
it
is
bounded.
If
it
stretches
outside
any
circle
w
e
w
ould
draw
a
round
it,
then
it
is
unb
ounded.
Figure:
−
2
≤
x
≤
2
and
−
3
≤
y
≤
3
is
b
ounded.
Figure:
−
2
≤
x
≤
2
is
unb
ounded.
Example
5.6.7
Finding
a
Global
Maximum
Consider
the
function
f
(
x,
y
)
=
x
2
+
2
y
2
−
x
2
y
on
the
domain
D
=
{
(
x,
y
)
|
{z
}
points
in
R
2
:
x
2
+
y
2
≤
16
,
x
≤
0
|
{z
}
conditions
}
a
Do
es
f
have
a
maximum
value
on
D
?
Ho
w
do
we
know?
383
Example
5.6.7
Finding
a
Global
Maximum
b
Find
the
critical
p
oints
of
f
.
c
Must
one
of
the
critical
p
oints
b
e
the
maximum?
d
Find
the
maximum
of
f
.
Rema
rk
The
set
notation
{
t
yp
e
of
objects
in
the
set
:
conditions
that
thoise
objects
must
satisfy
}
is
used
throughout
mathematics,
b
ecause
it
is
so
flexible.
It
can
denote
sets
of
numb
ers,
p
oints,
functions,
vecto
rs
o
r
any
other
objects.
Solution
a
f
is
a
p
olynomial,
so
it
is
continuous.
D
is
a
semi-disc
that
includes
its
boundary
,
so
it
is
closed
and
bounded.
The
extreme
value
theorem
guarantees
that
f
has
a
global
maximum
on
D
.
b
W
e
begin
by
computing
the
gradient
of
f
.
f
x
(
x,
y
)
=
2
x
−
2
xy
f
y
(
x,
y
)
=
4
y
−
x
2
384
These
are
never
undefined,
so
there
are
no
critical
p
oints
of
that
t
ype.
The
only
critical
points
will
be
where
b
oth
partial
derivatives
are
0
.
0
=
2
x
−
2
xy
0
=
4
y
−
x
2
0
=
2
x
(1
−
y
)
(
facto
r
2
x
−
2
xy
)
x
=
0
o
r
y
=
1
0
=
4
y
−
0
2
0
=
4(1)
−
x
2
(
examine
each
case
sep
erately
)
0
=
y
x
=
±
2
W
e
should
be
careful
not
to
lose
track
of
the
logic.
The
x
=
±
2
solution
go
es
with
the
y
=
1
case.
The
y
=
0
solution
go
es
with
the
x
=
0
case.
Mixing
these
up
will
give
invalid
solutions.
Y
ou
can
alwa
ys
plug
in
pair
of
(
x,
y
)
to
verify
they
satisfy
the
system
of
equations.
W
e
conclude
that
(0
,
0)
,
(2
,
1)
and
(
−
2
,
1)
are
the
critical
points,
but
(2
,
1)
is
not
in
the
domain,
so
w
e
disca
rd
it.
c
No.
Recall
our
metho
d
for
maximizing
single
va
riable
functions
on
a
closed
interval.
The
maximum
can
occur
at
the
endp
oint
of
the
interval
without
b
eing
detected
b
y
the
derivative.
The
same
is
true
here.
If
the
maximum
is
on
the
b
oundary
of
D
,
the
gradient
need
not
b
e
0
.
In
the
single-va
riable
case,
w
e
only
need
to
test
the
endp
oints
(by
evaluating
f
there).
There
are
infinitely
many
p
oints
on
the
b
oundary
of
D
.
Evaluating
f
on
all
of
them
is
not
an
option.
With
graphing
softw
are
w
e
can
see
that
the
maximum
o
ccurs
on
the
b
ounda
ry
somewhere
in
the
third
quadrant,
but
ho
w
can
we
solve
for
it
exactly?
385
Example
5.6.7
Finding
a
Global
Maximum
Click to Load Applet
Figure:
The
graph
of
y
=
f
(
x,
y
)
over
the
domain
D
d
T
o
narro
w
do
wn
the
sea
rch
for
a
maximum
on
the
b
oundary
of
D
,
w
e
will
use
the
boundary
equations
to
write
an
expression
for
f
that
is
valid
only
on
the
b
oundary
.
W
e
can
find
the
critical
p
oints
of
this
expression,
and
rule
out
any
p
oint
that
is
not
a
critical
p
oint.
Supp
ose
the
maximum
lies
on
x
=
0
.
The
function
on
x
=
0
is
f
(0
,
y
)
=
0
2
+
2
y
2
−
0
2
y
=
2
y
2
.
This
function
only
has
one
variable,
so
w
e
can
find
p
otential
maximums
by
lo
oking
for
its
critical
points.
f
′
(
y
)
=
4
y
This
is
never
undefined.
It
is
0
at
y
=
0
.
The
only
critical
p
oint
of
f
(
y
)
on
x
=
0
is
(0
,
0)
.
Ho
w
ever,
not
all
of
x
=
0
is
the
boundary
of
D
.
This
comp
onent
of
the
b
oundary
ends
at
(0
,
4)
and
(0
,
−
4)
.
Like
with
a
closed
interval,
the
derivative
of
f
(
y
)
cannot
detect
a
maximum
at
those
endp
oints.
Supp
ose
the
maximum
lies
on
x
2
+
y
2
=
16
.
On
this
graph,
w
e
can
similarly
reduce
f
(
x,
y
)
to
a
function
of
one
variable,
but
the
substitution
is
more
complicated.
We
solve
x
2
+
y
2
=
16
x
2
=
16
−
y
2
f
(
y
)
=
(16
−
y
2
)
+
2
y
2
−
(16
−
y
2
)
y
(
substitute
fo
r
x
2
)
=
y
3
+
y
2
−
16
y
+
16
f
′
(
y
)
=
3
y
2
+
2
y
−
16
0
=
3
y
2
+
2
y
−
16
(
solve
for
critical
p
oints
)
0
=
(3
y
+
8)(
y
−
2)
y
=
−
8
3
y
=
2
x
2
+
−
8
3
2
=
16
x
2
+
2
2
=
16
(
substituue
into
x
2
+
y
2
=
16)
x
2
=
16
−
64
9
x
2
=
16
−
4
386
x
=
−
r
80
9
x
=
−
√
12
(+
solutions
a
re
not
in
D
)
Our
critical
p
oints
are
−
q
80
9
,
−
8
3
and
−
√
12
,
2
.
This
comp
onent
of
the
boundary
also
ends
at
(0
,
4)
and
(0
,
−
4)
,
so
the
maximum
m
ight
lie
there.
W
e
can
no
w
argue
that
one
of
the
p
oints
we
have
found
is
the
maximum.
If
the
maximum
is
not
on
the
b
oundary
,
it
lies
at
(
−
2
,
1)
.
If
the
maximum
is
on
x
=
0
,
then
it
lies
at
(0
,
0)
,
(0
,
4)
or
(0
,
−
4)
.
If
the
maximum
is
on
x
2
+
y
2
=
16
,
then
it
lies
at
−
q
80
9
,
−
8
3
,
−
√
12
,
2
,
(0
,
4)
o
r
(0
,
−
4)
.
One
of
these
must
b
e
the
case.
T
o
figure
out
which
it
is,
we
can
evaluate
f
at
each
p
oint
and
see
which
p
roduces
the
largest
value.
f
(
−
2
,
1)
=
(
−
2)
2
+
2(1)
2
−
(
−
2)
2
(1)
=
2
f
(0
,
0)
=
(0)
2
+
2(0)
2
−
(0)
2
(0)
=
0
f
(0
,
4)
=
(0)
2
+
2(4)
2
−
(0)
2
(4)
=
32
f
(0
,
−
4)
=
(0)
2
+
2(
−
4)
2
−
(0)
2
(
−
4)
=
32
f
−
q
80
9
,
−
8
3
=
−
q
80
9
2
+
2
−
8
3
2
−
−
q
80
9
2
−
8
3
=
1264
27
(maximum)
f
−
√
12
,
2
=
(
−
√
12)
2
+
2(2)
2
−
(
−
√
12)
2
(2)
=
−
4
Main
Ideas
If
the
Extreme
Value
Theo
rem
applies,
then
all
we
need
to
do
is
find
the
critical
points
and
evaluate
f
at
each.
One
is
guaranteed
to
b
e
the
maximum,
and
one
is
guaranteed
to
b
e
the
minimum.
∇
f
=
0
will
detect
critical
p
oints
on
the
interior,
but
not
on
the
b
oundary
.
W
e
can
rewrite
the
function
on
a
b
oundary
comp
onent
using
substitution.
Set
the
derivative
equal
to
0
to
find
critical
p
oints.
Derivatives
will
not
detect
maximums
at
the
endpoints
of
a
boundary
curve.
These
must
be
included
in
y
our
set
of
critical
p
oints.
387
Section
5.6
Exercises
Summa
ry
Questions
Q1
Where
must
the
lo
cal
maximums
and
minimums
of
a
function
o
ccur?
Why
do
es
this
make
sense?
Q2
What
does
the
second
derivatives
test
tell
us?
Q3
What
hypotheses
do
es
the
Extreme
Value
Theorem
require?
What
do
es
it
tell
us?
Q4
Assuming
a
maximum
and
minimum
exist,
where
must
y
ou
lo
ok
in
a
domain
to
b
e
sure
you
find
them?
5.6.1
Q5
Raina
claims
that
(0
,
0)
is
the
maximum
of
f
(
x,
y
)
=
x
2
−
y
2
−
10
xy
.
Disprove
her
claim
without
using
calculus.
Q6
Is
a
global
maximum
also
a
lo
cal
maximum?
Explain.
Q7
Supp
ose
g
(
x,
y
)
=
e
f
(
x,y
)
.
If
(
a,
b
)
is
a
lo
cal
minimum
of
f
(
x,
y
)
,
is
it
also
a
lo
cal
minimum
of
g
(
x,
y
)
?
Explain.
Q8
Do
es
a
constant
function
have
any
lo
cal
maximums?
Justify
your
answer
with
the
definition
of
lo
cal
maximum.
388
5.6.2
Q9
Supp
ose
∇
f
(4
,
2)
=
⟨−
5
,
11
⟩
.
Where
would
you
travel
from
(4
,
2)
to
find
higher
values
of
f
?
Q10
The
function
f
(
x,
y
)
=
|
x
|
+
|
y
|
has
its
global
minimum
at
(0
,
0)
.
Is
this
a
critical
point?
Explain.
Q11
If
(
a,
b
)
produces
the
minimum
value
of
|∇
f
(
x,
y
)
|
,
must
(0
,
0)
must
b
e
a
critical
p
oint?
Explain.
Q12
Supp
ose
f
(
x
)
is
a
function
of
x
with
critical
p
oints
x
=
a
and
x
=
b
.
Suppose
g
(
y
)
is
a
function
of
y
with
critical
p
oints
y
=
c
and
y
=
d
.
What
are
the
critical
p
oints
of
h
(
x,
y
)
=
f
(
x
)
+
g
(
y
)
?
5.6.3
Q13
Find
the
critical
p
oints
of
f
(
x,
y
)
=
x
4
+
4
xy
+
y
4
.
Q14
Find
the
critical
p
oints
of
g
(
x,
y
)
=
x
2
+
y
2
−
3
xy
−
13
x
+
12
y
.
5.6.4
Q15
If
(
x
0
,
y
0
)
is
critical
point
and
f
(
xx
)(
x
0
,
y
0
)
=
0
,
can
(
x
0
,
y
0
)
b
e
a
local
maximum
of
f
?
What
must
be
the
value
of
f
xy
(
x
0
,
y
0
)
if
so?
Q16
F
o
r
what
values
of
a
do
es
f
(
x,
y
)
=
x
2
+
y
2
+
axy
have
a
lo
cal
minimum
at
the
o
rigin?
389
Section
5.6
Exercises
5.6.5
Q17
Find
the
critical
p
oints
of
h
(
x,
y
)
=
x
2
y
−
x
2
−
2
y
2
.
Classify
each
as
a
lo
cal
maximum,
local
minimum,
o
r
saddle
p
oint.
Q18
Find
all
critical
p
oints
of
f
(
x,
y
)
=
1
3
x
3
−
4
xy
+
2
y
2
.
Classify
them
as
local
maximums,
lo
cal
minimums,
o
r
saddle
p
oints.
Q19
Compute
the
critical
p
oints
of
f
(
x,
y
)
=
2
x
3
−
12
xy
+
3
y
2
and
classify
each
as
a
lo
cal
maximum,
lo
cal
minimum,
o
r
saddle
p
oint.
Q20
Let
h
(
x,
y
)
=
x
2
+
y
3
+
3
xy
.
Find
the
critical
p
oints
of
h
,
and
classify
each
as
a
lo
cal
maximum,
lo
cal
minimum
o
r
saddle
p
oint.
Q21
Let
f
(
x,
y
)
=
x
3
−
15
x
2
−
9
x
+
12
xy
−
3
y
2
−
18
y
.
Find
the
critical
p
oints
of
f
and
classify
each
one
as
local
maximum,
lo
cal
minimum
or
saddle
p
oint.
Q22
Let
f
(
x,
y
)
=
x
5
+
20
xy
+
5
y
2
.
Find
the
critical
points
of
f
and
classify
each
one
as
lo
cal
maximum,
local
minimum
or
saddle
p
oint.
Q23
Find
the
critical
p
oints
of
g
(
x,
y
)
=
e
x
3
+
y
2
−
12
x
+10
y
.
Classify
each
one
as
lo
cal
maximum,
lo
cal
minimum
o
r
saddle
p
oint.
Q24
Find
the
critical
p
oints
of
f
(
x,
y
)
=
1
x
4
−
x
2
y
+
y
2
+10
.
Classify
each
one
as
lo
cal
maximum,
lo
cal
minimum
o
r
saddle
p
oint.
5.6.6
Q25
Dra
w
a
sketch
of
D
=
{
(
x,
y
)
:
y
≥
x
2
,
y
≤
x
3
}
.
State
whether
D
is
closed
and
whether
D
is
b
ounded.
Q26
Dra
w
a
sketch
of
D
=
{
(
x,
y
)
:
y
≥
x,
y
≤
2
x,
xy
<
1
}
.
State
whether
D
is
closed
and
whether
D
is
b
ounded.
Q27
Dra
w
a
sketch
of
D
=
{
(
x,
y
)
:
x
>
0
,
y
≥
x
4
}
.
State
whether
D
is
closed
and
whether
D
is
b
ounded.
390
Q28
Dra
w
a
sk
etch
of
D
=
{
(
x,
y
)
:
−
1
<
x
2
+
y
2
≤
16
}
.
State
whether
D
is
closed
and
whether
D
is
bounded.
Q29
Let
D
=
{
(
x,
y
)
:
y
≥
x
2
}
.
Can
the
Extreme
V
alue
Theorem
gua
rantee
that
f
has
a
maximum
on
D
?
Explain.
Q30
Do
es
the
function
f
(
x,
y
)
=
1
x
2
+
y
2
have
a
maximum
and
minimum
value
on
the
domain
D
=
{
(
x,
y
)
:
−
3
≤
x
≤
3
,
−
4
≤
y
≤
4
}
?
If
y
es,
find
them.
If
not,
explain
why
the
extreme
value
theo
rem
does
not
apply
.
5.6.7
Q31
Dra
w
a
careful
diagram
of
D
=
{
(
x,
y
)
:
y
≥
x
2
,
x
2
+
y
2
≤
20
}
.
Where
w
ould
you
need
to
check
to
gua
rantee
you’d
find
the
maximum
value
of
a
continuous
function
f
on
D
?
Q32
Let
f
(
x,
y
)
b
e
a
differentiable
function
and
let
D
=
{
(
x,
y
)
:
y
≥
x
2
−
4
,
x
≥
0
,
y
≤
5
}
.
a
Sk
etch
the
domain
D
.
b
Do
es
the
Extreme
Value
Theo
rem
gua
rantee
that
f
has
an
absolute
minimum
on
D
?
Explain.
c
List
all
the
places
you
would
need
to
check
in
order
to
lo
cate
the
minimum.
Q33
Find
the
maximum
and
minimum
value
of
f
(
x,
y
)
=
e
x
+3
y
in
the
triangle
with
vertices
(0
,
0)
,
(6
,
0)
and
(0
,
3)
.
Q34
Find
the
maximum
and
minimum
value
of
f
(
x,
y
)
=
3
x
+
y
on
D
,
the
closed
region
b
ounded
by
y
=
x
2
and
y
=
16
.
Q35
Find
the
global
max
and
min
of
f
(
x,
y
)
=
x
3
−
12
x
+
y
3
−
3
y
on
the
rectangle
0
≤
x
≤
4
and
−
2
≤
y
≤
2
.
Q36
Consider
the
function
g
(
x,
y
)
=
x
4
−
2
x
2
+2
y
2
−
2
y
+2
on
the
rectangle
−
2
≤
x
≤
2
and
0
≤
y
≤
3
.
391
Section
5.6
Exercises
a
Do
es
the
extreme
value
theorem
apply
to
this
function?
Why
might
y
ou
b
e
concerned,
and
what
w
ould
y
ou
have
to
check?
b
Find
the
min
and
max
of
g
.
Synthesis
and
Extension
Q37
Consider
the
function
f
(
x,
y
)
=
x
2
−
4
xy
+
4
y
2
.
a
Find
the
critical
p
oint(s)
of
f
.
b
What
does
the
second
derivatives
test
say
ab
out
the
critical
p
oints
of
f
?
c
Can
y
ou
classify
the
critical
p
oints
using
algebra
instead?
Explain.
Q38
If
g
(
x
)
is
an
increasing
function,
explain
why
the
lo
cal
maximums
and
minimums
of
any
f
(
x,
y
)
a
re
the
same
as
the
lo
cal
maximums
and
minimums
of
g
(
f
(
x,
y
))
.
392
Section
5.7
Lagrange
Multipliers
Goals:
1
Find
minimum
and
maximum
values
of
a
function
subject
to
a
constraint
.
2
If
necessa
ry
,
use
Lagrange
multipliers
.
Many
of
the
functions
we
studied
do
not
have
maximum
values.
P
olynomials
and
exp
onential
functions
increase
without
b
ound.
Y
et
in
the
real
wo
rld,
w
e
never
see
co
rp
o
rations
p
roducing
infinite
quantities
of
go
o
ds.
W
e
never
see
infinite
p
opulations
of
animals.
Does
this
mean
that
p
olyonomials
and
exp
onentials
have
no
real-wo
rld
applications?
On
the
contrary
,
they
a
re
ubiquitous,
but
the
corporations
and
populations
that
opp
erate
under
these
mo
dels
also
have
constraints
on
their
inputs.
Co
rp
o
rations
do
not
have
infinite
money
to
invest.
Animals
do
not
have
infinite
foo
d
sources.
In
this
section
we
develop
the
to
ols
to
find
maximum
and
minimum
values
of
a
function,
when
our
inputs
a
re
constrained.
Question
5.7.1
What
Is
a
Constraint?
Sometimes
we
aren’t
interested
in
the
maximum
value
of
f
(
x,
y
)
over
the
whole
domain,
w
e
want
to
restrict
to
only
those
p
oints
that
satisfy
a
certain
constraint
equation.
The
maximum
on
the
constraint
is
unlik
ely
to
b
e
the
same
as
the
unconstrained
maximum
(where
∇
f
=
0
).
Can
w
e
still
use
∇
f
to
find
the
maximum
on
the
constraint?
Click to Load Applet
Figure:
Maximizing
f
such
that
x
+
y
=
1
W
e
explo
re
this
question
in
the
Maximums
on
a
Constraint
activity
.
Question
5.7.2
Ho
w
Do
W
e
Solve
a
Constrained
Optimization?
The
method
of
Lagrange
Multipliers
makes
use
of
the
following
theorem.
393
Question
5.7.2
Ho
w
Do
W
e
Solve
a
Constrained
Optimization?
Theo
rem
Supp
ose
an
objective
function
f
(
x,
y
)
and
a
constraint
function
g
(
x,
y
)
a
re
differentiable.
The
lo
cal
extremes
of
f
(
x,
y
)
given
the
constraint
g
(
x,
y
)
=
c
o
ccur
where
∇
f
=
λ
∇
g
fo
r
some
number
λ
,
or
else
where
∇
g
=
0
.
The
numb
er
λ
is
called
a
Lagrange
Multiplier
.
This
theo
rem
generalizes
to
functions
of
more
variables.
W
e
can
justify
the
theorem
visually
by
examining
the
relationship
∇
f
,
∇
g
and
the
constraint.
The
constraint
g
(
x,
y
)
=
c
is
b
y
definition
a
level
curve
of
g
.
It
is
normal
to
∇
g
.
Click to Load Applet
Figure:
Where
∇
f
is
not
pa
rallel
to
∇
g
,
we
can
travel
along
g
(
x,
y
)
=
c
and
increase
the
value
of
f
.
This
is
because
D
u
f
>
0
fo
r
some
u
along
the
constraint.
By
this
a
rgument,
the
only
place
a
maximum
or
minimum
of
the
objective
function
can
lie
of
the
contraint
is
where
D
u
f
w
ould
have
to
b
e
0
,
b
ecause
∇
f
is
parallel
to
∇
g
.
Rema
rk
When
∇
f
(
P
)
is
parallel
to
∇
g
(
P
)
(and
neither
of
these
vectors
is
0
),
the
level
curves
of
f
through
P
is
tangent
to
the
level
curve
g
(
x,
y
)
=
c
.
If
we
can
dra
w
the
level
curves
of
f
,
this
gives
us
a
visual
metho
d
of
identifying
the
p
otential
maximums
and
minimums.
Example
5.7.3
The
Maximum
on
a
Curve
Find
the
point(s)
on
the
ellipse
4
x
2
+
y
2
=
4
on
which
the
function
f
(
x,
y
)
=
xy
is
maximized.
394
The
EVT
and
constraints
Are
we
guaranteed
that
a
maximum
exists
at
all?
The
Extreme
Value
Theo
rem
can
still
b
e
applied
to
constraints.
Here
are
a
few
wa
ys
we
can
identify
that
a
constraint
is
closed:
1
A
curve
is
closed
if
it
includes
its
endp
oints
(or
none
exist).
2
A
surface
is
closed
if
it
includes
its
b
oundary
(or
none
exists).
3
The
level
set
of
a
continuous
function
is
alwa
ys
closed.
Even
a
rmed
with
these,
we
still
need
to
check
that
the
domain
is
b
ounded.
Solution
W
e’ll
check
the
conditions
of
the
Extreme
Value
Theorem
1
4
x
2
+
y
2
=
4
is
a
curve
with
no
endp
oints,
so
it
is
closed.
2
4
x
2
+
y
2
=
4
is
an
ellipse.
It
stays
within
a
b
ounded
distance
from
the
o
rigin.
3
f
is
continuous.
By
the
Extreme
V
alue
Theo
rem,
we
kno
w
that
a
maximum
exists.
We
will
use
Lagrange
multipliers
to
na
rro
w
do
wn
our
search
to
the
p
ossible
maximums.
We
set
g
(
x,
y
)
=
4
x
2
+
y
2
and
compute
the
gradient
vecto
rs
of
f
and
g
.
∇
f
(
x,
y
)
=
⟨
y
,
x
⟩
∇
g
(
x,
y
)
=
⟨
8
x,
2
y
⟩
The
theo
rem
allo
ws
t
w
o
p
ossibilities
at
a
maximum.
1
∇
g
(
x,
y
)
=
⟨
0
,
0
⟩
.
The
only
(
x,
y
)
that
satisfies
this
is
(0
,
0)
.
But
(0
,
0)
is
not
on
the
constraint,
so
it
is
not
a
valid
solution.
2
∇
f
=
λ
∇
g
.
We
can
facto
r
the
λ
across
each
comp
onent
of
the
vectors,
but
that
gives
us
tw
o
equations
and
three
variables
(
x
,
y
and
λ
).
We
need
another
equation,
and
fortunately
we
have
one.
x
and
y
must
satisfy
4
x
2
+
y
2
=
4
as
well.
Here
is
one
(but
not
the
only)
wa
y
to
solve
this
system
of
equations.
395
Example
5.7.3
The
Maximum
on
a
Curve
y
=
λ
8
x
x
=
λ
2
y
4
x
2
+
y
2
=
4
y
=
λ
8(
λ
2
y
)
0
=
λ
2
16
y
−
y
0
=
y
(4
λ
−
1)(4
λ
+
1)
either
0
=
y
x
=
λ
2(0)
4(0)
2
+
0
2
=
4
(
no
solution
)
o
r
λ
=
±
1
4
y
=
±
1
4
8
x
y
=
±
2
x
4
x
2
+
(
±
2
x
)
2
=
4
8
x
2
=
4
x
2
=
1
2
x
=
±
1
√
2
y
=
±
2
1
√
2
y
=
±
√
2
This
tells
us
the
only
p
ossible
lo
cations
for
the
maximum
are:
(
x,
y
)
=
±
1
√
2
,
±
√
2
W
e
identify
the
maximum
by
evaluating
f
at
each
p
oint.
f
1
√
2
,
√
2
=
1
f
−
1
√
2
,
√
2
=
−
1
f
−
1
√
2
,
−
√
2
=
1
f
1
√
2
,
−
√
2
=
−
1
W
e
conclude
that
the
maximum
o
ccurs
at
1
√
2
,
√
2
and
−
1
√
2
,
−
√
2
.
396
Figure:
The
four
p
oints
that
satisfy
∇
f
=
λ
∇
g
and
g
(
x,
y
)
=
c
.
Main
Idea
The
level
set
of
a
continuous
(constraint)
function
is
alw
a
ys
closed.
If
it
is
also
b
ounded
and
the
objective
function
is
differentiable,
then
one
of
the
p
oints
produced
by
Lagrange
multipliers
will
b
e
the
global
maximum
and
one
will
b
e
the
global
minimum
of
the
constrained
optimization.
Example
5.7.4
The
Maximum
on
a
Surface
Find
the
maximum
value
of
the
function
f
(
x,
y
,
z
)
=
x
4
y
4
z
on
the
sphere
x
2
+
y
2
+
z
2
=
36
.
Click to Load Applet
Figure:
The
gradient
vector
and
level
surface
of
a
constraint
function
and
the
gradient
vecto
r
of
the
objective
function
397
Example
5.7.4
The
Maximum
on
a
Surface
Solution
First
note
that
the
EVT
applies,
since
a
sphere
is
closed
and
b
ounded
and
f
is
continuous.
T
o
identify
p
otential
maximums,
w
e
app
eal
to
Lagrange
multipliers.
Set
g
(
x,
y
,
z
)
=
x
2
+
y
2
+
z
2
.
Then
∇
g
(
x,
y
,
z
)
=
⟨
2
x,
2
y
,
2
z
⟩
.
The
case
∇
g
(
x,
y
,
z
)
=
0
only
o
ccurs
at
the
o
rigin,
which
is
not
on
the
sphere.
The
critical
p
oints
must
b
e
only
the
p
oints
where
∇
f
=
λ
∇
g
.
∇
f
(
x,
y
,
z
)
=
4
x
3
y
4
z
,
4
x
4
y
3
z
,
x
4
y
4
.
Equating
each
co
ordinate
gives
us
three
equations,
and
the
constraint
is
a
fourth.
We
thus
have
a
system
of
four
equations
and
four
variables.
4
x
3
y
4
z
=
λ
2
x
4
x
4
y
3
z
=
λ
2
y
x
4
y
4
z
=
λ
2
z
x
2
+
y
2
+
z
2
=
36
The
most
obvious
w
a
y
to
solve
this
algebraically
is
to
solve
for
λ
,
but
this
requires
us
to
divide
by
x
,
y
and
z
.
We
would
need
to
rememb
er
that
another
p
ossible
solution
is
that
x
,
y
o
r
z
is
0
.
W
e
can
avoid
this
b
y
multiplying
and
factoring
instead.
4
x
3
y
4
z
=
λ
2
x
4
x
4
y
3
z
=
λ
2
y
x
4
y
4
=
λ
2
z
4
x
3
y
5
z
2
=
λ
2
xy
z
4
x
5
y
3
z
2
=
λ
2
xy
z
x
5
y
5
=
λ
2
xy
z
4
x
3
y
5
z
2
=
4
x
5
y
3
z
2
x
5
y
5
=
4
x
5
y
3
z
2
4
x
3
y
5
z
2
−
4
x
5
y
3
z
2
=
0
x
5
y
5
−
4
x
5
y
3
z
2
=
0
4
x
3
y
3
z
2
(
y
−
x
)(
y
+
x
)
=
0
x
5
y
3
(
y
−
2
z
)(
y
+
2
z
)
=
0
either
x
=
0
o
r
y
=
0
o
r
y
=
±
x
and
y
=
±
2
z
±
2
z
=
x
x
2
+
y
2
+
z
2
=
36
(
±
2
z
)
2
+
(
±
2
z
)
2
+
z
2
=
36
9
z
2
=
36
z
=
±
2
(
±
2)(
±
2)
=
x
y
=
(
±
2)(
±
2)
±
4
=
x
y
=
±
4
This
gives
us
8
critical
p
oints:
(
±
4
,
±
4
,
±
2)
.
In
addition
every
point
in
the
x
=
0
cross
section
of
the
sphere
is
a
critical
p
oint,
as
is
every
p
oint
in
the
y
=
0
cross-section.
This
is
infinitely
many
points
to
evaluate,
but
fortunately
the
algebra
of
our
objective
function
allows
us
to
evaluate
these
p
oints
in
la
rge
batches.
if
x
=
0
f
(
x,
y
,
z
)
=
0
4
y
4
z
=
0
if
y
=
0
f
(
x,
y
,
z
)
=
x
4
0
4
z
=
0
f
(
±
4
,
±
4
,
2)
=
(
±
4)
4
(
±
4)
4
(2)
=
2
17
f
(
±
4
,
±
4
,
−
2)
=
(
±
4)
4
(
±
4)
4
(
−
2)
=
−
2
17
Thus
the
maximum
value
is
2
17
.
It
o
ccurs
at
the
four
p
oints
(
±
4
,
±
4
,
2)
.
398
Rema
rk
If
we
hadn’t
seen
how
to
avoid
dividing
by
x
,
y
and
z
,
w
e
could
have
gone
ahead
and
done
the
division.
Rememb
er
that
when
you
divide
while
solving
an
equation,
you
obtain
an
extra
solution
where
the
divisor
is
0
.
This
would
lead
us
to
check
x
=
0
,
y
=
0
and
z
=
0
as
we
did
in
the
facto
ring
solution.
Synthesis
5.7.5
Using
the
Extreme
Value
Theo
rem
and
Lagrange
Multipliers
Ho
w
can
Lagrange
multipliers
help
us
find
the
maximum
of
f
(
x,
y
)
=
x
2
+
2
y
2
−
x
2
y
on
the
domain
D
=
{
(
x,
y
)
:
x
2
+
y
2
≤
16
,
x
≤
0
}
?
Solution
W
e
can
continue
Example
7.
After
finding
the
critical
p
oints
of
f
at
(0
,
0)
and
(
−
2
,
1)
,
we
turn
to
the
b
ounda
ries.
The
b
oundaries
are
level
curves.
F
o
r
x
2
+
y
2
=
16
,
set
g
(
x,
y
)
=
x
2
+
y
2
=
16
.
W
e
have
∇
f
(
x,
y
)
=
2
x
−
2
xy
,
4
y
−
x
2
∇
g
(
x,
y
)
=
⟨
2
x,
2
y
⟩
∇
g
(
x,
y
)
=
0
only
at
the
o
rigin,
which
isn’t
on
the
constraint.
So
we
solve
∇
f
(
x,
y
)
=
λ
∇
g
(
x,
y
)
and
g
(
x,
y
)
=
4
.
399
Synthesis
5.7.5
Using
the
Extreme
Value
Theo
rem
and
Lagrange
Multipliers
2
x
−
2
xy
=
λ
2
x
4
y
−
x
2
=
λ
2
y
x
2
+
y
2
=
16
2
x
−
2
xy
−
2
λx
=
0
2
x
(1
−
y
−
λ
)
=
0
if
x
=
0
0
2
+
y
2
=
16
y
=
±
4
if
1
−
y
−
λ
=
0
λ
=
1
−
y
4
y
−
x
2
=
(1
−
y
)2
y
2
y
2
+
2
y
=
x
2
(2
y
2
+
2
y
)
+
y
2
=
16
3
y
2
+
2
y
−
16
=
0
(3
y
+
8)(
y
−
2)
=
0
if
y
=
−
8
3
if
y
=
2
x
2
+
−
8
3
2
=
16
x
2
+
2
2
=
16
x
2
+
64
9
=
144
9
x
2
=
12
x
2
=
80
9
=
x
=
±
√
12
x
=
±
r
80
9
The
critical
p
oints
are
(0
,
±
4)
,
−
√
12
,
2
and
−
q
80
9
,
−
8
3
.
The
solutions
with
p
ositive
x
are
not
in
D
.
On
x
=
0
,
substitution
is
p
robably
the
easier
choice,
but
Lagrange
multipliers
a
re
still
p
ossible.
x
=
0
is
a
level
set
of
the
function
g
(
x,
y
)
=
x
.
∇
g
(
x,
y
)
=
⟨
1
,
0
⟩
∇
g
=
0
so
w
e
solve
∇
f
(
x,
y
)
=
λ
∇
g
(
x,
y
)
.
2
x
−
2
xy
=
λ
4
y
−
x
2
=
0
x
=
0
4
y
=
0
This
is
the
same
equation
we
obtained
by
substituting
x
=
0
into
f
and
differentiating.
400
Main
Idea
T
o
find
the
absolute
minimum
and
maximum
of
a
differentiable
function
f
(
x,
y
)
over
a
closed
and
b
ounded
domain
D
:
1
Compute
∇
f
and
find
the
critical
p
oints
inside
D
.
2
Identify
the
b
oundary
components.
Find
the
critical
p
oints
on
each
using
substitution
or
Lagrange
multipliers
.
3
Identify
the
endpoints
(intersections)
of
the
b
oundary
comp
onents.
4
Evaluate
f
(
x,
y
)
at
all
of
the
ab
ove.
The
minimum
is
the
low
est
numb
er,
the
maximum
is
the
highest.
Synthesis
5.7.6
The
Gradient
on
the
Boundary
Supp
ose
P
is
a
critical
p
oint
of
f
on
a
b
oundary
comp
onent
of
a
domain
D
.
What
do
es
the
direction
of
∇
f
(
P
)
tell
us
ab
out
whether
P
is
a
maximum
or
minimum?
Figure:
The
critical
p
oints
and
gradient
vectors
of
f
(
x,
y
)
on
a
closed
and
b
ounded
domain
Solution
First
supp
ose
∇
f
(
P
)
p
oints
into
D
.
Then
f
increases
as
we
travel
into
D
.
Thus
P
cannot
b
e
a
lo
cal
maximum.
401
Synthesis
5.7.6
The
Gradient
on
the
Boundary
P
ma
y
b
e
a
lo
cal
minimum
but
ma
y
not
b
e.
The
directional
derivative
along
the
b
oundary
is
0
,
so
f
could
curve
upw
a
rd
or
downw
a
rd
along
the
b
ounda
ry
.
If
f
curves
downw
a
rd
we
could
find
low
er
values
of
f
nearb
y
and
P
would
not
b
e
a
minimum.
If
f
curves
up
w
a
rd,
then
P
would
b
e
a
minimum.
We
could
compute
this
curvature
b
y
taking
the
substituted
version
of
f
that
w
e
used
to
solve
for
P
and
computing
its
second
derivative
at
P
.
On
the
other
hand,
if
we
supp
ose
that
∇
f
(
P
)
p
oints
out
of
D
,
then
D
decreases
as
we
travel
into
D
,
and
P
cannot
b
e
a
lo
cal
minimum.
It
may
or
ma
y
not
b
e
a
lo
cal
maximum.
Question
5.7.7
Can
This
Lagrange
Apply
to
More
Than
One
Constraint?
If
w
e
have
tw
o
constraints
in
three-space,
g
(
x,
y
,
z
)
=
c
and
h
(
x,
y
,
z
)
=
d
,
then
their
intersection
is
generally
a
curve.
Click to Load Applet
Figure:
The
intersection
of
the
constraints
g
(
x,
y
,
z
)
=
c
and
h
(
x,
y
,
z
)
=
d
Acco
rding
to
our
ea
rlier
argument
about
directional
derivatives,
at
a
maximum
P
on
the
constraint,
∇
f
(
P
)
must
b
e
normal
to
the
constraint.
There
a
re
mo
re
wa
ys
fo
r
this
to
happ
en
with
tw
o
constraint
equations.
1
∇
f
(
P
)
could
b
e
parallel
to
∇
g
(
P
)
.
2
∇
f
(
P
)
could
b
e
parallel
to
∇
h
(
P
)
.
3
∇
f
(
P
)
could
b
e
the
vector
sum
of
a
vector
parallel
to
∇
g
(
P
)
and
a
vecto
r
pa
rallel
to
∇
h
(
P
)
.
Y
ou
should
lo
ok
at
Figure
380
to
convince
y
ourself
that
these
∇
f
(
P
)
would
all
be
normal
to
the
constraint.
W
e
can
express
this
condition
algebraically
402
Theo
rem
If
f
(
x,
y
,
z
)
is
a
differentiable
function
and
g
(
x,
y
,
z
)
=
c
and
h
(
x,
y
,
z
)
=
d
are
tw
o
constraints.
If
P
is
a
maximum
of
f
(
x,
y
,
z
)
among
the
p
oints
that
satisfy
these
constraints
then
either
∇
f
(
P
)
=
λ
∇
g
(
P
)
+
µ
∇
h
(
P
)
fo
r
some
scala
rs
λ
and
µ
,
or
∇
g
(
P
)
and
∇
h
(
P
)
are
parallel.
This
system
of
equations
is
usually
difficult
to
solve
by
hand.
Rema
rk
Y
ou
can
check
the
reasonableness
of
this
metho
d
by
noting
that
it
gives
us
a
system
of
5
variables,
x
,
y
,
z
,
λ
,
µ
,
and
five
equations:
f
x
(
x,
y
,
z
)
=
λg
x
(
x,
y
,
z
)
+
µh
x
(
x,
y
,
z
)
g
(
x,
y
,
z
)
=
c
f
y
(
x,
y
,
z
)
=
λg
y
(
x,
y
,
z
)
+
µh
y
(
x,
y
,
z
)
h
(
x,
y
,
z
)
=
d
f
z
(
x,
y
,
z
)
=
λg
z
(
x,
y
,
z
)
+
µh
z
(
x,
y
,
z
)
W
e
therefo
re
generally
exp
ect
this
system
to
have
a
finite
numb
er
of
solutions,
though
there
are
plenty
of
counterexamples
to
this
exp
ectation.
Section
5.7
Exercises
Summa
ry
Questions
Q1
What
is
a
constraint?
Q2
What
equations
do
you
write
when
you
apply
the
metho
d
of
Lagrange
multipliers?
Q3
Is
the
set
of
p
oints
that
satisfies
a
constraint
closed
and
b
ounded?
Explain.
Q4
Ho
w
does
a
constraint
arise
when
finding
the
maximum
over
a
closed
and
b
ounded
domain?
403
Section
5.7
Exercises
5.7.1
Q5
Supp
ose
we
have
$230
to
sp
end
on
three
go
o
ds.
Go
o
d
1
costs
$13
p
er
unit.
Go
o
d
2
costs
$22
p
er
unit.
Go
o
d
3
costs
$11
p
er
unit.
W
rite
a
budget
constraint
that
expresses
what
purchases
(
x,
y
,
z
)
of
go
o
d
1,
goo
d
2
and
go
o
d
3
are
p
ossible,
if
you
sp
end
you
budget.
Q6
Supp
ose
the
maximum
value
of
f
(
x,
y
)
occurs
at
(3
,
−
4)
.
Where
is
the
maximum
value
of
f
(
x,
y
)
that
satisfies
the
constraint
x
2
+
y
2
=
25
?
Explain.
5.7.2
Q7
Supp
ose
f
(
x,
y
,
z
)
is
a
smo
oth
function.
Supp
ose
the
maximum
value
of
f
on
the
sphere
x
2
+
y
2
+
z
2
=
25
o
ccurs
at
P
.
What
can
you
say
about
∇
f
(
P
)
and
the
tangent
plane
to
the
sphere
at
P
?
Q8
Supp
ose
the
curve
b
elow
is
the
graph
of
g
(
x,
y
)
=
k
.
Use
metho
ds
from
calculus
to
find
and
ma
rk
the
appro
ximate
lo
cation
of
the
p
oint
that
maximizes
the
function
f
(
x,
y
)
=
3
y
−
x
subject
to
the
constraint
g
(
x,
y
)
=
k
.
Justify
your
reasoning
in
a
few
sentences.
Q9
Supp
ose
that
(
a,
b
)
is
a
lo
cal
maximum
of
the
smooth
function
f
(
x,
y
)
which
also
happ
ens
to
satisfy
the
constraint
g
(
a,
b
)
=
k
.
a
Is
(
a,
b
)
also
a
lo
cal
maximum
of
f
among
the
p
oints
on
the
constraint?
Explain.
b
If
we
used
Lagrange
multipliers
to
detect
(
a,
b
)
,
what
would
we
exp
ect
λ
to
b
e
equal
to
at
that
point?
404
Q10
Sho
w
that
(3
,
3)
is
not
a
local
maximum
of
f
(
x,
y
)
=
2
x
2
−
4
xy
+
y
2
−
8
x
on
the
graph
x
3
+
y
3
=
6
xy
.
5.7.3
Q11
Compute
the
maximum
value
of
y
−
x
2
on
the
constraint
x
2
+
y
2
=
4
.
Q12
Refer
to
y
our
“Maximums
on
a
Constraint”
wo
rksheet.
a
What
system
of
equations
would
you
set
up
to
find
the
critical
p
oints
of
f
on
the
constraint
p
(
x,
y
)
=
c
?
b
Can
y
ou
solve
it?
c
Which
w
as
easier,
using
Lagrange
or
using
substitution?
5.7.4
Q13
Find
the
maximum
value
of
f
(
x,
y
,
z
)
=
xy
z
on
the
sphere
x
2
+
y
2
+
z
2
=
36
.
Q14
Find
the
maximum
value
of
f
(
x,
y
,
z
)
=
xz
on
the
sphere
x
2
+
y
2
+
z
2
=
36
.
Q15
Find
the
maximum
value
of
f
(
x,
y
,
z
)
=
3
y
+
2
z
on
the
ellipsoid
25
x
2
+
y
2
+
4
z
2
=
100
.
Q16
The
function
h
(
x,
y
,
z
)
=
x
2
+
y
2
+
z
2
has
a
minimum
value
on
the
plane
3
x
+
5
y
−
2
z
=
30
.
Compute
it.
405
Section
5.7
Exercises
5.7.5
Q17
Supp
ose
f
(
x,
y
)
is
differentiable
but
has
no
critical
points.
Will
the
method
of
Lagrange
multipliers
detect
the
maximum
value
of
f
in
D
=
{
(
x,
y
)
:
x
2
+
y
2
≤
49
}
?
Explain.
Q18
Consider
the
follo
wing
tw
o
questions:
Find
the
maximum
value
of
f
(
x,
y
)
that
satisfies
x
2
+
y
2
≤
9
.
Find
the
maximum
value
of
f
(
x,
y
)
that
satisfies
x
2
+
y
2
=
9
.
a
Ho
w
a
re
the
ques
tions
different?
b
Which
question
tak
es
less
wo
rk
to
solve?
Explain
how
you
know.
c
Do
solutions
exist
to
b
oth
questions?
What
additional
info
rmation
would
guarantee
that
they
do?
Q19
Let
D
=
{
(
x,
y
)
:
x
2
+
y
2
≤
1
,
x
≥
0
,
y
≤
0
}
.
Find
the
maximum
and
minimum
values
of
f
(
x,
y
)
=
x
2
−
y
on
D
.
Q20
Consider
the
function
f
(
x,
y
)
=
x
2
+
6
xy
+
9
y
2
+
5
.
Find
the
maximum
and
minimum
values
of
f
on
the
domain
D
=
{
(
x,
y
)
:
y
≥
x,
x
≥
0
,
x
2
+
y
2
≤
10
}
Q21
Let
D
=
{
(
x,
y
)
:
x
2
+
y
2
≤
20
,
y
≥
−
x
}
.
Find
the
maximum
and
minimum
values
of
f
(
x,
y
)
=
x
4
y
on
D
.
Q22
Let
D
=
{
(
x,
y
)
:
x
2
+
y
2
≤
25
,
y
≥
x
+
1
,
y
≥
0
}
.
Find
the
maximum
and
minimum
values
of
f
(
x,
y
)
=
x
3
y
2
on
D
.
Q23
Let
D
=
{
(
x,
y
)
:
x
2
+
y
2
≤
20
,
y
≥
−
x
}
.
Find
the
maximum
and
minimum
values
of
f
(
x,
y
)
=
x
4
y
on
D
.
Q24
Let
D
=
(
x,
y
)
:
x
2
16
+
y
2
64
≤
1
,
x
≥
0
.
Find
the
p
oints
in
D
that
obtain
the
maximum
and
minimum
values
of
f
(
x,
y
)
=
2
x
+
3
y
.
406
5.7.6
Q25
Supp
ose
the
maximum
of
f
(
x,
y
)
on
D
=
{
(
x,
y
)
|
g
(
x,
y
)
≤
c
}
o
ccurs
at
P
on
the
boundary
of
D
.
We
know
that
∇
f
(
P
)
p
oints
out
of
D
.
What
do
es
this
tell
us
about
the
sign
of
λ
?
Q26
Explain
why
kno
wing
which
w
a
y
∇
f
points
is
not
useful
for
ruling
out
p
otential
maximums
given
a
domain
of
the
form
g
(
x,
y
)
=
c
.
5.7.7
Q27
Ho
w
does
the
metho
d
of
Lagrange
multipliers
suggest
we
solve
for
the
maximum
value
of
f
(
x,
y
)
on
the
constraints
x
+
y
=
1
and
x
−
y
=
0
?
Do
we
need
to
know
what
f
is
to
solve
this?
Why
shouldn’t
that
bother
us?
Q28
W
rite
a
system
of
equations
that
one
w
ould
solve
to
find
the
maximum
and
minimum
values
of
f
(
x,
y
,
z
)
=
x
on
the
tw
o
constraints
y
2
+
z
2
=
25
and
x
+
y
+
z
=
1
.
Synthesis
and
Extension
Q29
Consider
the
plane
p
with
normal
equation
7
x
+
6
y
−
3
z
−
42
=
0
a
Use
Lagrange
multipliers
to
find
the
p
oint
A
on
p
that
s
closest
to
the
o
rigin
O
.
b
Sho
w
that
−
→
O
A
is
a
no
rmal
vector
to
p
.
c
Sho
w
how
y
ou
can
use
the
observation
in
b
to
solve
fo
r
the
closest
p
oint
(
A
)
without
using
calculus.
Q30
Determine
the
smallest
rectangle
(pa
rallel
to
the
x
and
y
axes)
that
contains
the
ellipse
x
2
+
3
xy
+
4
y
2
−
4
x
−
13
y
+
4
=
0
.
407
Section
5.7
Exercises
Q31
An
aquarium
with
an
op
en
top
has
volume
20
m
3
.
Its
rectangular
base
is
made
of
slate,
and
its
sides
are
made
of
glass.
Slate
costs
five
times
as
much
(p
er
unit
area)
as
glass.
Set
up
and
solve
a
constrained
onstrained
optimization
problem
to
find
the
dimensions
(
ℓ,
w
,
h
)
of
the
aquarium
that
will
minimize
the
cost
of
materials.
Q32
Let
D
b
e
the
region
enclosed
by
2
x
+
y
=
8
,
y
=
8
and
x
=
4
.
Consider
the
function
f
(
x,
y
)
=
xy
−
3
y
−
6
x.
a
Do
es
f
have
a
maximum
and
minimum
value
on
D
?
What
to
ol
can
you
use
to
verify
this?
What
did
y
ou
need
to
check
b
efore
applying
this
to
ol?
b
Find
the
maximum
and
minimum
values
of
f
on
D
.
Demonstrate
in
your
w
o
rk
that
you’ve
check
ed
all
the
relevant
places
for
p
otential
maximums.
Q33
Find
the
maximum
and
minimum
values
of
f
(
x,
y
)
=
2
x
2
+
2
xy
+
5
y
2
on
the
ellipse
x
2
+
4
y
2
=
106
.
408
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