<
Chapter
4
Multiva
riable
F
unctions
This
chapter
intro
duces
functions
of
more
than
one
va
riable.
W
e
construct
the
higher
dimensional
spaces
needed
fo
r
their
domains,
w
e
p
roduce
to
ols
to
visualize
them,
and
w
e
compute
their
rates
of
change.
Contents
4.1
Three-Dimensional
Co
o
rdinate
Systems
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242
4.2
F
unctions
of
Several
V
a
riables
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259
4.3
Limits
and
Continuit
y
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275
4.4
P
a
rtial
Derivatives
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281
4.5
Linea
r
App
ro
ximations
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295
Section
4.1
Three-Dimensional
Co
o
rdinate
Systems
Goals:
1
Plot
p
oints
in
a
three-dimensional
co
o
rdinate
system.
2
Use
the
distance
fo
rmula.
3
Recognize
the
equation
of
a
sphere
and
find
its
radius
and
center.
4
Graph
an
implicit
function
with
a
free
va
riable.
Supp
ose
we
wanted
to
understand
the
growth
rate
of
a
sp
ecies
of
bacteria.
We
could
gro
w
several
dozen
cultures
and
tak
e
a
series
of
measurements
of
size
s
at
times
t
of
each.
Each
measurement
is
an
o
rdered
pair
(
t,
s
)
.
W
e
can
plot
these
pairs
in
a
co
ordinate
plane
to
get
a
visual
sense
of
ho
w
growth
o
ccurs
over
time.
We
might
even
fit
a
function
that
appro
ximates
s
as
a
function
of
t
.
What
if
we
w
anted
to
understand
the
role
of
some
other
measurement,
lik
e
temp
erature,
light,
or
the
availabilit
y
of
various
fo
o
d
sources?
We
could
gro
w
many
cultures
in
different
conditions.
No
w
a
single
measurement
has
three
o
r
mo
re
pieces
of
information.
While
we
could
strip
these
out
and
plot
our
data
on
a
temp
erature/size
co
o
rdinate
plane,
w
e
risk
missing
imp
ortant
relationships
with
the
other
variables.
In
order
to
tak
e
advantage
of
the
visual
and
computational
b
enefits
of
a
co
ordinate
system,
we
must
b
e
prepa
red
to
w
o
rk
with
a
coordinate
system
of
mo
re
than
t
w
o
va
riables.
Question
4.1.1
Ho
w
Do
Ca
rtesian
Coordinates
Extend
to
Higher
Dimensions?
The
b
est
wa
y
to
define
a
higher-dimensional
coordinate
system
is
to
extrap
olate
from
the
co
ordinate
plane.
This
wa
y
we
don’t
need
to
rememb
er
a
set
of
novel
and
arbitra
ry
rules,
and
our
tw
o-dimensional
exp
erience
will
b
e
a
guide
to
us
in
dimensions
where
we
have
no
visual
intuition.
Recall
ho
w
we
constructed
the
Ca
rtesian
plane.
x
y
−
4
−
4
−
3
−
3
−
2
−
2
−
1
−
1
1
1
2
2
3
3
4
4
(2
,
3)
1
Assign
o
rigin
and
tw
o
directions
(
x,
y
).
2
y
is
90
degrees
anticlo
ckwise
from
x
.
3
Axes
consist
of
the
p
oints
displaced
in
only
one
direction.
4
Co
o
rdinates
refer
to
displacement
from
the
o
rigin
in
each
direction.
5
Either
displacement
can
happ
en
first.
6
Each
p
oint
has
exactly
one
ordered
pair
that
refers
to
it.
In
a
three-dimensional
Ca
rtesian
co
ordinate
system.
We
can
extrapolate
from
tw
o
dimensions.
242
Click to Load Applet
1
Assign
o
rigin
and
three
directions
(
x,
y
,
z
).
2
Each
axis
makes
a
90
degree
angle
with
the
other
t
wo.
3
The
z
direction
is
determined
by
the
right-
hand
rule.
Question
4.1.2
Ho
w
Do
We
Establish
Which
Direction
Is
P
ositive
in
Each
Axis?
The
choice
of
which
direction
is
p
ositive
is
arbitra
ry
.
How
ever,
it
is
imp
ortant
that
we
all
make
the
same
choice,
or
our
visualizations
will
be
incompatible.
In
t
wo
dimensions,
w
e
agree
that
the
p
ositive
y
-axis
is
counterclo
ckwise
from
the
positive
x
-axis.
This
will
not
w
ork
in
three
dimensions.
Supp
ose
the
p
ositive
y
-axis
is
counterclo
ckwise
from
the
p
ositive
x
-axis
in
three-space.
If
you
rotate
your
p
oint
of
view
to
see
the
axes
from
the
other
side,
the
p
ositive
y
-axis
is
no
w
clo
ckwise
from
the
p
ositive
x
-axis.
Thus
the
relative
orientation
of
the
p
ositive
x
and
y
directions
do
es
not
matter.
Y
ou
could
pick
a
different
o
rientation,
and
just
b
e
lo
oking
at
three-space
from
a
different
viewp
oint.
The
z
direction
is
different.
Once
we’ve
chosen
a
p
ositive
x
and
y
direction,
there
are
tw
o
equally
valid
p
ossible
directions
for
p
ositive
z
,
p
ointing
in
opp
osite
directions
from
each
other.
The
choice
here
matters,
but
it
will
b
e
a
rbitrary
.
We
agree
to
define
the
p
ositive
z
direction
by
the
right
hand
rule
.
The
right
hand
rule
sa
ys
that
if
you
make
the
fingers
of
your
right
hand
follow
the
(counterclo
ckwise)
unit
circle
in
the
xy
-plane,
then
your
thumb
indicates
the
direction
of
the
p
ositive
z
-axis.
Click to Load Applet
Figure:
The
counterclo
ckwise
unit
circle
in
the
xy
-plane
243
Example
4.1.3
Dra
wing
a
Lo
cation
in
Three-Dimensional
Co
ordinates
The
p
oint
(2
,
3
,
5)
is
the
p
oint
displaced
from
the
origin
by
2
in
the
x
direction
3
in
the
y
direction
5
in
the
z
direction.
Ho
w
do
we
dra
w
a
reasonable
diagram
of
where
this
p
oint
lies?
Solution
W
e
can
b
egin
by
finding
the
p
oints
(2
,
0
,
0)
which
lies
on
the
x
-axis
tw
o
units
from
the
origin
and
(0
,
3
,
0)
which
lies
on
the
y
-axis
three
units
from
the
o
rigin.
Along
with
the
origin
itself,
these
p
oints
and
(2
,
3
,
0)
form
a
parallelogram.
Now
we
need
a
displacement
of
5
in
the
z
direction.
W
e
can
copy
the
length
and
direction
of
this
displacementof
the
segment
from
(0
,
0
,
0)
to
(0
,
0
,
5)
on
the
z
-axis.
We
dra
w
a
segment
of
that
length
and
direction
from
(2
,
3
,
0)
.
The
top
of
this
segment
is
(2
,
3
,
5)
.
Click to Load Applet
Rema
rk
The
extra
lines
we
used
to
construct
(2
,
3
,
5)
are
not
just
useful
for
guaranteeing
accuracy
,
they
also
help
our
audience
to
correctly
visualize
the
lo
cation
we
mean
to
plot.
When
we
project
three-space
onto
a
flat
page,
each
p
oint
on
the
page
represents
infinitely
many
p
oints
stretching
into
the
background.
If
we
only
draw
a
isolated
p
oint,
which
of
these
are
we
rep
resenting?
Lines
like
the
ones
we
p
ro
duced
in
this
example
trick
a
view
ers
brain
into
visualizing
co
rrect
three-dimensional
lo
cation
in
our
flat
diagram.
Ho
w
can
we
dra
w
a
reasonable
diagram
of
(
−
5
,
1
,
−
4)
?
244
Solution
The
p
ro
cedure
here
is
the
same,
except
that
the
displacements
in
the
x
and
z
directions
are
negative.
Thus
when
p
ro
ducing
these
displacements,
we
travel
backw
ard
along
their
axes.
Click to Load Applet
Question
4.1.4
Ho
w
Do
We
Measure
Distance
in
Three-Space?
Since
co
o
rdinate
displacements
in
tw
o-space
a
re
p
erp
endicula
r,
w
e
compute
the
distance
to
a
p
oint
using
the
Pythagorean
theo
rem.
This
reasoning
extends
to
higher
dimensions,
but
w
e
need
to
build
the
co
rrect
length
using
tw
o
or
mo
re
right
triangles.
Theo
rem
The
distance
from
the
o
rigin
to
the
p
oint
(
x,
y,
z
)
is
given
by
the
Pythagorean
Theorem
D
=
p
x
2
+
y
2
+
z
2
245
Question
4.1.4
Ho
w
Do
We
Measure
Distance
in
Three-Space?
Click to Load Applet
W
e
first
compute
the
distance
from
the
o
rigin
to
(
x,
y
,
0)
using
a
right
triangle
in
the
xy
-plane.
The
right
triangle
with
the
vertices
(0
,
0
,
0)
,
(
x,
y
,
0)
and
(
x,
y
,
z
)
allows
us
to
apply
the
Pythago
rean
theo
rem
again.
D
2
=
p
x
2
+
y
2
2
+
z
2
If
neither
of
the
p
oints
is
the
origin,
we
can
compute
the
displacements
by
subtraction.
This
is
a
natural
extension
of
the
t
wo-space
distance
fo
rmula.
Theo
rem
The
distance
from
the
p
oint
(
x
1
,
y
1
,
z
1
)
to
the
p
oint
(
x
2
,
y
2
,
z
2
)
is
given
b
y
D
=
p
(
x
1
−
x
2
)
2
+
(
y
1
−
y
2
)
2
+
(
z
1
−
z
2
)
2
Question
4.1.5
What
Is
a
Graph?
A
well-p
repared
calculus
student
has
learned
to
understand
the
graphs
of
many
equations:
lines,
circles,
pa
rab
olas.
The
definition
of
a
graph,
on
the
other
hand,
is
often
discarded
after
a
few
exercises
of
plotting
p
oints
b
y
hand.
The
definition
is
wo
rth
recalling.
It
applies
to
a
space
of
any
dimension.
Definition
The
graph
of
an
implicit
equation
is
the
set
of
p
oints
whose
co
ordinates
satisfy
that
equation.
In
other
w
ords,
the
t
wo
sides
are
equal
when
w
e
plug
the
co
ordinates
in
fo
r
x
,
y
and
z
.
This
definition
allows
us
to
immediately
understand
the
graphs
of
some
equations.
The
graph
of
the
following
equation
consists
of
the
p
oints
that,
when
plugged
into
a
sp
ecific
distance
formula
and
squa
red,
give
a
result
of
9
.
This
is
a
sphere.
246
Example
The
graph
of
x
2
+
(
y
−
4)
2
+
(
z
+
1)
2
=
9
is
the
set
of
p
oints
that
are
distance
3
from
the
p
oint
(0
,
4
,
−
1)
Click to Load Applet
Example
4.1.6
Graphing
an
Equation
with
Tw
o
Free
V
ariables
Sk
etch
the
graph
of
the
equation
y
=
3
.
Solution
The
naive
approach
would
have
us
seek
out
the
p
oint
mark
ed
with
3
on
the
y
-axis.
How
ever,
in
tw
o-
space,
w
e
know
that
the
graph
would
b
e
a
horizontal
line,
not
just
the
p
oint
(0
,
3)
.
Why
is
this?
Any
p
oint
of
the
form
(
x,
3)
satisfies
the
equation
y
=
3
.
Simila
rly
,
any
p
oint
of
the
form
(
x,
3
,
z
)
in
three-
space
satisfies
y
=
3
.
These
are
all
the
points
that
can
b
e
reached
from
(0
,
3
,
0)
by
displacements
in
the
x
and
z
directions.
They
create
a
plane
through
(0
,
3
,
0)
parallel
to
the
x
and
z
axes.
Much
as
lines
are
the
simplest
and
most
fundamental
one-dimensional
objects,
planes
are
the
simplest
and
most
fundamental
t
wo-dimensional
objects.
In
addition
to
co
ordinate
axes,
3
-dimensional
space
has
3
co
o
rdinate
planes
.
1
The
graph
of
z
=
0
is
the
xy
-plane.
2
The
graph
of
x
=
0
is
the
y
z
-plane.
3
The
graph
of
y
=
0
is
the
xz
-plane.
247
Example
4.1.6
Graphing
an
Equation
with
Tw
o
Free
V
ariables
Click to Load Applet
Figure:
The
co
o
rdinate
planes
in
3
-dimensional
space.
Rema
rk
Planes
extend
forever
but
our
pictures
of
them
cannot.
Notice
that
graphing
softw
are
cuts
them
off
pa
rallel
to
the
axes
they
contain.
The
resulting
images
are
parallelograms.
This
is
a
go
o
d
practice
when
dra
wing
planes
by
hand
to
o.
It
suggests
the
proper
orientation
to
the
viewer,
despite
the
limitations
of
a
flat
visualization.
Example
4.1.7
Graphing
an
Equation
with
One
F
ree
Va
riable
Sk
etch
the
graph
of
the
equation
z
=
x
2
−
3
.
Solution
W
e
should
recognize
this
as
the
equation
of
a
parabola.
If
we
ignore
the
variable
y
,
we
can
graph
this
equation
in
the
xz
plane.
What
do
es
the
absence
of
absence
of
y
in
the
equation
mean?
If
we
follow
the
definition
of
a
graph,
the
value
of
y
has
no
effect
on
whether
a
p
oint
lies
on
the
graph
or
not.
We
can
tak
e
the
pa
rab
ola
in
the
xz
plane,
and
project
it
in
the
y
direction
to
obtain
a
surface
c
alled
a
pa
rab
olic
cylinder
.
248
Click to Load Applet
z
=
x
2
−
3
Question
4.1.8
What
Do
the
Graphs
of
Implicit
Equations
Lo
ok
Lik
e
Generally?
Notice
that
the
graph
of
an
implicit
equation
in
the
plane
is
generally
one-dimensional
(a
curve),
whereas
the
graph
of
an
implicit
equation
in
three-space
is
generally
tw
o-dimensional
(a
surface).
Figure:
The
curve
y
=
x
2
−
3
Click to Load Applet
Figure:
The
surface
z
=
x
2
−
3
Question
4.1.9
What
Is
the
Slop
e-Intercept
Equation
of
a
Plane?
Unlik
e
a
line,
a
non-vertical
plane
has
tw
o
slop
es.
One
measures
rise
over
run
in
the
x
-direction,
the
other
in
the
y
-direction.
249
Question
4.1.9
What
Is
the
Slop
e-Intercept
Equation
of
a
Plane?
Click to Load Applet
Figure:
A
plane
with
slop
es
in
the
x
and
y
directions.
Equation
A
plane
with
z
intercept
(0
,
0
,
b
)
and
slop
es
m
x
and
m
y
in
the
x
and
y
directions
has
equation
z
=
m
x
x
+
m
y
y
+
b.
Example
4.1.10
W
riting
the
Equation
of
a
Plane
W
rite
the
equation
of
a
plane
with
intercepts
(4
,
0
,
0)
,
(0
,
6
,
0)
and
(0
,
0
,
8)
.
Solution
F
rom
the
p
oint
(4
,
0
,
0)
to
the
p
oint
(0
,
0
,
8)
,
the
plane
rises
by
8
while
x
is
reduced
b
y
4
.
This
gives
a
slop
e
in
the
x
direction.
m
x
=
8
−
0
0
−
4
=
−
2
.
Simila
rly
,
m
y
=
8
−
0
0
−
6
=
−
4
3
.
The
p
oint
(0
,
0
,
8)
is
on
the
z
-axis,
and
so
indicates
that
the
z
-intercept
is
8
.
Combining
these,
we
conclude
the
plane
has
equation:
z
=
−
2
x
−
4
3
y
+
8
250
Main
Idea
Given
three
p
oints
in
a
plane
A
=
(
x
1
,
y
1
,
z
1
)
,
B
=
(
x
2
,
y
2
,
z
2
)
and
C
=
(
x
3
,
y
3
,
z
3
)
1
If
t
wo
points
share
an
x
-coordinate,
w
e
can
directly
compute
m
y
and
vice
versa.
2
F
ailing
that,
we
can
set
up
a
system
of
equations
and
solve
fo
r
m
x
,
m
y
and
b
.
Question
4.1.11
Ho
w
Do
We
Extrapolate
to
Even
Higher
Dimensions?
The
measurements
we
take
of
each
observation,
the
more
dimensions
we
need
to
plot
the
data
we
have
produced.
Extrap
olating
from
three-space
to
even
higher
dimensions
introduces
no
new
difficulties,
except
that
we
cannot
visualize
the
result.
We
can
use
a
co
ordinate
system
to
describ
e
a
space
with
mo
re
than
3
dimensions.
k
-dimensional
space
can
b
e
defined
as
the
set
of
p
oints
of
the
form
P
=
(
x
1
,
x
2
,
.
.
.
,
x
k
)
.
Theo
rem
The
distance
from
the
o
rigin
to
P
=
(
x
1
,
x
2
,
.
.
.
,
x
k
)
in
k
-space
is
q
x
2
1
+
x
2
2
+
·
·
·
+
x
2
k
There
is
no
right
hand
rule
fo
r
higher
dimensions,
b
ecause
w
e
can’t
draw
these
spaces
anyw
ay
.
251
Section
4.1
Exercises
Summa
ry
Questions
Q1
What
displacements
a
re
represented
b
y
the
notation
(
a,
b,
c
)
?
Q2
What
is
the
right
hand
rule
and
what
do
es
it
tell
you
ab
out
a
three-dimensional
co
o
rdinate
system?
Q3
In
three-space,
what
is
the
y
-axis?
What
are
the
co
ordinates
of
a
general
p
oint
on
it?
Q4
In
three
space,
what
is
the
xz
-plane?
What
are
the
co
ordinates
of
a
general
p
oint
on
it?
What
is
its
equation?
Q5
Ho
w
do
we
use
a
free
va
riable
to
sketch
a
graph?
Q6
Ho
w
do
we
recognize
the
equation
of
a
sphere?
4.1.1
Q7
Supp
ose
that
instead
of
denoting
each
p
oint
P
=
(
x,
y
)
in
R
2
b
y
its
displacements
from
the
o
rigin
in
the
x
-
and
y
-directions,
we
denote
it
by
P
=
(
d,
m
)
where
d
is
its
distance
from
the
o
rigin,
and
m
is
the
slop
e
of
the
line
through
P
and
the
origin.
What
problems
could
arise
from
adopting
this
convention?
Q8
Supp
ose
the
x
and
y
axes
were
not
p
erp
endicula
r.
Could
w
e
still
assign
co
ordinates
to
each
p
oint
b
y
its
x
and
y
displacements
from
the
o
rigin?
Demonstrate
with
a
diagram.
252
4.1.2
Q9
Which
of
the
following
depictions
of
the
xy
-plane
are
consistent
with
the
usual
orientation,
and
which
a
re
backwa
rds?
a
The
p
ositive
x
axis
p
oints
up,
and
the
p
ositive
y
-axis
p
oints
left.
b
The
p
ositive
x
axis
p
oints
do
wn,
and
the
negative
y
-axis
p
oints
right.
c
The
p
ositive
x
axis
p
oints
left,
and
the
p
ositive
y
-axis
p
oints
up.
d
The
negative
x
axis
p
oints
right,
and
the
p
ositive
y
-axis
p
oints
down.
e
The
p
ositive
x
axis
p
oints
up
and
to
the
right,
and
the
p
ositive
y
-axis
p
oints
down
and
to
the
right.
Q10
Supp
ose
we
draw
the
xy
plane
on
our
pap
er
in
the
standard
wa
y
,
and
our
pap
er
is
lying
on
a
table.
Do
es
the
z
-axis
p
oint
down
into
the
table
or
up
out
of
the
table?
4.1.3
Q11
Dra
w
diagrams
of
p
oints
with
the
following
coordinates.
a
(6
,
1
,
2)
b
(
−
3
,
0
,
0)
c
(2
,
−
1
,
4)
d
(0
,
3
,
5)
Q12
Dra
w
diagrams
of
p
oints
with
the
following
coordinates.
a
(
−
4
,
0
,
0)
b
(3
,
−
2
,
0)
253
Section
4.1
Exercises
c
(4
,
5
,
−
3)
d
(
−
1
,
3
,
4)
4.1.4
Q13
Compute
the
distance
b
et
ween
(3
,
6
,
2)
and
(7
,
3
,
−
10)
.
Q14
Compute
the
distance
b
et
ween
(0
,
3
,
2)
and
(5
,
1
,
0)
.
Q15
Compute
the
distance
b
et
ween
(10
,
12
,
109)
and
(11
,
9
,
105)
.
Q16
Compute
the
distance
b
et
ween
(53
,
42
,
9)
and
(43
,
78
,
2)
.
4.1.5
Q17
Do
es
the
p
oint
(4
,
3
,
8)
lie
on
the
graph
of
z
=
x
2
−
2
?
Explain
ho
w
you
kno
w.
Q18
Do
es
(2
,
2
,
1)
lie
on
the
graph
of
x
2
+
y
2
+
z
2
=
9
?
Explain
ho
w
you
kno
w.
Q19
What
is
the
graph
of
y
2
+
z
2
=
−
1
?
Explain
y
our
reasoning.
Q20
The
p
oint
(2
,
3
,
4)
lies
on
the
graph
ax
+
ay
−
z
=
26
.
What
is
the
value
of
the
numb
er
a
?
Q21
Olivia
says
that
the
graph
of
(
x
−
2)(
y
−
3)
=
0
in
the
xy
-plane
is
the
p
oint
(2
,
3)
.
Do
y
ou
agree?
Ho
w
would
y
ou
explain
it?
Q22
Ho
w
is
the
graph
of
f
(
x,
y
,
z
)
g
(
x,
y
,
z
)
=
0
related
to
the
graphs
of
f
(
x,
y
,
z
)
=
0
and
g
(
x,
y
,
z
)
=
0
?
254
4.1.6
Q23
Do
es
the
graph
of
z
=
4
intersect
the
graph
of
z
=
6
?
Explain
b
oth
using
geometry
and
algebra.
Q24
Do
es
the
graph
of
x
=
2
intersect
the
graph
of
z
=
1
?
Explain.
4.1.7
Q25
Sk
etch
the
graph
of
each
equation.
a
x
=
−
4
b
x
2
+
y
2
=
9
c
x
2
+
4
x
+
y
2
+
z
2
−
2
z
=
4
Q26
Sk
etch
a
graph
of
z
=
−
2
.
Q27
Sk
etch
a
graph
of
y
=
−
z
2
.
Q28
Sk
etch
a
graph
of
x
2
+
z
2
=
25
.
4.1.8
Q29
What
dimension
y
ou
we
expect
the
graph
of
an
equation
to
b
e
in
6
-dimensional
space?
Q30
What
is
the
graph
of
x
2
+
y
2
=
0
in
the
xy
-plane?
Is
this
an
exception
to
our
intuition
about
the
dimension
of
a
graph?
Q31
Zo
e
and
Muhammad
b
oth
sketch
the
graph
of
y
=
x
2
.
Zo
e’s
graph
is
a
curve.
Muhammad’s
is
a
surface.
Has
one
of
them
dra
wn
the
wrong
graph?
Explain.
Q32
In
R
3
,
what
is
the
dimension
of
the
intersection
of
the
graphs
x
2
+
y
2
=
25
and
z
=
1
?
Can
you
explain
this
in
terms
of
our
intuition
ab
out
the
dimension
of
a
graph.
255
Section
4.1
Exercises
4.1.9
Q33
Supp
ose
that
y
is
a
free
variable
in
the
equation
of
a
plane.
What
do
es
that
tell
us
ab
out
m
x
and
m
y
?
Q34
Gabb
y
is
trying
to
find
the
equation
of
a
plane
P
,
but
she
do
esn’t
kno
w
any
p
oints
on
the
xz
-plane
o
r
y
z
-plane.
Instead
she
knows
that
P
contains
the
p
oints:
A
=
(1
,
3
,
6)
B
=
(5
,
3
,
4)
C
=
(7
,
5
,
10)
Using
p
oints
A
and
B
,
she
decides
that
m
x
=
4
−
6
5
−
1
=
−
1
2
.
Using
p
oints
A
and
C
,
she
decides
that
m
y
=
10
−
6
5
−
3
=
2
.
a
Which
of
Gabb
y’s
conclusions
do
you
agree
with
and
which
do
y
ou
disagree
with?
Why?
b
Ho
w
could
you
fix
the
one
that
is
wrong?
Q35
Sup
o
ose
y
ou
intend
to
write
the
equation
of
the
plane
through
A
,
B
and
C
in
slope-intercept
fo
rm.
If
A
=
(3
,
5
,
7)
and
B
=
(3
,
2
,
4)
,
what
value(s)
of
the
y
co
o
rdinate
of
C
would
make
it
easiest
to
compute
m
x
?
Q36
Recall
that
w
e
can
write
the
equation
of
a
line
in
R
2
in
p
oint-slop
e
form:
y
−
y
0
=
m
(
x
−
x
0
)
where
m
is
the
slop
e
and
(
x
0
,
y
0
)
is
a
known
p
oint.
This
was
esp
ecially
useful
in
single-variable
calculus
fo
r
writing
equations
of
tangent
lines.
a
Ho
w
w
ould
you
exp
ect
to
write
the
equation
of
the
plane
P
through
(2
,
4
,
−
6)
with
slopes
m
x
=
1
2
and
m
y
=
−
3
?
b
Do
es
y
our
answer
to
a
actually
pass
through
(2
,
4
,
−
6)
?
Ho
w
do
y
ou
know?
c
Is
your
answer
to
a
actually
the
equation
of
a
plane?
Ho
w
do
you
know?
Do
es
it
have
the
co
rrect
slop
es?
d
W
rite
a
general
expression
fo
r
p
oint-slop
e
form
fo
r
a
plane.
Q37
The
plane
P
has
slop
es
m
x
=
3
and
m
y
=
−
1
and
passes
through
(2
,
5
,
−
1)
.
a
W
rite
the
equation
of
P
is
p
oint-slop
e
form.
256
b
What
is
the
z
-intercept
of
P
.
Q38
Given
a
plane
with
m
x
=
5
and
m
y
=
2
,
we
can
conclude
that
the
plane
is
steep
er
in
the
x
-direction
than
the
y
-direction.
Is
the
x
-direction
the
steep
est
direction
we
could
travel
in?
If
not,
what
is?
4.1.10
Q39
W
rite
the
equation
of
a
plane
through
(3
,
0
,
0)
,
(0
,
7
,
0)
,
and
(0
,
0
,
−
1)
.
Q40
W
rite
the
equation
of
a
plane
with
intercepts
(2
,
0
,
0)
,
(0
,
−
2
,
0)
,
and
(0
,
0
,
4)
.
Q41
W
rite
the
equation
of
a
plane
through
(6
,
4
,
1)
,
(6
,
7
,
−
2)
,
and
(8
,
7
,
1)
.
Q42
W
rite
the
equation
of
a
plane
through
(2
,
2
,
1)
,
(4
,
2
,
9)
,
and
(2
,
0
,
0)
.
Q43
W
rite
the
equation
of
a
plane
through
(3
,
4
,
2)
,
(5
,
5
,
6)
,
and
(7
,
4
,
6)
.
Q44
W
rite
the
equation
of
a
plane
through
(1
,
5
,
2)
,
(11
,
5
,
4)
,
and
(6
,
3
,
−
3)
.
4.1.11
Q45
Assuming
you
could
draw
in
4
dimensions,
describ
e
how
y
ou
might
construct
the
graph
of
x
2
1
+
x
2
3
+
x
2
4
=
25
in
R
4
.
Q46
Assuming
you
could
draw
in
4
dimensions,
describ
e
how
y
ou
might
construct
the
graph
of
x
2
=
x
2
3
in
R
4
.
Q47
What
equation(s)
w
ould
describ
e
the
x
2
x
4
-plane
in
R
4
?
Q48
What
w
ould
you
call
the
object
in
R
4
defined
b
y
x
1
=
0
?
257
Section
4.1
Exercises
Extension
and
Synthesis
Q49
The
p
oints
(1
,
0
,
3)
and
(1
,
4
,
0)
a
re
b
oth
on
the
sphere
S
.
What
are
the
p
ossible
values
for
the
radius
of
S
?
Q50
The
graph
of
x
2
+
y
2
=
0
in
R
2
is
a
p
oint,
not
a
curve.
Use
this
idea
to
write
an
equation
for
the
intersection
of
the
graphs
f
(
x,
y
,
z
)
=
c
and
g
(
x,
y
,
z
)
=
d
.
What
do
you
exp
ect
the
dimension
of
this
intersection
to
b
e?
Q51
Supp
ose
the
x
and
y
axes
in
R
2
w
ere
not
p
erp
endicula
r.
Would
the
distance
fo
rmula
still
hold?
Demonstrate.
258
Section
4.2
F
unctions
of
Several
V
a
riables
Goals:
1
Convert
an
implicit
function
to
an
explicit
function.
2
Calculate
the
domain
of
a
multiva
riable
function.
3
Calculate
level
curves
and
cross
sections
.
If
we
want
to
understand
the
relationship
b
et
ween
variables,
a
function
is
the
gold
standard.
Fo
r
example,
when
we
can
write
y
as
a
function
of
x
,
then
at
each
value
of
x
,
we
simply
need
plug
in
the
value
and
simplify
the
arithmetic.
There
is
no
chance
that
algebraic
manipulation
will
lead
us
to
multiple
values
of
y
,
or
to
an
equation
we
cannot
solve.
Naturally
,
w
e
want
to
understand
this
type
of
relationship
b
etw
een
more
than
tw
o
variables.
Much
like
our
investigation
of
n
-space,
we’ll
b
egin
by
adding
one
va
riable.
After
this
initial
step,
extrap
olating
to
more
va
riables
will
b
e
straightforw
ard.
Question
4.2.1
What
Is
a
F
unction
of
More
than
One
V
ariable?
Definition
A
function
of
tw
o
variables
is
a
rule
that
assigns
a
numb
er
(the
output
)
to
each
ordered
pair
of
real
numb
ers
(
x,
y
)
in
its
domain
.
The
output
is
denoted
f
(
x,
y
)
.
Some
functions
can
b
e
defined
algeb
raically
.
If
f
(
x,
y
)
=
p
36
−
4
x
2
−
y
2
then
f
(1
,
4)
=
p
36
−
4
·
1
2
−
4
2
=
4
.
Example
4.2.2
The
Domain
of
a
F
unction
Identify
the
domain
of
f
(
x,
y
)
=
p
36
−
4
x
2
−
y
2
.
259
Example
4.2.2
The
Domain
of
a
F
unction
Solution
The
only
obstacle
to
evaluating
this
function
is
that
the
value
under
the
square
root
might
b
e
negative.
W
e
can
write
an
inequality
to
exp
ress
this
and
solve.
36
−
4
x
2
−
y
2
≥
0
36
≥
4
x
2
+
y
2
1
≥
x
2
9
+
y
2
36
These
a
re
the
p
oints
inside
an
ellipse
whose
intercepts
a
re
(
±
3
,
0)
and
(0
,
±
6)
.
Figure:
The
domain
of
a
function
Main
Idea
When
solving
for
the
domain
of
an
algebraic
function,
we
lo
ok
fo
r
the
same
obstacles
to
evaluating
the
function
that
w
e
do
for
one-va
riable
functions.
Exp
ressions
in
a
denominator
cannot
be
0
(including
built-in
fractions
lik
e
tan
x
=
sin
x
cos
x
)
Exp
ressions
in
a
square
root
must
b
e
greater
than
o
r
equal
to
0
.
Exp
ressions
in
a
logarithm
must
be
greater
than
0
.
The
conditions
these
produce
with
a
tw
o-va
riable
function
may
b
e
harder
to
visualize
o
r
simplify
than
with
a
function
of
one
va
riable.
260
Application
4.2.3
T
emperature
Maps
Many
useful
functions
cannot
b
e
defined
algebraically
.
There
is
a
function
T
(
x,
y
)
which
gives
the
temp
erature
at
each
latitude
and
longitude
(
x,
y
)
on
earth.
No
pair
(
x,
y
)
has
more
than
one
temp
erature,
and
no
pair
fails
to
have
a
temp
erature.
Still
there
is
no
hop
e
of
producing
an
expression
that
computes
T
for
any
x
and
y
.
Mathematically
(though
p
erhaps
not
meteorologically)
this
function
is
a
rbitrary
.
T
(
−
71
.
06
,
42
.
36)
=
50
T
(
−
84
.
38
,
33
.
75)
=
59
T
(
−
83
.
74
,
42
.
28)
=
41
Figure:
A
temp
erature
map
This
function
is
rep
resented
graphically
by
using
colo
r
to
p
ortra
y
the
value
of
T
at
each
p
oint.
Application
4.2.4
Digital
Images
A
digital
image
is
made
up
of
pixels,
each
with
a
different
colo
r.
In
many
mo
dern
images,
these
pixels
a
re
to
o
small
to
see.
The
colo
r
of
each
pixel
is
a
function
of
that
pixel’s
lo
cation.
Since
colors
a
re
ha
rder
to
define
numerically
,
we
can
consider
the
simpler
case:
where
each
pixel
is
a
different
shade
of
gra
y
.
In
this
case
we
have
a
brightness
function
B
(
x,
y
)
where
the
output
is
a
numb
er
that
represents
the
b
rightness
of
the
pixel
at
the
co
o
rdinates
(
x,
y
)
.
y
x
687
1024
B
(339
,
773)
=
158
B
(340
,
773)
=
127
Figure:
An
image
rep
resented
as
a
brightness
function
B
on
each
pixel
261
Application
4.2.4
Digital
Images
Rema
rk
The
b
rightness
function
differs
from
other
functions
w
e’ve
studied
in
one
k
ey
wa
y
.
It
is
only
defined
for
(
x,
y
)
where
x
and
y
are
integers.
Other
examples
can
take
any
real
numb
ers
as
co
ordinates.
This
makes
our
usual
calculus
metho
ds
impossible.
We
cannot
get
arbitra
rily
close
to
a
point
in
order
to
compute
a
limit.
All
other
p
oints
are
at
least
1
unit
a
w
ay
.
How
ever,
if
we
a
re
willing
to
settle
for
appro
ximations,
w
e
can
apply
calculus
and
get
useful
results.
Question
4.2.5
What
Is
the
Graph
of
a
Tw
o-Va
riable
Function?
A
graph
is
our
most
important
wa
y
to
visualize
a
function.
The
graph
of
a
one
variable
functions
is
an
object
in
t
wo-space.
One
dimension
measures
the
input
variable.
The
other
measures
the
output.
F
or
a
t
wo
variable
function,
the
graph
lies
in
three-space.
Definition
The
graph
of
a
function
f
(
x,
y
)
is
the
set
of
all
p
oints
(
x,
y
,
z
)
that
satisfy
z
=
f
(
x,
y
)
.
The
height
z
ab
ove
a
p
oint
(
x,
y
)
represents
the
value
of
the
function
at
(
x,
y
)
.
In
this
figure,
f
(1
,
4)
is
equal
to
the
height
of
the
graph
ab
ove
(1
,
4
,
0)
.
Click to Load Applet
Figure:
The
graph
z
=
p
36
−
4
x
2
−
y
2
262
Question
4.2.6
Ho
w
Do
We
Visualize
a
Graph
in
Three-Space?
Three-space
is
ha
rder
to
visualize
than
tw
o-space.
What’s
more,
plotting
points
is
more
arduous
with
t
wo
dimensions
of
inputs.
In
the
absence
of
computer
graphics,
mathematicians
have
used
a
variet
y
of
visualization
to
ols.
Definition
A
level
set
of
a
function
f
(
x,
y
)
is
the
graph
of
the
equation
f
(
x,
y
)
=
c
for
some
constant
c
.
Fo
r
a
function
of
t
wo
va
riables
this
graph
lies
in
the
xy
-plane
and
is
called
a
level
curve
.
Example
Consider
the
function
f
(
x,
y
)
=
p
36
−
4
x
2
−
y
2
.
The
level
curve
p
36
−
4
x
2
−
y
2
=
4
simplifies
to
4
x
2
+
y
2
=
20
.
This
is
an
ellipse.
Other
level
curves
have
the
form
p
36
−
4
x
2
−
y
2
=
c
o
r
4
x
2
+
y
2
=
36
−
c
2
.
These
a
re
larger
o
r
smaller
ellipses.
Level
curves
take
their
shap
e
from
the
intersection
of
z
=
f
(
x,
y
)
and
z
=
c
.
Seeing
many
level
curves
at
once
can
help
us
visualize
the
shap
e
of
the
graph.
Click to Load Applet
Figure:
The
graph
z
=
f
(
x,
y
)
,
the
planes
z
=
c
,
and
the
level
curves
263
Example
4.2.7
Dra
wing
Level
Curves
Where
a
re
the
level
curves
on
this
temp
erature
map?
Figure:
A
temp
erature
map
Solution
The
level
sets
are
the
p
oints
where
the
temp
erature
has
a
certain
value.
Since
the
colo
rs
represent
ranges
of
temp
eratures,
it’s
difficult
to
pick
out
the
level
sets
within
that
range.
Ho
wever,
at
the
transition
from
one
color
to
the
next,
we
know
that
the
temp
erature
is
equal
to
the
cutoff
temp
erature
b
etw
een
those
ranges.
The
picture
b
elow
shows
a
reasonable
attempt
to
sketch
three
level
curves
in
white.
Notice
that
the
level
curves
(esp
ecially
the
one
b
etw
een
green
and
yello
w)
are
not
connected,
and
that
drawing
them
in
p
erfect
detail
is
b
ey
ond
the
ability
of
a
human.
264
Example
4.2.8
Using
Level
Curves
to
Describ
e
a
Graph
What
features
can
w
e
discern
from
the
le
vel
curves
of
this
top
ographical
map?
Figure:
A
top
ographical
map
265
Example
4.2.8
Using
Level
Curves
to
Describ
e
a
Graph
Solution
A
B
C
D
D
D
There
a
re
many
features
we
could
describe.
Here
is
a
sample.
The
p
oint
A
is
surrounded
by
relatively
flat
terrain.
There
are
not
many
level
curves
here,
which
means
the
altitude
is
not
increasing
o
r
decreasing
to
higher
o
r
low
er
levels.
The
p
oints
B
and
C
are
on
slop
es.
If
we
travel
north
and
south
we
cross
level
curves,
meaning
our
altitude
is
increasing
or
decreasing.
The
slop
e
is
steep
er
at
B
than
at
C
,
b
ecause
traveling
no
rth
from
B
we
cross
more
level
curves
than
traveling
no
rth
from
C
The
p
oints
mark
ed
D
a
re
in
the
middle
of
a
series
of
rings
of
leve
l
curves.
These
are
either
the
tops
of
hills
o
r
(less
likely
given
the
context)
the
b
ottoms
of
valleys.
Example
4.2.9
A
Cross
Section
Definition
The
intersection
of
a
plane
with
a
graph
is
a
cross
section
.
A
level
curve
is
a
type
of
cross
section,
but
not
all
cross
sections
a
re
level
curves.
266
Find
the
cross
section
of
z
=
p
36
−
4
x
2
−
y
2
at
the
plane
y
=
1
.
Click to Load Applet
Figure:
The
y
=
1
cross
section
of
z
=
p
36
−
4
x
2
−
y
2
Example
4.2.10
Converting
an
Implicit
Equation
to
a
F
unction
Definition
W
e
sometimes
call
an
equation
in
x
,
y
and
z
an
implicit
equation
.
Often
in
o
rder
to
graph
these,
we
convert
them
to
explicit
functions
of
the
form
z
=
f
(
x,
y
)
W
rite
the
equation
of
a
pa
rab
oloid
x
2
−
y
+
z
2
=
0
as
one
o
r
more
explicit
functions
so
it
can
b
e
graphed.
Then
find
the
level
curves.
Click to Load Applet
Figure:
Level
curves
of
x
2
−
y
+
z
2
=
0
267
Question
4.2.11
Ho
w
Do
es
this
Apply
to
Functions
of
Mo
re
Va
riables?
W
e
can
define
functions
of
three
va
riables
as
well.
Denoting
them
f
(
x,
y,
z
)
.
Fo
r
even
mo
re
variables,
w
e
use
x
1
through
x
n
.
The
definitions
of
this
section
can
b
e
extrap
olated
as
follo
ws.
V
ariables
2
3
n
F
unction
f
(
x,
y
)
f
(
x,
y
,
z
)
f
(
x
1
,
.
.
.
,
x
n
)
Domain
subset
of
R
2
subset
of
R
3
subset
of
R
n
Graph
z
=
f
(
x,
y
)
in
R
3
w
=
f
(
x,
y,
z
)
in
R
4
x
n
+1
=
f
(
x
1
,
.
.
.
,
x
n
)
in
R
n
+1
Level
Sets
level
curve
in
R
2
level
surface
in
R
3
level
set
in
R
n
Observation
W
e
might
hop
e
to
solve
an
implicit
equation
of
n
variables
to
obtain
an
explicit
function
of
n
−
1
va
riables.
How
ever,
we
can
also
treat
it
as
a
level
set
of
an
explicit
function
of
n
variables
(whose
graph
lives
in
n
+
1
dimensional
space).
x
2
+
y
2
+
z
2
=
25
F
(
x,
y
,
z
)
=
x
2
+
y
2
+
z
2
F
(
x,
y
,
z
)
=
25
f
(
x,
y
)
=
±
p
25
−
x
2
−
y
2
Both
viewp
oints
will
b
e
useful
in
the
future.
268
Section
4.2
Exercises
Summa
ry
Questions
Q1
What
do
es
the
height
of
the
graph
z
=
f
(
x,
y
)
represent?
Q2
What
is
the
distinction
b
et
ween
a
level
set
and
a
cross
section?
Q3
What
a
re
level
sets
in
R
2
and
R
3
called?
Q4
What
is
the
difference
b
et
ween
an
implicit
equation
and
explicit
function?
4.2.1
Q5
If
f
(
x,
y
)
=
13
x
+
y
x
,
compute
f
(2
,
−
8)
.
Q6
If
f
(
x,
y
)
=
cos(
π
xy
)
,
compute
f
4
,
1
3
.
Q7
Is
f
(
x,
y
)
=
±
√
4
x
−
y
a
function?
Explain.
Q8
Is
the
follo
wing
a
function?
Explain.
f
(
x,
y
)
=
(
√
y
if
y
≥
0
√
x
if
x
≥
0
269
Section
4.2
Exercises
4.2.2
Q9
Compute
the
domain
of
f
(
x,
y
)
=
1
x
+
y
.
Q10
What
is
the
domain
of
f
(
x,
y
)
=
1
x
2
+
y
2
?
Q11
What
is
the
domain
of
g
(
x,
y
)
=
x
3
+
p
y
2
−
25
?
Q12
What
is
the
domain
of
g
(
x,
y
)
=
15
+
ln(
y
−
2
x
)
?
Q13
What
is
the
domain
of
f
(
x,
y
)
=
√
x
+3
y
2
−
x
?
Q14
Compute
the
domain
of
h
(
x,
y
)
=
4
x
y
−
ln
x
4.2.3
Q15
On
the
temp
erature
map,
we
saw
T
(
−
84
.
38
,
33
.
75)
=
59
.
Is
T
(
−
84
.
38
,
35
.
75)
greater
than
or
less
than
59
?
Q16
On
the
temp
erature
map,
we
saw
T
(
−
83
.
74
,
42
.
28)
=
41
.
Is
T
(
−
93
.
74
,
42
.
28)
greater
than
or
less
than
41
?
Q17
What
range
of
temp
eratures
are
found
in
South
Dakota?
In
which
parts
of
the
state
are
the
extreme
temp
eratures
found?
Q18
Can
y
ou
use
this
diagram
to
appro
ximate
T
(
−
61
.
06
,
42
.
36)
?
Explain.
4.2.4
Q19
In
our
image
of
Mona
Lisa,
what
is
the
domain
of
B
?
Q20
In
our
blow-up
of
the
digital
image,
we
see
Mona
Lisa’s
eye
is
near
the
co
ordiante
(369
,
800)
.
Where
is
her
other
ey
e?
270
4.2.5
Q21
Can
the
p
oints
(1
,
3
,
5)
and
(1
,
3
,
7)
b
oth
b
e
on
the
graph
of
z
=
f
(
x,
y
)
?
Explain.
Q22
If
the
graph
z
=
f
(
x,
y
)
is
b
elow
the
xy
-plane,
what
do
es
that
tell
us
ab
out
f
(
x,
y
)
?
Q23
If
f
(
x,
y
)
has
a
z
-intercept
of
c
,
what
do
es
that
tell
us
ab
out
f
?
Q24
What
is
the
significance
of
the
p
oints
where
the
graph
z
=
f
(
x,
y
)
intersects
the
xy
-plane?
4.2.6
Q25
Describ
e
the
level
curves
of
f
(
x,
y
)
=
(
x
−
2)
2
+
(
y
+
1)
2
.
Q26
Describ
e
the
level
curves
of
f
(
x,
y
)
=
x
2
−
3
y
+
5
.
Q27
Describ
e
the
level
curves
of
x
2
y
.
Q28
Describ
e
the
level
curves
of
g
(
x,
y
)
=
y
e
x
.
Q29
Give
the
equation
of
the
level
curve
of
f
(
x,
y
)
=
x
3
+
y
3
that
passes
through
(4
,
2)
.
Q30
Give
the
equation
of
the
level
curve
of
g
(
x,
y
)
=
17
x
2
−
3
xy
+
y
3
that
contains
the
p
oint
(1
,
2)
.
Q31
Given
a
function
f
(
x,
y
)
,
how
many
level
curves
might
pass
through
(3
,
7)
?
Q32
If
the
p
oints
(
x
1
,
y
1
)
and
(
x
2
,
y
2
)
lie
on
the
same
level
curve
of
h
(
x,
y
)
,
what
are
the
p
ossible
values
of
the
exp
ression
h
((
x
1
,
y
1
)
−
h
(
x
2
,
y
2
)
?
271
Section
4.2
Exercises
4.2.7
Q33
In
our
level
curves
on
the
temp
erature
map,
what
physical
meaning
can
we
tak
e
from
the
fact
that
the
green-y
ellow
and
red-orange
level
curves
are
closer
together
in
Kansas
than
they
are
fa
rther
east?
Q34
Explain
why
it
makes
sense
physically
that
level
curves
of
a
temp
erature
function
would
b
e
complicated
and
disconnected.
4.2.8
Q35
In
the
top
ographical
map,
what
can
w
e
deduce
from
the
fact
that
no
level
curves
cross
the
farm
fields
in
the
lo
wer
center
of
the
map?
Q36
Explain
why
it
makes
physical
sense
that
there
are
level
curves
alongside
the
creeks
in
this
map.
4.2.9
Q37
Give
an
equation
fo
r
the
y
=
2
cross-section
of
the
graph
z
=
f
(
x,
y
)
where
f
(
x,
y
)
=
x
3
+
y
3
.
Q38
Consider
the
plane
P
whose
equation
is
f
(
x,
y
)
=
3
x
−
5
y
+
7
.
i.
Give
the
equation
of
the
y
=
0
cross
section
of
P
.
What
is
this
graph?
What
is
the
significance
of
the
va
rious
parts
of
its
equation?
ii.
Give
the
equation
of
the
x
=
0
cross
section
of
P
.
What
is
the
significance
of
the
various
pa
rts
of
its
equation?
iii.
Give
the
equation
and
describ
e
the
set
of
all
level
curves
of
f
.
Q39
If
the
cross
sections
of
z
=
f
(
x,
y
)
in
the
planes
y
=
b
are
identical
for
all
values
of
b
,
what
do
es
that
tell
us
ab
out
f
?
Q40
If
f
(
x,
y
)
is
a
function
that
satisfies
f
(
x,
y
)
=
f
(
x,
−
y
)
for
all
x
and
y
,
how
will
this
be
refelected
in
the
cross
sections
of
z
=
f
(
x,
y
)
?
272
4.2.10
Q41
Rewrite
y
=
x
2
+
z
2
as
one
o
r
more
explicit
functions
z
=
f
(
x,
y
)
.
Q42
Rewrite
ln
x
+
ln
y
+
ln
z
=
0
as
one
or
more
explicit
functions
z
=
f
(
x,
y
)
.
Q43
Rewrite
x
2
+
y
2
+
z
2
+
xy
z
=
20
as
one
o
r
more
explicit
functions
z
=
f
(
x,
y
)
.
Q44
Explain
why
it
would
b
e
difficult
to
write
ln
y
z
−
√
xz
=
5
+
x
as
an
explicit
function
of
the
fo
rm
z
=
f
(
x,
z
)
.
Cho
ose
a
b
etter
dep
endent
and
variable
and
write
that
variable
as
a
function
of
the
other
t
wo.
4.2.11
Q45
Consider
the
function
f
(
x
1
,
x
2
,
x
3
,
x
4
,
x
5
)
.
a
What
space
do
es
the
graph
of
f
lie
in?
b
What
space
do
es
a
level
set
of
f
lie
in?
Q46
W
rite
xy
z
=
1
as
a
A
level
set
of
a
function
b
An
explicit
function
z
=
f
(
x,
y
)
Q47
Consider
a
one-va
riable
function
f
(
x
)
.
a
What
space
do
es
the
graph
of
f
(
x
)
lie
in?
b
Where
do
es
a
level
set
of
f
lie
in?
What
do
es
a
typical
level
set
lo
ok
lik
e?
Q48
Sho
w
how
the
graph
of
an
explicit
function
x
n
+1
=
f
(
x
1
,
x
2
,
.
.
.
,
x
n
)
can
b
e
converted
to
the
level
set
of
an
n
+
1
-va
riable
function.
273
Section
4.2
Exercises
Synthesis
&
Extension
Q49
Let
f
(
x,
y
)
=
x
2
.
Sk
etch
the
graph
of
z
=
f
(
x,
y
)
.
What
is
the
role
of
y
in
this
graph?
Q50
Consider
the
implicit
equation
z
x
=
y
a
Rewrite
this
equation
as
an
explicit
function
z
=
f
(
x,
y
)
.
b
What
is
the
domain
of
f
?
c
Solve
fo
r
and
sketch
a
few
level
sets
of
f
.
d
What
do
the
level
sets
tell
y
ou
ab
out
the
graph
z
=
f
(
x,
y
)
?
Q51
Consider
the
implicit
equation:
x
=
sin
z
.
a
Sk
etch
a
graph
of
the
equation.
b
Describ
e
(in
w
ords)
what
the
cross
section
of
the
graph
in
the
x
=
1
2
plane
lo
oks
lik
e.
274
Section
4.3
Limits
and
Continuit
y
Goals:
1
Understand
the
definition
of
a
limit
of
a
multiva
riable
function.
2
Use
the
Squeeze
Theo
rem
3
Apply
the
definition
of
continuit
y
.
Limits
of
multivariable
functions
are
conceptually
similar
to
one-va
riable
functions.
How
ever,
even
though
the
requirement
is
the
same,
it
is
a
much
harder
to
satisfy
.
Since
there
are
so
many
more
wa
ys
to
approach
a
given
p
oint
in
a
higher
dimensional
space,
there
a
re
more
nearb
y
p
oints
to
check
to
see
whether
the
function
is
actually
app
roaching
the
proposed
limit.
Question
4.3.1
What
Is
the
Limit
of
a
F
unction?
Definition
W
e
write
lim
(
x,y
)
→
(
a,b
)
f
(
x,
y
)
=
L
if
we
can
mak
e
the
values
of
f
stay
a
rbitrarily
close
to
L
by
restricting
to
a
sufficiently
small
neighb
orhoo
d
of
(
a,
b
)
.
Proving
a
limit
exists
requires
a
fo
rmula
or
rule.
Fo
r
any
amount
of
closeness
required
(
ϵ
),
you
must
b
e
able
to
p
ro
duce
a
radius
δ
a
round
(
a,
b
)
sufficiently
small
to
k
eep
|
f
(
x,
y
)
−
L
|
<
ϵ
.
Fo
r
this
reason,
w
e
will
not
prove
that
any
limits
exist.
We
will
present
three
examples
of
functions
whose
limit
does
not
exist.
Example
4.3.2
A
Limit
That
Do
es
Not
Exist
Sho
w
that
lim
(
x,y
)
→
(0
,
0)
x
2
−
y
2
x
2
+
y
2
do
es
not
exist.
275
Example
4.3.2
A
Limit
That
Do
es
Not
Exist
Solution
Let’s
define
f
(
x,
y
)
=
x
2
−
y
2
x
2
+
y
2
.
We
will
approach
the
p
oint
(0
,
0)
from
t
wo
different
directions.
If
w
e
app
roach
along
the
x
-axis,
then
the
p
oints
on
our
path
have
the
form
(
x,
0)
.
When
we
plug
these
into
the
function,
the
value
is
f
(
x,
0)
=
x
2
−
0
x
2
+0
.
This
is
equal
to
1
fo
r
all
values
of
x
except
0
,
so
as
x
app
roaches
0
,
the
values
of
f
are
arbitra
rily
close
(in
fact
exactly
equal)
to
1
.
On
the
other
hand,
if
we
approach
1
along
the
y
-axis,
then
the
p
oints
have
the
form
(0
,
y
)
.
When
w
e
plug
these
into
the
function,
the
value
is
f
(0
,
y
)
=
0
−
y
2
0+
y
2
.
This
is
equal
to
−
1
for
all
values
of
y
except
0
,
so
as
y
approaches
0
,
the
values
of
f
are
arbitra
rily
close
(in
fact
exactly
equal)
to
−
1
.
What
do
es
this
say
ab
out
the
limit
of
f
?
The
lim
(
x,y
)
→
(0
,
0)
f
(
x,
y
)
=
1
b
ecause
there
are
p
oints
on
the
y
-axis
do
not
give
values
close
to
1
,
but
any
neighb
orhoo
d
of
(0
,
0)
includes
some
points
on
the
y
-axis.
Simila
rly
,
lim
(
x,y
)
→
(0
,
0)
f
(
x,
y
)
=
−
1
.
If
w
e
tried
to
argue
that
the
limit
had
any
other
value,
the
x
-axis
and
y
-axis
would
b
oth
present
a
problem.
This
this
limit
do
es
not
exist.
W
e
can
identify
the
problem
b
ehavio
r
in
the
graph
of
z
=
f
(
x,
y
)
.
As
the
graph
app
roaches
the
o
rigin,
there
are
p
oints
of
all
heights
betw
een
−
1
and
1
.
Sp
ecifically
we
can
see
the
line
ab
ove
the
x
-axis
and
b
elo
w
the
y
-axis.
No
amount
of
closeness
can
exclude
this
range
of
values.
Click to Load Applet
Figure:
A
function
with
no
limit
at
(0
,
0)
W
e
might
tak
e
aw
a
y
the
idea
that
checking
limits
of
tw
o-variable
functions
requires
checking
in
b
oth
the
x
-direction
and
the
y
-direction.
Unfortunately
,
even
that
is
not
sufficient.
Example
4.3.3
Another
Limit
That
Do
es
Not
Exist
Sho
w
that
lim
(
x,y
)
→
(0
,
0)
xy
x
2
+
y
2
do
es
not
exist.
276
Solution
Let
f
(
x,
y
)
=
xy
x
2
+
y
2
.
We
can
check
the
values
of
this
function
on
the
x
-
and
y
-axes.
Except
at
(0
,
0)
,
f
(
x,
0)
=
0
and
f
(0
,
y
)
=
0
.
Ho
wever,
not
all
the
p
oints
close
to
(0
,
0)
lie
on
an
axis.
Supp
ose
we
wo
rk
with
the
p
oints
on
another
line:
y
=
mx
.
These
p
oints
have
the
form
(
x,
mx
)
.
W
e
can
evaluate
f
on
this
line.
f
(
x,
xm
)
=
(
x
)(
mx
)
x
2
+
(
mx
)
2
=
mx
2
(
m
2
+
1)
x
2
=
m
m
2
+
1
(
except
at
(0
,
0))
Thus
there
a
re
point
arbitra
rily
close
to
(0
,
0)
on
which
f
is
valued
as
lo
w
as
−
0
.
5
(
m
=
−
1
)
and
as
high
as
0
.
5
(
m
=
1
).
The
limit
do
es
not
exist.
Click to Load Applet
Figure:
The
graph
z
=
xy
x
2
+
y
2
and
the
line
of
height
1
2
over
x
=
y
.
W
e
might
take
aw
ay
the
idea
that
checking
limits
of
tw
o-variable
functions
requires
checking
along
each
line
through
the
p
oint
in
question.
Unfo
rtunately
,
even
that
is
not
sufficient.
Example
4.3.4
Y
et
Another
Limit
That
Do
es
Not
Exist
Sho
w
that
lim
(
x,y
)
→
(0
,
0)
xy
2
x
2
+
y
4
do
es
not
exist.
Solution
Let
f
(
x,
y
)
=
xy
2
x
2
+
y
4
.
We
can
check
the
values
of
this
function
on
the
x
-
and
y
-axes.
Except
at
(0
,
0)
,
277
Example
4.3.4
Y
et
Another
Limit
That
Do
es
Not
Exist
f
(
x,
0)
=
0
and
f
(0
,
y
)
=
0
.
We
can
also
check
the
values
along
a
line
of
the
form
y
=
mx
.
f
(
x,
xm
)
=
(
x
)(
mx
)
2
x
2
+
(
mx
)
4
=
m
2
x
3
x
2
(1
+
m
4
x
2
)
lim
x
→
0
f
(
x,
xm
)
=
lim
x
→
0
m
2
x
3
x
2
(1
+
m
4
x
2
)
=
lim
x
→
0
m
2
x
1
+
m
4
x
2
=
0
Thus
along
each
line,
the
values
of
f
approach
0
as
we
approach
the
o
rigin.
How
ever,
w
e
have
not
considered
paths
that
a
re
not
line.
Consider
the
parabola
x
=
y
2
.
Points
on
this
parabola
have
the
form
(
y
2
,
y
)
.
We
compute
the
values
on
this
parabola.
f
(
y
2
,
y
)
=
(
y
2
)(
y
)
2
(
y
2
)
2
+
y
4
=
y
4
2
y
4
F
or
any
point
on
this
parabola
except
the
origin
f
has
a
value
of
1
2
.
Thus
f
takes
values
of
1
2
and
0
in
any
neighb
o
rho
o
d
of
(0
,
0)
,
meaning
the
limit
do
es
not
exist.
Click to Load Applet
Figure:
The
graph
z
=
xy
2
x
2
+
y
4
,
which
limits
to
0
along
any
line
through
the
origin,
but
has
height
1
2
over
the
pa
rab
ola
x
=
y
2
W
e
take
a
wa
y
from
these
exercises
that
establishing
the
value
of
a
multi-variable
limit
cannot
b
e
reduced
to
computing
a
single-variable
limit,
or
even
a
family
of
single-variable
limits.
The
formal
a
rguments
that
establish
multi-variable
limits
a
re
more
advanced
and
beyond
the
scope
of
this
text.
278
Question
4.3.5
What
T
o
ols
Apply
to
Multi-Va
riable
Limits?
The
limit
la
ws
from
single-variable
limits
transfer
comfo
rtably
to
multi-variable
functions.
1
Sum/Difference
Rule
2
Constant
Multiple
Rule
3
Pro
duct/Quotient
Rule
These
rules
allo
w
us
to
compute
limits
of
complicated
functions
from
simpler
ones.
Ho
w
do
we
come
b
y
those
simpler
limits
in
the
first
place?
We
can
apply
the
kind
of
advanced
arguments
we
alluded
to
ea
rlier.
Another
to
ol
is
the
squeeze
theorem.
The
Squeeze
Theo
rem
If
g
<
f
<
h
in
some
neighb
orhoo
d
of
(
a,
b
)
and
lim
(
x,y
)
→
(
a,b
)
g
(
x,
y
)
=
lim
(
x,y
)
→
(
a,b
)
h
(
x,
y
)
=
L,
then
lim
(
x,y
)
→
(
a,b
)
f
(
x,
y
)
=
L.
Question
4.3.6
What
Is
a
Continuous
F
unction?
Definition
W
e
say
f
(
x,
y
)
is
continuous
at
(
a,
b
)
if
lim
(
x,y
)
→
(
a,b
)
f
(
x,
y
)
=
f
(
a,
b
)
.
In
a
rigorous
development
of
calculus,
we
compute
limits
and
use
them
to
show
that
functions
are
continuous.
Given
that
evaluating
limits
is
b
ey
ond
our
current
means,
w
e
will
reverse
the
process.
Rather
than
w
orrying
ab
out
how
to
prove
the
follo
wing
theorem,
we
will
assume
it
is
true
and
use
it
to
evaluate
limits.
279
Question
4.3.6
What
Is
a
Continuous
F
unction?
Theo
rem
P
olynomials,
ro
ots,
trig
functions,
exp
onential
functions
and
logarithms
are
continuous
on
their
domains.
Sums,
differences,
products,
quotients
and
comp
ositions
of
continuous
functions
are
continuous
on
their
domains.
The
limit
of
a
continuous
function
is
equal
to
the
value
of
the
function.
When
we
need
to
compute
a
limit
of
these
functions,
w
e’ll
just
evaluate
them
instead.
Why
didn’t
this
wo
rk
in
our
examples?
In
each
of
our
examples,
the
function
w
as
a
quotient
of
p
olynomials,
but
(0
,
0)
was
not
in
the
domain.
Rema
rk
Limits,
continuit
y
and
these
theorems
can
all
b
e
extrap
olated
to
functions
of
mo
re
variables.
Section
4.3
Exercises
Summa
ry
Questions
Q1
Why
is
it
ha
rder
to
verify
a
limit
of
a
multiva
riable
function?
Q2
What
do
y
ou
need
to
check
in
order
to
determine
whether
a
function
is
continuous?
280
Section
4.4
P
a
rtial
Derivatives
Goals:
1
Calculate
pa
rtial
derivatives
.
2
Realize
when
not
to
calculate
pa
rtial
derivatives.
The
first
task
in
developing
calculus
is
to
understand
rates
of
change.
In
the
single-variable
case,
we
ask
ho
w
the
dep
endent
variable
changes
p
er
unit
of
increase
in
the
indep
endent
va
riable.
With
mo
re
than
one
indep
endent
variable
we
must
ask:
what
kind
of
increase
do
we
mean?
There
is
more
than
one
p
ossible
answ
er.
Pa
rtial
derivatives
are
the
simplest
and
most
intuitive
rate
of
change.
Question
4.4.1
What
Is
the
Rate
of
Change
of
a
Multivariab
le
F
unction?
Motivational
Example
The
force
due
to
gravity
b
etw
een
tw
o
objects
dep
ends
on
their
masses
and
on
the
distance
b
etw
een
them.
Supp
ose
at
a
distance
of
8
,
000
km
the
force
b
etw
een
t
wo
particula
r
objects
is
100
newtons
and
at
a
distance
of
10
,
000
km,
the
force
is
64
newtons.
Ho
w
much
do
we
exp
ect
the
force
betw
een
these
objects
to
increase
o
r
decrease
p
er
kilometer
of
distance?
Solution
The
change
in
fo
rce
divided
by
the
change
in
distance
is
64
N
−
100
N
10
,
000
km
−
8
,
000
km
=
−
0
.
018
N
k
m
Notice
that
the
change
in
force
is
entirely
attributable
to
the
change
in
distance.
That
is
because
the
masses
of
the
objects
did
not
change.
The
only
change
in
the
dep
endent
variables
is
the
2
,
000
km
increase
in
distance.
Our
goals
in
understanding
multi-va
riable
rates
of
change
are
guided
b
y
what
we
accomplished
with
one
va
riable.
Derivatives
of
a
single-va
riable
function
were
a
wa
y
of
measuring
the
change
in
a
function.
Recall
the
follo
wing
facts
ab
out
f
′
(
x
)
.
1
Average
rate
of
change
is
realized
as
the
slop
e
of
a
secant
line:
f
(
x
)
−
f
(
x
0
)
x
−
x
0
281
Question
4.4.1
What
Is
the
Rate
of
Change
of
a
Multivariable
Function?
2
The
derivative
f
′
(
x
)
is
defined
as
a
limit
of
slop
es:
f
′
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
3
The
derivative
is
the
instantaneous
rate
of
change
of
f
at
x
.
4
The
derivative
f
′
(
x
0
)
is
realized
geometrically
as
the
slop
e
of
the
tangent
line
to
y
=
f
(
x
)
at
x
0
.
5
The
equation
of
that
tangent
line
can
b
e
written
in
p
oint-slop
e
form:
y
−
y
0
=
f
′
(
x
0
)(
x
−
x
0
)
In
the
physics
example
ab
ove,
the
rate
of
change
w
as
easier
to
understand
b
ecause
only
one
inde-
p
endent
va
riable
is
changing.
That
was
an
average
rate
of
change,
tak
en
b
etw
een
tw
o
p
oints.
We
now
develop
a
corresponding
instantaneous
rate
of
change.
A
pa
rtial
derivative
measures
the
rate
of
change
of
a
multiva
riable
function
as
one
variable
changes,
but
the
others
remain
constant.
Definition
The
pa
rtial
derivatives
of
a
t
wo-va
riable
function
f
(
x,
y
)
a
re
the
functions
f
x
(
x,
y
)
=
lim
h
→
0
f
(
x
+
h,
y
)
−
f
(
x,
y
)
h
and
f
y
(
x,
y
)
=
lim
h
→
0
f
(
x,
y
+
h
)
−
f
(
x,
y
)
h
.
W
e
can
see
the
idea
of
each
pa
rtial
derivative
in
the
formula.
f
x
compa
res
the
values
of
f
at
(
x
+
h,
y
)
and
(
x,
y
)
.
The
x
values
change
betw
een
these
tw
o
p
oints,
but
the
y
values
remain
constant.
The
opp
osite
is
true
in
the
fo
rmula
for
f
y
.
Notation
The
pa
rtial
derivative
of
a
function
can
b
e
denoted
a
va
riety
of
wa
ys.
Here
are
some
equivalent
notations
f
x
∂
f
∂
x
∂
z
∂
x
∂
∂
x
f
D
x
f
282
Example
4.4.2
Computing
a
P
artial
Derivative
Find
∂
∂
y
(
y
2
−
x
2
+
3
x
sin
y
)
.
Main
Idea
T
o
compute
a
pa
rtial
derivative
f
y
,
p
erform
single-variable
differentiation.
T
reat
y
as
the
indep
endent
va
riable
and
x
as
a
constant.
Solution
W
e
take
an
ordina
ry
derivative,
treating
y
as
the
variable
and
x
as
a
constant.
The
familiar
rules
of
derivatives
apply
.
The
sum
rule
means
we
can
differentiate
term-b
y-term.
∂
∂
y
y
2
=
2
y
∂
∂
y
x
2
=
0
,
since
the
x
2
term
is
treated
as
constant.
∂
∂
y
3
x
sin
y
=
3
x
cos
y
,
since
3
x
is
treated
as
constant
multiple
of
the
function
sin
y
.
T
ogether
this
gives
the
partial
derivative
∂
∂
y
(
y
2
−
x
2
+
3
x
sin
y
)
=
2
y
+
3
x
cos
y
.
Synthesis
4.4.3
Interp
reting
Derivatives
from
Level
Sets
Belo
w
are
the
level
curves
f
(
x,
y
)
=
c
for
some
values
of
c
.
Can
we
tell
whether
f
x
(
−
4
,
1
.
25)
and
f
y
(
−
4
,
1
.
25)
a
re
p
ositive
or
negative?
283
Synthesis
4.4.3
Interp
reting
Derivatives
from
Level
Sets
Figure:
Some
level
curves
of
f
(
x,
y
)
Solution
As
x
increases
and
y
remains
constant,
we
travel
to
the
right
in
the
co
ordinate
plane.
Based
on
the
lab
eling
of
the
level
curves,
this
takes
f
from
the
value
40
to
values
b
etw
een
40
and
50
,
meaning
f
increases.
Thus
f
x
>
0
.
Simila
rly
,
as
y
increases
and
x
remains
constant,
we
travel
upw
ards
in
the
co
o
rdinate
plane.
This
tak
es
f
from
the
value
40
to
values
b
etw
een
30
and
40
,
meaning
f
decreases.
Thus
f
y
<
0
.
Question
4.4.4
What
Is
the
Geometric
Significance
of
a
Pa
rtial
Derivative?
The
partial
derivative
f
x
(
x
0
,
y
0
)
is
realized
geometrically
as
the
slop
e
of
the
line
tangent
to
z
=
f
(
x,
y
)
at
(
x
0
,
y
0
,
z
0
)
and
traveling
in
the
x
direction.
Since
y
is
held
constant,
this
tangent
line
lives
in
y
=
y
0
,
a
plane
p
erp
endicula
r
to
the
y
-axis.
The
line
is
tangent
to
the
cross
section
of
the
graph
with
that
plane.
Click to Load Applet
Figure:
The
tangent
line
to
z
=
f
(
x,
y
)
in
the
x
direction
284
Example
4.4.5
Derivative
Rules
and
P
artial
Derivatives
Find
f
x
fo
r
the
following
functions
f
(
x,
y
)
:
a
f
=
√
xy
(
on
the
domain
x
>
0
,
y
>
0)
b
f
=
y
x
c
f
=
√
x
+
y
d
f
=
sin
(
xy
)
Solution
a
W
e
can
rewrite
this
as
f
(
x,
y
)
=
√
x
√
y
.
In
this
setting,
√
y
is
a
constant
multiple.
Thus
f
x
(
x,
y
)
=
1
2
√
x
√
y
b
W
e
can
rewrite
this
as
f
(
x,
y
)
=
1
x
y
.
We
treat
y
as
a
constant
multiple.
f
x
(
x,
y
)
=
−
1
x
2
y
.
c
W
e
cannot
rewrite
this
as
f
(
x,
y
)
√
x
+
√
y
,
b
ecause
that
is
not
a
valid
algebraic
manipulation.
Instead
w
e
use
the
chain
rule.
The
outer
function
is
√
x
.
Its
derivative
is
1
2
√
x
.
The
inner
function
is
x
+
y
.
Its
derivative
is
1
.
By
the
chain
rule
∂
∂
x
√
x
+
y
=
1
2
√
x
+
y
(1)
=
1
2
√
x
+
y
d
W
e
do
not
have
an
easy
trig
rule
to
b
reak
up
products.
We’ll
use
the
chain
rule
again.
The
outer
function
is
sin
x
.
Its
derivative
is
cos
x
.
The
inner
function
is
xy
.
Its
derivative
is
y
.
By
the
chain
rule
∂
∂
x
sin(
xy
)
=
cos(
xy
)
y
Main
Idea
Sometimes
we
can
detach
the
va
riable
held
constant
from
the
changing
variable
using
the
rules
of
algeb
ra.
When
we
can’t,
w
e’ll
often
need
a
differentiation
rule
(usually
the
chain
rule).
285
Question
4.4.6
What
If
W
e
Have
More
than
Tw
o
Va
riables?
W
e
can
also
calculate
partial
derivatives
of
functions
of
more
variables.
All
variables
but
one
are
held
to
b
e
constants.
:
Example
If
f
(
x,
y
,
z
)
=
x
2
−
xy
+
cos(
y
z
)
−
5
z
3
,
then
∂
f
∂
y
=
0
−
x
−
sin(
y
z
)
z
−
0
=
−
x
−
z
sin(
y
z
)
Example
4.4.7
A
F
unction
of
Three
Va
riables
F
or
an
ideal
gas,
we
have
the
law
P
=
nRT
V
,
where
P
is
pressure,
n
is
the
numb
er
of
moles
of
gas
molecules,
T
is
the
temp
erature,
and
V
is
the
volume.
a
Calculate
∂
P
∂
V
.
b
Calculate
∂
P
∂
T
.
c
(Science
Question)
Supp
ose
we’re
heating
a
sealed
gas
contained
in
a
glass
container.
Do
es
∂
P
∂
T
tell
us
ho
w
quickly
the
pressure
is
increasing
p
er
degree
of
temp
erature
increase?
Solution
a
W
e
can
write
P
=
nr
T
1
V
and
treat
nr
T
as
a
constant
multiple.
Then
∂
P
∂
V
=
nr
T
−
1
V
2
.
b
In
this
case,
nr
1
V
is
a
constant
multiple.
∂
P
∂
T
=
nr
1
V
(1)
.
c
No.
∂
P
∂
T
assumes
n
and
V
are
constant,
but
glass
expands
as
it
heats.
The
volume
of
b
oth
the
container
and
the
gas
is
increasing,
not
constant.
286
Question
4.4.8
Ho
w
Do
Higher
Order
Derivatives
Wo
rk?
T
aking
a
partial
derivative
of
a
partial
derivative
gives
us
a
higher
order
partial
derivative.
We
use
the
follo
wing
notation.
Notation
(
f
x
)
x
=
f
xx
=
∂
2
f
∂
x
2
W
e
need
not
use
the
same
variable
each
time
Notation
(
f
x
)
y
=
f
xy
=
∂
∂
y
∂
∂
x
f
=
∂
2
f
∂
y
∂
x
Rema
rk
Notice
the
subscript
notation
and
the
∂
notation
express
higher
order
derivatives
in
opp
osite
order.
Subscripts
a
re
added
to
the
right
of
f
,
which
the
differential
op
eration
is
applied
on
the
left
of
f
.
Example
4.4.9
A
Higher
Order
P
artial
Derivative
If
f
(
x,
y
)
=
sin(3
x
+
x
2
y
)
calculate
f
xy
.
287
Example
4.4.9
A
Higher
Order
P
artial
Derivative
Solution
First
w
e
compute
f
x
.
W
e’ll
need
the
chain
rule.
The
outer
function
is
sin
x
.
Its
derivative
is
cos
x
.
The
inner
function
is
3
x
+
x
2
y
.
Its
derivative
is
3
+
2
xy
.
f
x
=
cos(3
x
+
x
2
y
)(3
+
2
xy
)
.
Computing
(
f
x
)
y
will
require
the
p
ro
duct
rule.
∂
∂
y
cos(3
x
+
x
2
y
)
requires
the
chain
rule.
The
outer
function
is
cos
x
.
Its
derivative
is
−
sin
x
.
The
inner
function
is
3
x
+
x
2
y
.
Its
derivative
is
x
2
.
∂
∂
y
cos(3
x
+
x
2
y
)
=
−
sin(3
x
+
x
2
y
)(
x
2
)
.
No
w
we
apply
the
p
ro
duct
rule.
∂
∂
y
cos(3
x
+
x
2
y
)(3
+
2
xy
)
=
∂
∂
y
cos(3
x
+
x
2
y
)
(3
+
2
xy
)
+
cos(3
x
+
x
2
y
)
∂
∂
y
(3
+
2
xy
)
=
−
sin(3
x
+
x
2
y
)(
x
2
)(3
+
2
xy
)
+
cos(3
x
+
x
2
y
)(2
x
)
Question
4.4.10
Do
es
Differentiation
Order
Matter?
No.
Sp
ecifically
,
the
following
is
due
to
Clairaut:
Theo
rem
If
f
is
defined
on
a
neighb
orhoo
d
of
(
a,
b
)
and
the
functions
f
xy
and
f
y
x
a
re
b
oth
continuous
on
that
neighb
o
rho
o
d,
then
f
xy
(
a,
b
)
=
f
y
x
(
a,
b
)
.
This
readily
generalizes
to
larger
numb
ers
of
va
riables,
and
higher
o
rder
derivatives.
F
or
example
f
xy
yz
=
f
z
y
xy
.
288
Section
4.4
Exercises
Summa
ry
Questions
Q1
What
is
the
role
of
each
va
riable
when
w
e
compute
a
partial
derivative?
Q2
What
do
es
the
pa
rtial
derivative
f
y
(
a,
b
)
mean
geometrically?
Q3
Can
you
think
of
an
example
where
the
partial
derivative
do
es
not
accurately
mo
del
the
change
in
a
function?
Q4
What
is
Clairaut’s
Theo
rem?
4.4.1
Q5
Give
the
equation
of
the
line
that
lies
in
the
plane
x
=
2
and
is
tangent
to
the
graph
z
=
xe
3
xy
+
x
at
the
p
oint
(2
,
0
,
4)
.
Y
ou
may
give
y
our
equation
in
any
notation
that
wo
rks
in
2
dimensions.
Q6
Alexander
p
erforms
an
exp
eriment
with
his
wireless
netw
orking
router.
A
t
each
level
of
p
o
wer
output
(in
miliwatts)
and
distance
from
his
computer
(in
meters),
he
measures
T
(
p,
d
)
,
the
maximum
transfer
sp
eed
of
data
(in
megabits
p
er
second).
Here
is
a
table
of
his
observations.
0
mW
100
mW
200
mW
300
mW
400
mW
500
mW
600
mW
700
mW
10
m
6
15
40
100
300
800
800
800
20
m
0
2
15
30
90
300
800
800
30
m
0
0
2
10
50
100
400
800
40
m
0
0
0
0
5
20
50
100
50
m
0
0
0
0
2
5
20
45
a
Use
this
data
to
appro
ximate
T
p
(300
,
20)
.
Sho
w
what
values
you
used.
There
is
more
than
one
reasonable
w
ay
to
do
this.
b
What
do
es
the
derivative
in
a
mean
in
physical
terms?
Be
p
recise
and
include
units.
289
Section
4.4
Exercises
c
Use
this
data
to
appro
ximate
T
d
(500
,
30)
.
Sho
w
what
values
you
used.
There
is
more
than
one
reasonable
w
ay
to
do
this.
d
What
app
ears
to
b
e
true
ab
out
the
sign
of
T
d
(
p,
d
)
?
What
do
es
this
mean
in
physical
terms,
and
why
do
es
it
mak
e
sense?
4.4.2
Q7
Let
f
(
x,
y
)
=
7
x
2
+
5
y
cos
x
+
e
y
.
Compute
f
x
(
x,
y
)
.
Explain
the
role
of
y
in
each
term
where
it
is
p
resent.
Q8
Let
f
(
x,
y
)
=
sin
x
sin
y
.
Show
how
to
compute
f
y
(
x,
y
)
using
the
product
rule,
then
suggest
a
mo
re
efficient
approach.
4.4.3
Q9
In
the
diagram
from
this
example,
is
f
x
(3
,
0)
p
ositive
o
r
negative?
Explain.
Q10
In
the
diagram
from
this
example,
use
a
p
oint
on
the
c
=
30
level
set
to
app
roximate
f
y
(4
,
−
1
.
25)
.
290
Q11
In
the
diagram
from
this
example,
use
a
p
oint
on
the
c
=
50
level
set
to
appro
ximate
f
x
(4
,
−
1
.
25)
.
Q12
In
the
diagram
from
this
example,
what
is
f
y
(0
,
0)
?
Explain
your
reasoning.
4.4.4
Q13
Find
f
x
and
f
y
fo
r
the
following
functions
f
(
x,
y
)
a
f
(
x,
y
)
=
x
2
−
y
2
b
f
(
x,
y
)
=
p
y
x
(assume
x
>
0
and
y
>
0
)
c
f
(
x,
y
)
=
y
e
xy
Q14
Find
g
x
(
x,
y
)
and
g
y
(
x,
y
)
for
the
following
functions
g
(
x,
y
)
a
g
(
x,
y
)
=
e
x
2
+
y
2
b
g
(
x,
y
)
=
y
ln(
y
−
x
)
c
g
(
x,
y
)
=
3
x
2
+4
x
−
2
e
(
y
3
)
4.4.5
Q15
Extrap
olate
from
the
limit
defintion
of
f
x
(
x,
y
)
to
give
a
limit
definition
of
f
x
(
x,
y,
z
)
.
Explain
why
this
limit
rep
resents
a
change
in
f
where
only
x
is
changing.
Q16
Let
f
(
x,
y
,
z
)
=
e
3
x
y
+
3
√
y
z
+
x
3
z
7
.
Compute
∂
f
∂
z
.
Q17
Let
g
(
u,
v,
w
)
=
e
uv
+
w
2
.
Compute
∂
g
∂
v
.
Q18
Let
p
(
r
,
s,
t
)
=
e
r
+
e
s
+
e
t
rst
.
Compute
∂
p
∂
r
.
291
Section
4.4
Exercises
4.4.6
Q19
In
this
example,
do
es
the
fact
that
glass
expands
as
it
is
heated
suggest
that
∂
P
∂
T
overstates
or
understates
the
actual
rate
of
p
ressure
increase
as
T
increases?
Q20
Supp
ose
Jinteki
Co
rp
oration
makes
widgets
which
is
sells
fo
r
$100
each.
It
commands
a
small
enough
portion
of
the
mark
et
that
its
production
level
do
es
not
affect
the
demand
(p
rice)
for
its
p
ro
ducts.
If
W
is
the
numb
er
of
widgets
produced
and
C
is
their
op
erating
cost,
Jinteki’s
profit
is
mo
deled
b
y
P
=
100
W
−
C.
Since
∂
P
∂
W
=
100
do
es
this
mean
that
increasing
p
ro
duction
can
b
e
exp
ected
to
increase
profit
at
a
rate
of
$100
p
er
widget?
4.4.7
Q21
Supp
ose
g
(
s,
t
)
is
the
partial
derivative
of
f
(
s,
t
)
with
resp
ect
to
t
,
and
h
(
s,
t
)
is
the
partial
derivative
for
g
(
s,
t
)
with
resp
ect
to
s
.
Write
h
in
terms
of
f
using
b
oth
subscript
and
∂
notation.
Q22
Physicists
note
that
velo
city
is
the
derivative
of
position
with
resp
ect
to
time,
and
acceleration
is
the
derivative
of
velo
city
with
resp
ect
to
time.
If
s
(
t,
f
)
is
the
p
osition
of
a
ro
ck
et
with
f
kilograms
of
fuel
after
t
seconds,
what
is
the
physical
meaning
of
∂
3
s
∂
2
t∂
f
?
4.4.8
Q23
If
f
(
x,
y
)
=
sin(3
x
+
x
2
y
)
calculate
f
y
x
.
V
erify
that
you
get
the
same
answer
that
we
did
for
f
xy
.
Q24
Let
f
(
x,
y
)
=
ln(
x
2
+
y
)
.
Compute
f
xy
(
x,
y
)
.
Q25
Let
g
(
x,
y,
z
)
=
2
x
3
z
+
y
e
xy
2
.
a
Compute
∂
g
∂
y
.
292
b
Compute
∂
2
g
∂
x
2
.
Q26
Compute
the
follo
wing
partial
derivatives
of
g
(
x,
y,
z
)
=
x
3
sin(
xz
)
y
a
∂
g
∂
y
b
∂
2
g
∂
z
2
c
∂
2
g
∂
z
∂
x
4.4.9
Q27
If
f
(
x,
y
,
z
)
is
a
smo
oth
function,
which
of
the
following
are
equavalent
to
f
xy
yz
y
?
i.
f
xz
z
y
z
ii.
f
z
y
yxy
iii.
f
y
yy
z
x
iv.
f
xxxy
z
v.
f
xy
z
y
vi.
f
xy
z
vii.
f
y
xxz
x
Q28
Ho
w
many
third
pa
rtial
derivatives
does
a
t
wo-va
riable
function
have?
Assuming
these
derivatives
a
re
continuous,
which
of
them
are
equal
acco
rding
to
Clairaut’s
theorem?
293
Section
4.4
Exercises
Synthesis
&
Extension
Q29
Let
f
(
x,
y
)
=
e
xy
x
+
y
.
Is
∂
f
∂
x
=
∂
f
∂
y
?
If
so,
why?
If
not,
ho
w
are
they
related?
Q30
The
function
f
(
x,
y
)
=
e
x
+
y
has
the
strange
property
that
f
x
x,
y
=
f
y
(
x,
y
)
at
every
p
oint
(
x,
y
)
.
What
do
es
this
mean
geometrically
ab
out
the
function
f
?
Q31
Do
w
e
know
that
f
x
(
x,
y
)
is
in
fact
a
function?
What
fact
ab
out
limits
is
relevant
to
this
question?
294
Section
4.5
Linea
r
App
ro
ximations
Goals:
1
Calculate
the
equation
of
a
tangent
plane
.
2
Rewrite
the
tangent
plane
fo
rmula
as
a
linea
rization
or
differential
.
3
Use
linea
rizations
to
estimate
values
of
a
function.
4
Use
a
differential
to
estimate
the
erro
r
in
a
calculation.
In
single-variable
calculus,
the
tangent
line
was
one
of
the
great
applications
of
the
derivative.
It
solves
a
difficult
geometry
problem,
but
it
also
gives
a
metho
d
of
appro
ximating
a
difficult
to
compute
function.
The
height
of
the
tangent
line
is
close
to
the
height
of
the
graph
near
the
p
oint
of
tangency
.
This
means
the
value
of
the
tangent
line
function
appro
ximates
the
value
of
the
function,
close
to
the
p
oint
of
tangency
.
The
tw
o-variable
analogue
of
a
tangent
line
is
a
tangent
plane.
Question
4.5.1
What
Is
a
T
angent
Plane?
Definition
A
tangent
plane
at
a
p
oint
P
=
(
x
0
,
y
0
,
z
0
)
on
a
surface
is
a
plane
containing
the
tangent
lines
to
the
surface
through
P
.
Click to Load Applet
Figure:
The
tangent
plane
to
z
=
f
(
x,
y
)
at
a
p
oint
295
Question
4.5.1
What
Is
a
T
angent
Plane?
Equation
If
the
graph
z
=
f
(
x,
y
)
has
a
tangent
plane
at
(
x
0
,
y
0
)
,
then
it
has
the
equation:
z
−
z
0
=
f
x
(
x
0
,
y
0
)(
x
−
x
0
)
+
f
y
(
x
0
,
y
0
)(
y
−
y
0
)
.
Rema
rks
1
This
is
the
p
oint-slop
e
fo
rm
of
the
equation
of
a
plane.
f
x
(
x
0
,
y
0
)
and
f
y
(
x
0
,
y
0
)
a
re
the
slop
es.
2
x
0
and
y
0
a
re
numb
ers,
so
f
x
(
x
0
,
y
0
)
and
f
y
(
x
0
,
y
0
)
are
numb
ers.
The
variables
in
this
equation
a
re
x
,
y
and
z
.
The
cross
sections
of
the
tangent
plane
give
the
equation
of
the
tangent
lines
we
learned
in
single
va
riable
calculus.
y
=
y
0
x
=
x
0
z
−
z
0
=
f
x
(
x
0
,
y
0
)(
x
−
x
0
)
+
0
z
−
z
0
=
0
+
f
y
(
x
0
,
y
0
)(
y
−
y
0
)
This
sho
ws
that
the
tangent
plane
do
es
contain
these
t
wo
tangent
lines.
296
Example
4.5.2
W
riting
the
Equation
of
a
T
angent
Plane
Give
an
equation
of
the
tangent
plane
to
f
(
x,
y
)
=
√
xe
y
at
(4
,
0)
Solution
W
riting
the
formula
requires
us
to
fill
in
5
values.
1
x
0
=
4
is
given.
2
y
0
=
0
is
given.
3
z
0
is
the
height
of
the
graph
at
(4
,
0)
which
is
√
4
e
0
=
2
.
4
T
o
compute
f
x
(
x
0
,
y
0
)
w
e
compute
the
partial
derivative
function
f
x
(
x,
y
)
=
1
2
√
x
√
e
y
.
Then
w
e
evaluate
at
(4
,
0)
.
f
x
(4
,
0)
=
1
2
√
4
√
e
0
=
1
4
.
5
f
y
(
x
0
,
y
0
)
is
simila
r
though
we
will
use
the
chain
rule.
f
y
(
x,
y
)
=
√
x
1
2
√
e
y
e
y
f
y
(4
,
0)
=
√
4
1
2
√
e
0
e
0
=
1
W
e
plug
these
values
into
the
tangent
plane
fo
rmula.
z
−
2
=
1
4
(
x
−
4)
+
1(
y
−
0)
which
simplifies
to
z
−
2
=
1
4
(
x
−
4)
+
y
.
297
Question
4.5.3
Ho
w
Do
We
Rewrite
a
T
angent
Plane
as
a
Function?
Definition
If
w
e
write
z
as
a
function
L
(
x,
y
)
,
we
obtain
the
linea
rization
of
f
at
(
x
0
,
y
0
)
.
L
(
x,
y
)
=
f
(
x
0
,
y
0
)
+
f
x
(
x
0
,
y
0
)(
x
−
x
0
)
+
f
y
(
x
0
,
y
0
)(
y
−
y
0
)
If
the
graph
z
=
f
(
x,
y
)
has
a
tangent
plane,
then
L
(
x,
y
)
appro
ximates
the
values
of
f
near
(
x
0
,
y
0
)
.
Notice
f
(
x
0
,
y
0
)
just
calculates
the
value
of
z
0
.
This
formula
is
equivalent
to
the
tangent
plane
equation
after
w
e
solve
for
z
by
adding
z
0
to
b
oth
sides.
Example
4.5.4
App
roximating
a
F
unction
Use
a
linea
rization
to
appro
ximate
the
value
of
√
4
.
02
e
0
.
05
.
Solution
W
e
don’t
know
√
4
.
02
e
0
.
05
,
but
we
can
think
of
this
as
the
value
of
the
function
f
(
x,
y
)
=
√
xe
y
.
W
e
don’t
know
the
value
of
this
function
at
(4
.
02
,
0
.
05)
,
but
the
point
(4
,
0)
is
nea
rby
,
and
w
e
can
evaluate
it
there.
This
is
where
we’ll
produce
our
linea
rization.
We
already
produced
the
equation
of
the
tangent
plane
in
Example
.4.5.2.
z
−
2
=
1
4
(
x
−
4)
+
y
W
e
write
z
as
the
function
L
(
x,
y
)
and
solve
fo
r
it:
L
(
x,
y
)
=
2
+
1
4
(
x
−
4)
+
y
F
or
points
near
(4
,
0)
,
L
(
x,
y
)
is
close
to
f
(
x,
y
)
.
This
is
the
basis
of
our
appro
ximation.
√
4
.
02
e
0
.
05
=
f
(4
.
02
,
0
.
05)
≈
L
(4
.
02
,
0
.
05)
≈
2
+
1
4
(4
.
02
−
4)
+
0
.
05
≈
2
+
0
.
005
+
0
.
05
≈
2
.
055
298
Question
4.5.5
Ho
w
Do
es
Differential
Notation
Wo
rk
in
More
V
ariables?
The
one-va
riable
differential
is
a
shorthand
wa
y
to
exp
ress
change
in
the
linearization
of
a
function.
The
differential
dx
is
an
indep
endent
variable.
It
can
tak
e
on
any
value.
The
differential
dy
dep
ends
on
b
oth
x
0
and
dx
.
dy
=
f
′
(
x
0
)
dx
Once
we’ve
chosen
x
0
and
dx
,
dy
is
the
amount
that
the
tangent
line
to
y
=
f
(
x
)
at
x
0
rises
when
we
increase
x
b
y
dx
.
Click to Load Applet
Figure:
The
differentials
dx
and
dy
on
the
tangent
line
to
y
=
f
(
x
)
The
differential
dz
measures
the
change
in
the
linearization
of
f
(
x,
y
)
given
pa
rticular
changes
in
the
inputs:
dx
and
dy
.
It
is
a
useful
shorthand
when
one
is
estimating
the
error
in
an
initial
computation.
Definition
F
or
z
=
f
(
x,
y
)
,
the
differential
or
total
differential
dz
is
a
function
of
a
p
oint
(
x
0
,
y
0
)
and
tw
o
indep
endent
va
riables
dx
and
dy
.
dz
=
f
x
(
x
0
,
y
0
)
dx
+
f
y
(
x
0
,
y
0
)
dy
=
∂
z
∂
x
dx
+
∂
z
∂
y
dy
Rema
rk
The
differential
fo
rmula
is
just
the
tangent
plane
fo
rmula
with
dz
=
z
−
z
0
dx
=
x
−
x
0
dy
=
y
−
y
0
.
An
old
trigonometry
application
is
to
measure
the
height
of
a
p
ole
b
y
standing
at
some
distance.
W
e
then
measure
the
angle
θ
of
incline
to
the
top,
as
well
as
the
distance
b
to
the
base.
The
height
is
h
=
b
tan
θ
.
299
Question
4.5.5
Ho
w
Do
es
Differential
Notation
Wo
rk
in
More
V
ariables?
a
If
the
distance
to
the
base
is
13
m
and
the
angle
of
incline
is
π
6
,
what
is
the
height
of
the
p
ole?
b
Human
measurement
is
never
perfect.
If
our
measurement
of
b
is
off
b
y
at
most
0
.
1
m
and
our
measurement
of
θ
is
off
b
y
at
most
π
120
,
use
a
differential
to
appro
ximate
the
maximum
p
ossible
erro
r
in
our
h
.
Solution
a
The
height
is
13
tan
π
6
=
13
√
3
.
b
T
o
compute
the
differential,
we
need
to
kno
w
the
partial
derivatives
of
h
:
∂
h
∂
b
=
tan
θ
∂
h
∂
θ
=
b
sec
2
θ
∂
h
∂
b
(
13
,
π
6
)
=
1
√
3
∂
h
∂
θ
(
13
,
π
6
)
=
56
3
W
e
can
now
compute
the
differential.
dh
=
∂
h
∂
b
db
+
∂
h
∂
θ
dθ
=
1
√
3
db
+
56
3
dθ
dh
is
la
rgest
when
db
=
0
.
1
and
dθ
is
π
120
.
max
dh
=
1
√
3
(0
.
1)
+
56
3
π
120
=
1
10
√
3
+
13
π
90
300
Section
4.5
Exercises
Summa
ry
Questions
Q1
What
do
you
need
to
compute
in
o
rder
to
write
the
equation
of
a
tangent
plane
to
z
=
f
(
x,
y
)
at
(
x
0
,
y
0
,
z
0
)
?
Q2
F
or
what
kinds
of
functions
a
re
linear
app
roximations
useful?
Q3
Ho
w
are
the
tangent
plane
and
the
linea
rization
related?
Q4
Ho
w
is
the
differential
defined
for
a
tw
o
va
riable
function?
What
do
es
each
variable
in
the
formula
mean?
4.5.1
Q5
Let
p
(
x,
y
)
=
3
x
+
5
y
−
2
.
a
What
is
the
graph
z
=
p
(
x,
y
)
?
What
is
the
significance
of
3
,
5
and
−
2
?
b
Give
the
equation
of
the
tangent
plane
to
z
=
p
(
x,
y
)
at
(1
,
4
,
21)
c
Ho
w
is
the
tangent
plane
equation
related
to
z
=
p
(
x,
y
)
?
Why
do
es
this
make
sense?
Q6
Olivia
computes
the
tangent
plane
of
z
=
x
2
+
y
2
at
(4
,
3
,
25)
.
Her
answer
is
z
−
25
=
2
x
(
x
−
4)
+
2
y
(
y
−
3)
.
a
Is
this
the
equation
of
a
plane?
Explain.
b
What
do
es
Olivia
need
to
do
to
fix
her
answer?
Q7
If
the
equation
of
the
tangent
plane
of
z
=
f
(
x,
y
)
do
es
not
have
a
y
in
it,
do
es
that
mean
that
y
is
a
free
variable
of
f
?
Explain.
Q8
Can
our
tangent
plane
formula
ever
give
us
a
plane
pa
rallel
to
the
xy
-plane?
The
xz
-plane?
The
z
y
-plane?
Explain.
301
Section
4.5
Exercises
4.5.2
Q9
Compute
the
equation
of
the
tangent
plane
to
z
=
p
36
−
4
x
2
−
y
2
at
(2
,
2
,
4)
.
Q10
Let
g
(
x,
y
)
=
3
x
2
+4
x
−
2
e
(
y
3
)
.
W
rite
the
equation
of
the
tangent
plane
to
z
=
g
(
x,
y
)
at
(0
,
1)
.
Q11
Let
f
(
x,
y
)
=
p
y
x
.
W
rite
the
equation
of
the
tangent
plane
to
z
=
f
(
x,
y
)
at
(4
,
36
,
3)
.
Q12
Let
f
(
x,
y
)
=
ln(
x
2
+
y
)
.
Write
the
equation
of
the
tagent
plane
to
z
=
f
(
x,
y
)
at
(
e
3
,
0
,
6)
.
4.5.3
Q13
W
rite
a
linearization
of
f
(
x,
y
)
=
y
e
xy
at
(3
,
2)
.
Q14
W
rite
a
linearization
of
g
(
x,
y
)
=
e
x
2
+
y
2
at
(3
,
−
4)
.
4.5.4
Q15
Supp
ose
y
ou
want
to
appro
ximate
√
5
.
5
e
0
.
3
b
y
hand.
Would
using
the
linearization
of
f
(
x,
y
)
=
√
xe
y
at
(5
,
0)
b
e
a
go
o
d
strategy?
Explain.
Q16
Sho
w
how
to
use
an
app
ropriate
linea
rization
to
appro
ximate
1
5
.
12
sin
31
π
30
.
Q17
Let
g
(
x,
y
)
=
x
2
y
.
Supp
ose
you
don’t
rememb
er
how
to
divide
decimals.
Show
ho
w
you
can
use
a
linea
rization
of
g
to
app
roximate
3
.
97
2
1
.
05
.
Q18
Sho
w
how
to
use
a
linea
rization
to
appro
ximate
the
value
of
q
(4
.
02)
2
+
√
80
.
93
b
y
hand.
302
4.5.5
Q19
Let
f
(
x,
y
)
=
y
x
2
+
y
2
.
W
rite
the
differential
of
f
at
(4
,
3)
.
Q20
Let
g
(
p,
q
)
=
p
ln
q
.
Write
the
differential
of
g
at
(3
,
e
2
)
.
Q21
Bo
ris
is
measuring
the
area
of
a
rectangular
field,
so
he
can
decide
how
much
grass
seed
to
buy
.
Acco
rding
to
his
measurements,
the
field
is
30
m
by
50
m
,
giving
an
area
of
1500
m
2
.
If
we
accept
that
each
of
his
measurements
has
an
error
no
larger
than
0
.
2
m
,
use
a
differential
to
appro
ximate
the
maximum
erro
r
in
his
area
computation.
Q22
Supp
ose
I
decide
to
invest
$10
,
000
expecting
a
6%
annual
rate
of
return
fo
r
12
yea
rs,
after
which
I’ll
use
it
to
purchase
a
house.
The
fo
rmula
for
compound
interest
P
=
P
0
e
rt
indicates
that
when
I
w
ant
to
buy
a
house,
I
will
have
P
=
10
,
000
e
0
.
72
.
I
accept
that
my
e
xpected
rate
of
return
might
have
an
error
of
up
to
dr
=
2%
.
Also,
I
ma
y
decide
to
buy
a
house
up
to
dt
=
3
yea
rs
b
efore
or
after
I
exp
ected.
a
W
rite
the
formula
fo
r
the
differential
dP
at
(
r
0
,
t
0
)
=
(0
.
06
,
12)
.
b
Given
my
assumptions,
what
is
the
maximum
estimated
error
dP
in
my
initial
calculation?
c
What
is
the
actual
maximum
erro
r
in
P
?
Q23
Let
z
=
2
x
−
y
3
.
At
the
p
oint
(
x,
y
)
=
(5
,
2)
,
what
is
the
maximum
value
of
the
differential
dz
?
Q24
Let
f
(
x,
y
)
b
e
a
function.
What
differential
and
what
inputs
into
that
differential
w
ould
you
use
to
app
roximate
f
(5
.
5
,
3
.
2)
−
f
(4
.
7
,
3
.
8)
.
Synthesis
&
Extension
Q25
Let
L
(
x,
y
)
b
e
the
linearization
of
f
(
x,
y
)
at
(3
,
2)
.
If
f
y
y
(
x,
y
)
<
0
for
all
(
x,
y
)
,
at
which
p
oints
can
w
e
guarantee
that
L
(
x,
y
)
either
under
or
overestimates
the
value
of
f
(
x,
y
)
?
Explain.
Q26
Let
f
(
x,
y
)
=
25
−
(
x
+
1)
2
−
(
y
−
3)
2
.
Describ
e
the
set
of
p
oints
(
a,
b
)
such
that
the
tangent
plane
to
z
=
f
(
x,
y
)
at
(
a,
b,
f
(
a,
b
))
passes
through
the
origin.
303
Section
4.5
Exercises
Q27
Here
is
a
table
of
selected
values
for
a
function
f
(
x,
y
)
y
x
0
2
4
6
8
10
0
2
5
8
10
11
11
2
6
9
12
14
15
15
4
9
12
15
17
18
18
6
12
15
18
20
21
21
8
14
17
20
22
23
23
10
17
20
23
25
23
23
a
Using
any
reasonable
app
roximation
metho
d,
show
how
to
p
ro
duce
a
linearization
of
f
(
x,
y
)
at
(4
,
2)
.
b
Do
es
y
our
linearization
over
or
underestimate
f
(10
,
2)
?
Explain
what
that
suggests
ab
out
one
o
r
more
derivatives
of
f
(
x,
y
)
.
Q28
a
Give
an
equation
of
the
plane
that
passes
through
the
p
oints
(3
,
4
,
2)
,
(5
,
5
,
1)
and
(6
,
2
,
6)
.
b
Supp
ose
there
is
a
function
f
(
x,
y
)
and
the
plane
in
pa
rt
a
is
tangent
to
the
graph
z
=
f
(
x,
y
)
at
(3
,
4
,
2)
.
What
pa
rtial
derivatives
of
f
can
y
ou
compute
exactly
(b
e
sp
ecific)?
Compute
them.
304
>
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