3.3 - MATH
1081
 Wednesday,
February
2
 
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Unformatted text preview: MATH
1081
 Wednesday,
February
2
 
 Chapter
3
–
Section
3
 
 RATES
OF
CHANGE
 
 Homework
#3
(due
2/7):

 Section
3.2
#6,
20,
26
 
 
 
 
 
 
 
 
 Section
3.3
#8,
30,
34
 
 Clicker
Check­in:

Choose
any
letter
to
check
in
now.
 Before
we
begin
Section
3.3,
let’s
look
more
closely
at
the
last
 example
from
last
lecture.
 
 ⎧ x 2 + 1 if x < 0 ⎪ Is
the
function
 g ( x ) = ⎨ −1 if x = 0 
continuous
at
 x = 0 ?
 ⎪ ex if x > 0 ⎩ 
 If
not,
what
is
happening
at
the
point
 x = 0 ?
 • Is
the
point
just
in
the
“wrong”
place
(removable
 discontinuity)?
 • Do
the
2
one‐sided
limits
not
agree
(jump
discontinuity)?
 • Do
either
of
the
one‐sided
limits
go
to
infinity
(asymptote)?
 
 

 Suppose
you
are
travelling
along
a
highway
and
at
1pm
you
are
 at
mile
marker
37.

Later
at
2:30pm,
you
see
that
you
are
at
 mile
marker
142
(on
the
same
highway
in
the
same
state).
 
 What
was
your
average
rate
of
velocity
for
that
leg
of
the
trip?
 
 
 
 Now,
suppose
that
you
are
in
a
car
at
a
stoplight.

When
the
 light
turns
green,
you
begin
to
drive
along
a
road
so
that
your
 distance
(in
feet)
from
the
light
 t 
seconds
later
is
given
by
 s ( t ) = 3t 2 .
 
 What
is
your
average
rate
of
velocity
over
the
first
15
seconds
 after
you
leave
the
stoplight?
 Definition:

Average
Rate
of
Change
 The
average
rate
of
change
of
 f ( x ) 
with
respect
to
 x 
from
 x = a 
to
 x = b 
is
 f (b ) − f ( a ) 






 .
 b−a 
 
 Sometimes
a
ratio
such
as
this
is
called
a
difference
quotient.
 Note
that
this
is
the
same
as
the
slope
of
the
line
through
the
 points
( a, f ( a )) 
and
( b, f ( b )) .
 
 f (b ) − f ( a ) 
 
 b−a 
 
 Example:

Determine
the
average
rate
of
change
of
the

 
function
over
the
given
interval.
 
 y = 7 x − 9 

between
 x = −3
and
 x = 5 
 1. 
 
 
 
 2. f ( x ) = −4 x 2 − 6 
between
 x = 2 
and
 x = 6 
 
 
 Back
to
the
car
at
the
stoplight…
 Suppose
that
you
are
in
a
car
at
a
stoplight.

When
the
light
 turns
green,
you
begin
to
drive
along
a
road
so
that
your
 distance
(in
feet)
from
the
light
 t 
seconds
later
is
given
by
 s ( t ) = 3t 2 .
 
 
 How
can
we
determine
the
speed
of
the
car
at
one
precise
 moment
(like
10
seconds
after
leaving
the
stoplight)?
 
 Certainly
we
can
estimate
it
with
an
average
velocity
over
 some
time
interval
containing
 t = 10 ,
but
which
estimate
is
the
 best
one?
 
 Here
is
where
the
idea
of
a
limit
comes
in.

For
the
best
 estimate,
we
can
find
the
average
velocity
of
smaller
and
 smaller
intervals
of
time;

 like
from
 t = 10 
to
 t = 10.1
and
then
 t = 10 
to
 t = 10.05 
and
so
on.
 
 We
can’t
compute
the
rate
of
change
if
we
only
have
one
point,
 so
the
length
of
the
interval
never
equals
0,
but
we
want
it
to
 become
smaller
and
smaller.
 
 So,
we
want
to
compute
 f (b ) − f ( a ) 
 
 
 
 
 
 
 lim .
 b→ a b−a 
 Looking
at
the
graph,
we
are
looking
for
the
limiting
value
of
 the
slope
of
the
line
as
the
two
points
( a, f (a )) 
and
( b, f (b )) 
get
 closer
and
closer
together.
 
 
 
 For
the
purpose
of
computation,
we
will
make
a
slight
 adjustment
to
the
expression
a
write
the
following:
 
 Definition:
Instantaneous
Rate
of
Change
 The
instantaneous
rate
of
change
of
a
function
 f ( x ) 
at
 x = a 
 is
given
by
 f (a + h) − f (a) 
 
 
 
 
 lim 
 h→ 0 h provided
that
the
limit
exists.
 
 
 If
we
visualize
the
graph
of
 f ( x ) ,
we
are
computing
the
slope
 between
the
points
( a, f ( a )) 
and
( a + h, f ( a + h )) 
and
then
 making
two
points
get
closer
and
closer
by
 h → 0 .
 
 
 
 
 Example:

Determine
the
instantaneous
rate
of
change
of
the

 
function
at
the
given
point.
 
 y = 7 x − 9 

at

 x = −3
 1. 
 
 
 2. f ( x ) = −4 x 2 − 6 

at

 x = 2 
 
 
 
 3. g ( x ) = 3x − 2 

at

 x = 1
 
 Clicker
Check­out:
Choose
any
letter
to
check
out
now.
 
 
 Tomorrow
in
recitation:

Workshop
#2
 
 Next
time:

Section
3.4
 
 ...
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