Jim Lambers
Math 2B
Fall Quarter 2004-05
Leture 25 Notes
These notes orrespond to Setion 9.1 in the text.
Ar Length
In this leture, we will learn how to use alulus to ompute the length of a urve that is desribed
by an equation of the form
y
=
f
(
x
), for some given funtion
f
(
x
). Just as we learned how to
ompute the area under suh a urve as the limit of a sum of areas of simpler regions (namely,
retangles), we an ompute the length of the urve by interpreting the length as a limit of a sum
of lengths of the simplest urves kno
wn, whih are line segments.
Suppose that we wish to ompute the length of the urve
y
=
f
(
x
) between
x
=
a
and
x
=
b
. We
an approximate this length by dividing the interval [
a; b
℄ into subintervals of length
x
= (
b
a
)
=n
,
just as we did when we were trying to ompute the approximate area under
y
=
f
(
x
). Consider
any subinterval [
x
i
1
; x
i
℄. Then, if
x
is hosen to be suÆiently small, the length of the urve
y
=
f
(
x
) between
x
=
x
i
1
and
x
=
x
i
an be well approximated by the length of the line segment
between the points (
x
i
1
; f
(
x
i
1
)) and (
x
i
; f
(
x
i
)). This line segment is the hypotenuse of a right
triangle with legs of length
x
and
j
f
(
x
i
)
f
(
x
i
1
)
j
, and therefore the length
L
i
of the urve
y
=
f
(
x
) between
x
=
x
i
1
and
x
=
x
i
is approximately
L
i
p
x
2
+ (
f
(
x
i
)
f
(
x
i
1
))
2
=
x
s
1 +
f
(
x
i
)
f
(
x
i
1
)
x
2
:
(1)
It follows that the length
L
of the urve between
x
=
a
and
x
=
b
is approximated by
L
n
X
i
=1
L
i
=
n
X
i
=1
s
1 +
f
(
x
i
)
f
(
x
i
1
)
x
2
x
=
n
X
i
=1
s
1 +
f
(
x
i
)
f
(
x
i
1
)
x
i
x
i
1
2
x:
(2)
As
n
, the number of line segments, approahes
1
,
x
approahes zero, so the length of eah
subinterval [
x
i
1
; x
i
℄ tends to zero. It follows from the denition of the derivative that
lim
x
!
0
f
(
x
i
)
f
(
x
i
1
)
x
i
x
i
1
= lim
x
!
0
f
(
x
i
1
+
x
)
f
(
x
i
1
)
x
=
f
0
(
x
i
1
) (3)
and therefore the sum onverges to the denite integral
L
= lim
n
!1
n
X
i
=1
L
i
=
Z
b
a
p
1 + [
f
0
(
x
)℄
2
dx:
(4)
1
