Calculus II
MAT 146
Integration Applications: Arc Length
Again we use a definite integral to sum an infinite number of measures, each
infinitesimally small. We seek to determine the length of a curve that represents
the graph of some real-valued function f, measuring from the point ( a, f ( a)) on the
curve to the point ( b, f ( b)) on the curve.
The graph of y = f is shown. We seek to determine the length of the curve, known as arc length, from the point
on the curve to the point
(b, f (b)) .
Together with endpoints
( a, f ( a))
and
(a, f (a))
(b, f (b)) , we select additional points ( xi, f ( xi )) on the curve. Here, we’ve
created n line segments. The length of the ith line segment is denoted by Li, with 1 ≤ i ≤ n.
Concentrate on the segment whose length is Li. Determine the slope of that segment:
slope of segment whose length is Li = mi
=
f ( xi ) ! f ( xi!1 )
xi ! xi!1
=
"yi
"xi
# mi =
"yi
"xi
# mi $ "xi = "yi
We know by the Mean Value Theorem that there exists an xi*, with xi–1 ≤ xi* ≤ xi-1, such
that mi = f ! ( xi* ) . We use this fact, and the distance formula, to calculate the typical
segment length Li:
Li =
2
( xi ! xi!1 ) + ( f ( xi ) ! f ( xi!1 ))
2
2
2
2
=
("xi ) + ("yi )
=
("xi ) + ("yi )
=
("xi ) + ( mi "xi )
=
("xi ) + ( f # ( xi* ) "xi )
=
("xi )
2
2
2
2
2
$1+ f # x 2 '
&% ( ( i* )) )(
= "xi 1+ ( f # ( xi* ))
2
2
We now let the number of segments between (a, f(a)) and (b, f(b)) grow without bound.
That is, we let n—> ∞. We can now write a definite integral to add all the lengths Li:
total length of the curve from x = a to x = b :
b
&
a
2
1+ "# f ! ( x )$% dx
Example #1: Determine the length of the curve y = x2 from x = 0 to x = 3.
b
2
We use the arc length formula & 1+ "# f ! ( x )$% dx . Here, a = 0 and b = 3. The function
a
2
y = f ( x ) = x has derivative f ! ( x ) = 2x , so
b
arc length =
2
& 1+ "# f ! ( x )$% dx
a
3
=
2
& 1+ [2x ] dx
0
3
=
& 1+ 4x 2 dx
0
1
3
= ln 37 + 6 +
37
4
2
' 9.747089
(
)
x3 1 1
+ , ! x !1?
6 2x 2
Example #2: What is the arc length for y =
b
2
Again, use the arc length formula & 1+ "# f ! ( x )$% dx . Here, a = ½ and b = 1, with
a
dy x 2
1
= ! 2.
dx 2 2x
b
arc length =
2
& 1+ "# f ! ( x )$% dx
a
2
" x2
1 $
1+ ( ' 2 ) dx
# 2 2x %
1
=
&
1
=
2
31
48
Example #3: For x = ln (1! y 2 ) , determine the length of arc x(y) for 0 ≤ y ≤ ½.
d
2
Here, we use the arc length formula with y as the independent variable: & 1+ "#h! ( y)$% dy .
c
Here, we have c = 0 and d = ½, with
dx " 1 %
!2y
.
=$
!2y) =
2 '(
dy # 1! y &
1! y 2
d
arc length =
1+ "# f ! ( y)$% dy
1
" '2y $
1+ (
dy
2)
#1' y %
c
=
2
&
2
&
0
* 0.5986123
2