1. No category

Section 7.4 Arc Length
Arc Length
Suppose that a curve C is defined by the equation y = f (x), where f is continuous and a ≤ x ≤ b. We
obtain a polygonal approximation to C by dividing the interval [a, b] into n subintervals with endpoints
x0 , x1 , . . . , xn and equal width ∆x. If yi = f (xi ), then the point Pi (xi , yi) lies on C and the polygon with
vertices P0 , P1 , . . . , Pn is an approximation to C.
The length L of C is approximately the length of this polygon and the approximation gets better as we
let n increase. Therefore, we define the length L of the curve C with equation y = f (x), a ≤ x ≤ b, as
the limit of the lengths of the inscribed polygons with vertices P0 , P1 , . . . (if the limit exists):
L = lim
n→∞
n
X
i=1
|Pi−1 Pi |
(1)
Notice that the procedure for defining arc length is very similar to the procedure we used for defining area
and volume: We divided the curve into a large number of small parts. We then found the approximate
lengths of the small parts and added them. Finally, we took the limit as n → ∞.
The definition of arc length given by Equation 1 is not very convenient for computational purposes, but
we can derive an integral formula for L in the case where f has a continuous derivative. [Such a function
f is called smooth because a small change in x produces a small change in f ′ (x).]
If we let ∆yi = yi − yi−1 , then
|Pi−1 Pi | =
p
(xi − xi−1 )2 + (yi − yi−1 )2 =
p
(∆x)2 + (∆yi )2
By applying the Mean Value Theorem to f on the interval [xi−1 , xi ], we find that there is a number x∗i
between xi−1 and xi such that
f (xi ) − f (xi−1 ) = f ′ (x∗i )(xi − xi−1 )
that is,
∆yi = f ′ (x∗i )∆x
Thus we have
p
p
p
(∆x)2 + [f ′ (x∗i )∆x]2 = (∆x)2 + [f ′ (x∗i )]2 (∆x)2
p
p
p
p
= (1 + [f ′ (x∗i )]2 )(∆x)2 = 1 + [f ′ (x∗i )]2 (∆x)2 = 1 + [f ′ (x∗i )]2 ∆x
|Pi−1 Pi | =
(∆x)2 + (∆yi )2 =
1
Section 7.4 Arc Length
Therefore, by the Definition above,
L = lim
n→∞
n
X
i=1
|Pi−1 Pi | = lim
n→∞
We recognize this expression as being equal to
Z bp
n
X
p
1 + [f ′ (x∗i )]2 ∆x
i=1
1 + [f ′ (x)]2 dx
a
by the definition of a definite integral. This integral exists because the function g(x) =
continuous. Thus we have proved the following theorem:
p
1 + [f ′ (x)]2 is
THEOREM: If f ′ is continuous on [a, b], then the length of the curve y = f (x), a ≤ x ≤ b, is
L=
Z bp
1 + [f ′ (x)]2 dx
(2)
a
If we use Leibniz notation for derivatives, we can write the arc length formula as follows:
L=
Z
a
b
s
1+
dy
dx
2
(3)
dx
EXAMPLE: Find the length of the segment of the horizontal line y = 1 between the points (a, 1) and
(b, 1).
Solution: Since f (x) = 1, we have f ′ (x) = 0, and (2) gives
Z bp
Z b
Z b
√
′
2
L=
1 + [f (x)] dx =
1 + 0dx =
dx = b − a
a
a
a
EXAMPLE: Find the length of the segment of the line y = x between the points (a, a) and (b, b).
Solution: Since f (x) = x, we have f ′ (x) = 1, and (2) gives
Z b
Z bp
√ Z b
√
√
′
2
1 + [f (x)] dx =
1 + 1dx = 2
dx = 2(b − a)
L=
a
a
a
EXAMPLE: Find the length of the segment of the line y = mx + n between the points (a, f (a)) and
(b, f (b)).
Solution: Since f (x) = mx + n, we have f ′ (x) = m, and (2) gives
L=
Z bp
a
1+
[f ′ (x)]2 dx
=
Z b√
1+
m2 dx
a
=
√
1+
m2
Z
a
b
dx =
√
1 + m2 (b − a)
EXAMPLE: Find the length of the arc of the semicubical parabola y 2 = x3 between the points (1, 1) and
(4, 8).
2
Section 7.4 Arc Length
EXAMPLE: Find the length of the arc of the semicubical parabola y 2 = x3 between the points (1, 1) and
(4, 8).
Solution: For the top half of the curve we have
dy
3
= x1/2
dx
2
y = x3/2
and so the arc length formula gives
Z
L=
4
1
s
1+
dy
dx
2
dx =
Z
r
4
9
1 + xdx
4
1
9
9
13
If we substitute u = 1 + x, then du = dx. When x = 1, u = ; when x = 4, u = 10. Therefore
4
4
4
"
3/2 #
10
Z
√
√
8
4 10 √
13
1
4 2 3/2
=
103/2 −
L=
= (80 10 − 13 13)
udu = · u
9 13/4
9 3
27
4
27
13/4
If a curve has the equation x = g(y), c ≤ y ≤ d, and g ′ (y) is continuous, then by interchanging the roles
of x and y in (2) and (3), we obtain the following formula for its length:
L=
Z
c
d
p
1 + [g ′ (y)]2dy =
Z
c
d
s
1+
dx
dy
2
dy
EXAMPLE: Find the length of the arc of the parabola y 2 = x from (0, 0) to (1, 1).
3
(4)
Section 7.4 Arc Length
EXAMPLE: Find the length of the arc of the parabola y 2 = x from (0, 0) to (1, 1).
Solution: Since x = y 2, we have dx/dy = 2y, and (4) gives
s
2
Z 1
Z 1p
dx
L=
1+
1 + 4y 2dy
dy =
dy
0
0
p
1
1
We make the trigonometric substitution y = tan θ, which gives dy = sec2 θdθ and 1 + 4y 2 =
2
2
√
1 + tan2 θ = sec θ. When y = 0, tan θ = 0, so θ = 0; when y = 1, tan θ = 2, so θ = tan−1 2 = α,
say. Thus
Z α
Z
1
1 α 3
2
L=
sec θ · sec θdθ =
sec θdθ
2
2 0
0
iα
1 1h
= ·
sec θ tan θ + ln | sec θ + tan θ|
2 2
0
1
= (sec α tan α + ln | sec α + tan α|)
4
2
2
(We could have
√ used Formula 21 in the Table of Integrals.) Since tan α = 2, we have sec α = 1+tan α = 5,
so sec α = 5 and
√
√
5 ln( 5 + 2)
+
L=
2
4
Because of the presence of the square root sign in Formulas 2 and 4, the calculation of an arc length often
leads to an integral that is very difficult or even impossible to evaluate explicitly. Thus we sometimes have
to be content with finding an approximation to the length of a curve, as in the following example.
EXAMPLE:
(a) Set
up an integral for the length of the arc of the hyperbola xy = 1 from the point (1, 1) to the point
1
.
2,
2
(b) Use Simpson’s Rule with n = 10 to estimate the arc length.
Solution:
(a) We have
y=
1
x
1
dy
=− 2
dx
x
and so the arc length is
L=
Z
1
2
s
1+
dy
dx
2
dx =
Z 2r
1
1+
1
dx
x4
p
(b) Using Simpson’s Rule (see Section 7.7) with a = 1, b = 2, n = 10, ∆x = 0.1, and f (x) = 1 + 1/x4 ,
we have
Z 2r
∆x
1
L=
[f (1) + 4f (1.1) + 2f (1.2) + 4f (1.3) + . . . + 2f (1.8) + 4f (1.9) + f (2)] ≈ 1.1321
1 + 4 dx ≈
x
3
1
4
Section 7.4 Arc Length
The Arc Length Function
We will find it useful to have a function that measures the arc length of a curve from a particular starting
point to any other point on the curve. Thus if a smooth curve C has the equation y = f (x), a ≤ x ≤ b,
let s(x) be the distance along C from the initial point P0 (a, f (a)) to the point Q(x, f (x)). Then s is a
function, called the arc length function, and, by (2),
Z xp
s(x) =
1 + [f ′ (t)]2 dt
(5)
a
We can use Part 1 of the Fundamental Theorem of Calculus to differentiate (5) (since the integrand is
continuous):
s
2
p
dy
ds
′
2
= 1 + [f (x)] = 1 +
(6)
dx
dx
The last Equation shows that the rate of change of s with respect to x is always at least 1 and is equal to
1 when f ′ (x), the slope of the curve, is 0. The differential of arc length is
s
2
dy
dx
(7)
ds = 1 +
dx
and this equation is sometimes written in the symmetric form
(ds)2 = (dx)2 + (dy)2
(8)
The geometric interpretation ofZEquation 8 is shown in the Figure
on the right. If we write L =
ds, then from (8) either we can
solve to get (7), which gives (3), or we can solve to get
s
2
dx
ds = 1 +
dy
dy
which gives (4).
1
EXAMPLE: Find the arc length function for the curve y = x2 − ln x taking P0 (1, 1) as the starting point.
8
5
Section 7.4 Arc Length
1
EXAMPLE: Find the arc length function for the curve y = x2 − ln x taking P0 (1, 1) as the starting point.
8
1
Solution: If f (x) = x2 − ln x, then
8
1
8x
2
2
1
1
1
1
1
1
2
2
′
2
= 4x + +
= 2x +
= 1 + 4x − +
1 + [f (x)] = 1 + 2x −
8x
2 64x2
2 64x2
8x
f ′ (x) = 2x −
p
1 + [f ′ (x)]2 = 2x +
1
8x
Thus the arc length function is given by
Z
Z xp
′
2
1 + [f (t)] dt =
s(x) =
1
x
1
1
2t +
8t
x
1
1
dt = t + ln t = x2 + ln x − 1
8
8
1
2
For instance, the arc length along the curve from (1, 1) to (3, f (3)) is
s(3) = 32 +
1
ln 3
ln 3 − 1 = 8 +
≈ 8.1373
8
8
6