Download Section 8.1 Arc Length

Section 8.1 Arc Length
Ruipeng Shen
November 27
Definition 1. The length L of the curve C with equation y = f (x), a ≤ x ≤ b is defined to be
the limit of the lengths of inscribed polygons.
L = lim
n→∞
n
X
|Pi−1 Pi |.
i=1
y
Pi-1(xi-1,yi-1)
y=f(x)
Pi(xi,yi)
P1
Pn
Pn-1
P0
Δx
x
a=x0
x1
xi-1
xi
xn-1
xn=b
Formula for arc length We can calculate the distance |Pi−1 Pi | by
p
p
|Pi−1 Pi | = (yi − yi−1 )2 + (xi − xi−1 )2 = [f (xi ) − f (xi−1 )]2 + [xi − xi−1 ]2
1
By the mean value theorem, there exists a number x?i ∈ (xi−1 , xi ) such that f (xi ) − f (xi−1 ) =
f 0 (x?i ) · (xi − xi−1 ). Thus we have
q
q
|Pi−1 Pi | = [f 0 (x?i )]2 · [xi − xi−1 ]2 + [xi − xi−1 ]2 = |xi − xi−1 | 1 + [f 0 (x?i )]2
q
= 1 + [f 0 (x?i )]2 · ∆x.
Making a sum and taking the limit, we have
Z
n q
X
?
2
0
L = lim
1 + [f (xi )] · ∆x =
n→∞
b
p
1 + [f 0 (x)]2 dx.
a
i=1
We can rewrite this with Leibniz notation for derivatives
s
2
Z b
dy
dx.
1+
L=
dx
a
Example 2. Find the length of the arc of the semicubical parabola y 2 = x3 between the points
(1, 1) and (4, 8).
●
7.5
(4,8)
5
y2= x3
2.5
●
-5
-2.5
0
(1,1)
2.5
5
7.5
10
12.5
-2.5
Solution This piece of curve can be given by
derivative
dy
=
dx
the equation y = x3/2 , 1 ≤ x ≤ 4. We have the
3 1/2
x
2
2
Thus we have
4
Z
s
L=
1+
1
u(4)
Z
√
dy
dx
2
r
4
Z
dx =
9
9
9
1 + x let u = 1 + x, du = dx
4
4
4
1
3/2 10
4
4 u
u · du =
·
9
9 3/2 13/4
u(1)
"
3/2 #
√ i
8
13
1 h √
3/2
=
80 10 − 13 13 .
10 −
=
27
4
27
=
Arc Length by integration in y If a curve has the equation x = g(y), c ≤ y ≤ d, and g 0 (y)
is continuous, then by interchanging the roles of x and y, we obtain the following formula of arc
length
s
2
Z d
Z dp
dx
0
2
dy.
1 + [g (y)] dy =
1+
L=
dy
c
c
Example 3. Find the length of the arc of the parabola y 2 = x from (0, 0) to (1, 1).
Solution
By the formula above we have
s
2
Z 1
Z 1p
dx
L=
1+
dy =
1 + 4y 2 dy
dy
0
0
Applying the trigonometric substitution y = (1/2) tan θ, 0 ≤ θ ≤ α = arctan 2, we have
p
1 + 4y 2 = 1 + tan2 θ = sec2 θ ⇒ 1 + 4y 2 = sec θ;
dy = (1/2) sec2 θ.
Thus
Z
α
sec θ ·
L=
0
1
1
sec2 θ dθ =
2
2
Z
α
sec3 θ dθ
0
1 1
α
= · [sec θ tan θ + ln | sec θ + tan θ|]0
2 2
1
= [sec α tan α + ln | sec α + tan α| − sec 0 tan 0 − ln | sec 0 + tan 0|]
4
i √5
√
√
1 h√
=
+ ln |2 + 5|.
5 · 2 + ln | 5 + 2| =
4
2
√
√
Here we use tan α = 2 ⇒ sec α = 1 + tan2 α = 5.
The arc length function Assume that a smooth curve C has the equation y = f (x), a ≤
x ≤ b. If we define s(x) be the distance along C from the initial point (a, f (a)) to the point
(x, f (x)), then s is a function, called the arc length function, and
Z xp
1 + [f 0 (t)]2 dt.
s(x) =
a
3
We can rewrite this into
ds
=
dx
s
1+
dy
dx
2
ds2 = dx2 + dy 2 .
⇒
y
P(x,f(x))
B(b,f(b))
S(x)
A(a,f(a))
x
a
x
Example 4. Find the arc length function for the curve y = x2 −
starting point.
Solution
The curve is given by y = f (x) = x2 −
1
8
b
1
8
ln x taking P (1, 1) as the
ln x, we have
1
f 0 (x) =2x −
;
s 8x
2 r
q
1
1
1
2
0
1 + [f (x)] = 1 + 2x −
= 1 + 4x2 +
−
2
8x
64x
2
r
1
1
1
= 4x2 +
+ = 2x +
.
64x2
2
8x
The formula above gives
x
Z x
1
1 + [f 0 (t)]2 dt =
2t +
dt
8t
1
1
x
1
1
= t2 + ln t = x2 + ln x − 1.
8
8
1
Z
s(x) =
p
4