Section 8.2 Lecture Slides

1. Chapter 8: Further Applications of Integration
Section 8.1: Arc Length
Alea Wittig
SUNY Albany

2. Outline
Length of a Curve
The Arc Length Function

3. ▶ So far we have learned methods to compute the area under a
curve, the area between curves, and the volume of a solid of
revolution.
▶ In section 8.1 we will learn how to compute the length of a
curve.
▶ In section 8.2 we will learn how to compute the surface area of
a surface of revolution.

4. Length of a Curve

5. Length of a Curve
▶ How do we measure the length of a curve?
▶ Imagine fitting a string to the curve and measuring the string
with a ruler.
▶ To make that idea mathematically precise, we start by
approximating the curve with line segments as shown below.
▶ Denote the line segment connecting point Pi−1 to Pi by
Pi−1Pi .

6. Length of a Curve
▶ Now summing up the lengths of the line segments, and taking
n → ∞ we get the length of the curve:
L = lim
n→∞
n
X
i=1
|Pi−1Pi |
where |Pi−1Pi | is the distance between points Pi−1 and Pi .
|Pi−1Pi | =
q
(xi − xi−1)2 + (yi − yi−1)2
=
q
(△x)2 + (△yi )2
= △x
s
1 +
△yi
△x

7. Length of a Curve
By the mean value theorem for derivatives, there exists a number
x∗
i ∈ (xi−1, xi ) such that
f ′
(x∗
i ) =
f (xi ) − f (xi−1)
xi − xi−1
=
△yi
△x
Now we have
L = lim
n→∞
n
X
i=1
s
1 +
△yi
△x
△x
= lim
n→∞
n
X
i=1
q
1 + f ′(x∗
i )△x
=
Z b
a
q
1 + [f ′(x)]2dx

8. Arc Length Formula (1)
If f ′ is continuous on [a, b], then the length of the curve y = f (x),
a ≤ x ≤ b is
L =
Z b
a
q
1 + [f ′(x)]2dx
Or, in Leibniz notation
L =
Z b
a
r
1 +
dy
dx
2
dx

9. Example 1
1/2
Find the length of the arc of the semicubical parabola y2 = x3
between the points (1, 1) and (4, 8).
▶ Solving for y we have y = ±
√
x3, but since we are only
concerned with the length of the curve between y = 1 and
y = 8 we take y =
√
x3 = x3/2
▶ We have
dy
dx
=
3x1/2
2

10. Example 1
2/2
Z 4
1
r
1 +
3x1/2
2
2
dx =
Z 4
1
r
1 +
9
4
xdx
| {z }
u=1+9x
4
, du=9
4
dx→4
9
du=dx
u1=13/4, u2=10
=
4
9
Z 10
13/4
u1/2
du
=
4
9
2
3
u3/2

13. 10
13/4
=
8
27
(103/2
− (13/4)3/2
)
=
8
27
(10
√
10 −
1
8
· 13
√
13)
=
80
27
√
10 −
13
27
√
13

14. Arc Length Formula (2)
If a curve has equation x = g(y) where g′(y) is continuous on
c ≤ y ≤ d then we can interchange the role of x and y in the arc
length formula so
L =
Z d
c
q
1 + [g′(y)]2dy
Or in Leibniz notation
L =
Z b
a
s
1 +
dx
dy
2
dy

15. Example 2
1/4
Find the length of the arc of the parabola y2 = x from (0, 0) to
(1, 1).
▶ We use the arc length formula (2).
x = y2
,
dx
dy
= 2y

16. Example 2
2/4
L =
Z 1
0
q
1 + (2y)2dy =
Z 1
0
p
1 + 4y2dy
| {z }
u=2y, du=2dy→ 1
2
du=dy
u1=0, u2=2
=
1
2
Z 2
0
p
1 + u2du
| {z }
tan t=u, du=sec2 tdt
t1=0, t2=tan−1 (2)
=
1
2
Z tan−1 (2)
0
√
sec2 t sec2
tdt
=
1
2
Z tan−1 (2)
0
sec3
tdt

17. Example 2
3/4
▶ Recall from previous examples
Z
sec3
xdx =
1
2
(sec x tan x + ln | sec x + tan x|) + C
=⇒
1
2
Z tan−1 (2)
0
sec3
tdt =
1
4
(sec t tan t+ln | sec t + tan t|)

20. tan−1 (2)
0
where tan−1
(2) = θ ⇐⇒ tan θ = 2 =⇒
sec θ =
p
tan2 θ + 1 =
p
22 + 1 =
√
5 =⇒
sec (tan−1
2) =
√
5
tan (tan−1
2) = 2
sec 0 = 1
tan 0 = 0

21. Example 2
4/4
1
4
(sec x tan x + ln | sec x + tan x|)

24. tan−1 2
0
=
1
4
2
√
5 + ln (
√
5 + 2)
. . . . . . − (0 + ln 1)
=
√
5
2
+
1
4
ln (
√
5 + 2)

25. Example 3
1/2
Find the length of the curve y = ln sin x
on π/4 ≤ x ≤ π/2.
▶ Try this on your own before proceeding to the solution.

26. Example 3
2/2
dy
dx
=
1
sin x
cos x = cot x
L =
Z π/2
π/4
p
1 + cot2 xdx
=
Z π/2
π/4
csc xdx
= ln | cot x − csc x|

29. π/2
π/4
= ln |0 − 1| − ln |1 −
√
2|
= ln |
1
1 −
√
2
|
= ln |
1 +
√
2
−1
|
= ln (1 +
√
2)

30. The Arc Length Function

31. Arc Length Function
▶ The function that gives the length of the curve from a starting
point a to x, called the arc length function is
s(x) =
Z x
a
q
1 + [f ′(t)]2dt
▶ Differentiating with respect to x gives
ds
dx
=
q
1 + [f ′(x)]2
=
r
1 +
dy
dx
2
by FTC.
▶ Note that it is always greater than or equal to 1 with equality
when f ′(x) = 0, ie, when f (x) is constant.

32. Arc Length Function
▶ Multiplying both sides by differential dx we get the arc length
differential
ds =
r
1 +
dy
dx
2
dx (1)
▶ Squaring both sides yields
(ds)2
= (dx)2
+ (dy)2
▶ And rearranging we have
ds =
s
1 +
dx
dy
2
dy (2)
So we can write L =
R
ds using either eqn (1) or (2) for ds.

33. Example 4
1/3
Find the arc length function for the curve f (t) = t2 − 1
8 ln t, taking
t = 1 as the starting point.

34. Example 4
2/3
f (t) = t2
−
1
8
ln t
f ′
(t) = 2t −
1
8t
[f ′
(t)]2
= 4t2
− 2(
1
4
) +
1
64t2
1 + [f ′
(t)]2
= 4t2
+
1
2
+
1
64t2
s(x) =
Z x
1
r
4t2 +
1
2
+
1
64t2
dt
=
Z x
1
r
256t4 + 32t2 + 1
64t2
dt
=
Z x
1
1
8t
p
256t4 + 32t2 + 1dt

35. Example 4
3/3
=
Z x
1
1
8t
q
(16t2 + 1)2dt
=
Z x
1
1
8t
(16t2
+ 1)dt
=
Z x
1
2t +
1
8t
dt
= t2
+
1
8
ln |t|

38. x
1
= x2
+
1
8
ln x − 1 −
1
8
ln 1
= x2
+
1
8
ln x − 1