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Lecture 16 : Arc Length
In this section, we derive a formula for the length of a curve y = f (x) on an interval [a, b]. We will
assume that f is continuous and differentiable on the interval [a, b] and we will assume that
its derivative f
′ is also continuous on the interval [a, b]. We use Riemann sums to approximate the
length of the curve over the interval and then take the limit to get an integral.
We see from the picture above that
L = lim n→∞
∑^ n
i=
|Pi− 1 Pi|
Letting ∆x =
b−a n
= |xi− 1 − xi|, we get
|Pi− 1 Pi| =
(xi − xi− 1 )^2 + (f (xi) − f (xi− 1 ))^2 = ∆x
[
f (xi) − f (xi− 1 )
xi − xi− 1
] 2
Now by the mean value theorem from last semester, we have
f (xi)−f (xi− 1 ) xi−xi− 1
= f
′ (x
∗ i ) for some^ x
∗ i in the
interval [xi− 1 , xi]. Therefore
L = lim n→∞
n ∑
i=
|Pi− 1 Pi| = lim n→∞
n ∑
i=
1 + [f ′(x
∗ i )]
(^2) ∆x =
∫ (^) b
a
1 + [f ′(x)]^2 dx
giving us
L =
b
a
1 + [f
′ (x)]
2 dx or L =
b
a
[
dy
dx
] 2
dx
Example Find the arc length of the curve y =
2 x^3 /^2 3
from (1,
2 3
) to (2,
4
√ 2 3
Example Find the arc length of the curve y =
ex+e−x 2
, 0 ≤ x ≤ 2.
Example Set up the integral which gives the arc length of the curve y = e
x , 0 ≤ x ≤ 2. Indicate
how you would calculate the integral. (the full details of the calculation are included at the end of your
lecture).
For a curve with equation x = g(y), where g(y) is continuous and has a continuous derivative on the
interval c ≤ y ≤ d, we can derive a similar formula for the arc length of the curve between y = c and
y = d.
L =
∫ (^) d
c
1 + [g
′ (y)]
2 dy or L =
∫ (^) d
c
[
dx
dy
] 2
dy
Example Find the length of the curve 24xy = y
4
4 3
, 2) to (
11 4
We cannot always find an antiderivative for the integrand to evaluate the arc length. However, we can
use Simpson’s rule to estimate the arc length.
Example Use Simpson’s rule with n = 10 to estimate the length of the curve
x = y +
y, 2 ≤ y ≤ 4
dx/dy = 1 +
y
L =
2
[
dx
dy
] 2
dy =
2
[
y
] 2
dy =
2
y
4 y
dy
Worked Examples
Example Find the length of the curve y = e
x , 0 ≤ x ≤ 2.
L =
0
[
dy
dx
] 2
dx =
0
[
e
x
] 2
dx =
0
1 + e
2 x dx
Let u = e
x , du = udx or dx = du/u. u(0) = 1 and u(2) = e
2
. This gives
2
0
1 + e
2 x dx =
e^2
1
1 + u
2
u
du
Letting u = tan θ, where −π/ 2 ≤ θ ≤ π/2, we get
1 + u
2
1 + tan
2 θ =
sec
2 θ = sec θ and
du = sec
2 θdθ ∫ (^) tan− (^1) (e (^2) )
π 4
sec θ
tan θ
sec
2 θdθ
∫ (^) tan− (^1) (e (^2) )
π 4
sec
3 θ
tan θ
dθ =
∫ (^) tan− (^1) (e (^2) )
π 4
sec
3 θ tan θ
tan
2 θ
dθ
∫ (^) tan− (^1) (e (^2) )
π 4
sec
3 θ tan θ
sec
2 θ − 1
dθ
Letting w = sec θ, we have w(
π 4
2, w(tan
− 1 (e
2 )) =
1 + e
4 from a triangle and dw = sec θ tan θ.
Our integral becomes
√ 1+e^4
√ 2
w
2
w^2 − 1
dw =
√ 1+e^4
√ 2
w^2 − 1
dw =
√ 1+e^4
√ 2
w − 1
w + 1
dw
= w +
ln |w − 1 | −
ln |w + 1|
]
√ 1+e^4
√ 2
= w +
ln
w − 1
w + 1
]
√ 1+e^4
√ 2
1 + e
4 −
ln
1 + e^4 − 1 √ 1 + e^4 + 1
