Proof of the arc-length formula for smooth graphs
March 21, 2017
Let f W Œa; b ! R be a continuous function. For any partition
P D fx0 D a; x1 ; : : : ; xn 1 ; xn D bg
of the interval Œa; b, consider the points Ai D .xi ; f .xi // on the graph of f . The union of
the line segments
A0 A1 A1 A2 An 2 An 1 An 1 An
is the polygonal approximation to the graph of f given by the partition P. The length
`.f; P/ of this polygonal approximation is clearly
n
X
(1)
`.f; P/ D
distance.Ai 1 ; Ai /:
i D1
By definition, the length `.f / of the graph of f is the least upper bound of the lengths of
all such polygonal approximations:
`.f / D sup `.f; P/:
(2)
P
As we have seen, there are continuous functions f for which `.f / D C1 (a simple
variation of the von Koch snowflake would provide such an example). However, when f
is sufficiently smooth, `.f / is finite and can be expressed as an integral. To motivate this
formula, let us take a partition P as above and use the notations
xi D xi
xi
1
and
yi D f .xi /
f .xi
1 /:
Then, by the distance formula in the plane, (1) becomes
s
n
n
X
X
p
yi 2
2
2
`.f; P/ D
.xi / C .yi / D
1C
xi :
x
i
i D1
i D1
Assuming f is differentiable, the mean value theorem shows that
yi
D f 0 .ci /
for some ci in .xi 1 ; xi /:
xi
Hence,
n
X
p
(3)
`.f; P/ D
1 C .f 0 .ci //2 xi
i D1
which is a Riemann sum for the function
informally, suggests the following result:
p
1 C .f 0 /2 and the partition P. This, at least
Theorem. Suppose the derivative f 0 exists and is continuous on Œa; b. Then
Z bp
`.f / D
1 C .f 0 .x//2 dx:
a
In particular, `.f / is finite.
The proof depends on the following simple lemma which says that adding points to a
partition can only increase the length of the corresponding polygonal approximation.
Lemma. If a partition Q is a refinement of another partition P, then `.f; P/ `.f; Q/.
Proof. It suffices to verify the claim when Q has only one point more than P. The general
case will then follow by repeated application of this case. So, let
P D fx0 D a; x1 ; : : : ; xi
1 ; xi ; : : : ; xn 1 ; xn
D bg
and
Q D fx0 D a; x1 ; : : : ; xi 1 ; t; xi ; : : : ; xn 1 ; xn D bg;
with the extra point t in Q lying between xi 1 and xi . Calling T D .t; f .t//, we see that
the polygonal approximation given by Q is the same as the one given by P except that the
line segment Ai 1 Ai should be replaced by the union of the two segments Ai 1 T and TAi .
By the triangle inequality
distance.Ai
1 ; Ai /
distance.Ai
1; T /
C distance.T; Ai /;
2
p
Proof of the theorem. For convenience, let g D 1 C .f 0 /2 . By the assumption, g is
continuous, hence it is integrable on Œa; b. Moreover, (3) shows that
proving `.f; P/ `.f; Q/.
(4)
L.g; P/ `.f; P/ U.g; P/
for every P;
where, as usual, L.g; P/ and U.g; P/ are the lower and upper sums of g for the partition P.
Since `.f; P/ `.f / by the definition (2), the left inequality in (4) shows that
L.g; P/ `.f /
for every partition P. Taking the supremum over all P gives
Z b
(5)
g `.f /:
a
To prove the reverse inequality, take any two partitions P and Q of Œa; b and let S D P [ Q
be their common refinement. Then, `.f; P/ `.f; S/ by the above lemma, `.f; S/ U.g; S/ by (4), and U.g; S/ U.g; Q/ because upper sums decrease upon refining the
partition. Putting these three inequalities together, it follows that
`.f; P/ U.g; Q/
for any two partitions P and Q. Taking the infimum over all Q gives
Z b
`.f; P/ g
a
for every P. Finally, taking the supremum over all P shows
Z b
(6)
`.f / g:
a
The proof is now complete once we put (5) and (6) together.
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