Length of a Curve

Suppose one wanted to find the length of the curve above. Let us suppose also that this curve has aparametrization c : [a, b]→ R2 which we’ll abbreviate as c(t) = (x(t), y(t)). Suppose also that c(t) isdifferentiable i.e x(t), y(t) are differentiable. Let P = {x0 = a < t1 < · · · < tn = b} be a partition of [a, b].We will denote the distance between (x(ti−1), y(ti−1)) and (x(ti), y(ti)) as Li(t) where:

Li(t) =√

(x(ti−1)− x(ti))2 + (y(ti−1)− y(ti))2

Since x(t), y(t) are differentiable then there exists t∗i ∈ [ti−1, ti] s.t, x′(t∗i )(ti − ti−1) = x(t1)− x(ti−1). Letuse denote ∆t = ti − ti−1 then we can rewrite the above equation as:

Li(t) =√

(x′(t∗i )∆t)2 + (y′(t∗i )∆t)

2 =√

(x′(t∗i ))2 + (y′(t∗i ))

2 ∆t

We observe that Li(t) resembles a term in the Riemann sum for f(x) =√x′(t)2 + y′(t)2. In other words,

if we know that:

SN =N∑j=1

Lj(t) <∞

Then it follows from our definition of the Riemann Integral that we have:

limN→∞

SN =

∫ b

a

√x′(t)2 + y′(t)2 dt =

∫ b

a

‖c′(t)‖ dt

we take this to be the definition for the length of a parametrized curve. It can be easily shown that thedefinition above generalizes to arbitrary curves c : I → Rn.

1

Arc-Length Function

From now on we will assume γ′(0) 6= 0 i.e γ is regular. Suppose we wish to only look at the length of acurve segment for γ : [a, b]→ Rn where n = 2, 3. Then for t > a we can define the length of the segmentfor γ on [a, t] by:

s(t) :=

∫ t

a

‖γ′(u)‖ du

Observe that by the Fundamental Theorem of Calculus the speed of the particle at time t is given by:

ds

dt= ‖γ′(t)‖

From here, we can begin our journey on defining curvature. Intuitively, one would think that a goodmeasure for curvature would be ‖γ′′(t)‖ i.e observe how the tangent vector is changing along γ. However,this measure is speed dependent i.e not invariant under reparametrization. In the same vein, if we couldperhaps make it so that ‖γ′(t)‖ = 1 for all t then κ(t) = ‖γ′′(t)‖ becomes well-defined and is timeindependent. To see this, let t = g(s) and consider the reparametrization γ1(s) = γ(g(s)) then:

(γ ◦ g)′(s) = γ′(g(s)) · g′(s) = γ′(g(s)) · dtds

=γ′(g(s))

ds

dt

⇒ ‖r′′1(s)‖ =γ′′(t)

ds

dt

= ‖γ′′(t)‖

The big question now becomes, what is g(s)? How do we find it? Does it always exist? Consider s(t)given above and observe that it is continuous and strictly increasing. Thus, it admits an inverse g(s).Here g : [a, b]→ [a, b] is a continuous bijection (?) i.e γ1(s) = γ(g(s)) is a reparametrization of γ and:

(s ◦ g)(t) = t⇒ s′(g(t)) · g′(t) = 1⇒ g′(t) =1

(s ◦ g)′(t)=

1

γ′(g(s))

With this we can now show that (γ ◦ g)(s) has unit speed:

‖(γ ◦ g)′(s)‖ = ‖γ′(g(s)) · g′(s)‖ =

∥∥∥∥γ′(g(s)) · 1

γ′(g(s))

∥∥∥∥ = 1

2

Curvature

In order to stay on the path above, we must change direction. This change of direction is due to the pathbending. Since tangent vectors give the direction of a particles movement, then one might say that agood measure for curvature would be:

κ(t) = ‖γ′′(t)‖

i.e we look at the rate in which the tangent vector changes. We observe that such a definition is speeddependent in the sense that the tangent vector will change much quicker is someone is running asopposed to walking. However, we know that we can find a parameter s such that ‖γ′(s)‖ = 1. Thus is sis the unit speed parameter then we define:

curvature = κ(s) = ‖γ′′(s)‖

In practice, computing the inverse for s(t) is nearly impossible so we would like to get a definition whichescapes this process. The idea behind such a scheme is to suppose we have t = g(s) where g(s) is theunit-speed parameter. Define γ1(s) = (γ ◦ g)(s) then γ′1(s) is a unit vector at g(s). Observe that if wedefine the unit tangent vector:

T (t) =γ′(t)

‖γ′(t)‖= (T ◦ g)(s)

then T (t) is also a unit vector at g(s) i.e r′1(s) = T (t). To finish we differentiate and use chain rule:

r′′1(s) =d

ds(T ◦ g)(s) = T ′(g(s)) · g′(s) = T ′(g(s)) · dt

ds=T ′(t)

ds

dt

=T ′(t)

‖γ′(t)‖⇒ κ(s) = ‖r′′1(s)‖ =

‖T ′(t)‖‖γ′(t)‖

3

A More Suitable Definition

Even with the newly condensed definition of curvature for α : I → R2 or 3 there is a bit of work to do.Thankfully we can condense this definition even further and later pick which one is more suitable for acertain problem. We will now choose to drop (t) in the following calculations i.e T (t) will be abbreviatedas T and so forth.

T =γ′

‖γ′‖⇒ γ′ = ‖γ′‖T

⇒ γ′

= ‖γ′‖T ′ + T‖γ′‖′

⇒ r′ × r′′ = ‖γ′‖T × (‖γ′‖T ′ + T‖γ′‖′)= ‖γ′‖2

⇒ ‖r′ × r′′‖‖γ′‖3

=‖T ′‖‖γ′‖

= κ(t)

We know have a working definition for curvature. Since the above uses cross products, it is very usefulfor regular curves α : I → R3. For regular plane curves which are graphs of some function f : I → R wecan use this definition to derive another beautiful formula. We can parametrize graph(f) by v.v.functionγ(x) = 〈x, f(x)〉. Next, we view γ as a curve in 3-space by adding a zero in the last coordinate i.e wehave γ̃(x) = 〈x, f(x), 0〉. Now we run γ̃ through that curvature formula given above:

κ(x) =‖〈1, f ′(x), 0〉 × 〈0, f ′′(x), 0〉‖

(1 + (f ′(x))2)32

=|f ′′(x)|

(1 + (f ′(x))2)32

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Visualizing Curvature

If one would like to visualize curvature at a point p for γ : I → R2 or 3 then one could construct the“osculating circle” at p. To parametrize this circle we need two things, center and radius. We will allowthe radius to be r = 1

κ. That way if the curvature is large then our circle will be small, otherwise if small,

our circle will be very big. From the image above, the center is given by ~OP + 1κ~N(t), but what is ~N?

Here ~N is the unit normal vector i.e ~N ⊥ ~T and we choose ~N s.t it points in the direction in which thecurve is turning. To quantify ~N mathematically we have:

~N(t) =~T ′(t)

‖~T ′(t)‖

Choosing again to drop (t) we observe that T · T = 1 and so (T · T )′ = 2T · T ′ = 0⇒ T ′ ⊥ T i.e N ⊥ T .Now that we have our unit normal vector, we can get the center which is given by:

center of curvature = ~OP +1

κ~N(t)

To get a parametrization γ(t) for the osculating circle, we recall that for a circle of radius 1κ

centered at(0, 0), it’s parametrization is given by:

1

κ〈cos t, sin t〉 ⇒ γ(t) =

(~OP +

1

κ~N(t)

)+

1

κ〈cos t, sin t〉

5

Frenet- Serret Formulas

The Frenet-Serret Formulas were discovered independently by French mathematicians Jean Frdric Frenet(1816-1900) and Joseph Alfred Serret (1819-1885). The formulas describe the geometric properties of adifferentiable curve α in 3-space. It has been said that if one wants to know something about curves thenthe result is probably given by “tinkering” with the Frenet-Serret Formulas. In other words, theseformulas basically sealed the deal for curve theory (“the study of curves”). The construction basicallygoes as follows:

• Let α : I → R3 be a differentiable curve

• Given α(s) we consider a frame about α(s) i.e a local coordinate system

• The orthonormal frame is given by ~Tα(s), ~Nα(s) and ~B = ~Tα(s) × ~Nα(s)

• Given any orthonormal frame {e1p, e2p, e3p} and vp we have:

vp =∑j

(vp · ejp) ejp

• Now using orthonormal expansion (given above), express ~T ′α(s), ~N ′α(s) and ~B′ = ~Tα(s) × ~Nα(s) in

terms of ~Tα(s), ~Nα(s) and ~Bα(s)

Following the program given above we produce the Frenet-Serret formulas which state:

dT

ds= κN

dN

ds= −κT + τB

dB

ds= −τN

6

Total Curvature, Knots and Fary-Milnor

The total curvature for regular smooth space curve parametrized γ : I → R3 which is also parametrizedby arc-length is given by: ∫ b

a

κ(s) ds =

∫ b

a

‖γ′′(s)‖ ds

A knot is an embedding α : S1 → R3. Ignoring this maybe technical definition, we think of a knot as theproduct of starting with a finite piece of string, twist and turn as you may please (making sure to keepthe endpoints at your fingertips) then glue the endpoints together (a picture is given below). Two knotsbeing equivalent if there exists an ambient isotopy between them i.e seq. of allowable invertible-moves.

Question: What is the minimal total curvature for a knot? Well, this question was answeredindependently by John Milnor (1950) and Fary(1949). The exact statement is that any closed curve K inEuclidean space for which we can define the curvature κ at each of its points then:

∮K

|κ(s)| ds ≤ 4π ⇒ K is an unknot

7

Problems to Think About

1. If γ is a curve who’s endpoints lie at (±1, 0) on S1. Can you give bounds for the total curvature if weconsider the collection of γ with fixed endpoint given above and γ ⊂ S1?

2. Suppose two curves γ1, γ2 are homotopic relative to their endpoints. If H is a homotopy between theγi’s then are their total curvature related in any ‘interesting” way using H?

3. Now that we are done with curvature for curves, how can we being to think about curvature forsurfaces (images given below)?

4. Given α : I → R3, when can we guarantee that α always lies on some plane P?

5. Given a path γ : I → S2, how much does γ need to bend in order to stay on the surface?

8