Math 162A - Notes

Neil Donaldson

Winter 2018

Introduction and notation

Classical Differential Geometry is the study of curves and surfaces in the plane and three-dimensionalspace. The word differential indicates that we will be using calculus! Indeed the basic ingredients ofdifferential geometry are multi-variable calculus, linear algebra and differential equations. At grad-uate level, one also requires a great deal of topology, analysis and algebra, but none of this is appearsin our course! Our purposes is to use calculus to describe certain properties of curves and surfaces.

Of particular interest is the notion of curvature. For curves this will turn out to be fairly simple:curvature is a measure of the bendiness of a curve. A straight line should have zero curvature,whereas a circle should have curvature which increases as the radius decreases: a very large circleshould have a very small curvature.

Zero curvature Small curvature Larger curvature

Changing curvature

Understanding and quantifying this concept for more complicated curves is our first important goal.For instance, the curvature of the spiral increases as one travels towards the center. The rough ideais as follows: imagining a curve as a roller-coaster along which you travel at a constant speed, thecurvature should equal the force you feel, necessary to keep you travelling along the curve.

For surfaces, curvature is a more difficult concept. We will hunt for quantities which measure howmuch a surface appears to be dome- or saddle-shaped. Understanding this will occupy much of thesecond half of the course.

Dome-shaped Saddle-shaped More complicated

The third surface is saddle-shaped near the narrow neck and dome-shaped away from it.

Euclidean Space Classical differential geometry is usually conducted in Euclidean space En. It isimportant to distinguish this from Rn.

Rn Is the set of n-tuples of real numbers. These can be written as row- or column-vectors, yourchoice. Thus

Rn ={(x1, . . . , xn) : xi R

}In this course, Rn is nothing more than a co-ordinate space. It is not assumed to have any addi-tional vector-space structure (no scalar multiplaction or addition of vectors), and certainly nodot product!

En is the set Rn together with the usual notion of dot product. We usually write these as columnvectors, and denote elements bold face1. For instance,

x =

x1x2x3

E3This helps us distinguish vectors in Euclidean space from co-ordinates. The dot product allows

us to measure length and angle. To refresh; if a =(

a1a2

)and b =

(b1b2

)are two vectors in E2,

then:

The dot product of a and b is a b = a1b1 + a2b2. The length of a is |a| =

a a.

The angle between a and b satisfies cos =a b|a| |b| .

In this class, Rn will always have n = 1 (for curves) or n = 2 (for surfaces). Similarly En will alwayshave n = 2 or 3. If a definition or result is true in both E2 and E3 (and more generally in En), we willjust write En.

1Underline x or over-arrow ~x when hand-writing.

2

Examples

1. Let x(t) =(

cos tsin t

)where t R. This defines a function x : R E2. The components of x are

the functions

x1(t) = cos t, x2(t) = sin t

It is perfectly acceptable to use x and y for the components if you prefer:

x(t) =(

x(t)y(t)

)=

(cos tsin t

)Just make sure you dont get x and x confused!

2. Let x(u, v) =

uvu2 + v2

where u, v R (or (u, v) R2 if you prefer). This defines a functionx : R2 E3. Its components are

x(u, v) = u, y(u, v) = v, z(u, v) = u2 + v2

It isnt important at the moment, but these examples describe, respectively, a circle in the plane anda paraboloid in three-dimensions.

1 Curves

1.1 Parameterized Curves in Euclidean space

Definition 1.1. A parameterized differentiable curve in En is a smoothfunction x : I En of a subinterval I = (a, b) of the real line Rinto En.Its velocity, or tangent vector at t0 I is the vector x(t0). In termsof the components, we can write

x(t0) =

x1(t0)

...xn(t0)

O

x(t0)

x(t0)

Smooth means differentiable with continuous derivatives; that the quotient

x(t) = limh0

x(t + h) x(t)h

exists for every t I and that the function x(t) is continuous. The fact that we are differentiating avector-valued function is of no concern: we are not dividing by a vector in the above quotient!

3

Examples

Straight line Fix constant vectors a, b En. The parameterizedcurve defined by

x(t) = a + t(b a)

describes the line through the point A and B with position vec-tors, respectively, a and b.

AB

O

x(0) = ax(1) = b

b a

Helix Let x(t) =

cos tsin tt

. The function x : R E3 describesa helix. To help visualize the helix, you can try several tricks.Imaging sitting on top of the z-axis and looking down: youd only

see the horizontal projection of the curve t 7(

cos tsin t

)which de-

scribes the unit circle counter-clockwise. The projection onto thez-axis is simply t 7 t, so the curve moves upwards at a constantspeed. One could also project onto the xz- or yz-planes. All theseimages are plotted below

1

1y

1 1x

z

1 1x

2

3z

1 1y

2

3

Tangent Line Given a parameterized differentiable curve x : I En, we can easily define its tangent line at any fixed t0. This issimply the straight line through the point with position vectorx(t0), pointing in the direction of the tangent vector x(t0). It isitself a parameterized curve, y : R En:

y(s) = x(t0) + sx(t0)

For example, the tangent line to the helix at t0 = 3 has equation

y(s) =

12323

+ s

3

2121

4

Corners and Cusps

We will almost always think about parameterized curves whichare differentiable everywhere. It is, however, easy to constructcurves which are not. Consider the curve

x(t) =(

t1 |t|

), t (1, 1)

This has a corner at x(0) =(

01

)and is not differentiable there.

Indeed, this is not a parameterized curve in the language of Defi-nition 1.1.

1y

1 0 1x

A similar difficulty appears with a cusp: for example

x(t) =(

t|t|

), t (1, 1)

is not differentiable at t = 0.

1y

1 0 1x

In either case we call the non-differentiable point a singularity of the curve.

Self-intersections

Curves which cross themselves might seem difficult, but the pa-rameterization takes care of everything! The curve in the picturehas parameterization

x(t) =(

sin 2t3cos t

), t [0, 6)

The curve is differentiable everywhere, even though it intersectsitself. For instance, the curve passes through the origin at botht1 = 32 and t2 =

92 , with corresponding tangent vectors

x( 32 ) =( 23

1

), x( 92 ) =

( 231

) 1

1y

1 1x

The important thing to note about self-intersections is that we always talk about the tangent vectorto a parameterized curve at a co-ordinate t = t0. In this example, it is illegal to talk about the tangentvector to the curve at the origin. The parameterization takes care of the fact that there are two distincttangent vectors at the origin.

5

Speed

Definition 1.2. The speed of a curve x(t) is the norm/length of its velocity vector

v(t) :=x(t) = x1(t)2 + + xn(t)2

Note that the speed of a curve is a non-negative, scalar function, whereas the velocity is a vector. More-over, for a smooth parameterized curve, the speed must be a continuous function.

Examples

1. The helix parameterized by x(t) =

cos tsin tt

has speedv(t) =

( sin t)2 + (cos t)2 + 12 =

2

2. The curve x(t) =(

t3

t6

)has tangent vector x(t) =

(3t2

6t5

)and speed v(t) = 3t2

1 + 4t6.

3. The curve x(t) =(

t1 |t|

)drawn above, has tangent vector

x(t) =

(11

)if t < 0,(

11

)if t > 0,

and is undefined at t = 0. Its speed is therefore v(t) =

2 for t 6= 0 and undefined for t = 0.

4. The curve x(t) =(

sin 2t3cos t

)has tangent vector

x(t) =( 2

3 cos2t3

sin t

)

and speed v(t) =

49 cos

2 2t3 + sin

2 t.

The picture shows the curve and its tangent vector in blue: observe how the speed of the curve(length of the tangent vector) changes as we move round the curve.2

2Click the picture when open in a full-feature viewer such as Acrobat.

6

1.2 Regular curves and the arc-length parameterization

If we are going to do differential geometry, we need to be able to differentiate. Moreover, if we wantto do anything with the tangent vector to a curve, we had better make sure that it is non-zero. Wetherefore want to isolate those curves which have non-zero tangent vectors as our main objects ofstudy.

Definition 1.3. A curve x : I En is regular at t0 if x(t0) 6= 0. A curve is regular if it is regular for allt I. Equivalently, its speed is never zero.

Examples We consider our examples from the previous section:

1. The helix is regular since v(t) =

2 is always positive.

2. x : t 7(

t3

t6

)is non-regular at t = 0, since x(0) = 0.

3. x : t 7(

t1 |t|

)is non-regular at t = 0, since it is not differentiable there.

4. x : t 7(

sin 2t3cos t

)is regular, since x(t) is never zero.

Reparameterization

The second example above is non-regular only at t = 0. Consider now the curve

y(s) =(

ss2

)This results in the same physical curve in E2 via the correspondence

s = t3 ! x(t) = y(t3)

As parameterized curves however, these are distinct examples. In particular,

y(s) =(

12s

)whence y is a regular curve with non-zero speed

v(s) =

1 + 4s2

Regularity is not an intrinsic property of the curve as a subset of the plane, rather it is a property of thecurves parameterization.

Definition 1.4. A reparameterization of a parameterized curve x(t) is a curve y(s) := x((s)) where(s) is some (increasing or decreasing) differentiable function.For reasons seen in the next theorem, when talking about regular curves, the function (s) must besmooth (differentiable with continuous derivatives) and non-zero.

7

Theorem 1.5. Suppose that x : I En is a regular parameterized curve. If (s) is smooth and non-zero,then the reparameterization y(s) = x((s)) is also regular.

Proof. By the chain rule, dyds = (s) dxdt , whence the velocity of the reparameterization is dyds

= (s) x(t)Since (s) is non-zero, it follows that the velocity in the new parameterization is also non-zero.

To compare with the above example, the reparameterization (s) = 3

s is not differentiable at s = 0:this is the reason it was able to turn a non-regular parameterization into a regular one.

Example Let x(t) =(

cos tsin t

)where t [0, 2). This describes the unit circle. A simple reparame-

terization could be obtained using t = (s) = 2s. This yields the curve

y(s) = x(2s) =(

cos 2ssin 2s

)The domain is now s [0, ), and the new curve travels at twice the speed of the original:y(s) = 2x(2s) = 2 x(t)Notice that both parameterizations are regular.

Aside: a non-regularizable curveExample 3, above, cannot be made regular via any reparameterization. Indeed let (s) be an increas-ing, differentiable function such that (0) = 0, and set y(s) = x((s)). Then

y(s) = (s)x((s))

If y is to be regular at s = 0, we should be able to compute y(0). However this is impossible:x((0)) = x(0) is undefined, whence we would have to attempt to define

y(0) = lims0

(s)x((s)) = (0) lims0

x((s))

Since lims0

x((s)) is also undefined,3 we could only have (0) = 0, resulting in y(0) = 0, whence

the curve remains non-regular at t = 0.The difference between Examples 2 and 3 is simple: you cant regularize corners, but you can regu-larize smooth curves that have been parameterized so that the curve sometimes comes to a halt.

3limt0+ x(t) =(

11)6=(

11

)= limt0 x(t).

8

A special parameterization for regular curves

On of the difficulties of parameterized curves is that one might struggle to tell which properties ofa curve are truly inherent to the curve, and which depend on the parameterization. If a curve isregular, then we may choose to travel along it in such a way that our speed is always 1. This allowssomething of a uniform parameterization for regular curves.

Definition 1.6. The (signed) arc-length of a curve x(t) from t0 to t is the integral of the speed

s(t) = t

t0

x(T) dT = tt0

v(T)dT

A curve is unit-speed if |x(t)| = 1 for all t I.The arc-length is signed because, if t < t0 we have a negative length: we are measuring length againstthe direction (orientation) of the curve. To complete the picture we need the following proposition:

Theorem 1.7. Unit speed curves are parameterized by arc-length from some value t0. Conversely, any regularcurve x has a unit-speed parameterization.

Proof. Suppose first that x(t) is unit-speed. Then s(t) = t

t01 dt = t t0, and so t is the arc-length of

x from t = 0.

Conversely suppose that x(t) is a regular curve. Then x(t) has a well-defined arc-length s(t) fromany fixed value t = t0. It follows4 that s(t) = dsdt = v(t).Since x is regular, it follows that v(t) > 0, whence s(t) > 0 and so s(t) is an increasing function. Sucha function is necessarily invertible. Let the inverse be t(s) and define y(s) = x(t(s)). But then, by thechain rule,

y(s) =dds

x(t(s)) = x(t)dtds

=x(t)s(t)

=y(s) = |x(t)||s(t)| = v(t)v(t) = 1

Notice that we need regularity here or else we might end up dividing by zero!

There are many possible arc-length parameterizations of a curve. If x(t) is an arc-length parameter-ization, then so is x(t t0) for any t0. Since these parameterizations involving travelling along thecurve in the same direction, we say that they preserve orientation. Another collection of arc-lengthparameterizations is given by x(t0 t) for any t0: these have opposite orientation since they traversethe curve in reverse.

Computing an arc-length parameterization can be hard work:

It may not be possible to explicitly evaluate the integral s(t) = t

t0|x(T)| dT.

Even if you can compute s(t), you may not be able to find its inverse t(s).

However, by the Theorem, we can always assume that one exists. Computing in terms of it will likelymake things easier.

We now try to find arc-length parameterizations for our examples.

4Fundamental Theorem of Calculus: ddt t

t0v(T)dT = v(t) . . .

9

Examples

1. The helix x(t) =

cos tsin tt

has constant speed 2. The arc-length from t0 = 0 is therefores(t) =

2t, with inverse function t(s) = s

2. It follows that

y(s) =

cos(s/

2)sin(s/

2)

s/

2

is a unit speed parameterization of the helix.

2. Let x(t) =(

tt2

)(weve modified the example to make it regular). The arc-length from t0 = 0 is

s(t) = t

0

1 + 4T2 dT

This can be computed with some judicious use of substitutions,5 though it isnt fun!

s(t) =12

t

1 + 4t2 +14

ln2t +1 + 4t2

Now you need an inverse! This problem is essentially impossible to do explicitly.

3. At least where it is regular, x(t) =(

t1 |t|

)has speed v(t) =

2 (t 6= 0). Thus x(s) =(

s/

21

s/2)

is a unit speed parameterization, away from s = 0.

4. Explicitly finding a unit speed parameterization of x(t) =(

sin 2t3cos t

)is also far too difficult. It

would involve computing the integral

s(t) = t

0

49

cos22T3

+ sin2 T dT

As you can see, it is generally pointless to hunt for an explicit arc-length parameterization. Whatthen is the purpose of such? They are generally used abstractly: if a parameterized curve is regular,then we know that an arc-length parameterization exists. We can therefore assume that an abstractcurve comes with such a parameterization. This makes certain calculations nad theorems much eas-ier. In practice, however, it makes work with explicit examples harder: we are hiding behind anincomputable integral!

5Try tan = 2T . . .

10

The Catenary We finish with one final example of a regular curve with explicit unit-speed param-eterization, though it uses some potentially unfamiliar functions. A catenary is the curve made by achain hanging between two fixed points. It has equation y = cosh x or,6 as a parameterized curve,

x(t) =(

tcosh t

)The curve has speed

v(t) =x = 1 + sinh2 t = cosh t

The arc-length measured from t0 = 0 is

s(t) = t

0cosh T dT = sinh t

Hence y(s) = x(t(s)) has unit speed when t(s) = sinh1 s:

y(s) =(

sinh1 s1 + s2

)It is easily checked that y has unit speed:

y(s) =

(1

1+s2s

1+s2

)=

y(s)2 = 11 + s2

+s2

1 + s2= 1

1

y

1 0 1x

Curvature

Now that we know about unit speed curves, we can properly define our principal notion of bendi-ness.

Definition 1.8. If x : I En is a regular, twice-differentiable, unit-speed curve, its curvature is thefunction

(s) =x(s)

For curves in the plane (n = 2), we modify this slightly: curvature is given a positive (negative) signif the direction of travel bends the curve to the left (right).

Examples

1. A straight line has curvature zero.

6The basic hyperbolic functions are cosh x = ex+ex

2 and sinh x =exex

2 . They satisfyd

dx cosh x = sinh x andd

dx sinh x = cosh x. If youve never seen hyperbolic functions before, you can still do the entire example using expnentials

and logarithms. Indeed a little calculus should convince you that sinh1 s = lns +1 + s2.

11

2. The circle of radius r has unit-speed parameterization

x(s) = r(

cos srsin sr

)and so

x(s) = 1r

(cos srsin sr

)= (s) = 1

r

Since we are bending to the left as we traverse the curve, the curvature should be kept positive.Observe that the curvature is inversely proportional to the radius: smaller circles have largercurvature.

3. The standard helix with unit-speed parameterization

x(s) =

cos(s/

2)sin(s/

2)

s/

2

has

x(s) = 12

cos(s/

2)sin(s/

2)

0

= (s) = 12

4. The catenary above has

y(s) =1

(1 + s2)3/2

(s1

)= (s) = 1

1 + s2

Compare with the picture: y(0) is the point of maximum curvature = 1 at the bottom of thecurve. As we move away, the curvature decreases and the curve becomes straighter.

We shall return to curvature and its meaning later, after we have developed a litte of the theory ofmoving frames.

1.3 Products of Vectors and Moving Frames

The vector (cross) product (in E3) and scalar product (on E2, E3) will be of use to us throughout. Herewe review several useful properties, and put them to work to create families of axes fitted to a givencurve.

Definition 1.9. The scalar product of two vectors x, y En is defined by

x y =n

i=1

xiyi

The norm or length of a vector x is |x| := x xThe angle [0, ] between two non-zero vectors x, y is = cos1 x y|x| |y|Vectors x, y are orthogonal or perpendicular if x y = 0 (if x, y 6= 0 this is equivalent to = 2 )

12

Note that is bilinear (linear in both slots) and symmetric: that is

(ax + by) z = ax z + by + z, x (cy + dz) = cx y + dx z, and x y = y x

Definition 1.10. The vector product, or cross product, of two vec-tors x, y E3 is defined by

x y =

x2 y2x3 y3

x1 y1x3 y3

x1 y1x2 y2

=e1 e2 e3x1 x2 x3y1 y2 y3

where e1, e2, e3 form the standard basis of E3. Otherwise said,x y is the unique vector in E3 such that, for all z E3,

(x y) z = det(x, y, z) =

x1 x2 x3y1 y2 y3z1 z2 z3

This last expression is the scalar triple product of x, y, z.

Example

123

121

=82

4

Theorem 1.11. The vector product satisfies the following:

1. (x y) x = 0 = (x y) y. Therefore x y is orthogonal to the plane spanned by x, y.

2. x y = 0 x, y are parallel (at least one of x, y is a scalar multiple of the other).

3. x y = y x (skew-symmetry).

4. is bilinear.

5. (x y) (zw) =x z x wy z y w

. It follows that|x y| = |x| |y| |sin |

which is the area of the parallelogram spanned by x, y.

6. The scalar triple product is the (signed) volume of the paral-lelepiped (see picture) spanned by x, y, z. The sign is positive ifand only if z and x y lie on the same side of the plane spannedby x, y.

13

Example The area of the parallelogram spanned by

123

and121

is 82 + 22 + 42 = 221.Proof. Everything follows from the basic properties of determinant as seen in a linear algebra course.We prove 1, 2 and 5 and leave the remainder as exercises.

1. (x y) x =

x1 x2 x3y1 y2 y3x1 x2 x3

= 0, since two rows of the matrix are the same. (x y) y = 0 issimilar.

2. If x and y are parallel, then one column in each of the 2 2 determinants in the definition is ascalar multiple of the. The determinant are therefore zero.Now for the converse. If either of x, y is zero, then the vectors are parallel and x y = 0. Thuswe assume that both x, y 6= 0 and that x y = 0. We also assume, WLOG, that x1 6= 0 (if not,reindex what follows).By definitionx1 y1x2 y2

= 0 = x1y2 = x2y1 = (y1y2)=

y1x1

(x1x2

)Similarlyx1 y1x3 y3

= 0 = x1y3 = x3y1 = (y1y3)=

y1x1

(x1x3

)In particular, y = y1x1 x, whence x and y are parallel.

5. Because of the bilinearity of , and the 2 2 determinant, we need only check the first formulawhen x, y, z are two of the standard basis vectors e1, e2, e3. This is easy, but tedious, to check.We then apply the formula with z = x and w = y to obtain

|x y|2 = |x|2 x yx y |y|2

= |x|2 |y|2 (x y)2 = |x|2 |y|2 (1 cos2 ) = (|x| |y| |sin |)2where is the angle between x, y. Since the area of the triangle spanned by x, y making angle is 12 |x| |y| sin , and this comprises half the parallelogram, we are done.

Calculus and moving frames

It is important to be able to differentiate products of vectors. Thankfully both satisfy the usual prod-uct rule and can be proved similarly.

Theorem 1.12 (Product Rule). If x(t), y(t) are curves, then

ddt

(x(t) y(t)) = x(t) y(t) + x(t) y(t)ddt

(x(t) y(t)) = x(t) y(t) + x(t) y(t)

14

Examples

1. Suppose that x(t), y(t) are curves such that x(t) is orthogonal to both y(t) and y(t) for every t.Then,

x(t) y(t) = ddt

(x(t) y(t)) x(t) y(t) = 0

so that x is orthogonal to y also.

2. Suppose that x(t) x(t) = k is a constant vector. Then

0 =ddt

(x x) = x x + x x = x x

and so x is parallel to x.

Aside: Central ForcesThis second example is what happens in the case of a particle being acted on by a central force.According to Newton, forces cause acceleration, so the action of a central force (always pointingtowards some central point such as the center of the earth) may be modelled by the equation x = x,where is some scalar function and x is the position of the particle relative to the center. But then

ddt

(x x) = x x = x x = 0

and so x x = k is constant. Without knowing anything about the magnitude of the force, we knowthat the particle is constrained to live in the constant plane orthogonal to k.

We now use our two products to define the crucial notion of a moving frame.

Definition 1.13. A frame7 of E3 is an orthonormal basis {e1, e2, e3} oriented such that e3 = e1 e2.A frame along a curve x(t), or a moving frame, is a family of frames E(t) = (e1(t) e2(t) e3(t)), one foreach t.For a moving frame along a curve in E2, simply ignore e3(t).

When dealing with a moving frame along a regular curve, we will assume that the functions ei(t) arealso regular.

e1(t)e2(t)

e1(t)

e2(t)

e1(t)

e2(t)

A frame in E2 A frame in E3

7Strictly this is an orthonormal frame. These are the only frames we shall consider. The notation E(t) = (e1(t) e2(t) e3(t))is a useful abuse: think of it as a 1 3 matrix whose entries are vector-valued functions; if the vectors are explicit then E(t)becomes a 3 3 matrix.

15

Example Suppose that x(t) =(

cos tsin t

)parameterizes the

unit circle. We can define a moving frame along the circleas follows:

e1(t) =(

cos 2tsin 2t

)e2(t) =

( sin 2tcos 2t

)The frame rotates twice as rapidly as one travels roundthe circle! The frame can be expressed as a matrix-valuedfunction:

E(t) =(

cos 2t sin 2tsin 2t cos 2t

)

The Structure Equations for a Moving Frame

The main idea of moving frames is that there are certain natural choices of frame with respect towhich fundamental properties of the curve become clear (we shall construct such frames in the nextsection). We pay a price to gain this advantage: we give up having a fixed, unmoving, system ofreference and we now have to understand how a moving frame moves. As the following resultsshow, the movement of a moving frame is determined by three scalar functions.

Lemma 1.14. If (e1 e2 e3) is a moving frame in E3, then

ei ej = ei ej, and so ei ei = 0

Proof. Just differentiate each of the expressions ei ej = constant.

Theorem 1.15. If E = (e1 e2 e3) is a moving frame, then there exist unique functions w12, w13, w23

(e1 e

2 e

3)=(e1 e2 e3

) 0 w12 w13w12 0 w23w13 w23 0

Moreover, each function is simply wij = ei ej.

Proof. Multiply out and apply Lemma 1.14.

The equations in Theorem 1.15 are known as the structure equations of the moving frame. In E2 thereis only the single function w12.

16

Examples

1. Without bothering to worry about an explicit curve, consider the moving frame

e1(t) =

cos tsin t0

, e2(t) = sin tcos t

0

, e3(t) =00

1

This frame rotates about the z-axis. We compute

e1(t) =

sin tcos t0

= e2(t), e2(t) = cos t sin t

0

= e1(t), e3(t) = 0Hence w12 = e1 e2 = 1, w13 = 0 = w23 and

(e1 e

2 e

3)=(e1 e2 e3

)0 1 01 0 00 0 0

2. Here is a second moving frame

e1(t) =

cos2 tcos t sin tsin t

, e2(t) = sin t cos t

0

, e3(t) =sin t cos tsin2 t cos t

.We compute

e1(t) = cos te2(t) e3(t), e2(t) = cos te1(t) + sin te3(t), e3(t) = e1(t) sin te2(t).

from which w12 = cos t, w13 = 1 and w23 = sin t. The structure equations can be written

E = E

0 cos t 1 cos t 0 sin t1 sin t 0

3. For the moving frame

e1(t) =(

cos 2tsin 2t

)e2(t) =

( sin 2tcos 2t

)along the unit circle, we have

e1(t) =(2 sin 2t

cos 2t

)= 2e2(t) and e2(t) =

(2 cos 2t2 sin 2t

)= 2e1(t)

Therefore w12(t) = 2 and the structure equations can be written

E = (e1 e2) = (e1 e2)

(0 22 0

)17

1.4 Local theory of curves: the Frenet Frame

Part of the our purpose is to attempt to answer the following question: how independet can twocurves be? A more stringent version of the same question might ask how many arbitrary functionsare required in order to define a curve. We are essentially attempting to classify curves in the sensethat two curves are considered the same provided they differ only by a rigid motion (rotation ortranslation).A nave approach says that a regular curve in E3 requires three (essentially8) arbitrary scalar functions

x(t) =( x(t)

y(t)z(t)

)but this doesnt take into account equivalence under rigid motions. We can also

parameterize the same curve in many ways. In order to separate out the essential properties of acurve we do two things:

1. Choose a unit-speed parameterization. The only two vagaries that can now depend on theposition are the orientation (direction of travel) along the curve, and a choice fixed point (usuallywhen t = 0).

2. Choose a moving frame which is adapted to the curve in such a way that the structure equationsare very simple. We shall then see that the structure equations of the frame in fact define thecurve up to rigid motions.

First we rehash Definition 1.8.

Definition 1.16. Suppose that x : I E3 is a twice-differentiable, oriented, regular curve, parame-terized by arc-length. The vector field T(t) := x(t) is the unit tangent vector field along the curve. Thecurvature of x is the scalar function

(t) =x(t) = T(t)

Note that T(t) is a unit vector field. Similarly to Lemma 1.14, we note that

T T = 1 = T T + T T = 0 = T T = 0

whence T is orthogonal to its derivative. The curvature is therefore measuring how rapidly the unittangent vector changes direction. Curvature is therefore one measure of the deviation of a curve from astraight line. It doesnt tell the whole story however. Recall that a circle of radius 2 and the standardhelix both have the same curvature9 = 12 :

Circle x(t) =

2 cos t22 sin t20

= (t) = 12 cos t2 12 sin t2

0

= 12Helix x(t) =

cost2

sin t2

t2

= (t) =

12 cos

t2

12 sin t20

=

12

To classify curves more completely, we need to restrict our field of study a little.

8The components x(t), y(t), z(t) have to be differentiable with continuous derivatives.9Remember that both curves must be parameterized at unit-speed!

18

Definition 1.17. A regular, twice-differentiable curve is biregular if is defined everywhere and isnever zero.

We could equivalently say that a curve x : I E3 is biregular if and only if it is twice-differentiableand the vectors x(t) and x(t) are everywhere linearly independent.

Theorem 1.18. Let x : I E3 be biregular and unit speed. Then there exists a unique moving frame(T, N, B) along x such that x = T and

(T N B

)=(T N B

)0 0 0 0 0

where > 0 is the curvature and is a function called the torsion10 of x.

The structure equations can be written explicitly as

T = N, N = T + B, B = N

Proof. Setting T = x fixes T as the unit velocity vector.Since x is unit speed, we have T T (see Definition 1.16).Since x is biregular we have = |T| 6= 0, so we may define N = 1 T to be the unit vector pointing inthe same dierection as T.The final vector in the moving frame is fixed via B := T N is then fixed by orientation and theexistence of = N B follows from Theorem 1.15.

Definition 1.19. The frame (T, N, B) is the Frenet frame of x and the three vectors are, respec-tively, the unit tangent, principal normal and binormal vector fields of the curve.

The equations of Theorem 1.18 are known as the FrenetSerret equations or simply the structureequations for a unit-speed space-curve.

The plane spanned by T and N is called the osculating plane of x. The plane spanned by T, B isthe rectifying plane, and the plane spanned by N, B is the normal plane.

Example 1 For a horizontal helix x(t) =

t2

cos t2

sin t2

parameterized by arc-length, we have

T(t) = x(t) =12

1 sin t2cos t

2

But then

T(t) = 12

0cos t2sin t

2

= N(t) = 0cos t2

sin t2

and (t) = 1210Several authors take to be the negative of ours.

19

Finally,

B(t) = T(t)N(t) = 12

1sin t2 cos t

2

gives us the binormal vector field, whence the torsion is

(t) = N(t) B(t) = 12

0sin t2 cos t

2

12

1sin t2 cos t

2)

= 12The FrenetSerret equations for the helix are therefore

(T N B

)=(T N B

)0 12 012 0 120 12 0

Example 2 Let x(t) =

13 (1 + t)3/212 t13 (1 t)3/2

for t (1, 1). Then

x(t) =

12 (1 + t)1/212 12 (1 t)1/2

= v(t) = x(t) = 14 (1 + t) + 12 + 14 (1 t) = 1so that we have have a unit-speed parameterization. It follows that T = x. Now differentiate again:

T(t) =

14 (1 + t)1/2014 (1 t)1/2

= (t) = T(t) = 14

1

1 + t+

11 t =

2

4

1 t2

20

whence

N(t) =1

(t)T(t) =

4

1 t22

14 (1 + t)1/2014 (1 t)1/2

= 12

(1 t)1/20(1 + t)1/2

We finally have

B(t) = T(t)N(t) = 12

(1 + t)1/22(1 + t)1/2

= (t) = N(t) B(t) = 12

2

1 t2

Calculation of curvature and torsion in any parameterization

As weve discussed, there are relatively few curves for which an explicit unit-speed parameterizationcan be found. We want, therefore, to be able to find the Frenet frame for any biregular curve, re-gardless of its parameterization. There is nothing stopping us from doing this already, the difficultycomes mainly from a profusion of square-roots. . .

Example Consider the exponential spiral x(t) =

et cos tet sin tet

. We have

x(t) = et

cos t sin tsin t + cos t1

= v(t) = 3et= T(t) = 1

3

cos t sin tsin t + cos t1

= T = 13

sin t cos tcos t sin t0

21

= N(t) = 12

sin t cos tcos t sin t0

= B(t) = TN = 1

6

cos t + sin t sin t cos t2

This example looks a bit like a helix, but it in fact it spirals up the surface of the cone x2 + y2 = z2,rather than up the surface of a cylinder. It might be tempting to say that the curvature of this example

is

23 , since T

=

23 N, but this is not the case: since the curve is not unit speed, we need to think a

little harder.

Theorem 1.20. The speed, curvature and torsion of a biregular spacecurve x(t) with arbitrary parameteriza-tion are

v(t) =x(t) = x x, (t) = |x x|

v3, (t) =

(x x) xv62

The Frenet frame with respect to an arbitrary parameterization is

T(t) =1v

x, N(t) =vx vx

v3, B(t) =

x xv3

Moreover in this parameterization, the structure equations are

(T N B

)=(T N B

) 0 v 0v 0 v0 v 0

The proof is a (simple!) application of the chain-rule, the difficulty is primarily with notation.

Proof. First we clarify some notation. If you are confident using the chain rule, some of the followingis unnecessary.

The arc-length of x(t) measured from some fixed value of t is s(t). The inverse function is t(s).The speed is v(t) = dsdt .

The original curve parameterized by arc-length is y(s) = x(t(s)) (exists by Theorem 1.7).

Quantities (Frenet frame, curvature, etc.) with respect to the unit-speed curve will be denotedT(s), (s), etc. The corresponding expressions in terms of t may be found using, for instance,T(t) = T(s(t)) = T(s).

The last part is then straightforward:

ddt(T N B

)=

dsdt

dds(T N B

)(Chain Rule)

= v(T N B

)0 0 0 0 0

(Theorem 1.18)22

=(T N B

) 0 v 0v 0 v0 v 0

We now compute each of the terms explicitly:

T(t) = T(s) = y(s) =dtds

dxdt

=1v

x

= (s)N(s) = T(s) = y(s) = dtds

ddt

(1v

x)=

1v

(vx vx

v2

)()

= N(t) = N(s) = vx vxv3

Taking cross products with T we obtain

B(t) = B(s) = T(s) N(s) = x xv3

()

Since B(t) is a unit vector, we immediately have

(t) = (s) =|x x|

v3

Now differentiate () with respect to s (more chain rule!) to obtain

(s)N(s) + (s)N(s) =dtds

ddt

(vx vx

v3

)=

v2x (vv + 3v2)xv5

Finally, take the scalar product of both sides with B = T N using the expression in () to obtain

N B = = = x x xv6

Dividing by yields the result.

Examples

1. Our example x(t) =

et cos tet sin tet

from earlier has T = 23 N = vN = = 23 et.Similarly B = 1

3N = vN = = 13 et. One can also use the above expressions for

and to compute them directly from the derivatives.

2. The helix with radius r and vertical separation 2h between spirals has the form x(t) =

r cos tr sin tht

.We compute the Frenet frame:

x(t) =

r sin tr cos th

, x(t) =r cos tr sin t

0

, x(t) = r sin tr cos t

0

,23

whence

v =

r2 + h2, =r

r2 + h2, =

hr2 + h2

and

T(t) =1

r2 + h2

r sin tr cos th

, N(t) = cos t sin t

0

, B(t) = 1r2 + h2

h sin th cos tr

3. Consider the curve x(t) =

tt2t3

. This has

x(t) =

12t3t2

, x(t) = 02

6t

, x(t) =00

6

,from which

v(t) =

1 + 4t2 + 9t4, (t) = 2

1 + 9t2 + 9t4

(1 + 4t2 + 9t4)3, (t) =

31 + 9t2 + 9t4

Calculating the curvature and torsion of even simple curves can be extremely messy.

1.5 The Fundamental Theorem of Biregular Spacecurves

Weve already seen that straight lines have curvature zero. The converse is true, as is a similar state-ment connecting plane curves and torsion.

Theorem 1.21. 1. A twice-differentiable regular spacecurve has curvature identically zero if and only if itis a straight line.

2. A biregular spacecurve has torsion identically zero if and only if it is contained in a fixed plane (theunmoving osculating plane of the curve).

Proof. 1. We may suppose that x(t) is unit speed. Then (t) = |x(t)| = 0 x(t) = 0. Thusx is a straight line.

2. Suppose that x(t) is a unit-speed biregular curve which lives in a fixed plane. Then x, x areparallel to this plane, and so T, N are also. But then B is a continuous unit vector orthogonal tothe plane and thus constant. Hence N = B = 0, so that x has zero torsion.Conversely, if = 0, then B is constant. But then

(x B) = x B = T B = 0,

hence x B is constant. x thus lives in a fixed plane normal to B.

24

Theorem 1.21 says that curvature measures the de-viation of a curve from a straight line, while tor-sion measures how much a curve is trying to leapout of its osculating plane. For this reason, curva-ture is often referred to as bending and torsion astwisting of a curve.In the pictures, the Frenet frame and osculatingplane of a helix are drawn at a given time. Thesecond picture is a still of the first, rotated to beside-on to the osculating plane. The non-zerotorsion ( = 12 for this curve) may be observedin how the curve crosses the osculating plane: itlooks a little like how the cubic function y = x3

crosses the x-axis.

We will say a little more regarding the geometricmeaning of curvature later on, when we discussosculating circles.

The Equivalence Problem for Spacecurves

We wish to classify all biregular curves up to equivalence by rigid motions. The idea is that and tell you everything about the shape of a curve. In fact these functions are enough to classify biregularcurves up to a choice of initial position and direction. The phrase used is up to isometry: we need tosee what is meant by an isometry.

Definition 1.22. A 3 3 matrix A is orthogonal if AT A = I, where I is the identity matrix. The set oforthogonal matrices with real number entries is denoted O3(R).

Several facts are immediate:

Lemma 1.23. 1. A is invertible, with inverse AT.

2. The product of two orthogonal matrices is orthogonal.

3. A is orthogonal if and only if (Ax) (Ay) = x y for all vectors x, y E3.

4. If A = (e1 e2 e3) so that e1, e2, e3 are the columns of A, then A is orthogonal if and only if the vectorsform an orthonormal set.

5. An orthogonal matrix A has determinant det A = 1. If det A = 1 we call the matrix a rotation, andif det A = 1 we call it a reflection. The set of rotations is denoted11 SO3(R).

6. A moving frame is a function E : I SO3(R).

Definition 1.24. An isometry of E3 is a map x 7 Ax+b where A is a constant 3 3 orthogonal matrixand b E3 is a constant vector. An isometry is either direct (det A = +1) or indirect (det A = 1). IfA is the identity matrix, the isometry x 7 x + b is known as a translation.

11This is the set of special orthogonal matrices.

25

Isometries can be thought of as all possible combinations of translations, rotations and reflections.These are all the rigid motions of E3 that will preserve the distance between points and angles betweenvectors. Since both of these concepts are measured using the dot product, this is precisely what part3 of the Lemma is saying. We now consider how curve transforms under the action of an isometry.

Theorem 1.25. Under an isometry x = Ax+b, the curvature and torsion of a biregular spacecurve transformas follows:

Direct isometry: = , = .

Indirect isometry: = , = .

Proof. Suppose WLOG that x(t) is unit-speed. Then

x(t) = Ax(t) = v(t) =x(t) = x(t) = 1

since orthogonal matrices preserve the scalar product (Lemma 1.23, part 3). We can now compute:

x = Ax = T = AT, x = Ax = =x2 = Ax2 = x2 = .

Therefore curvature is invariant under any isometry.For torsion, observe that the second formula above implies N = AN. Since A is orthogonal, itpreserves angles, whence AB is orthogonal to T, N. It remains only to decide if AB is B. For this,consider the fact that the moving frame E = (T N B) is a special orthogonal matrix. We now have

AE = (AT AN AB) = (T N AB)

It follows that AE = E AB = B det A = 1. It follows that

B = (det A)AB =

{AB if the isometry is direct,AB if the isometry is indirect.

Now

= B N = (det A)(AB) (AN) = det AB N = (det A)

as required.

Example The simplest indirect isometry is S(x) = Ax =

1 0 00 1 00 0 1

x. If we apply this to theunit-speed helix x(t) =

cos(t/

2)sin(t/

2)

t/

2

, we obtain x(t) =cos(t/

2)

sin(t/

2)t/

2

which spirals down ratherthan up. Indeed

T = x =12

sin(t/

2)cos(t/

2)

1

= AT = T =12 cos(t/

2)

sin(t/

2)0

= 12 = 26

N =

cos(t/

2) sin(t/

2)

0

= N = AN B = T N = 12

sin(t/

2)cos(t/

2)

1

= AB = B N = 1

2

To visualize, here are pictures of the two curves with their Frenet frames at t = 0.

Observe that the first curve is twisting up (in the B direction) out of its osculating plane, while thesecond twists down (in the B direction).

Existence and Uniqueness of Solutions to Initial Value Problems

Our classification of spacecurves depends on the usual existence and unique result for ODEs. In-deed the relationship of curves and surfaces to the existence of solutions to differential equations is acornerstone of modern differential geometry research.

Theorem 1.26 (Picard, Lindelof, etc.). Let a and c be fixed elements of R and Rn respectively. Let A(x, t)be a continuous vector-valued function defined on some closed region

|x c| K and |t a| Tand suppose that the partial derivatives of A with respect to the co-ordinates of x are bounded.12 Suppose alsothat M = sup |A(x, t)| on the closed region. Then the initial value problem

dFdt

= A(F, t)

F(a) = c

has a unique solution F : I Rn on the interval defined by |t a| min(T, KM

).

12The requirement that the partial derivatives be bounded may be relaxed. The technical term is that the function A(x, t)be Lipshitz continuous which essentially says that |A(x, t)A(y, t)| B |x y| for some constant B.

27

The proof is beyond the difficulty of this course as it depends crucially on ideas from topology. Therough idea is that we produce a sequence of functions

F1(t) = c, F2(t) = c + t

aA(F1(t), t)dt, F3(t) = c +

ta

A(F2(t), t)dt, . . .

which are then shown to converge to the required solution.

Example Consider the initial value problem dFdt = 2tF(t) with initial condition F(0) = 1. We obtain

F1(t) = 1, F2(t) = 1 + t

02t dt = 1 + t2,

F3(t) = 1 + t

02t(1 + t2)dt = 1 + t2 +

12

t4, . . .

The Picard iteration is building up the correct solution as a power series

F(t) = et2=

n=0

t2n

n!= 1 + t2 +

12

t4 +16

t6

Corollary 1.27. If the ODE in Picards Theorem is linear that is if A(x, t) = A(t)x + b(t) for some matrixvalued-function A(t) and vector-valued b(t), then the unique solution is defined on |t a| T.Proof. Since the only dependence of A on x is through multiplication, it follows that A is continuousfor all x Rn. Essentially K can be taken to be arbitrarily large: K = MT is sufficient for the result.

Corollary 1.28. Let I = [t0 T, t0 + T] be an interval and W : I M3(R) a continuous matrix-valuedfunction such that each W(t) is skew-symmetric.13 Suppose that O is an orthogonal matrix. Then:

1. There exists a unique solution E : I O3(R) to the initial value problem{dEdt = EW

E(t0) = O

2. If detO = 1, then E : I SO3(R).Proof. 1. The initial value problem consists of a system of nine linear first order ODEs in the entries

of the 3 3 matrix E. By the above, we know there exists a unique solution E : I M3(R).Now differentiate:

ddt

(EET) = EET + EET = EWET + EWTET = 0

since W is skew-symmetric. Thus EET is constant. However E(t0)E(t0)T = I, whence thisconstant is the identity matrix. It follows that E(t) is always orthogonal.

2. Determinant is a continuous function of the entries of a matrix (it is a polynomial. . . ). Since Eis differentiable, it follows that det E : I R is continuous. However det E = 1 since E isorthogonal. Since E is defined on an interval, it must therefore be the constant 1. E is thereforespecial orthogonal.

13I.e. WT = W.

28

We have now essentially proved the main result!

Theorem 1.29 (Fundamental theorem of biregular curves in E3). Let I = [t0 T, t0 + T] be an interval,c E3 a constant vector and suppose that (T0 N0 B0) is a fixed oriented orthonormal basis of E3. Givenfunctions14 > 0 and from I R, there exists a unique unit-speed biregular spacecurve x : I E3with these curvature and torsion functions such that x(t0) = c and the Frenet frame at t = t0 is the basis(T0 N0 B0).

Proof. The structure equations E = E( 0 0

0 0 0

)put us in the situation of Corollary 1.28, whence

there exists a unique solution E = (T N B) : I SO3(R). To recover the curve x simply integratethe unit tangent vector:

x(t) = c + t

t0T(t)dt

It is clear that x his the unique curve with the required initial conditions, curvature and torsion.

An alternative way of stating the same theorem is to say that a biregular curve is determined up todirect isometry by its curvature and torsion. Indeed:

Corollary 1.30. Given two biregular spacecurves with the same curvature and torsion, there exists a directisometry between them.

Proof. Suppose that x1 : I E3 and x2 : I E3 have the same curvature and torsion functions.Choose some t0 I and suppouse that their Frenet frames are E1, E2 respectively. The isometry werequire must satisfy the conditions at t0. Indeed we need an direct isometry S : x 7 Ax + b such that

S(x1(t0)

)= x2(t0)

AE1(t0) = E2(t0)

This exists and is unique: we are forced to choose A = E2(t0)E1(t0)1 and b = x2(t0) Ax1(t0).Note that A necessarily has determinant 1 since both E1 and E2 do so.Now apply Theorem 1.25) x3 := S(x1) is a spacecurve with the same initial conditions (at t0), curva-ture and torsion as x2. The Fundamental Theorem says that x2 = x1.

Compare what weve done with elementary integration. The distance-speed problem from (one-dimensional) calculus states that if you know your starting position and your speed as a function oftime, you may compute how far you have travelled. Position x(t) is thus computed from the pair(x(t0), v(t)).The Fundamental Theorem provides a three-dimensional analogue: if you know your starting loca-tion, your initial orientation (direction and direction of curvature), your speed (for us its always unitspeed, but theres nothing stopping us working with arbitrary speed) and how much you bend andtwist, you can determine the entire motion. The required data is the sextuple

(x(t0), T(t0), N(t0), v(t), (t), (t))

As a final application of the Fundamental Theorem, consider the following.

14 must be differentiable with continuous derivatives and be continuous.

29

Corollary 1.31. Every biregular curve with , both constant is a circular helix (circle if 0).Proof. We have already seen that the general helix (example 2, page 23) has this property. Nowsuppose that , are constant. The unit-speed helix

x(t) =

cost

2+2

sin t2+2

t2+2

has these curvature and torsion. By Corollary 1.30, this is only such curve up to direct isometry.

1.6 Plane curves

We can follow the same course with curves in E2 as we did in E3. The situation is a lot simplerin that we can make a common choice of N simply by rotating T counter-clockwise, rather thandifferentiating. In particular, we dont need to be able to differentiate curve twice in order to find theFrenet frame.

Definition 1.32. Let x : I E2 : t 7(

x(t)y(t)

)be a regular curve with speed v(t). The unit tangent

vector is T(t) = 1v(t)x(t). Define the unit normal vector by

N(t) = JT(t) where J =(

0 11 0

)Observe that N is simply T rotated counter-clockwise by 2 radians: this is the same as imaginingthat the triple (T N k) is an oriented orthonormal basis of E3. We now have the following analogueof Theorem 1.18.

Theorem 1.33. Given a regular curve x : I E2 of speed v(t), there exists a function : I R (the(signed) curvature of x) satisfying,

(T N

)=(T N

) ( 0 vv 0

)These are the structure equations for x.

Recall Definition 1.8. In contrast to the case of curves in E3, the curvature of a planar curve may benegative.{

> 0 x bends towards N < 0 x bends away from N

Observe the pictures. If these were curves in E3, thefirst would have B pointing out of the page. For thesecond picture we would have to reverse the directionof N and have B pointing into the page.Also observe that orienting a planar curve in the oppo-site direction changes the sign of the curvature.

T(t)N(t)

> 0

T(t)

N(t) < 0

30

Theorem 1.34. If v(t) is the speed of a twice-differentiable regular curve x : I E2, then its signed curvatureis given by

=x Jx

v3

This reduces to = |x(t)| if x is unit-speed, so we recover the same notion as in Definition 1.8.

Proof. From the structure equations,

v = T N = vx vx

v2 1

vJx =

1v2

x Jx

since Jx is perpendicular to x.

Examples

1. For a circle of radius a parameterized counter-clockwise, the curvature is = 1a . Parameterizedclockwise, the curvature is 1a .

2. Any function y = f (x) may be parameterized by x(t) =(

tf (t)

). Its curvature is then

(t) =f (t)

(1 + f (t)2)3/2

as you might have seen in an elementary calculus course. In particular, the catenary (page 11)

x(t) =(

tcosh t

)has

(t) =cosh t

(1 + sinh2 t)3/2=

11 + sinh2 t

which is the same formula as before via the change of parameterization s = sinh t.

What does tell you? Suppose that x is unit speed and that (s) describes the direction of T: thus

T(s) =(

cos (s)sin (s)

), T =

( sin cos

), N = JT =

( sin cos

)= (s) = (s)

Curvature is therefore the rate of change (counter-clockwise) of with respect to arc-length.In E2 the curvature is enough to completely determine a regular curve up to rigid motions, as thefollowing analogue to Theorem 1.29 shows.

Theorem 1.35. Given a countinuous function : I R, there exists a unique unit-speed curve x(t) in E2

with curvature (t) such that x(0) is the unit vector(

10

)and x(0) = 0.

31

The proof proceeds in the same way as Theorem 1.29: the existence theorem from ODEs gives the

vectors T(t), N(t) satisfying the structure equations and T(0) =(

10

), N(0) =

(01

). Integrating

T = x yields the curve.

Aside: Curves in Arbitrary DimensionThere is a general theorem (and Frenet frame) for curves in En: for a unit-speed curve x in En whosefirst n derivatives at each point form a basis of En, we may Gram-Schmidt orthogonalize the basis{x, x, . . . , x(n)} to obtain a moving frame {e1, . . . , en} and functions 1, . . . , n1 (the generalizedcurvatures of x) satisfying

(e1, . . . , en) = (e1, . . . , en)

0 1 0 0 01 0 2 0 00 2 0 0 0...

. . ....

0 0 0 0 n10 0 0 n1 0

Conversely, the n 1 generalized curvatures then determine the curve.

1.7 Radii of curvature

We have already seen that curvature and torsion measure how far a curve is from being a straightline and from being planar respectively. There is a little more geometry contained in the notion ofcurvature. We have seen that the only planar curve with constant curvature k is the circle of radius1k . We could have started with this as our definition of curvature, and said that a curve has curvaturek at a certain point, if the circle which best approximates the curve at that point has radius 1k . Ofcourse, we have to define what is meant by best approximation.

Definition 1.36. Two curves x, y have nth order contact at an intersection point x(t0) = y(s0), if thefirst n derivatives agree there: i.e. x(i)(t0) = y(i)(s0) for all 1 i n.

Let x(t) be a unit-speed curve in E2 and consider a unit-speed circle c(s) of (signed) radius r satisfying the fol-lowing:

c(0) = x(t0),

c(0) = x(t0),

r > 0 the circle lies on the same side of thecurve as N.

It should be clear that there is precisely one circle c ofeach singed radius r R.

T(t0)N(t0)

r > 0r < 0

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Now think out how to parameterize such a circle: a little thought shows that

c(s) = x(t0) + rN(t0) center

+ r sin(s/r)T(t0) r cos(s/r)N(t0) rotation

is the unit-speed parameterization for which c(0) = x(t0). Moreover

c(s) = cos(s/r)T(t0) + sin(s/r)N(t0) = c(0) = T(t0)

The circles all have first-order contact with the curve at x(t0). Moreover,

c(s) = 1r

sin(s/r)T(t0) +1r

cos(s/r)N(t0) = c(0) =1r

N(t0)

The circle has second-order contact with the curve at t0 if and only if

c(0) = x(t0) 1r= (t0)

Definition 1.37. The radius r = 1(t0)

is the radius of curvature of the curve x at t0. The circle parame-terized by

c(s) = x(t0) +1

(t0)(sin(s/r)T(t0) + (1 cos(s/r))N(t0)

)is called the osculating circle at t0. Its center is called the center of curvature.

There is nothing stopping us from calculating the radius of curvature and the osculating circles foran arbitrary speed curve: all we need is the curvature and the Frenet frame at the relevant point.

Example We calculate the osculating circles for the parabola y = x2 parameterized in the obvious

way x(t) =(

tt2

). The relevant ingredients are

x(t) =(

12t

)= v(t) =

1 + 4t2 = T(t) = 1

1 + 4t2

(12t

)x(t) =

(02

), (t) =

2(1 + 4t2)3/2

, N(t) =1

1 + 4t2

(2t1

)The center of the osculating circle at x(t) has position vector

x(t) +1

(t)N(t) =

(tt2

)+

1 + 4t2

2

(2t1

)=

( 4t312 + 3t

2

)

Its radius is 1 =(1+4t2)3/2

2 . An animation of several of the osculating circles is given below.

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The idea of second-order contact can be used to obtain local approximations to a curve: indeed recallthat the nth Taylor polynomial of a function y at t0 is the unique degree n polynomial which hasnth order contact with y at t0.

Theorem 1.38. Let x be a regular curve in E2 which passes horizontally through the origin at t = 0. Then(locally) the graph of the curve is given by,

y =(0)

2x2 + higher order terms.

Proof. Such a curve may be parameterized by x(t) =(

ty(t)

), where y(0) = 0 = y(0). The curvature

at t = 0 is simply y(0). The rest is Taylor series.

Obvious modifications may be made to describe a curve near other points.

Examples

1. The parabola y = x2 passes horizontally through the origin, where its curvature is = 2. Thetheorem tells us what we expect to see, that

y =22

x2 +

The higher order terms are thus all zero.

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2. The catenary y = cosh x has initial curvature = 1 at the point (0, 1), where its gradient is zero.We therefore see that

y = 1 +12

x2 +

This is simply the start of the Taylor series of cosh x.

Evolutes

Definition 1.39. The evolute e(t) of a planar curve x(t) is the curve defined by the centers of curvatureof x:

e(t) = x(t) +1

(t)N(t)

Example We have already calculated the evolute of the parabola x(t) =(

tt2

)when we found its

osculating circles:

e(t) = x(t) +1

(t)N(t) =

( 4t312 + 3t

2

)which may be written as y = 12 + 3

( x4

)2/3. The animation has been updated to include the evolute.

The gray lines in the animation are the normal lines to the parabola. If you want a little challengewith calculus definitions, try to prove the following result:

Proposition 1.40. The limit of the intersections of nearby normal lines of a curve x describe the evolute e.

This is often described by saying that the evolute is a focal curve for the normal lines.

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Involutes

A related notion to the evolute is the involute of a curve. This is obtained by rolling a line along acurve and seeing what curve the end of the line traces out.

Definition 1.41. Suppose that x(t) is unit speed. The involute of x(t) is the curve i(t) := x(t) tx(t).Note that the involute depends on where the zero of the parameterization is located: at t = 0 theinvolute intersects the original curve.

Example The involute of the catenary x(t) =(

sinh1 t1 + t2

)is a curve called the tractrix:

i(t) =(

sinh1 t t(1 + t2)1/2(1 + t2)1/2

)This is the curve obtained when a mass (the green dot) at (0, 1) is towed by attaching a (orange) ropeof length 1 to a vehicle (orange dot/tractor!) moving along the x-axis.

Observe from the picture that the tangent lines to the catenary are being wrapped round the curveand then unwrapped as we move to the right. It is as if a string has been wrapped tight around thecurve, then cut at t = 0, before being unwrapped while still under tension.

Evolutes and involutes are inverse concepts to each other as the following theorem shows.

36

Theorem 1.42. The evolute of any involute of a curve is the original curve, minus those points where the t = 0or = 0.

Proof. Let x(t) be unit speed, then

i(t) = x(t) tx(t) = x(t) tT(t)is the involute, where T is the unit tangent vector of x. Let N, denote the unit normal and curvatureof x, and v, T, N, the speed, Frenet frame and curvature of i. Then

i = T T tT = tN = v(t) = |t|= T = sgn(t)N= N = sgn(t)JN = sgn(t)T

By the structure equations, we have

vN = T = sgn(t)N = sgn(t)T = Nand so = v =

|t| . The evolute of the involute of x is therefore

i +|t|

sgn(t)T = x tT + t

T = x

Note that the proof fails precisely when v(t) = t(t) = 0.

1.8 Further thoughts on curves

Several of the ideas described earlier may be generalized, and there are many further topics in clas-sical curve theory that we ignore.

Osculating circles may be defined for a curve in E3 (or higher dimension). At each point of a curvex(t), the osculating circle is the circle in the osculating plane with center x(t) + 1

(t)N(t) and

radius 1(t) : the radius of curvature. It can be shown that this is the unique circle in E

3 havingsecond order contact with x.

Involutes of a curve in E3 may also be defined similarly: if x(t) is an arc-length parameterization,then i(t) = x(t) tx(t) is an involute of x(t).

Evolutes are somewhat more difficult: we want an evolute to be a curve e = x + N + B forsome functions , such that x is an involute of e. It can be seen that, if t is an arc-lengthparameterization,

e(t) = x(t) +1

(t)

(N(t) + cot

((t)dt

)B(t)

)where

(t)dt is an anti-derivative of the torsion. There are thus an infinite family of evolutes.

If we want to restrict to an evolute that lying in the osculating plane of x at each point y insistingthat

t, cot(

(t)dt)= 0

(t)dt =

2

This is clearly if the anti-derivative is the constant c = 2 and (t) 0: the only curves with anevolute always in the osculating plane are curves which are already planar!

37

Global properties of curves can also be considered, in contrast to the local theory weve developed(differentiation is a local concept since only the behavior of a curve near a point matters).In particular, questions can be asked about simple closed curves (closed curves without self-intersection). For example:

The Isoperimetric Inequality states that the area in the plane bounded by a simple closedcurve x of length l satisfies the inequality

4A l2 with equality x is a circle.

The Four-Vertex theorem states that every simple closed curve in the plane with differentiablecurvature has at least four vertices (points where = 0). For an easy example of this last,consider an ellipse

x(t) =(

a cos tb sin t

)Its curvature is

(t) =ab

a2 sin2 t + b2 cos2 t

Therefore

(t) =ab(b2 a2) cos t sin t[a2 sin2 t + b2 cos2 t]3/2

= 0 t = 0, 2

, ,32

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CurvesParameterized Curves in Euclidean spaceRegular curves and the arc-length parameterizationProducts of Vectors and Moving FramesLocal theory of curves: the Frenet FrameThe Fundamental Theorem of Biregular SpacecurvesPlane curvesRadii of curvatureFurther thoughts on curves

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