Curves and Surfaces in the Plane and Three-Dimensional Space

Home , Arc length, Center of curvature, Differentiable curve, Involute, Radius of curvature

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-dimensional space. The word differential indicates that we will be using calculus! Indeed the basic ingredients of differential 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 appears in 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 circle should 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 idea is as follows: imagining a curve as a roller-coaster along which you travel at a constant speed, the curvature 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 how much a surface appears to be dome- or saddle-shaped. Understanding this will occupy much of the second 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 is important to distinguish this from Rn.

Rn Is the set of n-tuples of real numbers. These can be written as row- or column-vectors, your choice. Thus Rn = (x ,..., x ) : x R 1 n i ∈ 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 no dot product! En is the set Rn together with the usual notion of dot product. We usually write these as column vectors, and denote elements bold face1. For instance,   x1 3 x = x2 E ∈ x3 This helps us distinguish vectors in Euclidean space from co-ordinates. The dot product allows a b us to measure length and angle. To refresh; if a = 1 and b = 1 are two vectors in E2, a2 b2 then: • The dot product of a and b is a b = a b + a b . · 1 1 2 2 • The length of a is a = √a a. | | · a b • The angle θ between a and b satisﬁes cos θ = · . a b | | | | In this class, Rn will always have n = 1 (for curves) or n = 2 (for surfaces). Similarly En will always have n = 2 or 3. If a deﬁnition or result is true in both E2 and E3 (and more generally in En), we will just write En.

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

2 Examples

cos t 1. Let x(t) = where t R. This deﬁnes a function x : R E2. The components of x are sin t ∈ → 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) cos t x(t) = = y(t) sin t

Just make sure you don’t get x and x confused!

 u  2. Let x(u, v) =  v  where u, v R (or (u, v) R2 if you prefer). This deﬁnes a function ∈ ∈ u2 + v2 x : R2 E3. It’s components are → x(u, v) = u, y(u, v) = v, z(u, v) = u2 + v2

It isn’t important at the moment, but these examples describe, respectively, a circle in the plane and a paraboloid in three-dimensions.

1 Curves

1.1 Parameterized Curves in Euclidean space

A in En is a Deﬁnition 1.1. parameterized differentiable curve smooth x0(t0) function x : I En of a subinterval I = (a, b) of the real line R → into En. It’s velocity, or tangent vector at t I is the vector x (t ). In terms 0 ∈ 0 0 of the components, we can write

  x(t0) x10 (t0)  .  x0(t0) =  .  xn0 (t0) O Smooth means differentiable with continuous derivatives; that the quotient

x(t + h) x(t) x0(t) = lim − h 0 h → exists for every t I and that the function x (t) is continuous. The fact that we are differentiating a ∈ 0 vector-valued function is of no concern: we are not dividing by a vector in the above quotient!

3 Examples B b a Straight line Fix constant vectors a, b En. The parameterized A − ∈ curve deﬁned by x(1) = b x(t) = a + t(b a) − x(0) = a describes the line through the point A and B with position vec- O tors, respectively, a and b.

cos t Helix Let x(t) = sin t. The function x : R E3 describes → t a helix. To help visualize the helix, you can try several tricks. Imaging sitting on top of the z-axis and looking down: you’d only cos t see the horizontal projection of the curve t which de- 7→ sin t scribes the unit circle counter-clockwise. The projection onto the z-axis is simply t t, so the curve moves upwards at a constant 7→ speed. One could also project onto the xz- or yz-planes. All these images are plotted below

y z z 1 3π 3π

2π 2π

π π 11 − x 11 1 1 1 − π x − π y − − − Tangent Line Given a parameterized differentiable curve x : I n → E , we can easily deﬁne its tangent line at any ﬁxed t0. This is simply the straight line through the point with position vector x(t0), pointing in the direction of the tangent vector x0(t0). It is itself a parameterized curve, y : R En: →

y(s) = x(t0) + sx0(t0)

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

 1   √3  2 2 y(s) = √3 + s −1  2   2  π 3 1

4 Corners and Cusps

We will almost always think about parameterized curves which y are differentiable everywhere. It is, however, easy to construct 1 curves which are not. Consider the curve

t x(t) = , t ( 1, 1) 1 t ∈ − − | | 0 101 This has a corner at x(0) = and is not differentiable there. − 1 x Indeed, this is not a parameterized curve in the language of Deﬁ- nition 1.1.

A similar difﬁculty appears with a cusp: for example y 1 t x(t) = p , t ( 1, 1) t ∈ − | | is not differentiable at t = 0.

101 − x 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- y rameterization takes care of everything! The curve in the picture 1 has parameterization

sin 2t x(t) = 3 , t [0, 6π) cos t ∈ 1 1 The curve is differentiable everywhere, even though it intersects − itself. For instance, the curve passes through the origin at both x 3π 9π t1 = 2 and t2 = 2 , with corresponding tangent vectors

2 2 1 3π 9π − x0( ) = − 3 , x0( ) = − 3 2 1 2 1 − The important thing to note about self-intersections is that we always talk about the tangent vector to a parameterized curve at a co-ordinate t = t0. In this example, it is illegal to talk about the tangent vector to the curve at the origin. The parameterization takes care of the fact that there are two distinct tangent vectors at the origin.

5 Speed Deﬁnition 1.2. The speed of a curve x(t) is the norm/length of its velocity vector q 2 2 v(t) := x0(t) = x (t) + + x (t) 10 ··· n0 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 cos t 1. The helix parameterized by x(t) = sin t has speed t q v(t) = ( sin t)2 + (cos t)2 + 12 = √2 − t3 3t2 2. The curve x(t) = has tangent vector x (t) = and speed v(t) = 3t2√1 + 4t6. t6 0 6t5 t 3. The curve x(t) = drawn above, has tangent vector 1 t − | |  !  1  if t < 0,  1 x0(t) = !  1  if t > 0,  1 − and is undeﬁned at t = 0. Its speed is therefore v(t) = √2 for t = 0 and undeﬁned for t = 0. 6 sin 2t 4. The curve x(t) = 3 has tangent vector cos t

2 cos 2t x (t) = 3 3 0 sin t − q 4 2 2t 2 and speed v(t) = 9 cos 3 + sin 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 want to do anything with the tangent vector to a curve, we had better make sure that it is non-zero. We therefore want to isolate those curves which have non-zero tangent vectors as our main objects of study.

Deﬁnition 1.3. A curve x : I En is regular at t if x (t ) = 0. A curve is regular if it is regular for all → 0 0 0 6 t 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.

t3 2. x : t is non-regular at t = 0, since x (0) = 0. 7→ t6 0

t 3. x : t is non-regular at t = 0, since it is not differentiable there. 7→ 1 t − | | sin 2t 4. x : t 3 is regular, since x (t) is never zero. 7→ cos t 0

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

s y(s) = s2

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

3 3 s = t ! x(t) = y(t )

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

1 y (s) = 0 2s whence y is a regular curve with non-zero speed p 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 the curve’s parameterization.

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

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

dy dx Proof. By the chain rule, ds = α0(s) dt , whence the velocity of the reparameterization is

dy = α0(s) x0(t) ds

Since α0(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 α(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.

cos t Example Let x(t) = where t [0, 2π). This describes the unit circle. A simple reparame- sin t ∈ terization could be obtained using t = α(s) = 2s. This yields the curve

cos 2s y(s) = x(2s) = sin 2s

The domain is now s [0, π), and the new curve travels at twice the speed of the original: ∈

y0(s) = 2x0(2s) = 2 x0(t)

Notice that both parameterizations are regular.

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

y0(s) = α0(s)x0(α(s))

If y is to be regular at s = 0, we should be able to compute y0(0). However this is impossible: x0(α(0)) = x0(0) is undeﬁned, whence we would have to attempt to deﬁne

y0(0) = lim α0(s)x0(α(s)) = α0(0) lim x0(α(s)) s 0 s 0 → → 3 Since lim x0(α(s)) is also undeﬁned, we could only have α0(0) = 0, resulting in y0(0) = 0, whence s 0 the curve→ remains non-regular at t = 0. The difference between Examples 2 and 3 is simple: you can’t regularize corners, but you can regularize smooth curves that have been parameterized so that the curve sometimes comes to a halt.

3 1 1 limt 0+ x0(t) = = = limt 0 x0(t). → 1 6 1 → − −

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

Z t Z t s(t) = x0(T) dT = v(T) dT t0 t0 A curve is unit-speed if x (t) = 1 for all t I. | 0 | ∈ The arc-length is signed because, if t < t0 we have a negative length: we are measuring length against the 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 regular curve x has a unit-speed parameterization. R t Proof. Suppose first that x(t) is unit-speed. Then s(t) = 1 dt = t t0, and so t is the arc-length of t0 − x from t = 0. Conversely suppose that x(t) is a regular curve. Then x(t) has a well-defined arc-length s(t) from 4 ds any fixed value t = t0. It follows that s0(t) = dt = v(t). Since x is regular, it follows that v(t) > 0, whence s0(t) > 0 and so s(t) is an increasing function. Such a function is necessarily invertible. Let the inverse be t(s) and define y(s) = x(t(s)). But then, by the chain rule,

d dt x0(t) x0(t) v(t) y0(s) = x(t(s)) = x0(t) = = y0(s) = | | = = 1 ds ds s (t) ⇒ s (t) v(t) 0 | 0 | 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 parameterization, then so is x(t t ) for any t . Since these parameterizations involving travelling along the − 0 0 curve in the same direction, we say that they preserve orientation. Another collection of arc-length parameterizations is given by x(t t) for any t : these have opposite orientation since they traverse 0 − 0 the curve in reverse.

Computing an arc-length parameterization can be hard work: R t • It may not be possible to explicitly evaluate the integral s(t) = x0(T) dT. t0 | | • 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 likely make things easier.

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

R t 4Fundamental Theorem of Calculus: d v(T) dT = v(t) ... dt t0

9 Examples

cos t 1. The helix x(t) = sin t has constant speed √2. The arc-length from t0 = 0 is therefore t s(t) = √2t, with inverse function t(s) = s . It follows that √2

cos(s/√2) y(s) = sin(s/√2) s/√2

is a unit speed parameterization of the helix.

t 2. Let x(t) = (we’ve modiﬁed the example to make it regular). The arc-length from t = 0 is t2 0

Z t p s(t) = 1 + 4T2 dT 0

This can be computed with some judicious use of substitutions,5 though it isn’t fun!

1 p 1 p s(t) = t 1 + 4t2 + ln 2t + 1 + 4t2 2 4 Now you need an inverse! This problem is essentially impossible to do explicitly.

t 3. At least where it is regular, x(t) = has speed v(t) = √2 (t = 0). Thus x(s) = 1 t 6 ! − | | s/√2 is a unit speed parameterization, away from s = 0. 1 s/√2 − sin 2t 4. Explicitly ﬁnding a unit speed parameterization of x(t) = 3 is also far too difﬁcult. It cos t would involve computing the integral r Z t 4 2T s(t) = cos2 + sin2 T dT 0 9 3

As you can see, it is generally pointless to hunt for an explicit arc-length parameterization. What then 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 abstract curve comes with such a parameterization. This makes certain calculations nad theorems much easier. In practice, however, it makes work with explicit examples harder: we are hiding behind an incomputable integral!

5Try tan θ = 2T ...

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

t x(t) = cosh t

The curve has speed y q 2 v(t) = x0 = 1 + sinh t = cosh t

The arc-length measured from t0 = 0 is 1

Z t s(t) = cosh T dT = sinh t 0

1 Hence y(s) = x(t(s)) has unit speed when t(s) = sinh− s: 101 1 − sinh− s x y(s) = √1 + s2

It is easily checked that y has unit speed:

1 ! 2 √1+s2 2 1 s y0(s) = = y0(s) = + = 1 s ⇒ 1 + s2 1 + s2 √1+s2

Curvature Now that we know about unit speed curves, we can properly deﬁne our principal notion of bendiness.

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

κ(s) = x00(s)

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

Examples

1. A straight line has curvature zero.

x x x x 6 e +e− e e− d The basic hyperbolic functions are cosh x = 2 and sinh x = −2 . They satisfy dx cosh x = sinh x and d sinh x = cosh x. If you’ve never seen hyperbolic functions before, you can still do the entire example using expnentials dx 1 √ 2 and logarithms. Indeed a little calculus should convince you that sinh− s = ln s + 1 + s .

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

s cos r x(s) = r s sin r and so s 1 cos r 1 x00(s) = s = κ(s) = − r sin r ⇒ 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 larger curvature. 3. The standard helix with unit-speed parameterization

cos(s/√2) x(s) = sin(s/√2) s/√2 has cos(s/√2) 1 1 x00(s) = sin(s/√2) = κ(s) = −2 ⇒ 2 0

4. The catenary above has

1 s 1 y00(s) = − = κ(s) = (1 + s2)3/2 1 ⇒ 1 + s2 Compare with the picture: y(0) is the point of maximum curvature κ = 1 at the bottom of the curve. 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 of moving 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. Here we review several useful properties, and put them to work to create families of axes fitted to a given curve. Definition 1.9. The scalar product of two vectors x, y En is defined by ∈ n x y = ∑ xiyi · i=1 The norm or length of a vector x is x := √x x | | · 1 x y The angle θ [0, π] between two non-zero vectors x, y is θ = cos− · ∈ x y | | | | Vectors x, y are orthogonal or perpendicular if x y = 0 (if x, y = 0 this is equivalent to θ = π ) · 6 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 vectors x, y E3 is deﬁned by ∈   x2 y2

 x3 y3    e1 e2 e3  x1 y1  x y =   = x1 x2 x3 × − x3 y3    y1 y2 y3  x y  1 1 x2 y2

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

x1 x2 x3

(x y) z = det(x, y, z) = y y y × · 1 2 3 z1 z2 z3

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

1  1  8 − − Example 2  2  =  2 × − 3 1 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. × x z x w 5. (x y) (z w) = · · . It follows that × · × y z y w · · x y = x y sin θ | × | | | | | | | 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 if and only if z and x y lie on the same side of the plane spanned × by x, y.

13 1  1 − Example The area of the parallelogram spanned by 2 and  2  is √82 + 22 + 42 = 2√21. 3 1 − 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.

x1 x2 x3

1. (x y) x = y y y = 0, since two rows of the matrix are the same. (x y) y = 0 is × · 1 2 3 × · x1 x2 x3 similar.

2. If x and y are parallel, then one column in each of the 2 2 determinants in the deﬁnition is a × scalar 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. Thus × we assume that both x, y = 0 and that x y = 0. We also assume, WLOG, that x = 0 (if not, 6 × 1 6 reindex what follows). By deﬁnition x1 y1 y1 y1 x1 = 0 = x1y2 = x2y1 = = x2 y2 ⇒ ⇒ y2 x1 x2 Similarly x1 y1 y1 y1 x1 = 0 = x1y3 = x3y1 = = x3 y3 ⇒ ⇒ y3 x1 x3

In particular, y = y1 x, whence x and y are parallel. x1 5. Because of the bilinearity of , and the 2 2 determinant, we need only check the ﬁrst formula · × × when 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 2 x y x y 2 = | | · = x 2 y 2 (x y)2 = x 2 y 2 (1 cos2 θ) = ( x y sin θ )2 | × | x y y 2 | | | | − · | | | | − | | | | | | · | | where θ is the angle between x, y. Since the area of the triangle spanned by x, y making angle θ is 1 x y sin θ, and this comprises half the parallelogram, we are done. 2 | | | |

Calculus and moving frames It is important to be able to differentiate products of vectors. Thankfully both satisfy the usual product rule and can be proved similarly.

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

d (x(t) y(t)) = x0(t) y(t) + x(t) y0(t) dt · · · d (x(t) y(t)) = x0(t) y(t) + x(t) y0(t) dt × × ×

14 Examples

1. Suppose that x(t), y(t) are curves such that x(t) is orthogonal to both y(t) and y0(t) for every t. Then, d x0(t) y(t) = (x(t) y(t)) x(t) y0(t) = 0 · dt · − ·

so that x0 is orthogonal to y also. 2. Suppose that x(t) x (t) = k is a constant vector. Then × 0 d 0 = (x x0) = x0 x0 + x x00 = x x00 dt × × × ×

and so x00 is parallel to x.

Aside: Central Forces This 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 pointing towards some central point such as the center of the earth) may be modelled by the equation x00 = λx, where λ is some scalar function and x is the position of the particle relative to the center. But then d (x x0) = x x00 = λx x = 0 dt × × × and so x x = k is constant. Without knowing anything about the magnitude of the force, we know × 0 that 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 e , e , e oriented such that e = e e . { 1 2 3} 3 1 × 2 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 for each t. 2 For a moving frame along a curve in E , simply ignore e3(t).

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

e2(t)

e1(t)

e1(t) e2(t) e2(t)

e1(t) A frame in E2 A frame in E3

7 Strictly 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 cos t Example Suppose that x(t) = parameterizes the sin t unit circle. We can deﬁne a moving frame along the circle as follows:

cos 2t sin 2t e1(t) = e2(t) = − sin 2t cos 2t

The frame rotates twice as rapidly as one travels round the circle! The frame can be expressed as a matrix-valued function:

cos 2t sin 2t E(t) = − sin 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 to which fundamental properties of the curve become clear (we shall construct such frames in the next section). We pay a price to gain this advantage: we give up having a ﬁxed, unmoving, system of reference and we now have to understand how a moving frame moves. As the following results show, the movement of a moving frame is determined by three scalar functions.

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

e0 e = e e0, and so e0 e = 0 i · j − i · j i · i Proof. Just differentiate each of the expressions e e = constant. i · j

Theorem 1.15. If E = (e1 e2 e3) is a moving frame, then there exist unique functions w12, w13, w23   0 w12 w13 e0 e0 e0 = e1 e2 e3  w12 0 w23 1 2 3 − w w 0 − 13 − 23 Moreover, each function is simply w = e e . ij i · 0j 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 there is only the single function w12.

16 Examples

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

cos t  sin t 0 − e1(t) = sin t , e2(t) =  cos t  , e3(t) = 0 0 0 1

This frame rotates about the z-axis. We compute

 sin t  cos t − − e0 (t) =  cos t  = e2(t), e0 (t) =  sin t = e1(t), e0 (t) = 0 1 2 − − 3 0 0

Hence w = e e = 1, w = 0 = w and 12 1 · 20 − 13 23 0 1 0 − e10 e20 e30 = e1 e2 e3 1 0 0 0 0 0

2. Here is a second moving frame

 cos2 t   sin t  sin t cos t 2 e1(t) = cos t sin t , e2(t) =  cos t , e3(t) =  sin t  . − sin t 0 cos t − We compute

e0 (t) = cos te (t) e (t), e0 (t) = cos te (t) + sin te (t), e0 (t) = e (t) sin te (t). 1 − 2 − 3 2 1 3 3 1 − 2 from which w = cos t, w = 1 and w = sin t. The structure equations can be written 12 13 23 −  0 cos t 1  E0 = E  cos t 0 sin t − − 1 sin t 0 − 3. For the moving frame

cos 2t sin 2t e1(t) = e2(t) = − sin 2t cos 2t

along the unit circle, we have

2 sin 2t 2 cos 2t e0 (t) = − = 2e2(t) and e0 (t) = − = 2e1(t) 1 cos 2t 2 2 sin 2t − − Therefore w (t) = 2 and the structure equations can be written 12 − 0 2 E0 = (e0 e0 ) = (e1 e2) − 1 2 2 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 two curves be? A more stringent version of the same question might ask how many arbitrary functions are required in order to deﬁne a curve. We are essentially attempting to classify curves in the sense that two curves are considered the same provided they differ only by a rigid motion (rotation or translation). A na¨ıve approach says that a regular curve in E3 requires three (essentially8) arbitrary scalar functions x(t) x(t) = y(t) but this doesn’t take into account equivalence under rigid motions. We can also z(t) parameterize the same curve in many ways. In order to separate out the essential properties of a curve we do two things:

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

2. Choose a moving frame which is adapted to the curve in such a way that the structure equations are very simple. We shall then see that the structure equations of the frame in fact deﬁne the curve 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) := x0(t) is the unit tangent vector field along the curve. The curvature of x is the scalar function

κ(t) = x00(t) = T0(t)

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

T T = 1 = T0 T + T T0 = 0 = T0 T = 0 · ⇒ · · ⇒ · whence T is orthogonal to its derivative. The curvature κ is therefore measuring how rapidly the unit tangent vector changes direction. Curvature is therefore one measure of the deviation of a curve from a straight line. It doesn’t tell the whole story however. Recall that a circle of radius 2 and the standard 9 1 helix both have the same curvature κ = 2 :  t   1 t  2 cos cos 2 − 2 2 1 Circlex (t) = 2 sin t  = κ(t) =  1 sin t  = 2 ⇒ − 2 2 2 0 0

 t    cos 1 cos t √2 − 2 √2 1 Helixx (t) = sin t  = κ(t) =  1 sin t  =  √2   2 √2  t ⇒ − 2 √2 0 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 κ is deﬁned everywhere and is never zero. We could equivalently say that a curve x : I E3 is biregular if and only if it is twice-differentiable → and the vectors x0(t) and x00(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 x0 = T and 0 κ 0  − T0 N0 B0 = TNB κ 0 τ − 0 τ 0 where κ > 0 is the curvature and τ is a function called the torsion10 of x. The structure equations can be written explicitly as

T0 = κN, N0 = κT + τB, B0 = τN − − Proof. Setting T = x0 fixes T as the unit velocity vector. Since x is unit speed, we have T T (see Definition 1.16). 0 ⊥ Since x is biregular we have κ = T = 0, so we may define N = 1 T to be the unit vector pointing in | 0| 6 κ 0 the same dierection as T0. The final vector in the moving frame is fixed via B := T N is then fixed by orientation and the × existence of τ = N B follows from Theorem 1.15. 0 · Definition 1.19. • The frame (T, N, B) is the Frenet frame of x and the three vectors are, respectively, the unit tangent, principal normal and binormal vector fields of the curve.

• The equations of Theorem 1.18 are known as the Frenet–Serret equations or simply the structure equations 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 is the rectifying plane, and the plane spanned by N, B is the normal plane.

 t  √2 ( ) = cos t  Example 1 For a horizontal helix x t  √2  parameterized by arc-length, we have sin t √2   1 1 t T(t) = x (t) =  sin  0 √ − √2  2 cos t √2 But then     0 0 1 t t 1 T0(t) = cos  = N(t) = cos  and κ(t) = −2  √2  ⇒ −  √2  2 sin t sin t √2 √2

10Several authors take τ to be the negative of ours.

19 Finally,   1 1 t B(t) = T(t) N(t) =  sin  × √  √2  2 cos t − √2 gives us the binormal vector ﬁeld, whence the torsion is     0 1 1 t 1 t 1 τ(t) = N0(t) B(t) =  sin   sin  = · √  √2  · √  √2  2 2 cos t 2 cos t ) − √2 − √2 The Frenet–Serret equations for the helix are therefore

 1  0 2 0 1 − 1 T0 N0 B0 = TNB  2 0 2  1 − 0 2 0

 1 3/2 3 (1 + t) Example 2 Let x(t) =  1 t  for t ( 1, 1). Then  √2  ∈ − 1 (1 t)3/2 3 −  1 1/2  2 (1 + t) 1 1 1 1 x0(t) =   = v(t) = x0(t) = (1 + t) + + (1 t) = 1  √2  ⇒ 4 2 4 − 1 (1 t)1/2 − 2 − so that we have have a unit-speed parameterization. It follows that T = x. Now differentiate again:

 1 1/2 (1 + t)− r 4 1 1 1 √2 T0(t) =  0  = κ(t) = T0(t) = + = 1 1/2 ⇒ 4 1 + t 1 t 4√1 t2 (1 t)− − 4 − − 20 whence

 1 (1 + t) 1/2 (1 t)1/2 1 4√1 t2 4 − 1 − N(t) = T(t) = −  0  =  0  κ(t) √ √ 2 1 (1 t) 1/2 2 (1 + t)1/2 4 − − We ﬁnally have

 (1 + t)1/2 1 − 1 B(t) = T(t) N(t) =  √2  = τ(t) = N0(t) B(t) = 2 × 2 1/2 ⇒ · 2√2√1 t (1 + t) −

Calculation of curvature and torsion in any parameterization As we’ve discussed, there are relatively few curves for which an explicit unit-speed parameterization can 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 difﬁculty comes mainly from a profusion of square-roots. . .

et cos t Example Consider the exponential spiral x(t) = et sin t. We have et

cos t sin t t − t x0(t) = e sin t + cos t = v(t) = √3e ⇒ 1 cos t sin t  sin t cos t 1 − 1 − − = T(t) = sin t + cos t = T0 =  cos t sin t  ⇒ √ ⇒ √ − 3 1 3 0

21  sin t cos t 1 − − = N(t) =  cos t sin t  ⇒ √ − 2 0  cos t + sin t 1 − = B(t) = T N =  sin t cos t ⇒ × √ − − 6 2

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 q q 2 2 is 3 , since T0 = 3 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 parameterization are

x0 x00 (x0 x00) x000 v(t) = x0(t) = √x x , κ(t) = | × |, τ(t) = × · 0 · 0 v3 v6κ2 The Frenet frame with respect to an arbitrary parameterization is

1 vx00 v0x0 x0 x00 T(t) = x0, N(t) = − , B(t) = × v v3κ v3κ Moreover in this parameterization, the structure equations are

 0 vκ 0  − T0 N0 B0 = TNB vκ 0 vτ − 0 vτ 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 following is unnecessary.

• The arc-length of x(t) measured from some ﬁxed value of t is s(t). The inverse function is t(s). ds The speed is v(t) = dt . • 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 denoted Tˆ (s), κˆ(s), etc. The corresponding expressions in terms of t may be found using, for instance, T(t) = Tˆ (s(t)) = Tˆ (s).

The last part is then straightforward:

d ds d TNB = Tˆ Nˆ Bˆ (Chain Rule) dt dt ds 0 κˆ 0  − = v Tˆ Nˆ Bˆ κˆ 0 τˆ (Theorem 1.18) − 0 τˆ 0

22  0 vκ 0  − = TNB vκ 0 vτ − 0 vτ 0

We now compute each of the terms explicitly:

dt dx 1 T(t) = Tˆ (s) = y0(s) = = x0 ds dt v dt d 1 1 vx00 v0x0 = κˆ(s)Nˆ (s) = Tˆ 0(s) = y00(s) = x0 = − ( ) ⇒ ds dt v v v2 ∗ vx v x = N(t) = Nˆ (s) = 00 − 0 0 ⇒ v3κ Taking cross products with T we obtain

x x B(t) = Bˆ (s) = Tˆ (s) Nˆ (s) = 0 × 00 (†) × v3κ Since B(t) is a unit vector, we immediately have

x x κ(t) = κˆ(s) = | 0 × 00| v3 Now differentiate ( ) with respect to s (more chain rule!) to obtain ∗ 2 2 dt d vx00 v0x0 v x000 (vv00 + 3v0 )x0 κˆ0(s)Nˆ (s) + κˆ(s)Nˆ 0(s) = − = − ds dt v3 v5

Finally, take the scalar product of both sides with Bˆ = Tˆ Nˆ using the expression in (†) to obtain × x0 x00 κˆNˆ 0 Bˆ = κˆτˆ = κτ = x000 × · · v6κ Dividing by κ yields the result.

Examples

 t  e cos t q t 2 √2 t 1. Our example x(t) = e sin t from earlier has T0 = N = vκN = κ = e− . 3 ⇒ 3 et Similarly B = 1 N = vτN = τ = 1 e t. One can also use the above expressions for κ 0 − √3 − ⇒ 3 − and τ to compute them directly from the derivatives.

r cos t 2. The helix with radius r and vertical separation 2πh between spirals has the form x(t) = r sin t. ht We compute the Frenet frame:

 r sin t  r cos t  r sin t  − − x0(t) =  r cos t  , x00(t) =  r sin t , x000(t) =  r cos t , − − h 0 0

23 whence p r h v = r2 + h2, κ = , τ = r2 + h2 r2 + h2 and

 r sin t  cos t  h sin t  1 − − 1 T(t) =  r cos t  , N(t) =  sin t , B(t) =  h cos t √ 2 2 − √ 2 2 − r + h h 0 r + h r

 t  3. Consider the curve x(t) = t2. This has t3

 1   0  0 x0(t) =  2t  , x00(t) =  2  , x000(t) = 0 , 3t2 6t 6

from which s p 1 + 9t2 + 9t4 3 v(t) = 1 + 4t2 + 9t4, κ(t) = 2 , τ(t) = (1 + 4t2 + 9t4)3 1 + 9t2 + 9t4

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

1.5 The Fundamental Theorem of Biregular Spacecurves We’ve 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 it is a straight line.

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

Proof. 1. We may suppose that x(t) is unit speed. Then κ(t) = x (t) = 0 x (t) = 0. Thus | 00 | ⇐⇒ 00 x is a straight line.

2. Suppose that x(t) is a unit-speed biregular curve which lives in a ﬁxed plane. Then x0, x00 are parallel to this plane, and so T, N are also. But then B is a continuous unit vector orthogonal to the plane and thus constant. Hence τN = B = 0, so that x has zero torsion. − 0 Conversely, if τ = 0, then B is constant. But then

(x B)0 = x0 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 deviation of a curve from a straight line, while torsion measures how much a curve is trying to leap out of its osculating plane. For this reason, curvature is often referred to as bending and torsion as twisting of a curve. In the pictures, the Frenet frame and osculating plane of a helix are drawn at a given time. The second picture is a still of the ﬁrst, rotated to be side-on to the osculating plane. The non-zero 1 torsion (τ = 2 for this curve) may be observed in how the curve crosses the osculating plane: it looks a little like how the cubic function y = x3 crosses the x-axis.

We will say a little more regarding the geometric meaning of curvature later on, when we discuss osculating circles.

The Equivalence Problem for Spacecurves We wish to classify all biregular curves up to equivalence by rigid motions. The idea is that κ and τ tell you everything about the shape of a curve. In fact these functions are enough to classify biregular curves up to a choice of initial position and direction. The phrase used is up to isometry: we need to see 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 of × orthogonal 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 vectors form an orthonormal set.

5. An orthogonal matrix A has determinant det A = 1. If det A = 1 we call the matrix a rotation, and ± if det A = 1 we call it a reﬂection. The set of rotations is denoted11 SO (R). − 3 6. A moving frame is a function E : I SO (R). → 3 Deﬁnition 1.24. An isometry of E3 is a map x Ax + b where A is a constant 3 3 orthogonal matrix 7→ × and b E3 is a constant vector. An isometry is either direct (det A = +1) or indirect (det A = 1). If ∈ − A is the identity matrix, the isometry x x + b is known as a translation. 7→ 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 between vectors. Since both of these concepts are measured using the dot product, this is precisely what part 3 of the Lemma is saying. We now consider how curve transforms under the action of an isometry.

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

Direct isometry: κˆ = κ, τˆ = τ.

Indirect isometry: κˆ = κ, τˆ = τ. − Proof. Suppose WLOG that x(t) is unit-speed. Then

xˆ 0(t) = Ax0(t) = vˆ(t) = xˆ 0(t) = x0(t) = 1 ⇒ since orthogonal matrices preserve the scalar product (Lemma 1.23, part 3). We can now compute:

2 2 2 xˆ 0 = Ax0 = Tˆ = AT, xˆ 00 = Ax00 = κˆ = xˆ 00 = Ax00 = x00 = κ. ⇒ ⇒ Therefore curvature is invariant under any isometry. For torsion, observe that the second formula above implies Nˆ = AN. Since A is orthogonal, it preserves angles, whence AB is orthogonal to Tˆ , Nˆ . It remains only to decide if AB is Bˆ . For this, ± consider the fact that the moving frame E = (TNB) is a special orthogonal matrix. We now have

AE = (AT AN AB) = (Tˆ Nˆ AB)

It follows that AE = Eˆ AB = Bˆ det A = 1. It follows that ⇐⇒ ⇐⇒ ( AB if the isometry is direct, Bˆ = (det A)AB = AB if the isometry is indirect. − Now

τˆ = Bˆ 0 Nˆ = (det A)(AB0) (AN) = det AB0 N = (det A)τ − · − · − · as required.

1 0 0  Example The simplest indirect isometry is S(x) = Ax = 0 1 0  x. If we apply this to the 0 0 1 − cos(t/√2) cos(t/√2) unit-speed helix x(t) = sin(t/√2), we obtain xˆ(t) = sin(t/√2) which spirals down rather t/√2 t/√2 than up. Indeed −     sin(t/√2) cos(t/√2) 1 − 1 − 1 Tˆ = xˆ 0 =  cos(t/√2)  = AT κˆ = Tˆ 0 =  sin(t/√2) = = κ √ 2 − 2 2 1 0 − 26  cos(t/√2)  sin(t/√2) − 1 − Nˆ =  sin(t/√2) = N = AN Bˆ = Tˆ Nˆ =  cos(t/√2)  = AB − × √ − 0 2 1 1 τˆ = Bˆ 0 Nˆ = − · −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 the second twists down (in the Bˆ 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 a cornerstone 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 T | − | ≤ | − | ≤ and suppose that the partial derivatives of A with respect to the co-ordinates of x are bounded.12 Suppose also that M = sup A(x, t) on the closed region. Then the initial value problem | |  dF  = A(F, t) dt F(a) = c has a unique solution F : I Rn on the interval defined by t a min T, K . → | − | ≤ M 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. The rough idea is that we produce a sequence of functions

Z t Z t F1(t) = c, F2(t) = c + A(F1(t˜), t˜) dt˜, F3(t) = c + A(F2(t˜), t˜) dt˜,... a a which are then shown to converge to the required solution.

dF Example Consider the initial value problem dt = 2tF(t) with initial condition F(0) = 1. We obtain Z t 2 F1(t) = 1, F2(t) = 1 + 2t˜dt˜ = 1 + t , 0 Z t 2 2 1 4 F3(t) = 1 + 2t˜(1 + t˜ ) dt˜ = 1 + t + t ,... 0 2 The Picard iteration is building up the correct solution as a power series

∞ 2n 2 t 1 1 F(t) = et = ∑ = 1 + t2 + t4 + t6 n=0 n! 2 6 ··· Corollary 1.27. If the ODE in Picard’s Theorem is linear that is if A(x, t) = A(t)x + b(t) for some matrix valued-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 continuous for all x Rn. Essentially K can be taken to be arbitrarily large: K = MT is sufficient for the result. ∈ Corollary 1.28. Let I = [t T, t + T] be an interval and W : I M (R) a continuous matrix-valued 0 − 0 → 3 function such that each W(t) is skew-symmetric.13 Suppose that is an orthogonal matrix. Then: O 1. There exists a unique solution E : I O (R) to the initial value problem → 3 ( dE dt = EW E(t ) = 0 O 2. If det = 1, then E : I SO (R). O → 3 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 M (R). × → 3 Now differentiate:

d T T T T T T (EE ) = E0E + EE0 = EWE + EW E = 0 dt T T since W is skew-symmetric. Thus EE is constant. However E(t0)E(t0) = I, whence this constant 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 E is differentiable, it follows that det E : I R is continuous. However det E = 1 since E is → ± orthogonal. Since E is deﬁned on an interval, it must therefore be the constant 1. E is therefore special 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 = [t T, t + T] be an interval, 0 − 0 c E3 a constant vector and suppose that (T N B ) is a ﬁxed oriented orthonormal basis of E3. Given ∈ 0 0 0 functions14 κ > 0 and τ from I R, there exists a unique unit-speed biregular spacecurve x : I E3 → → with these curvature and torsion functions such that x(t0) = c and the Frenet frame at t = t0 is the basis (T0 N0 B0).

0 κ 0 Proof. The structure equations E0 = E κ −0 τ put us in the situation of Corollary 1.28, whence 0 τ −0 there exists a unique solution E = (TNB) : I SO (R). To recover the curve x simply integrate → 3 the unit tangent vector:

Z t x(t) = c + T(t˜) dt˜ t0 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 to direct isometry by its curvature and torsion. Indeed:

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

Proof. Suppose that x : I E3 and x : I E3 have the same curvature and torsion functions. 1 → 2 → Choose some t I and suppouse that their Frenet frames are E , E respectively. The isometry we 0 ∈ 1 2 require must satisfy the conditions at t . Indeed we need an direct isometry S : x Ax + b such that 0 7→ • S x1(t0) = x2(t0)

• AE1(t0) = E2(t0) This exists and is unique: we are forced to choose A = E (t )E (t ) 1 and b = x (t ) Ax (t ). 2 0 1 0 − 2 0 − 1 0 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), curvature and torsion as x2. The Fundamental Theorem says that x2 = x1.

Compare what we’ve 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 of time, 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 it’s always unit speed, but there’s nothing stopping us working with arbitrary speed) and how much you bend and twist, you can determine the entire motion. The required data is the sextuple

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

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

14κ must be differentiable with continuous derivatives and τ be continuous.

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

κ cos t  √κ2+τ2 ( ) = κ sin t  x t  √κ2+τ2  τt √κ2+τ2 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 simpler in that we can make a common choice of N simply by rotating T counter-clockwise, rather than differentiating. In particular, we don’t need to be able to differentiate curve twice in order to find the Frenet frame. x(t) Definition 1.32. Let x : I E2 : t be a regular curve with speed v(t). The unit tangent → 7→ y(t) 1 vector is T(t) = v(t) x0(t). Define the unit normal vector by

0 1 N(t) = JT(t) where J = − 1 0

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

Theorem 1.33. Given a regular curve x : I E2 of speed v(t), there exists a function κ : I R (the → → (signed) curvature of x) satisfying, 0 vκ T0 N0 = TN − vκ 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 be negative. ( κ > 0 x bends towards N ⇐⇒ κ < 0 x bends away from N ⇐⇒ 3 Observe the pictures. If these were curves in E , the N(t) T(t) ﬁrst would have B pointing out of the page. For the κ < 0 second picture we would have to reverse the direction κ > 0 N(t) of N and have B pointing into the page. Also observe that orienting a planar curve in the oppo- T(t) site direction changes the sign of the curvature.

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

x Jx κ = 00 · 0 v3 This reduces to κ = x (t) if x is unit-speed, so we recover the same notion as in Deﬁnition 1.8. ± | 00 | Proof. From the structure equations,

vx00 v0x0 1 1 vκ = T0 N = − Jx0 = x00 Jx0 · v2 · v v2 · since Jx0 is perpendicular to x0.

Examples

1 1. For a circle of radius a parameterized counter-clockwise, the curvature is κ = a . Parameterized clockwise, the curvature is 1 . − a t 2. Any function y = f (x) may be parameterized by x(t) = . Its curvature is then f (t)

f (t) ( ) = 00 κ t 2 3/2 (1 + f 0(t) ) as you might have seen in an elementary calculus course. In particular, the catenary (page 11) t x(t) = has cosh t

cosh t 1 κ(t) = = (1 + sinh2 t)3/2 1 + sinh2 t

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

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

cos θ(s) sin θ sin θ T(s) = , T0 = θ0 − , N = JT = − = κ(s) = θ0(s) sin θ(s) cos θ cos θ ⇒

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

Theorem 1.35. Given a countinuous function κ : I R, there exists a unique unit-speed curve x(t) in E2 → 1 with curvature κ(t) such that x (0) is the unit vector and x(0) = 0. 0 0

31 The proof proceeds in the same way as Theorem 1.29: the existence theorem from ODEs gives the 1 0 vectors T(t), N(t) satisfying the structure equations and T(0) = , N(0) = . Integrating 0 1 T = x0 yields the curve.

Aside: Curves in Arbitrary Dimension There is a general theorem (and Frenet frame) for curves in En: for a unit-speed curve x in En whose ﬁrst n derivatives at each point form a basis of En, we may Gram-Schmidt orthogonalize the basis (n) x0, x00,..., x to obtain a moving frame e1,..., en and functions κ1,..., κn 1 (the generalized { } { } − curvatures of x) satisfying

 0 κ 0 0 0  − 1 ··· κ1 0 κ2 0 0   −   0 κ 0 0 0   2  (e10 ,..., en0 ) = (e1,..., en)  . . .   . .. .     0 0 0 0 κn 1 − − 0 0 0 κn 1 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 straight line and from being planar respectively. There is a little more geometry contained in the notion of curvature. We have seen that the only planar curve with constant curvature k is the circle of radius 1 k . We could have started with this as our deﬁnition of curvature, and said that a curve has curvature 1 k at a certain point, if the circle which best approximates the curve at that point has radius k . Of course, 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 the ﬁrst n derivatives agree there: i.e. x(i)(t ) = y(i)(s ) for all 1 i n. 0 0 ≤ ≤

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

• c(0) = x(t0), T(t0) N(t0) • c0(0) = x0(t0),

• r > 0 the circle lies on the same side of the ⇐⇒ curve as N. r > 0 It should be clear that there is precisely one circle c of r < 0 each singed radius r R. ∈

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

c(s) = x(t0) + rN(t0) + r sin(s/r)T(t0) r cos(s/r)N(t0) | {z } | −{z } center rotation is the unit-speed parameterization for which c(0) = x(t0). Moreover

c0(s) = cos(s/r)T(t ) + sin(s/r)N(t ) = c0(0) = T(t ) 0 0 ⇒ 0

The circles all have ﬁrst-order contact with the curve at x(t0). Moreover, 1 1 1 c00(s) = sin(s/r)T(t ) + cos(s/r)N(t ) = c00(0) = N(t ) − r 0 r 0 ⇒ r 0

The circle has second-order contact with the curve at t0 if and only if 1 c00(0) = x00(t ) = κ(t ) 0 ⇐⇒ r 0

1 Deﬁnition 1.37. The radius r = is the radius of curvature of the curve x at t0. The circle parame- κ(t0) terized by

1 c(s) = x(t0) + sin(s/r)T(t0) + (1 cos(s/r))N(t0) κ(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 for an 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 t way x(t) = . The relevant ingredients are t2 1 p 2 1 1 x0(t) = = v(t) = 1 + 4t = T(t) = 2t ⇒ ⇒ √1 + 4t2 2t 0 2 1 2t x00(t) = , κ(t) = , N(t) = − 2 (1 + 4t2)3/2 √1 + 4t2 1

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

1 t 1 + 4t2 2t 4t3 x(t) + N(t) = 2 + − = 1− 2 κ(t) t 2 1 2 + 3t

1 (1+4t2)3/2 Its radius is κ = 2 . An animation of several of the osculating circles is given below.

33 The idea of second-order contact can be used to obtain local approximations to a curve: indeed recall that the nth Taylor polynomial of a function y at t is the unique degree n polynomial which has 0 ≤ nth 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,

κ(0) y = x2 + higher order terms. 2 t Proof. Such a curve may be parameterized by x(t) = , where y(0) = 0 = y (0). The curvature y(t) 0 at t = 0 is simply y00(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 κ = 2. The theorem tells us what we expect to see, that

2 y = x2 + 2 ··· The higher order terms are thus all zero.

34 2. The catenary y = cosh x has initial curvature κ = 1 at the point (0, 1), where its gradient is zero. We therefore see that 1 y = 1 + x2 + 2 ··· 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 curvature of x: 1 e(t) = x(t) + N(t) κ(t)

t Example We have already calculated the evolute of the parabola x(t) = when we found its t2 osculating circles:

1 4t3 e(t) = x(t) + N(t) = 1− 2 κ(t) 2 + 3t

1 x 2/3 which may be written as y = 2 + 3 4 . 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 challenge with 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.

35 Involutes A related notion to the evolute is the involute of a curve. This is obtained by rolling a line along a curve 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). − 0 Note that the involute depends on where the zero of the parameterization is located: at t = 0 the involute intersects the original curve.

1 sinh− t Example The involute of the catenary x(t) = is a curve called the tractrix: √1 + t2

1 2 1/2 sinh− t t(1 + t )− i(t) = − 2 1/2 (1 + t )− This is the curve obtained when a mass (the green dot) at (0, 1) is towed by attaching a (orange) rope of 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 curve and then unwrapped as we move to the right. It is as if a string has been wrapped tight around the curve, 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 = 0 or κ = 0. Proof. Let x(t) be unit speed, then

i(t) = x(t) tx0(t) = x(t) tT(t) − − is the involute, where T is the unit tangent vector of x. Let N, κ denote the unit normal and curvature of x, and ve, Te, Ne , κe the speed, Frenet frame and curvature of i. Then

i0 = T T tT0 = tκN = v˜(t) = tκ − − − ⇒ | | = Te = sgn(tκ)N ⇒ − = Ne = sgn(tκ)JN = sgn(tκ)T ⇒ − By the structure equations, we have

vκNe = Te0 = sgn(tκ)N0 = sgn(tκ)κT = κNe ee − κ κ and so κe = v = tκ . The evolute of the involute of x is therefore e | | tκ tκ i + | |sgn(tκ)T = x tT + T = x κ − κ Note that the proof fails precisely when ve(t) = tκ(t) = 0.

1.8 Further thoughts on curves Several of the ideas described earlier may be generalized, and there are many further topics in classical curve theory that we ignore. Osculating circles may be defined for a curve in E3 (or higher dimension). At each point of a curve 1 x(t), the osculating circle is the circle in the osculating plane with center x(t) + κ(t) N(t) and 1 E3 radius κ(t) : the radius of curvature. It can be shown that this is the unique circle in having second 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). − 0 Evolutes are somewhat more difficult: we want an evolute to be a curve e = x + λN + µB for some functions λ, µ such that x is an involute of e. It can be seen that, if t is an arc-length parameterization, 1 Z e(t) = x(t) + N(t) + cot τ(t) dt B(t) κ(t) R where τ(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 insisting that Z Z π t, cot τ(t) dt = 0 τ(t) dt = ∀ ⇐⇒ 2 This is clearly if the anti-derivative is the constant c = π and τ(t) 0: the only curves with an 2 ≡ evolute 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 we’ve 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 closed curve x of length l satisﬁes the inequality

4πA l2 with equality x is a circle. ≤ ⇐⇒ The Four-Vertex theorem states that every simple closed curve in the plane with differentiable curvature has at least four vertices (points where κ0 = 0). For an easy example of this last, consider an ellipse

a cos t x(t) = b sin t

Its curvature is ab κ(t) = p a2 sin2 t + b2 cos2 t Therefore ab(b2 a2) cos t sin t π 3π κ0(t) = − = 0 t = 0, , π, [a2 sin2 t + b2 cos2 t]3/2 ⇐⇒ 2 2