APPENDIX A

CURVES

A.1 Parametrization of Curves and Arclength

For n = 2 or 3, a smooth parametrized curve is a subset of W1 of the form

{C(t)\te(a,b)}

where C(f ) : (a,b) Rn is assumed to be at least twice differentiable, and C'(f) 7^ 0 for any in (a, b). It is quite common to think of the function C(t) itself as the curve, and such will be the case here as well. Thus, Ci (t) = (i2,cosf) is a curve in R2, and C2(i) = (cosf,sini,i) isacurveinR3 for any domain (,fe), but 03(f) = (2,||) and C4(i) = (3,2) fail to be curves for any domain (a,b) that contains 0. Of course, both C3(i) and C^t) are smooth parametrized curves over any domain (a,b) that does not contain 0. In general, the domain (a,b) will be left unspecified, and it will be assumed that every curve is in fact smooth and parametrized.

An arc of the curve C(t) is any portion that corresponds to a closed subinterval [a, b] of the curve's domain. The length of such an arc is defined as

fb / \C'{t)\dt. Ja

Introduction to Topology and Geometry, Second Edition. 4 2 9 By Saul Stahl and Catherine Stenson Copyright 2013 John Wiley & Sons, Inc.

4 3 0 APPENDIX A

Figure A.l Arclength.

This definition is justified by the following informal argument. In Figure A.l let ds denote the length of a small arc of the curve C(i) = (x(t),y(t)) in R2. Then

ds Vdx2 + dy2 y/[x?(t)dt]2 + \/(t)]2dt

= ^[x'{t)]2 + \y

CURVES 431

3

2

1

f = 0

- 1

- 2

- 3

Figure A.2 A parametrized curve.

Theorem A.1.2 Let C(t) be a curve. Then C'(to) is parallel to the tangent to C(t) at C(to), and |C;(io)| is the instantaneous rate with which this parametrization traces the curve at C(to).

PROOF: By definition

C ' ( , o ) = l i m C ( * + A f ) - C ( f o ) . /-) At

Since (l/At)[C(t0 + ) - C(/0)] is parallel to the chord from C(t0) to C(i0 + ), and since the tangent line at C(io) is the limiting position of this chord, it follows that C'(io) is parallel to the tangent at C(fo)

Let As denote the length of the arc of C(f ), to 0 At

which limit, by the Mean Value Theorem for integrals, has value |C'(io) |. Q.E.D.

Example A.1.3 The curve C(i) = (t2,t) of Figure A.2 has the points correspond-ing to t 3, 2, 1, 0, 1, 2, 3 marked on it. The different lengths of the arcs determined by these points are roughly proportional to the corresponding values of

|C'(0| = \ / l + 4 i 2 .

The parameter s of the curve C(s) is said to be an arclengthparameter if |C'(.s) | = 1 for all s in the domain.

432 APPENDIX A

C(i) = D(s)

Figure A.3 The arclength parameter of a semicircle.

Example A.1.4 In the curve C(s) = (I/y/2)(coss,sins,s),s is an arclength param-eter, because

\C'(s)\ = - ^ \ / s i n 2 s + cos 2 s+ l = =y/l=\. 1 W l V2 \/2

Theorem A.1.5 Every curve has an arclength parametrization.

PROOF: Let C(t) be a curve with a point A on it. The point A divides the curve into two parts of which one (it doesn't matter which one) is designated as positive and the other as negative. If s is a positive (negative) number, let D(s) denote that point P on the positive (negative) part such that the arc of O(s) from A to P has length \s\. Define the function t = t(s) by the equation

C(f(s))=D(5),

and let s = s(t) be its inverse. Then

ID'WI CM |C(0 ds/dt |C(0I- C'(0I Q.E.D.

Example A. 1.6 For the semicircle C(r) = (\/l 2,), - 1 < t < 1, set A = (1,0), and designate the upper part as the positive direction (Fig. A.3). Then

CURVES 433

O

Curve

Unit tangent

Figure A.4 Tangents.

Tangent line

and hence s(t)=f\C(T)\dx=f

Jo Jo

Consequently, t(s) = sin s, and hence

dx

VT = sin xt.

O(s) =C(t(s)) = (v 1 sin2 A, sin ) = (cos s, sin s).

For any curve C(f ), the unit tangent is

C'(f) T(f) :

|C(0I" We note in passing that while the tangent line to a curve is visualized at its point of contact with the curve, the unit tangent is anchored at the origin (Fig. A.4).

Example A.1.7 If C(t) = (a\ + b\t,a2 + b2t,a3 + bit), then

^ + ^ + ' C/(f) = (*i,&2,&3) and T(0 =

which, as expected, is a constant.

Example A.1.8 If C(t) = (cosi,sini,i), then

C'(f) = (-sini,cosi, l) and ( = - r=(-cosi ,s ini , l ) . V2

Here C is a helix, and T rotates about the origin while keeping a constant angle to the xy-plane.

Example A.1.9 If C(f) = (t,t2,t3), then

C'(i) = (l,2i,3i2) and T(i) = (l,2f,3f2)

> / l+4i 2 + 9i4"

4 3 4 APPENDIX A

The curvature of a curve C(t) is a measure of the rate at which it is changing di-rection (relative to the traversed distance). Since the direction of a curve is specified by its unit tangent, it follows from Theorem A. 1.2 that the curvature of C should equal

ds = -

dT\ dt

\ds\ \dt \

1T'(Q| lew

Example A.1.10 For the curve C(t) = (a\ -\-b\t,ci2 + ^2^,3 H-^3i), wehaveT^f) = 0, whereas C'(f) 0 and hence : = 0.

Example A.1.11 For the curve C(f) = (cosi,sini,/), wehaveC'(i) = (sini,cosi, 1), T'(f) = (cosi,sin/,0), and hence

|T'(Q |C(0

_1_ V2'

Example A.1.12 For the curve C(i) = (,2,3),

( -4 -18 3 , 2 -18 4 , 6 +12 3 ) T(t) :

|'

( l + 4 i 2 + 9f4)3/2

/4 + 52;2 + 216r4 + 468r6-+ 324;8

(1+42 + 94)3/2 '

|C'(0| = |(1,2,32)| = Vl+4 / 2 + 9i4,

fC = V4 + 52i2 + 216f4 + 468r6 + 324;8

(1+92 + 94)2 '

In the interest of brevity the derivative dT(t{s))/ds will be abbreviated to dT/ds, where s denotes the arclength parameter of C(t). Note that since T is a unit vector,

= 1 .

When this equation is differentiated with respect to s, we get

dT ^ ^ dT .T + T. = 0 ds ds

and hence JT ds

T = 0.

It follows that the vector dT/ds is orthogonal to the direction of C. For this reason, the principal normal to C(t) is defined as the unit vector

~ds =l d T

K ds ' N =

dT "57

(2)

CURVES 435

Example A.1.13 For the curve C(t) = (cosi,sini,f), set A = (1,0,0), so that

s(t) = / Vcos2T + s in 2 T+ ld r = Vt, Jo

Hence D ( , ) = ( c o s - ^ s i n - ^

is the arclength parametrization of C(f ). Consequently,

T W = D ' ( i ) ^ ( - s i ^ , c o s - ^ )

and

dT 1 / s . s

^ = H~ C 0 S 7 i '~ s i n 7rV

N = ( cos= , sinT=,0 1 y/l y/l

Theorem A.1.14 Let C(t) be a curve, and let ' and " denote first and second deriva-tives with respect to t. Then

_ \a x c"i

PROOF: Let D(s) be the arclength parametrization of C(t). By the chain rule,

dC dDds , *- = ~ r ~ ~i~~r =s *i at as at

C" - / T + s'T = s"T + s,(^-(^= s"T + S'2KN, ds dt

and hence C' x C" = " x T + / 3 icTx N.

However, since T and N are othogonal unit vectors, it follows that T x N is also a unit vector. Hence

| C ' X C " | = S ' 3 K : ,

from which it follows that

[CxC'l ic'xC'i * - ? 5 - H e p - QED

Example A.1.15 For the curve C(f) = (t,t ,t ) we have

C(f) = (l,2f,3f2), C"(i) = (0,2,6i),

4 3 6 APPENDIX A

and hence

Consequently,

K = \C x C"|

|C'p

C ' x C " = (6i2,-6i,2).

V36r4 + 36i2 + 4 / 94 + 92 + 1 ~ (1+42 + 94)3/2 - y (94 + 4 2 + 1 ) 3

The reader should take note of the fact that both the derivation and the final answer here are simpler than those in Example A. 1.12.

Example A.1.16 The plane graph of y f(x) can be placed in R3 as the curve C(f) - (i,/(i),0) with C'(f) = ( l , / ( i ) ,0 ) and C"(t) = (0,/"(i),0). It follows that its curvature is

IC 'xC ' l |(0,0,/") | _ \f"\ "/13 (x/+71)3 (i+/'2)3/2'

The signed curvature of a plane curve is *, where the positive value is used if and only if the curve is turning counterclockwise.

For a given curve C(f ), the vector

B(f ) = T(0 x N(0

is called the binormal of C(f ). It is clear that the vectors T, N, B form an orthonormal triple. The arclength derivatives of these vectors are used to describe how the curve twists through space. Let s be an arclength parameter of C. Since B B = 1, it follows by differentiation with respect to the arclength s that

. B = 0. ds

In addition,

dB dT dN = - - + = : + 0 = 0. ds ds ds

The two equations above imply that dB/ds is orthogonal to both B and T. Since N is also orthogonal to these two vectors, it follows that there exists a scalar function , called the torsion of C, such that

= - . ds

We will describe how this function is computed and used after the above considera-tions are summarized.

CURVES 437

Theorem A.1.17 (Frenet-Serret formulas) Let s be an arclength parameter of the curve C. Then

as

= - + , as dB

= - . as PROOF: It only remains to prove the second of the three equations. This, however,

follows by differentiating the equation N = B x T with respect to s:

-B ds ds ds

= B x : - = -- + . Q.E.D.

Proposition A.1.18 The curve C is a straight line if and only ifK = 0. It is a plane curve if and only ifx = 0.

PROOF: If C is a straight line, then , and it follows from the first of the Frenet-Serret equations that = 0.

If C is a plane curve, then both T and N lie in a plane parallel to that of C. The binormal B must then be one of the two normals to this plane, and it is therefore constant. It follows from the third equation that .

If s is arclength and K = 0, then by the first equation dT/ds = 0 and hence T = dC/ds is constant. It follows that

C(s )=sE + F

for some constant vectors E and F. In other words, C is a straight line. If = 0, then, by the third equation, B is constant. Set

f(s) = (C(J ) - C(0)) B for all s.

Then df dC -4- = ^ B = T B = 0. ds ds

In other words, f(s) is constant. Since /(0) = 0, it follows that

( C ( J ) - C ( 0 ) ) - B = 0 for all s

implying that C(s) is contained in the plane orthogonal to B and containing the point C(0). Q.E.D.

Proposition A.1.18 makes it reasonable to regard the torsion as a measure of the nonplanarity of the curve. There is a straightforward analog of Theorem A. 1.14 for the computation of the torsion.

438 APPENDIX A

Theorem A.1.19 If ' denotes differentiation with respect to t, then the torsion of C(t) is given by

C'xC"-C"' T~ |C'xC"|2 '

PROOF: The following equations are easily verified, and the details are relegated to Exercise 7:

C " = / T + K: ( / ) 2 N, (3)

C x C" = K{S')% (4)

C'" = [s'"- C V ) 3 ] T + [3icV' + )f'(/)2]N+ ^, (5) c'xc-r = \axcr ' D(6)

A curve C is said to be a cylindrical helix provided there is a fixed unit vector A such that the angle from T to A is a constant . Equivalently, T A = cos# is constant. For example, the helix

C(t) = (acost,asint,bt)

is a cylindrical helix with A = (0,0,1).

Proposition A.1.20 The curve C with K>0is a cylindrical helix if and only ifx/K is a constant

PROOF: Let ' denote differentiation with respect to an arclength parameter of the cylindrical helix C with T A = cos . Then

0 = ( ) / = , = .

Since : > 0, it follows that N A = 0, and consequently A lies in the plane spanned by T and B. Since all these vectors have unit length,

A = cos0T + sin0B.

Differentiation and the first and third Frenet-Serret formulas yield

0 = KTcosN-TsinN.

or / cot = constant

Conversely, suppose > 0 and / is constant. Let = cot-1 (/). If we set

A = cos0T + sin0B,

CURVES 439

then, again by the Frenet-Serret formulas,

A' = ( K-COS - sin 0)N = 0.

It follows that A is a constant unit vector such that T A is also constant. Q.E.D.

Proposition A.1.21 The curve C is part of a circle if and only if it has constant nonzero curvature and its torsion vanishes everywhere.

PROOF: Suppose C is part of a circle. After a rotation and/or a translation it may be assumed that C lies in the xy-plane and is centered at the origin, so that it has arclength parametrization

C(s) = r(cos(s/r),sin(s/r),0),

where r is the radius of the circle. It is then easily verified by means of Theorems A. 1.14 and A. 1.18 that = \/r = constant, and is identically 0.

Conversely, suppose 0 is constant and vanishes everywhere. It follows from Proposition A. 1.18 that C is planar. We next show that if s is an arclength parameter of C, then the point

O(s)=C(s) + K(s)

is constant. When the equation above is differentiated with respect to s, we get

D ' ( ) = C'(i) + - N ' ( J ) = T + -(-JCT + xB) = T - T = 0.

Hence, O(s) is a constant point, say D. Since the distance of C(.s) from this point D is the constant l/fc, and since C is known to be planar, it follows that it is also part of a circle. Q.E.D.

Exercises A

1. Suppose

( ,^-'*--) -1

4 4 0 APPENDIX A

3. Let (x(t),y(t)) be a parametrized plane curve. Show that

_ \x'y"-x"y'\ K- (y2+y2)3/2-

4. The curve C(i) = (coshi, sinht,t) is known as the hyperbolic helix. Show that its curva-ture is l/(2cosh2i).

5. Let denote the inclination of the tangent line to the graph of y = f(x) to the x-axis. Prove that /ds equals the signed curvature.

6. Suppose f(x) = Ax2 + g(*), where 0 = g'(0) = g"(0). Then the signed curvature of y = f(x) at;c = 0is2A.

7. Prove Equations (3)-(6). 8. Suppose C(i) is a curve in M3, A is a fixed 3-vector, and B is a fixed 3 x 3 rotation

matrix. Set D(i) = C(i) + A, E(r) = BC(f). Prove that

9. Evaluate the curvature and torsion of the following curves:

(a) C(t) = {3t-t3,3t2,3t + t3).

(b) C(i) = (i sini, 1 -cos i , i ) .

(c) C(i) = (l+cos2r,sin2i,2sini).

(d) C(t) = (t,t2,t3).