$1.EXAMPLES,ARCLENGTH PARAMETRIZATION 9 7.Find the arclength of the tractrix,given in Example 2,starting at(0.1)and proceeding to an arbitrary point #8.Let P.ER3 and let a:[a.b]R3 be any parametrized curve with a(a)=P,a(b)=Q.Let v=-P.Prove that length()lvll.so that the line segment from P to gives the shortest possible path.(Hint:Consider '(r).vdt and use the Cauchy-Schwarz inequality u Of course,with the altemative definition on p.6,it's even easier.) 9.Consider a uniform cable with density8 hanging in equilibrium.As shown in Figure 1.12.the tension forces T(x+Ax).-T(x),and the weight of the piece of cable lying over [x.x+Ax]all balance. If the bottom of the cable is atx =0.To is the magnitude of the tension there,and the cable is T(x+△x) 1B+△9 -Tx0-g8△s x+△x FIGURE1.12 du make the substitution u=sinh v.) 10.As shown in Figure 1.13,Freddy Flintstone wishes to drive his car with square wheels along a strange road.How should you design the road so that his ride is perfectly smooth,ie.so that the center of his wheel travels in a horizontal line?(Hints:Start with a square with vertices at (1.+1).with center FIGURE1.13 C at the origin.If(s)=(x(s).y(s))is an arclength parametrization of the road,starting at(0.-1). consider the vector O=p+P+C,where P=a(s)is the point of contact and is the midpoint of the edge ofthe square.Use=s(s)and the fact that isaunit vector orthogonal to
÷1. EXAMPLES, ARCLENGTH PARAMETRIZATION 9 7. Find the arclength of the tractrix, given in Example 2, starting at .0; 1/ and proceeding to an arbitrary point. ]8. Let P; Q 2 R 3 and let ˛W Œa; b ! R 3 be any parametrized curve with ˛.a/ D P, ˛.b/ D Q. Let v D Q P. Prove that length.˛/ kvk, so that the line segment from P to Q gives the shortest possible path. (Hint: Consider Z b a ˛ 0 .t / vdt and use the Cauchy-Schwarz inequality u v kukkvk. Of course, with the alternative definition on p. 6, it’s even easier.) 9. Consider a uniform cable with density ı hanging in equilibrium. As shown in Figure 1.12, the tension forces T.x C x/, T.x/, and the weight of the piece of cable lying over Œx; x C x all balance. If the bottom of the cable is at x D 0, T0 is the magnitude of the tension there, and the cable is FIGURE 1.12 the graph y D f .x/, show that f 00.x/ D gı T0 p 1 C f 0 .x/2. (Remember that tan D f 0 .x/.) Letting C D T0=gı, show that f .x/ D C cosh.x=C /Cc for some constant c. (Hint: To integrate Z du p 1 C u2 , make the substitution u D sinh v.) 10. As shown in Figure 1.13, Freddy Flintstone wishes to drive his car with square wheels along a strange road. How should you design the road so that his ride is perfectly smooth, i.e., so that the center of his wheel travels in a horizontal line? (Hints: Start with a square with vertices at .˙1; ˙1/, with center C P Q O FIGURE 1.13 C at the origin. If ˛.s/ D .x.s/; y.s// is an arclength parametrization of the road, starting at .0; 1/, consider the vector ! OC D ! OP C ! PQ C ! QC, where P D ˛.s/ is the point of contact and Q is the midpoint of the edge of the square. Use ! QP D s˛ 0 .s/ and the fact that ! QC is a unit vector orthogonal to
CHAPTER 1.CURVES unexpectedly.Now use the re of Exercise4tond()the hint for Exereise) 11.Show that the curve a(t)= (L,1sin(π/)），t≠0 has infinite length on [0.1].(Hint:Consider 0.0). 1=0 (a,P)withP={0.1/N.2/2N-1),1/N-1).,1/2,2/3.1) 12.Prove that no four distinct points on the twisted cubic (see Example 1(e))lie on a plane 13.(a special case of a recent American Mathematical Monthly problem)Suppose [a.bR2 is a smooth parametrized plane curve (perhaps not arclength-parametrized).Prove that if the chord length l(s)-(t)depends only on s-,then must be a (subset of)a line or a circle.(How many derivatives of do you need to use?) 2.Local Theory:Frenet Frame What distinguishes a circle or a helix from a line is their curvature,i.e.,the tendency of the curve to change direction.We shall now see that we can associate to each smooth()arclength-parametrized curve a natural"m noving frame"(an orthonormal basis forchosen at each point on the curve.adapted to the geometry of the curve as much as possible). We begin with a fact from vector calculus which will appear throughout this course Lemma 2.1.Suppose f.g:(a.b)R3 are differentiable and satisfy f(r)g(t)=const for all t.Then f)·gt) -f(r)g'(t).In particular, lf(r)l=const if and only if f(r).f (r)=0 for all t. Proof.Since a function is constant on an interval if and only if its derivative is everywhere zero,we deduce from the product rule, (f·g'0=f)gt)+f)g) that if f.g is constant,then f.g=-f.g.In particular,fl is constant if and only iflf2=f.f is constant. and this occurs if and only if f.f=0. ◇ Remark.This result is intuitively clear.If a particle moves on a sphere centered at the origin,then its velocity vector must be orthogonal to its position vector,any component in the direction of the position vector would move the particle off the sphere.Similarly,suppose fand g have constant length and a constant angle between them.Then in order to maintain the constant angle.as fturs towards g.we see that g mus turn away from fat the same rate. Using Lemma 2.1 repeatedly,we now construct the Frenet frame of suitable regular curves.We assume throughout that the curve a is parametrized by arclength.Then,for starters,(s)is the unit tangent vector to the curve.which we denote by T(s).Since T has constant length.T(s)will be orthogonal to T(s Assuming T'(s)0,define the principal normal vector N(s)=T'(s)/T'(s)and the curvature K(s)= IT'(s)ll.So far,we have T(s)=k(s)N(s)
10 CHAPTER 1. CURVES ! QP. Express the fact that C moves horizontally to show that s D y 0 .s/ x 0 .s/; you will need to differentiate unexpectedly. Now use the result of Exercise 4 to find y D f .x/. Also see the hint for Exercise 9.) 11. Show that the curve ˛.t / D 8 < : t; t sin.=t / ; t ¤ 0 .0; 0/; t D 0 has infinite length on Œ0; 1. (Hint: Consider `.˛;P/ with P D f0; 1=N; 2=.2N 1/; 1=.N 1/; : : : ; 1=2; 2=3; 1g.) 12. Prove that no four distinct points on the twisted cubic (see Example 1(e)) lie on a plane. 13. (a special case of a recent American Mathematical Monthly problem) Suppose ˛W Œa; b ! R 2 is a smooth parametrized plane curve (perhaps not arclength-parametrized). Prove that if the chord length k˛.s/ ˛.t /k depends only on js tj, then ˛ must be a (subset of) a line or a circle. (How many derivatives of ˛ do you need to use?) 2. Local Theory: Frenet Frame What distinguishes a circle or a helix from a line is their curvature, i.e., the tendency of the curve to change direction. We shall now see that we can associate to each smooth (C 3 ) arclength-parametrized curve ˛ a natural “moving frame” (an orthonormal basis for R 3 chosen at each point on the curve, adapted to the geometry of the curve as much as possible). We begin with a fact from vector calculus which will appear throughout this course. Lemma 2.1. Suppose f; gW.a; b/ ! R 3 are differentiable and satisfy f.t / g.t / D const for all t. Then f 0 .t / g.t / D f.t / g 0 .t /. In particular, kf.t /k D const if and only if f.t / f 0 .t / D 0 for all t: Proof. Since a function is constant on an interval if and only if its derivative is everywhere zero, we deduce from the product rule, .f g/ 0 .t / D f 0 .t / g.t / C f.t / g 0 .t /; that if f g is constant, then f g 0 D f 0 g. In particular, kfk is constant if and only if kfk 2 D f f is constant, and this occurs if and only if f f 0 D 0. Remark. This result is intuitively clear. If a particle moves on a sphere centered at the origin, then its velocity vector must be orthogonal to its position vector; any component in the direction of the position vector would move the particle off the sphere. Similarly, suppose f and g have constant length and a constant angle between them. Then in order to maintain the constant angle, as f turns towards g, we see that g must turn away from f at the same rate. Using Lemma 2.1 repeatedly, we now construct the Frenet frame of suitable regular curves. We assume throughout that the curve ˛ is parametrized by arclength. Then, for starters, ˛ 0 .s/ is the unit tangent vector to the curve, which we denote by T.s/. Since T has constant length, T 0 .s/ will be orthogonal to T.s/. Assuming T 0 .s/ ¤ 0, define the principal normal vector N.s/ D T 0 .s/=kT 0 .s/k and the curvature .s/ D kT 0 .s/k. So far, we have T 0 .s/ D .s/N.s/:
$2.LOCAL THEORY:FRENET FRAME If(s)=0,the principal normal vector is not defined.Assuming,we continue.Define the binormal vector B(s)=T(s)xN(s).ThenT(s).N(s).B(s)form a right-handed orthonormal basis for3 Now,N'(s)must be a linear combination of T(s),N(s),and B(s).But we know from Lemma 2.1 that N'(s)-N(s)=0and N'(s).T(s)=-T'(s).N(s)=-K(s).We define the torsion t(s)=N'(s).B(s).This gives us N'(s)=-k(s)T(s)+t(s)B(s). Finally,B'(s)must be a linear combination of T(s),N(s),and B(s).Lemma2.1 tells us that B'(s)-B(s)=0, B'(s).T(s)=-T'(s).B(s)=0,and B'(s).N(s)=-N'(s).B(s)=-r(s).Thus, B(s)=-t(s)N(s). In summary,we have: Frenet formulas T'(s)= k(s)N(s) N'(s)=-k(s)T(s) +t(s)B() B'S)= -t(s)N(s) The skew-symmetry of these equations is made clearest when we state the Frenet formulas in matrix form: 10 0 T(s)N(s)B'(s) T(s)N(s)B(s) K(S)0 -t(S) 1110(s)0 Indeed,note that the coefficient matrix appearing on the right is skew-symmetric.This is the case whenever we differentiate an orthogonal matrix depending on a parameter(s in this case).(See Exercise A.1.4.) Note that,by definition,the curvature,is always nonnegative;the torsion,.however,has a sign,as we shall now see Example 1.Consider the helix,given by its arclength parametrization (see Exercise 1.1.2)a(s)= (a cos(s/c).a sin(s/c).bs/c),where c=a2+b2.Then we have Ts)-(-asin(s/e).acos(s/e).b) T(s)=c(-a cos(s/e).-a sin(s/e).)=(-cos(s/e).-sin(s/).0) N Summarizing. k0=是=2+F a and N(s)=(-cos(s/c).-sin(s/c).0). Now we deal with B and the torsion B(s)=T(s)x N(s)=(bsin(s/c).-bcos(s/c).a) )(bcots/e).bsin(s/e).))
÷2. LOCAL THEORY: FRENET FRAME 11 If .s/ D 0, the principal normal vector is not defined. Assuming ¤ 0, we continue. Define the binormal vector B.s/ D T.s/ N.s/. Then fT.s/; N.s/;B.s/g form a right-handed orthonormal basis for R 3 . Now, N 0 .s/ must be a linear combination of T.s/, N.s/, and B.s/. But we know from Lemma 2.1 that N 0 .s/ N.s/ D 0 and N 0 .s/T.s/ D T 0 .s/ N.s/ D .s/. We define the torsion .s/ D N 0 .s/B.s/. This gives us N 0 .s/ D .s/T.s/ C .s/B.s/: Finally, B 0 .s/ must be a linear combination of T.s/, N.s/, and B.s/. Lemma 2.1 tells us that B 0 .s/B.s/ D 0, B 0 .s/ T.s/ D T 0 .s/ B.s/ D 0, and B 0 .s/ N.s/ D N 0 .s/ B.s/ D .s/. Thus, B 0 .s/ D .s/N.s/: In summary, we have: Frenet formulas T 0 .s/ D .s/N.s/ N 0 .s/ D .s/T.s/ C .s/B.s/ B 0 .s/ D .s/N.s/ The skew-symmetry of these equations is made clearest when we state the Frenet formulas in matrix form: 2 6 4 j j j T 0 .s/ N 0 .s/ B 0 .s/ j j j 3 7 5 D 2 6 4 j j j T.s/ N.s/ B.s/ j j j 3 7 5 2 6 4 0 .s/ 0 .s/ 0 .s/ 0 .s/ 0 3 7 5 : Indeed, note that the coefficient matrix appearing on the right is skew-symmetric. This is the case whenever we differentiate an orthogonal matrix depending on a parameter (s in this case). (See Exercise A.1.4.) Note that, by definition, the curvature, , is always nonnegative; the torsion, , however, has a sign, as we shall now see. Example 1. Consider the helix, given by its arclength parametrization (see Exercise 1.1.2) ˛.s/ D a cos.s=c/; a sin.s=c/; bs=c , where c D p a 2 C b 2. Then we have T.s/ D 1 c a sin.s=c/; a cos.s=c/; b T 0 .s/ D 1 c 2 a cos.s=c/; a sin.s=c/; 0 D a c 2 „ƒ‚. cos.s=c/; sin.s=c/; 0 „ ƒ‚ . N : Summarizing, .s/ D a c 2 D a a 2 C b 2 and N.s/ D cos.s=c/; sin.s=c/; 0 : Now we deal with B and the torsion: B.s/ D T.s/ N.s/ D 1 c b sin.s=c/; b cos.s=c/; a B 0 .s/ D 1 c 2 b cos.s=c/; b sin.s=c/; 0 D b c 2 N.s/;
CHAPTER I.CURVES h b so we infer that t(s)= .h2 Note that both the curvature and the torsion are constants.The torsion is positive when the helix is "right-handed"(b>0)and negative when the helix is"left-handed"(b<0)It is interesting to observe that,.asb0,the helix becomes very tightly wound and almost planar,and0;as boo,the helix twists extremely slowly and looks more and more like a straight line on the cylinder and, once again,0.As the reader can check,the helix has the greatest torsion when b=a;why does this seem plausible? In Figure 2.I we show the Frenet frames of the helix at some sample points.(In the latter two pictures. FIGURE 2.1 the perspective is misleading.T.N.B still form a right-handed frame:In the third,T is in front of N,and in the last,B is pointing upwards and out of the page.) We stop for a moment to contemplate what happe ens with the frenet formulas when we are dealing with anon-arclength-parametrized,regular curve.As we did in Section 1,we can(theoretically)reparametrize by arclength,obtaining B(s).Then we have()=B(s(t)),so,by the chain rule, ( a'(t)=B'(s(r))s'(t)=v(r)T(s(r)) where v(t)=s'(t)is the speed.3 Similarly.by the chain rule.once we have the unit tangent vector as a function ofdifferentiating with respect to,we have (Tos)'(t)=T'(s(t))s'(t)=v(t)K(s())N(s()). Using the more casual-but convenient-Leibniz notation for derivatives, T or dt dvdn
12 CHAPTER 1. CURVES so we infer that .s/ D b c 2 D b a 2 C b 2 . Note that both the curvature and the torsion are constants. The torsion is positive when the helix is “right-handed” (b > 0) and negative when the helix is “left-handed” (b < 0). It is interesting to observe that, fixing a > 0, as b ! 0, the helix becomes very tightly wound and almost planar, and ! 0; as b ! 1, the helix twists extremely slowly and looks more and more like a straight line on the cylinder and, once again, ! 0. As the reader can check, the helix has the greatest torsion when b D a; why does this seem plausible? In Figure 2.1 we show the Frenet frames of the helix at some sample points. (In the latter two pictures, T N B T N B T N B T N B FIGURE 2.1 the perspective is misleading. T; N;B still form a right-handed frame: In the third, T is in front of N, and in the last, B is pointing upwards and out of the page.) O We stop for a moment to contemplate what happens with the Frenet formulas when we are dealing with a non-arclength-parametrized, regular curve ˛. As we did in Section 1, we can (theoretically) reparametrize by arclength, obtaining ˇ.s/. Then we have ˛.t / D ˇ.s.t //, so, by the chain rule, () ˛ 0 .t / D ˇ 0 .s.t //s0 .t / D .t /T.s.t //; where .t / D s 0 .t / is the speed.3 Similarly, by the chain rule, once we have the unit tangent vector as a function of t, differentiating with respect to t, we have .Tıs/0 .t / D T 0 .s.t //s0 .t / D .t /.s.t //N.s.t //: Using the more casual—but convenient—Leibniz notation for derivatives, dT dt D dT ds ds dt D N or N D dT ds D dT dt ds dt D 1 dT dt : 3 is the Greek letter upsilon, not to be confused with , the Greek letter nu.
$2.LOCAL THEORY:FRENET FRAME 3 Example 2.Let's calculate the curvature of the tractrix (see Example 2 in Section 1).Using the first parametrization,we have a'()=(-sin +csc 0,cos )and so v(e)=lla'(0)Il =(-sin +csc)2+cos2 0=csc20-1=-cot 0 (Note the negative sign becauseTherefore. 1 T)=-cd←sn+cc8.cos)=-lameotcos.cos)=（←cos.-sn8. Of course,looking at Figure 1.9.the formula for T should come as no surprise.Then,to find the curvature. we calculate -cot Since-tan>and (sin0.-cos)is a unit vector we conclude that x(0)=-tan0 and N(0)=(sin0,-cos0). Later on we will see an interesting geometric consequence of the equality of the curvature and the (absolute value of)the slope.V Example 3.Let's calculate the"Frenet apparatus"for the parametrized curve )=(31-13,32,31+13 We begin by calculatingand determining the unit tangent vector T and speed v: )=31-t2,21,1+t2).s0 v)=Ia)川=3V1-122+(2)2+(1+t22=3V2(1+t22=3V21+2and 11 T(t)= 方0-21+=方 Now N=r 1 dT 1 1+内a+网+号 1 -4121-2) 2t西立（品 2 N Here we have factored out the length of the derivative vector and left ourselves with a unit vector in it direction,which must be the principal normal N:the magnitude that is left must be the curvature K.In summary,so far we have 0=0+原时N0=(片o） 1
÷2. LOCAL THEORY: FRENET FRAME 13 Example 2. Let’s calculate the curvature of the tractrix (see Example 2 in Section 1). Using the first parametrization, we have ˛ 0 . / D . sin C csc ; cos /, and so . / D k˛ 0 . /k D q . sin C csc /2 C cos2 D p csc2 1 D cot : (Note the negative sign because 2 < .) Therefore, T. / D 1 cot . sin C csc ; cos / D tan .cot cos ; cos / D . cos ; sin /: Of course, looking at Figure 1.9, the formula for T should come as no surprise. Then, to find the curvature, we calculate N D dT ds D dT d ds d D .sin ; cos / cot D . tan /.sin ; cos /: Since tan > 0 and .sin ; cos / is a unit vector we conclude that . / D tan and N. / D .sin ; cos /: Later on we will see an interesting geometric consequence of the equality of the curvature and the (absolute value of) the slope. O Example 3. Let’s calculate the “Frenet apparatus” for the parametrized curve ˛.t / D .3t t 3 ; 3t2 ; 3t C t 3 /: We begin by calculating ˛ 0 and determining the unit tangent vector T and speed : ˛ 0 .t / D 3.1 t 2 ; 2t; 1 C t 2 /; so .t / D k˛ 0 .t /k D 3 q .1 t 2/ 2 C .2t /2 C .1 C t 2/ 2 D 3 q 2.1 C t 2/ 2 D 3 p 2.1 C t 2 / and T.t / D 1 p 2 1 1 C t 2 .1 t 2 ; 2t; 1 C t 2 / D 1 p 2 1 t 2 1 C t 2 ; 2t 1 C t 2 ; 1 : Now N D dT ds D dT dt ds dt D 1 .t / dT dt D 1 3 p 2.1 C t 2/ 1 p 2 4t .1 C t 2/ 2 ; 2.1 t 2 / .1 C t 2/ 2 ; 0 D 1 3 p 2.1 C t 2/ 1 p 2 2 1 C t 2 „ ƒ‚ . 2t 1 C t 2 ; 1 t 2 1 C t 2 ; 0 „ ƒ‚ . N : Here we have factored out the length of the derivative vector and left ourselves with a unit vector in its direction, which must be the principal normal N; the magnitude that is left must be the curvature . In summary, so far we have .t / D 1 3.1 C t 2/ 2 and N.t / D 2t 1 C t 2 ; 1 t 2 1 C t 2 ; 0 :