14 CHAPTER 1.CURVES Next we find the binormal B by calculating the cross product 助=mx0=方（品高 And now,at long last,we calculate the torsion by differentiating B: -N== 1 dB 两(p)】 1 1 4t 1 211-t2 N s0t0)=k0=30+r 1 Now we see that curvature enters naturally when we compute the acceleration of a moving particle Differentiating the formula(*)on p.12,we obtain "(t)=v'(t)T(s())+v(t)T'(s())s'(t) =v'()T(st)+v)2(k(s(t)N(st)). Suppressing the variables for a moment,we can rewrite this equation as (*) a"=U'T+Kv2N The tangential component of acceleration is the derivative of speed;the normal component(the"centripetal acceleration"in the case of circular motion)is the product of the curvature of the path and the square of the speed.Thus,from the physics of the motion we can recover the curvature of the path Propositio.For ary regular parametrized curvehave 3 Proof.Sincea'×a"=(wT)×(U'T+kv2N)=kv3T×N and Kv3>0,we obtain Kv3=la'×a"l and so=la'x"/3,as desired. We next proceed to study various theoretical consequences of the Frenet formulas. Proposition 2.3.A space curve is a line if and only if its curvature is everywhere 0. Proof.The general line is given by (s)=sv+efor some unit vector v and constant vector e.Then a'(s)=T(s)=v is constant,so K=0.Conversely,if =0.then T(s)=To is a constant vector, and,integrating,we obtain (s)= Td().This is,once again,the parametrie equation of a line. Example 4.Suppose all the tangent lines of a space curve pass through a fixed point.What can we say about the curve Without loss of generality,we take the fixed point to be the origin and the curve to be
14 CHAPTER 1. CURVES Next we find the binormal B by calculating the cross product B.t / D T.t / N.t / D 1 p 2 1 t 2 1 C t 2 ; 2t 1 C t 2 ; 1 : And now, at long last, we calculate the torsion by differentiating B: N D dB ds D dB dt ds dt D 1 .t / dB dt D 1 3 p 2.1 C t 2/ 1 p 2 4t .1 C t 2/ 2 ; 2.t2 1/ .1 C t 2/ 2 ; 0 D 1 3.1 C t 2/ 2 „ ƒ‚ . 2t 1 C t 2 ; 1 t 2 1 C t 2 ; 0 „ ƒ‚ . N ; so .t / D .t / D 1 3.1 C t 2/ 2 . O Now we see that curvature enters naturally when we compute the acceleration of a moving particle. Differentiating the formula () on p. 12, we obtain ˛ 00.t / D 0 .t /T.s.t // C .t /T 0 .s.t //s0 .t / D 0 .t /T.s.t // C .t /2 .s.t //N.s.t // : Suppressing the variables for a moment, we can rewrite this equation as () ˛ 00 D 0T C 2N: The tangential component of acceleration is the derivative of speed; the normal component (the “centripetal acceleration” in the case of circular motion) is the product of the curvature of the path and the square of the speed. Thus, from the physics of the motion we can recover the curvature of the path: Proposition 2.2. For any regular parametrized curve ˛, we have D k˛ 0 ˛ 00k k˛0k 3 . Proof. Since ˛ 0 ˛ 00 D . T/. 0TC 2N/ D 3TN and 3 > 0, we obtain 3 D k˛ 0 ˛ 00k, and so D k˛ 0 ˛ 00k= 3 , as desired. We next proceed to study various theoretical consequences of the Frenet formulas. Proposition 2.3. A space curve is a line if and only if its curvature is everywhere 0. Proof. The general line is given by ˛.s/ D sv C c for some unit vector v and constant vector c. Then ˛ 0 .s/ D T.s/ D v is constant, so D 0. Conversely, if D 0, then T.s/ D T0 is a constant vector, and, integrating, we obtain ˛.s/ D Z s 0 T.u/du C ˛.0/ D sT0 C ˛.0/. This is, once again, the parametric equation of a line. Example 4. Suppose all the tangent lines of a space curve pass through a fixed point. What can we say about the curve? Without loss of generality, we take the fixed point to be the origin and the curve to be
$2.LOCAL THEORY:FRENET FRAME 6 arclength-parametrized by Then there is a scalar functionso that for everys we have(s)=(s)T(s). Differentiating.we have T(s)=a'(s)=λ'()T(s)+(s)T'(s)=2'(s)T(s)+A(s)k(s)Ns). Then ('(s)-1)T(s)+(s)(s)N(s)=0,so,since T(s)and N(s)are linearly independent,we infer that A(s)=s+c for some constant c and k(s)=0.Therefore.the curve must be a line through the fixed point Somewhat more challenging is the following Proposition 2.4.A space curve is planar if and only if its torsion is everywhere 0.The only planar curves with nonzero constant curvature are(portions of)circles. Proof.If a curve lies in a plane,then T(s)and N(s)span the plane Po parallel toand passing through the origin.Therefore.B Tx N is a constant vector(the normal to Po),and so B'=-N =0, from which we conclude that=0.Conversely,if=0.the binormal vector B is a constant vector Bo. Now,consider the function f(s)=a(s).Bo:we have f(s)=a'(s).Bo =T(s)-B(s)=0,and so f(s)=c for some constant c.This means that a lies in the plane x.Bo =c. We leave it to the reader to check in Exercise 2a.that a circle of radius a has constant curvature 1/a. (This can also be deduced as a special case of the calculation in Example 1.)Now suppose a planar curve has constant curvature Ko.Consider the auxiliary function B(s)=(s)+ -N(s).Then we have B'(s)= a'(s)+ (-Ko(s)T(s))=T(s)-T(s)=0.Therefore B is a constant function,say B(s)=P for all s. Now we ciaim that a is a(subset of a)circle centered at P.for l(s)-P=ll(s)-B(s)=1/ko. We have already seen that a circular helix has constant curvature and torsion.We leave it to the reader to check in Exercise 10 that these are the only curves with constant curvature and torsion.Somewhat more interesting are the curves for whichis a constant FIGURE2.2 A generalized helix is a space curve with0all of whose tangent vectors make a constant angle with a fixed direction.As shown in Figure 2.2,this curve lies on a generalized cylinder,formed by taking the
÷2. LOCAL THEORY: FRENET FRAME 15 arclength-parametrized by ˛. Then there is a scalar function so that for every s we have ˛.s/ D .s/T.s/. Differentiating, we have T.s/ D ˛ 0 .s/ D 0 .s/T.s/ C .s/T 0 .s/ D 0 .s/T.s/ C .s/.s/N.s/: Then .0 .s/ 1/T.s/ C .s/.s/N.s/ D 0, so, since T.s/ and N.s/ are linearly independent, we infer that .s/ D s C c for some constant c and .s/ D 0. Therefore, the curve must be a line through the fixed point. O Somewhat more challenging is the following Proposition 2.4. A space curve is planar if and only if its torsion is everywhere 0. The only planar curves with nonzero constant curvature are (portions of) circles. Proof. If a curve lies in a plane P, then T.s/ and N.s/ span the plane P0 parallel to P and passing through the origin. Therefore, B D T N is a constant vector (the normal to P0), and so B 0 D N D 0, from which we conclude that D 0. Conversely, if D 0, the binormal vector B is a constant vector B0. Now, consider the function f .s/ D ˛.s/ B0; we have f 0 .s/ D ˛ 0 .s/ B0 D T.s/ B.s/ D 0, and so f .s/ D c for some constant c. This means that ˛ lies in the plane x B0 D c. We leave it to the reader to check in Exercise 2a. that a circle of radius a has constant curvature 1=a. (This can also be deduced as a special case of the calculation in Example 1.) Now suppose a planar curve ˛ has constant curvature 0. Consider the auxiliary function ˇ.s/ D ˛.s/ C 1 0 N.s/. Then we have ˇ 0 .s/ D ˛ 0 .s/ C 1 0 .0.s/T.s// D T.s/ T.s/ D 0. Therefore ˇ is a constant function, say ˇ.s/ D P for all s. Now we claim that ˛ is a (subset of a) circle centered at P, for k˛.s/ Pk D k˛.s/ ˇ.s/k D 1=0. We have already seen that a circular helix has constant curvature and torsion. We leave it to the reader to check in Exercise 10 that these are the only curves with constant curvature and torsion. Somewhat more interesting are the curves for which = is a constant. FIGURE 2.2 A generalized helix is a space curve with ¤ 0 all of whose tangent vectors make a constant angle with a fixed direction. As shown in Figure 2.2, this curve lies on a generalized cylinder, formed by taking the
CHAPTER 1.CURVES union of the lines(rulings)in that fixed direction through each point of the curve.We can now characterize generalized helices by the following Proposition 25.A curve isa generalized helix if and only if/is constant. Proof.Suppose is an arclength-parametrized generalized helix.Then there is a(constant)unit vector A with the property that T.A =cos for some constant 0.Differentiating,we obtain KN.A =0,whence N.A=0.Differentiating yet again,we have () (-KT+tB)·A=0. Now,note that A lies in the plane spanned by T and B,and thus B.A =+sin0.Thus,we infer from cquation (that r/cot which is indeed constant. Conversely,if/k is constant,set/=cot0 for some angle (0.).Set A(s)=cos0T(s)+ sinB(s).Then A'(s)=(k cos0-r sin)N(s)=0,so A(s)is a constant unit vector A.and T(s).A= cos is constant,as desired. Example 5.In Example we saw a curve with=,so from the proof of Proposition 25 we see that the curve should make a constant angle =/4 with the vector A=-(T+B)=(0.0.1)(as should have been obvious from the formula for T alone).We verify this in Figure 2.3 by drawing along with the vertical cylinder built on the projection of onto the xy-plane.V FIGURE 2.3 The Frenet formulas actually characterize the local picture of a space curve. (Local canonical fomm).Letbe a smooth(orbetter)arclength-parametrized curve Ifa(0)=0,then for s near0,we have a⊙=气-3+T0+(2+会3+）No+(0+)o
16 CHAPTER 1. CURVES union of the lines (rulings) in that fixed direction through each point of the curve. We can now characterize generalized helices by the following Proposition 2.5. A curve is a generalized helix if and only if = is constant. Proof. Suppose ˛ is an arclength-parametrized generalized helix. Then there is a (constant) unit vector A with the property that T A D cos for some constant . Differentiating, we obtain N A D 0, whence N A D 0. Differentiating yet again, we have () .T C B/ A D 0: Now, note that A lies in the plane spanned by T and B, and thus B A D ˙ sin . Thus, we infer from equation () that = D ˙ cot , which is indeed constant. Conversely, if = is constant, set = D cot for some angle 2 .0; /. Set A.s/ D cos T.s/ C sin B.s/. Then A 0 .s/ D . cos sin /N.s/ D 0, so A.s/ is a constant unit vector A, and T.s/ A D cos is constant, as desired. Example 5. In Example 3 we saw a curve ˛ with D , so from the proof of Proposition 2.5 we see that the curve should make a constant angle D =4 with the vector A D 1 p 2 .T C B/ D .0; 0; 1/ (as should have been obvious from the formula for T alone). We verify this in Figure 2.3 by drawing ˛ along with the vertical cylinder built on the projection of ˛ onto the xy-plane. O FIGURE 2.3 The Frenet formulas actually characterize the local picture of a space curve. Proposition 2.6 (Local canonical form). Let ˛ be a smooth (C 3 or better) arclength-parametrized curve. If ˛.0/ D 0, then for s near 0, we have ˛.s/ D s 2 0 6 s 3 C : : :! T.0/ C 0 2 s 2 C 0 0 6 s 3 C : : : N.0/ C 0 0 6 s 3 C : : : B.0/:
$2.LOCAL THEORY:FRENET FRAME (Hereo.toand denote,respectively,the values of,andat and lim.3=0.) Proof.Using Taylor's Theorem,we write a0=a0+sa0+20+石a”0+ where lim./s3 =0.Now,()=0.()=T().and a"()=T(0)=KoN(0).Differentiating again,we have a"(0)=(KN)(0)=KoN(0)+Ko(-KoT(0)+to B(0)).Substituting.we obtain a0=sT0+2xoN0)+名3（←T0)+6N0+ooB0)+. =（-经，3+T0+(92+总3+）0+(03+）B0. as required.☐ We now introduce three fundamental planes at P=a(0): (i)the osculating plane,spanned by T(0)and N(0), (ii)the rectifving plane,spanned by T(0)and B(0),and (iii)the normal plane,spanned by N(0)and B(0). We see that,locally,the projections of into these respective planes look like (0(u,Ko/2)u2+(0/6u3+.) (间）(u,(Koto/u3+.),and (）(2.(2)）3+) where=0.Thus,the projections of into these planes look locally as shown in Figure 2.4. The osculating("kissing")plane is the plane that comes closest to containing near P(see also Exercise osculating plane rectifying plane normal plane FIGURE 2.4 23);the rectifying ("straightening")plane is the one that comes closest to flattening the curve near P:the normal plane is normal (perpendicular)to the curve at P.(Cf.Figure 1.3.) EXERCISES 1.2 1.Compute the curvature of the following arclength-parametrized curves:
÷2. LOCAL THEORY: FRENET FRAME 17 (Here 0, 0, and 0 0 denote, respectively, the values of , , and 0 at 0, and lim s!0 : : : =s3 D 0.) Proof. Using Taylor’s Theorem, we write ˛.s/ D ˛.0/ C s˛ 0 .0/ C 1 2 s 2˛ 00 .0/ C 1 6 s 3˛ 000.0/ C : : : ; where lim s!0 : : : =s3 D 0. Now, ˛.0/ D 0, ˛ 0 .0/ D T.0/, and ˛ 00.0/ D T 0 .0/ D 0N.0/. Differentiating again, we have ˛ 000.0/ D .N/ 0 .0/ D 0 0N.0/ C 0.0T.0/ C 0B.0//. Substituting, we obtain ˛.s/ D sT.0/ C 1 2 s 2 0N.0/ C 1 6 s 3 2 0T.0/ C 0 0N.0/ C 0 0B.0/ C : : : D s 2 0 6 s 3 C : : :! T.0/ C 0 2 s 2 C 0 0 6 s 3 C : : : N.0/ C 0 0 6 s 3 C : : : B.0/; as required. We now introduce three fundamental planes at P D ˛.0/: (i) the osculating plane, spanned by T.0/ and N.0/, (ii) the rectifying plane, spanned by T.0/ and B.0/, and (iii) the normal plane, spanned by N.0/ and B.0/. We see that, locally, the projections of ˛ into these respective planes look like (i) .u; .0=2/u2 C .0 0 =6/u3 C : : :/ (ii) .u; .0 0=6/u3 C : : :/, and (iii) .u2 ; p 2 0 3 p0 u 3 C : : :/, where lim u!0 : : : =u3 D 0. Thus, the projections of ˛ into these planes look locally as shown in Figure 2.4. The osculating (“kissing”) plane is the plane that comes closest to containing ˛ near P (see also Exercise osculating plane rectifying plane normal plane T T N N B B FIGURE 2.4 23); the rectifying (“straightening”) plane is the one that comes closest to flattening the curve near P; the normal plane is normal (perpendicular) to the curve at P. (Cf. Figure 1.3.) EXERCISES 1.2 1. Compute the curvature of the following arclength-parametrized curves:
CHAPTER 1.CURVES a.a(s)=(coss.coss.sins) b.x(s)=(个+s2n(s+√个+s2) 2.Calculate the unit tangent vector,principal normal,and curvature of the following curves: a a circle of radius a:(t)=(acost.a sint) b.a(t)=(t,cosht) c.a(t)=(cos3t,sin3t)t∈(0，π/2) 3.Calculate the Frenet a aratus (T,N,B,and)of the following curves b. a(t)=(ef (sint +cost).e!(sint -cost).et) a0)=(+z,ln+√+ d.a(t)=(e'cost,e'sint.e) e.a(t)=(cosht,sinht.t) Prove that the curvature ofthe plane curvey()is given by 1f" 5.Use sn of the tractrix given in Example recompute the curvature *6.By differentiating the equation B=Tx N,derive the equation B'=-rN. 7.Suppose is an arclength-parametrized space curve with the property that (s)()=Rfo all s sufficiently close to so.Prove that k(so)1/R.(Hint:Consider the function f(s)=lla(s)2 What do you know about f"(so)?) 8.Let be a regular(arclength-parametrized)curve with nonzero curvature.The normal line toat(s) is the line through (s)with direction vector N(s).Suppose all the normal lines to pass through a fixed point.What can you say about the curve? 9.a.Prove that if all the normal planes of a curve pass through a particular point,then the curve lies on a sphere.(Hint:Apply Lemma 2.1.) b.Prove that if all the osculating planes of a curve pass through a particular point,then the curve is planar. 10.Prove that if=Ko and=to are nonzero constants,then the curve is a(right)circular helix.(Hint The only solutions of the differential equation y"+k2y=0 are y=ci cos(kt)+c2 sin(kt).) Remark.It is an amusing exercise to give a and b(in our formula for the circular helix)in terms of Ko and to. 11.Proceed as in the derivation of Proposition 2.2 to show that t=(axam) '×"2
18 CHAPTER 1. CURVES a. ˛.s/ D p 1 2 cos s; p 1 2 cos s;sin s b. ˛.s/ D p 1 C s 2; ln.s C p 1 C s 2/ *c. ˛.s/ D 1 3 .1 C s/3=2 ; 1 3 .1 s/3=2 ; p 1 2 s , s 2 .1; 1/ 2. Calculate the unit tangent vector, principal normal, and curvature of the following curves: a. a circle of radius a: ˛.t / D .a cost; a sin t / b. ˛.t / D .t; cosh t / c. ˛.t / D .cos3 t;sin3 t /, t 2 .0; =2/ 3. Calculate the Frenet apparatus (T, , N, B, and ) of the following curves: *a. ˛.s/ D 1 3 .1 C s/3=2 ; 1 3 .1 s/3=2 ; p 1 2 s , s 2 .1; 1/ b. ˛.t / D 1 2 e t .sin t C cost /; 1 2 e t .sin t cost /; et *c. ˛.t / D p 1 C t 2; t; ln.t C p 1 C t 2/ d. ˛.t / D .et cost; et sin t; et / e. ˛.t / D .cosh t;sinh t; t / ]4. Prove that the curvature of the plane curve y D f .x/ is given by D jf 00j .1 C f 02/ 3=2 . ]*5. Use Proposition 2.2 and the second parametrization of the tractrix given in Example 2 of Section 1 to recompute the curvature. *6. By differentiating the equation B D T N, derive the equation B 0 D N. ]7. Suppose ˛ is an arclength-parametrized space curve with the property that k˛.s/k k˛.s0/k D R for all s sufficiently close to s0. Prove that .s0/ 1=R. (Hint: Consider the function f .s/ D k˛.s/k 2 . What do you know about f 00.s0/?) 8. Let ˛ be a regular (arclength-parametrized) curve with nonzero curvature. The normal line to ˛ at ˛.s/ is the line through ˛.s/ with direction vector N.s/. Suppose all the normal lines to ˛ pass through a fixed point. What can you say about the curve? 9. a. Prove that if all the normal planes of a curve pass through a particular point, then the curve lies on a sphere. (Hint: Apply Lemma 2.1.) *b. Prove that if all the osculating planes of a curve pass through a particular point, then the curve is planar. 10. Prove that if D 0 and D 0 are nonzero constants, then the curve is a (right) circular helix. (Hint: The only solutions of the differential equation y 00 C k 2y D 0 are y D c1 cos.kt / C c2 sin.kt /.) Remark. It is an amusing exercise to give a and b (in our formula for the circular helix) in terms of 0 and 0. *11. Proceed as in the derivation of Proposition 2.2 to show that D ˛ 0 .˛ 00 ˛ 000/ k˛0 ˛00k 2 :