CHAPTER 1.CURVES 2π1 FIGURE1.6 Brief review.Just as the circle=1 is parametrized by (cos.sin),the portion of the hyperbolax2-y2=1 lying to the right of the y-axis,as shown in Figure 1.7,is parametrized by (cosht,sinht),where cosht=e +e- and 2 sinht =ef -e- 2 sinht By analogy with circular trigonometry,we set tanht= (cosh t.sinh) FIGURE1.7 formulas are easy to check cosh2 t-sinh2 t=1. tanh2 t +scch2t=I sinh'(r)=cosht cosh'(t)=sinht tanh'(t)=sech2t. sech'(t)=-tanht secht
4 CHAPTER 1. CURVES FIGURE 1.6 Brief review of hyperbolic trigonometric functions. Just as the circle x 2Cy 2 D 1 is parametrized by .cos ;sin /, the portion of the hyperbola x 2 y 2 D 1 lying to the right of the y-axis, as shown in Figure 1.7, is parametrized by .cosh t;sinh t /, where cosh t D e t C e t 2 and sinh t D e t e t 2 : By analogy with circular trigonometry, we set tanh t D sinh t cosh t and secht D 1 cosh t . The following (cosh t, sinh t) FIGURE 1.7 formulas are easy to check: cosh2 t sinh2 t D 1; tanh2 t C sech2 t D 1 sinh0 .t / D cosh t; cosh0 .t / D sinh t; tanh0 .t / D sech2 t; sech0 .t / D tanh t secht:
$1.EXAMPLES,ARCLENGTH PARAMETRIZATION 5 (h)When a uniform and flexible chain hangs from two its length.The shape it takes is called a catenary.As we ask the reader to check in Exercise 9. the catenary is the graph of f(x)=C cosh(x/C).for any constant C>0.This curve will appear FIGURE 1.8 numerous times in this course. Example 2.One of the more interesting curves that arises"in nature"is the traditional story is this:A dog is at the end of a l-unit leash and buries a bone at (0.1)as his owner begins to walk down the x-axis,starting at the origin.The dog tries to get back to the bone,so he always pulls the leash taut as he is dragged along the tractrix by his owner.His pulling the leash taut means that the leash will be tangent to the curve.When the master is at (1.).let the dog's position be (x(r).()).and let the leash 4(0.1) (x,y) FIgUre 1.9 make angle 0(t)with the positive x-axis.Then we havex(r)=t +cos(t),y(t)=sin(t),so m0=贵-0=品 cos8()8') Therefore,0'(r)=sin().Separating variables and integrating.we have fd0/sin=fdr,and so t =-In(csc 6 cot 0)+c for some constant c.Since =/2 when t =0,we see that c =0.Now, 2c0s2(8/2) since cse +cotcicot(2),we can rewrite this asIn tan(). sin Thus,we can parametrize the tractrix by x(0)=(cos6+In tan(0/2),sin8),π/2≤0<元
÷1. EXAMPLES, ARCLENGTH PARAMETRIZATION 5 (h) When a uniform and flexible chain hangs from two pegs, its weight is uniformly distributed along its length. The shape it takes is called a catenary. 1 As we ask the reader to check in Exercise 9, the catenary is the graph of f .x/ D C cosh.x=C /, for any constant C > 0. This curve will appear FIGURE 1.8 numerous times in this course. O Example 2. One of the more interesting curves that arises “in nature” is the tractrix. 2 The traditional story is this: A dog is at the end of a 1-unit leash and buries a bone at .0; 1/ as his owner begins to walk down the x-axis, starting at the origin. The dog tries to get back to the bone, so he always pulls the leash taut as he is dragged along the tractrix by his owner. His pulling the leash taut means that the leash will be tangent to the curve. When the master is at .t; 0/, let the dog’s position be .x.t /; y.t //, and let the leash FIGURE 1.9 make angle .t / with the positive x-axis. Then we have x.t / D t C cos .t /, y.t / D sin .t /, so tan .t / D dy dx D y 0 .t / x 0 .t / D cos .t /0 .t / 1 sin .t /0 .t /: Therefore, 0 .t / D sin .t /. Separating variables and integrating, we have R d= sin D R dt, and so t D ln.csc C cot / C c for some constant c. Since D =2 when t D 0, we see that c D 0. Now, since csc Ccot D 1 C cos sin D 2 cos2 .=2/ 2 sin.=2/ cos.=2/ D cot.=2/, we can rewrite this as t D ln tan.=2/. Thus, we can parametrize the tractrix by ˛. / D cos C ln tan.=2/;sin ; =2 < : 1From the Latin catena, chain. 2From the Latin trahere, tractus, to pull
6 CHAPTER 1.CURVES Alternatively,since tan(0/2)=e,we have 2e' 2 sin0 =2sin(@/2)cos(/2)=sech e-t-er 匹0=cos02-02片ah and so we can parametrize the tractrix instead by B()=(t-tanh.secht),t≥0.7 The fundamental concept underlying the geometry of curves is the arclength of a parametrized curve. Definition.If [a,b]R3 is a parametrized curve,then for any ab,we define its arclength from a to t to be s(t)= ()du.That is,the distance a particle travels-the arclength of its trajectory-is the integral of its speed. An alternative approach is to start with the following Definition.Let:[a.b]R3 be a(continuous)parametrized curve.Given a partition P=a=to< <.<t=b)of the interval [a.b],let (a,P)=Ia(G)-at-1). i=1 That is,is the length of the inscribed polygon with vertices at ),i=0.k,as indicated in Given this partition.of] the length of this polygonal path is (o.) FIGURE 1.10 Figure 1.10.We define the arclength of to be length()=sup(.):a partition of [a.] provided the set of polygonal lengths is bounded above. Now,using this definition,we can prove that the distance a particle travels is the integral of its speed. We will need to use the result of Exercise A.2.4
6 CHAPTER 1. CURVES Alternatively, since tan.=2/ D e t , we have sin D 2 sin.=2/ cos.=2/ D 2et 1 C e 2t D 2 e t C et D secht cos D cos2 .=2/ sin2 .=2/ D 1 e 2t 1 C e 2t D e t e t e t C et D tanh t; and so we can parametrize the tractrix instead by ˇ.t / D t tanh t;secht /; t 0: O The fundamental concept underlying the geometry of curves is the arclength of a parametrized curve. Definition. If ˛W Œa; b ! R 3 is a parametrized curve, then for any a t b, we define its arclength from a to t to be s.t / D Z t a k˛ 0 .u/kdu. That is, the distance a particle travels—the arclength of its trajectory—is the integral of its speed. An alternative approach is to start with the following Definition. Let ˛W Œa; b ! R 3 be a (continuous) parametrized curve. Given a partition P D fa D t0 < t1 < < tk D bg of the interval Œa; b, let `.˛;P/ D X k iD1 k˛.ti/ ˛.ti1/k: That is, `.˛;P/ is the length of the inscribed polygon with vertices at ˛.ti/, i D 0; : : : ; k, as indicated in a b FIGURE 1.10 Figure 1.10. We define the arclength of ˛ to be length.˛/ D supf`.˛;P/ W P a partition of Œa; bg; provided the set of polygonal lengths is bounded above. Now, using this definition, we can prove that the distance a particle travels is the integral of its speed. We will need to use the result of Exercise A.2.4
$1.EXAMPLES,ARCLENGTH PARAMETRIZATION Proposition 1.1.Leta:[a,b]R3 be a piecewise-l parametrized curve.Then length(a)=l()dt. Proof.For any partitionof [a,b],we have so length(a)≤ (ldt.The same holds on any interval. Now,for b,define s(r)to be the arclength of the curve a on the interval [a.].Then forh> we have h since s(+h)-s(r)is the arelength of the curve a on the interval [(See Exercise 8 for the first inequality.)Now Therefore,by the squeeze principle, 思0+》-0=w0m h A similar argument works for h,and we conclude thats)=().Therefore s)=la'(u)ldu,a≤tsb, and,in particular,s(b)=length(@)= 。le()ldi,as desired. We say the curve a is parametrized by arclength if (=1 for all t,so that s(t)=t-a.In this event,we usually use the parameter s and write (s). Example 3. (a)The standard parametrization of the circle of radius a is ( (a cost,a sint),t [0.2].so a'(t)=(-a sint.a cost)and llo'(t)ll a.It is easy to see from the chain rule that if we reparametrize the curve by B(s)=(a cos(s/a).a sin(s/a)),s [0.2ma], then B'(s)=(-sin(s/a).cos(s/a))and B'(s)=1 for all s.Thus,the curve B is parametrized by arclength. (b)Let a(s)=((1+s)3/2.(1-s)32.s),s e(-1.1).Then we have a'(s)=((1+)12. -l-sy/2,方》，and'sl=1 for all.s.Thus is parametried by arclength
÷1. EXAMPLES, ARCLENGTH PARAMETRIZATION 7 Proposition 1.1. Let ˛W Œa; b ! R 3 be a piecewise-C 1 parametrized curve. Then length.˛/ D Z b a k˛ 0 .t /kdt: Proof. For any partition P of Œa; b, we have `.˛;P/ D X k iD1 k˛.ti/ ˛.ti1/k D X k iD1 Z ti ti1 ˛ 0 .t /dt X k iD1 Z ti ti1 k˛ 0 .t /kdt D Z b a k˛ 0 .t /kdt; so length.˛/ Z b a k˛ 0 .t /kdt. The same holds on any interval. Now, for a t b, define s.t / to be the arclength of the curve ˛ on the interval Œa; t. Then for h > 0 we have k˛.t C h/ ˛.t /k h s.t C h/ s.t / h 1 h Z tCh t k˛ 0 .u/kdu; since s.t C h/ s.t / is the arclength of the curve ˛ on the interval Œt; t C h. (See Exercise 8 for the first inequality.) Now lim h!0C k˛.t C h/ ˛.t /k h D k˛ 0 .t /k D lim h!0C 1 h Z tCh t k˛ 0 .u/kdu: Therefore, by the squeeze principle, lim h!0C s.t C h/ s.t / h D k˛ 0 .t /k: A similar argument works for h < 0, and we conclude that s 0 .t / D k˛ 0 .t /k. Therefore, s.t / D Z t a k˛ 0 .u/kdu; a t b; and, in particular, s.b/ D length.˛/ D Z b a k˛ 0 .t /kdt, as desired. We say the curve ˛ is parametrized by arclength if k˛ 0 .t /k D 1 for all t, so that s.t / D t a. In this event, we usually use the parameter s and write ˛.s/. Example 3. (a) The standard parametrization of the circle of radius a is ˛.t / D .a cost; a sin t /, t 2 Œ0; 2, so ˛ 0 .t / D .a sin t; a cost / and k˛ 0 .t /k D a. It is easy to see from the chain rule that if we reparametrize the curve by ˇ.s/ D .a cos.s=a/; a sin.s=a//, s 2 Œ0; 2a, then ˇ 0 .s/ D . sin.s=a/; cos.s=a// and kˇ 0 .s/k D 1 for all s. Thus, the curve ˇ is parametrized by arclength. (b) Let ˛.s/ D 1 3 .1 C s/3=2 ; 1 3 .1 s/3=2 ; p 1 2 s , s 2 .1; 1/. Then we have ˛ 0 .s/ D 1 2 .1 C s/1=2 ; 1 2 .1 s/1=2 ; p 1 2 , and k˛ 0 .s/k D 1 for all s. Thus, ˛ is parametrized by arclength. O
CHAPTER 1.CURVES An important observation from a theoretical standpoint is that any regular parametrized curve can be reparametrized by arclength.For if is regular,the arclength functions()= l(u)ldu is an increas- ing function (since s'()=(0for all )and therefore has an inverse function=(s).Then we can consider the parametrization B(s)=a(t(s)). Note that the chain rule tells us that B'(s)=a(t(s))t'(s)=a'(t(s))/s'(t(5))=a'(t(s))/lla'(t(s))Il is everywhere a unit vector,in other words,B moves with speed 1. EXERCISES 1.1 *1.Parametrize the unit circle(less the point (-1.0))by the lengtht indicated in Figure 1.11. (-1,0の FIGURE 1.11 Consider the helix()=(acost.asint.br).Calculate).and reparametrize by ar- clength. 3.Let0）=(3os1+方sin.方eos.方cos-方sn).Calculateo,lal. by arclength *4.Parametrize the graph y=f(x),a sxsb,and show that its arclength is given by the traditional formula 5.a.Show that the arclength of the catenary)=(亿，cosh)for0≤t≤is sinh b. b.Reparametrize the catenary by arclength.(Hint:Find the inverse of sinh by using the quadratic formula.) 6.Consider the curve()=(e.e-)Calculate a(),(),and reparametrize by arclength, starting at =0
8 CHAPTER 1. CURVES An important observation from a theoretical standpoint is that any regular parametrized curve can be reparametrized by arclength. For if ˛ is regular, the arclength function s.t / D Z t a k˛ 0 .u/kdu is an increasing function (since s 0 .t / D k˛ 0 .t /k > 0 for all t), and therefore has an inverse function t D t .s/. Then we can consider the parametrization ˇ.s/ D ˛.t .s//: Note that the chain rule tells us that ˇ 0 .s/ D ˛ 0 .t .s//t0 .s/ D ˛ 0 .t .s//=s0 .t .s// D ˛ 0 .t .s//=k˛ 0 .t .s//k is everywhere a unit vector; in other words, ˇ moves with speed 1. EXERCISES 1.1 *1. Parametrize the unit circle (less the point .1; 0/) by the length t indicated in Figure 1.11. t (−1,0) (x,y) FIGURE 1.11 ]2. Consider the helix ˛.t / D .a cost; a sin t; bt /. Calculate ˛ 0 .t /, k˛ 0 .t /k, and reparametrize ˛ by arclength. 3. Let ˛.t / D p 1 3 costCp 1 2 sin t; p 1 3 cost; p 1 3 costp 1 2 sin t . Calculate ˛ 0 .t /, k˛ 0 .t /k, and reparametrize ˛ by arclength. *4. Parametrize the graph y D f .x/, a x b, and show that its arclength is given by the traditional formula length D Z b a q 1 C f 0 .x/2 dx: 5. a. Show that the arclength of the catenary ˛.t / D .t; cosh t / for 0 t b is sinh b. b. Reparametrize the catenary by arclength. (Hint: Find the inverse of sinh by using the quadratic formula.) *6. Consider the curve ˛.t / D .et ; et ; p 2t /. Calculate ˛ 0 .t /, k˛ 0 .t /k, and reparametrize ˛ by arclength, starting at t D 0