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regular curve with speed function, ity and acceleration of a are given by (Fig. 2.15.), , 4.2 Lemma, , If a is, , a, , velocv, then the, , a'= VT, a"=:dy-T+ wN., dt, , Since a, a[s), where s is the, Lemma 4.5 of Chapter 1, that, , Proof., , using, , =, , arc, , length function of, , a,, , we, , find,, , ds, , vT(s)= vT., a'='(s)=, dt, Then a second differentiation yields, , aT+Tr-T+niN,, dt, dt, where, , we use, , Lemma 4.1., , vT is to be expected since a and T are each tangent, The formula a', v. The formula for, while || | |, to the curve and T has a unit length,, &is the rate of change of the, acceleration is more interesting. By definition,, =, , =, , a =v?T, , T, , Kv'N, FIG. 2.15
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velocity a', and in general both the length and the direction of a' are changing. The tangential component (dvldt)T of a" measures the rate of change of, the length of a' (that is, of the speed of a). The normal component Kv'N measures the rate of change of the direction of a'. Newton's laws of motion show, that these components may be experienced as forces. For example, in a car, that is specding up or slowing down on a straight road, the only force one, feels is due to (dv/dt) T. If one takes an unbanked curve al speed v, the result, , ing sideways force is due to av'N. Here *'measures how sharply the road turns;, the effect of speed is given by r, so 60 miles per hour is four times as unset, , tling as 30., We now find etlectively computable expressions for the Frenet apparatus., , 4.3 Theorem, , Let abe a regular curve in R'. Then, , T=a'la'|., N, , =, , Bx, , * =la'xa" |/|«'|., , T, , B=a'xa"|a'xt" |. r=(a'xa").a"/|a'xa"I, Proof., a, , Since v = | a' ||> 0, the formula T = a'l|a"| is equivalent to, , = vT. From the preceding lemma we get, , a'xa" =(vT)xT+*N, =, , N =Kv°'B,, TxT+Kv'Tx, dt, , since Tx T = 0. Taking norms we find, , a'xa"|=| xv'B||=v, because| | B| = 1, *, 0, and v> 0. Indeed, this equation shows that for, regular curves, || a x a | > 0 is equivalent to the usual condition K> 0., , (Thus for x>0, @' and a" are linearly independent and determine the osculating plane at each point, as do Tand N.) Then, B, , Xa", , x, , a", , a'xa" l, Since N= B x Tis true for any Frenet frame field (Exercise 4 of Section, 3), only the formula for torsion remains to be proved., To find the dot product (ax a)° o" we express, everything in terms of, , T, N, B. We already know that o, , x a" = v°'B. Thus, since 0, , N B, we need only find the B component of a". But, , =T, , B=
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24, , T+v'N, , =, , = A 'tB+*'., we, , use Lemma, , a'xa"|, , ple, , =, , A', we have the required, , Example, , W'+, , a", (a'x a")*, , formula, , * ' v , and, , =, , find, , (a, , x, , Exercise, , But, , a") " ., , Frenet apparatus, We compute the, , at), , of the, , 3-curve, , (31 -t', 3t2, 3t +'), , =, , The derivatives are, , a 0), , =, , 3(1 -,21, 1+)., , a "t)= 6(-t, 1, /),, a ) =6(-1,0, 1), Now,, , a')*a') = 18(1 +2+r'),, sO, , vt)=|a0)|=V18(1+1), Applying the definition of cross product yields, , U, , a'0)xa")=181- 2, , Us, , 1+=18(-1 +, -2, 1+1), , Dotting this vector with itself, we get, , (18(-1+) +4t+(1+|=2(18(1 +t., Hence, , at)x a"t)I=18/2(1 +), The, , expressions above for ar x, , a" and o", , yield, axa") a" =6 18 2., , since, , 1ort, , could (by, formula for T, anyway,, we need a i a, , this, scalar product in, a"xa", be written a', , ection I)also, more efficient to, 4.4, , 4.1., , Conscquently,, , 73, , Arbitrary-Speed Curves, , 4 of, , so Tt
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74, , 2., , It remains, , N, , Frame Fields, , only, , substitute this data into the formulas in Theorem 4.3, with, being computed by another cross product. The final results are, to, , T, , 2, 1+2, , ,, , N2(1+1), N, , 2,1-t2,0), 1+t2, , t,, , -21, 1+12), , 21+12), K= =, , 31+
