DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

14 Chapter 1. Curvescurvature κ. Insummary, so far we haveκ(t) =13(1 + t 2 ) 2 and N(t) =(−2t1+t 2 , 1 − )t21+t 2 , 0 .Next we find the binormal B by calculating the cross productB(t) =T(t) × N(t) = √ 1 (− 1 − t22 1+t 2 , − 2t )1+t 2 , 1 .And now, at long last, we calculate the torsion by differentiating B:so τ(t) =κ(t) =−τN = dB dBds = dtds13(1 + t 2 ) 2 . ▽dt= 1 dBυ(t) dt()1 4t√2 (1 + t 2 ) 2 , 2(t2 − 1)(1 + t 2 ) 2 , 01=3 √ 2(1 + t 2 )(1= −3(1 + t 2 )} {{ 2 −2t1+t}2 , 1 − )t21+t 2 , 0 ,} {{ }τNNow we see that curvature enters naturally when we compute the acceleration of a movingparticle. Differentiating the formula (∗) onp.12, we obtainα ′′ (t) =υ ′ (t)T(s(t)) + υ(t)T ′ (s(t))s ′ (t)= υ ′ (t)T(s(t)) + υ(t) 2( κ(s(t))N(s(t)) ) .Suppressing the variables for a moment, we can rewrite this equation as(∗∗)α ′′ = υ ′ T + κυ 2 N.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 pathand the square of the speed. Thus, from the physics of the motion we can recover the curvature ofthe path:Proposition 2.2. Forany regular parametrized curve α, wehave κ = ‖α′ × α ′′ ‖‖α ′ ‖ 3 .Proof. Since α ′ × α ′′ =(υT) × (υ ′ T + κυ 2 N)=κυ 3 T × N and κυ 3 > 0, we obtain κυ 3 =‖α ′ × α ′′ ‖, and so κ = ‖α ′ × α ′′ ‖/υ 3 ,asdesired. □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 α(s) =sv + c for some unit vector v and constant vectorc. Then α ′ (s) =T(s) =v is constant, so κ =0. Conversely, if κ =0,then T(s) =T 0 is a constant∫ svector, and, integrating, we obtain α(s) = T(u)du = sT 0 + c for some constant vector c. This0is, once again, the parametric equation of a line. □