15.1. SPACE CURVES 277

Thus using 15.2 and this parametrization,

v×a=

∣∣∣∣∣∣∣i j k

−r sin t r cos t 0−r cos t −r sin t 0

∣∣∣∣∣∣∣= kr2

Thus

κ =r2

r3 =1r

Since v is constant, it follows from 15.1 that

a=1r|v|2 N =

1r

v2N

Example 15.1.4 Let R(t) =(cos(t) , t, t2

)for t ∈ [0,3]. Find the speed, velocity, curva-

ture, and write the acceleration in terms of normal and tangential components.

First of all, v (t) = (−sin t,1,2t) and so the speed is given by

|v|=√

sin2 (t)+1+4t2.

Therefore,

aT =ddt

(√sin2 (t)+1+4t2

)=

sin(t)cos(t)+4t√(2+4t2− cos2 t)

.

It remains to find aN . To do this, you can find the curvature first if you like.

a(t) =R′′ (t) = (−cos t,0,2) .

Then

κ =|(−cos t,0,2)× (−sin t,1,2t)|(√

sin2 (t)+1+4t2

)3 =

√4+(−2sin(t)+2(cos(t)) t)2 + cos2 (t)(√

sin2 (t)+1+4t2

)3

Then aN = κ |v|2

=

√4+(−2sin(t)+2(cos(t)) t)2 + cos2 (t)(√

sin2 (t)+1+4t2

)3

(sin2 (t)+1+4t2)

=

√4+(−2sin(t)+2(cos(t)) t)2 + cos2 (t)√

sin2 (t)+1+4t2.

You can observe the formula a2N +a2

T = |a|2 holds. Indeed a2N +a2

T =

√

4+(−2sin(t)+2(cos(t)) t)2 + cos2 (t)√sin2 (t)+1+4t2

2

+

(sin(t)cos(t)+4t√(2+4t2− cos2 t)

)2

=4+(−2sin t +2(cos t) t)2 + cos2 t

sin2 t +1+4t2+

(sin t cos t +4t)2

2+4t2− cos2 t= cos2 t +4 = |a|2