Kenneth Kuttler

346 CHAPTER 16. SPACE CURVES

R(t)

Lemma 16.6.1 Define T (t)≡R′ (t)/∣∣R′ (t)

∣∣. Then |T (t)|= 1 and if T ′ (t) ̸= 0, thenthere exists a unit vector N (t) perpendicular to T (t) and a scalar valued function κ (t),with T ′ (t) = κ (t) |v|N (t).

Proof: It follows from the definition that |T | = 1. Therefore, T ·T = 1 and so, upondifferentiating both sides, T ′ ·T +T ·T ′ = 2T ′ ·T = 0. Therefore, T ′ is perpendicular toT . Let N (t)

∣∣T ′∣∣≡ T ′. Note that if∣∣T ′∣∣= 0, you could let N (t) be any unit vector. Then

letting κ (t) be defined such that∣∣T ′∣∣ ≡ κ (t) |v (t)|, it follows T ′ (t) =

∣∣T ′ (t)∣∣N (t) =

κ (t) |v (t)|N (t) .

Definition 16.6.2 The vector T (t) is called the unit tangent vector and the vectorN (t) is called the principal normal. The function κ (t) in the above lemma is called thecurvature. The radius of curvature is defined as ρ = 1/κ . The plane determined by thetwo vectors T and N in the case where T ′ ̸= 0 is called the osculating1 plane. It identifiesa particular plane which is in a sense tangent to this space curve.

The important thing about this is that it is possible to write the acceleration as the sumof two vectors, one perpendicular to the direction of motion and the other in the directionof motion.

Theorem 16.6.3 For R(t) the position vector of a space curve, the acceleration isgiven by the formula

a=d |v|dt

T +κ |v|2N ≡ aTT +aNN . (16.11)

Furthermore, a2T +a2

N = |a|2.

Proof: a= dvdt =

ddt (R

′)= ddt (|v|T )= d|v|

dt T + |v|T ′ = d|v|dt T + |v|2 κN. This proves

the first part.For the second part,

|a|2 = (aTT +aNN) · (aTT +aNN)

= a2TT ·T +2aNaTT ·N +a2

NN ·N = a2T +a2

because T ·N = 0.From 16.11 and the geometric properties of the cross product,

a×v = κ |v|2N ×v

1To osculate means to kiss. Thus this plane could be called the kissing plane. However, that does not soundformal enough so we call it the osculating plane.

346 CHAPTER 16. SPACE CURVESR(t)xLemma 16.6.1 Define T(t) = R(t) /|R' (t)|. Then |T (t)| = 1 and if T’ (t) £0, thenthere exists a unit vector N (t) perpendicular to T (t) and a scalar valued function « (t),with T' (t) = K(t) |v|_N (0).Proof: It follows from the definition that |T'| = 1. Therefore, T-T' = 1 and so, uponev OT both sides, T’-T+7T-T’ = 2T’ -T =0. Therefore, T’ is perpendicular toT. Let N (t)|T"| = 7". Note that if |T"| = i yon could let N (t) be uy unit vector. Thenletting K(t) be defined such that |Z’| = K(t)|v(t)|, it follows T" (t) = |T"(t)|.N (t) =K(t)|v(@)|N(). OfDefinition 16.6.2 The vector T (t) is called the unit tangent vector and the vectorN (t) is called the principal normal. The function « (t) in the above lemma is called thecurvature. The radius of curvature is defined as p = 1/«. The plane determined by thetwo vectors T and N in the case where T' £0 is called the osculating' plane. It identifiesa particular plane which is in a sense tangent to this space curve.The important thing about this is that it is possible to write the acceleration as the sumof two vectors, one perpendicular to the direction of motion and the other in the directionof motion.Theorem 16.6.3 for R(t) the position vector of a space curve, the acceleration isgiven by the formula_ dle!a IT +«|\v? N =arT +ayN. (16.11)Furthermore, a3. +>, = \a|*.Proof: a= @ = £(R’)= 4 (|v|T) = del p+ ly |T’ = del T+ ||? KN. This provesthe first part.For the second part,|a|? = (arpT +ayN) -(arT +ayN)=a;T -T +2ayarT-N +ayN-N =a; +aybecause T-N=0. JJFrom 16.11 and the geometric properties of the cross product,axv=klvlNxv'To osculate means to kiss. Thus this plane could be called the kissing plane. However, that does not soundformal enough so we call it the osculating plane.