Kenneth Kuttler

16.7. GEOMETRY OF SPACE CURVES∗ 349

16.7 Geometry of Space Curves∗

If you are interested in more on space curves, you should read this section. Otherwise,proceed to the exercises. Denote by R(s) the function which takes s to a point on this curvewhere s is arc length. Thus R(s) equals the point on the curve which occurs when you havetraveled a distance of s along the curve from one end. This is known as the parametrizationof the curve in terms of arc length. Note also that it incorporates an orientation on the curvebecause there are exactly two ends you could begin measuring length from. In this section,assume anything about smoothness and continuity to make the following manipulationsvalid. In particular, assume that R′ exists and is continuous.

Lemma 16.7.1 Define T (s) ≡ R′ (s). Then |T (s)| = 1 and if T ′ (s) ̸= 0, then thereexists a unit vector N (s) perpendicular to T (s) and a scalar valued function κ (s) withT ′ (s) = κ (s)N (s).

Proof: First, s =∫ s

∣∣R′ (r)∣∣ dr because of the definition of arc length. Therefore, from

the fundamental theorem of calculus, 1 =∣∣R′ (s)

∣∣ = |T (s)|. Therefore, T ·T = 1 and soupon differentiating this on both sides, yields T ′ ·T +T ·T ′ = 0 which shows T ·T ′ = 0.Therefore, the vector T ′ is perpendicular to the vector T . In case T ′ (s) ̸= 0, let N (s) =T ′(s)|T ′(s)| and so T ′ (s) =

∣∣T ′ (s)∣∣N (s), showing the scalar valued function is κ (s) =

∣∣T ′ (s)∣∣.

The radius of curvature is defined as ρ = 1κ

. Thus at points where there is a lot ofcurvature, the radius of curvature is small and at points where the curvature is small, theradius of curvature is large. The plane determined by the two vectors T and N is calledthe osculating plane. It identifies a particular plane which is in a sense tangent to this spacecurve. In the case where

∣∣T ′ (s)∣∣= 0 near the point of interest, T (s) equals a constant and

so the space curve is a straight line which it would be supposed has no curvature. Also, theprincipal normal is undefined in this case. This makes sense because if there is no curvinggoing on, there is no special direction normal to the curve at such points which could bedistinguished from any other direction normal to the curve. In the case where

∣∣T ′ (s)∣∣= 0,

κ (s) = 0 and the radius of curvature would be considered infinite.

Definition 16.7.2 The vector T (s) is called the unit tangent vector and the vectorN (s) is called the principal normal. The function κ (s) in the above lemma is calledthe curvature. When T ′ (s) ̸= 0 so the principal normal is defined, the vector B (s) ≡T (s)×N (s) is called the binormal.

The binormal is normal to the osculating plane and B′ tells how fast this vector changes.Thus it measures the rate at which the curve twists.

Lemma 16.7.3 Let R(s) be a parametrization of a space curve with respect to arclength and let the vectors T,N, and B be as defined above. Then B′ = T ×N ′ and thereexists a scalar function τ (s) such that B′ = τN.

Proof: From the definition of B = T ×N, and you can differentiate both sides and getB′ =T ′×N +T ×N ′. Now recall that T ′ is a multiple called curvature multiplied by Nso the vectors T ′ and N have the same direction, so B′ = T ×N ′. Therefore, B′ is eitherzero or is perpendicular to T. But also, from the definition of B,B is a unit vector and soB (s) ·B (s) = 1. Differentiating this, B′ (s) ·B (s)+B (s) ·B′ (s) = 0 showing that B′ is

16.7. GEOMETRY OF SPACE CURVES* 34916.7 Geometry of Space Curves*If you are interested in more on space curves, you should read this section. Otherwise,proceed to the exercises. Denote by R(s) the function which takes s to a point on this curvewhere s is arc length. Thus R(s) equals the point on the curve which occurs when you havetraveled a distance of s along the curve from one end. This is known as the parametrizationof the curve in terms of arc length. Note also that it incorporates an orientation on the curvebecause there are exactly two ends you could begin measuring length from. In this section,assume anything about smoothness and continuity to make the following manipulationsvalid. In particular, assume that R’ exists and is continuous.Lemma 16.7.1 Define T (s) = R'(s). Then \T(s)| = 1 and if T'(s) #0, then thereexists a unit vector IN (s) perpendicular to T'(s) and a scalar valued function K(s) withT’ (s) = k(s) N(s).Proof: First, s = {j |’ (r)| dr because of the definition of arc length. Therefore, fromthe fundamental theorem of calculus, 1 = |’ (s)| = |T'(s)|. Therefore, T'-T' = 1 and soupon differentiating this on both sides, yields T’-T + T'-T’ = 0 which shows T'- T’ = 0.Therefore, the vector T’ is perpendicular to the vector T’. In case T’ (s) 4 0, let N(s) =oa and so I” (s) = |T" (s)| N (s), showing the scalar valued function is « (s) = |" (s)].|The radius of curvature is defined as p = i, Thus at points where there is a lot ofcurvature, the radius of curvature is small and at points where the curvature is small, theradius of curvature is large. The plane determined by the two vectors J’ and WN is calledthe osculating plane. It identifies a particular plane which is in a sense tangent to this spacecurve. In the case where |r" (s)| = 0 near the point of interest, T (s) equals a constant andso the space curve is a straight line which it would be supposed has no curvature. Also, theprincipal normal is undefined in this case. This makes sense because if there is no curvinggoing on, there is no special direction normal to the curve at such points which could bedistinguished from any other direction normal to the curve. In the case where |T" (s)| =0,« (s) = 0 and the radius of curvature would be considered infinite.Definition 16.7.2 The vector T (s) is called the unit tangent vector and the vectorN (s) is called the principal normal. The function «K(s) in the above lemma is calledthe curvature. When T'(s) 4 0 so the principal normal is defined, the vector B(s) =T (s) x N (s) is called the binormal.The binormal is normal to the osculating plane and B’ tells how fast this vector changes.Thus it measures the rate at which the curve twists.Lemma 16.7.3 Let R(s) be a parametrization of a space curve with respect to arclength and let the vectors T, N, and B be as defined above. Then B’ = T x N' and thereexists a scalar function t(s) such that B'= tN.Proof: From the definition of B = T x N, and you can differentiate both sides and getB'=T'xN+T x N’. Nowrecall that T” is a multiple called curvature multiplied by Nso the vectors T” and N have the same direction, so B’ = T x N’. Therefore, B’ is eitherzero or is perpendicular to T’. But also, from the definition of B, B is a unit vector and soB(s)-B(s) =1. Differentiating this, B’ (s)- B(s)+ B(s)-B’(s) =0 showing that B’ is