15.2. GEOMETRY OF SPACE CURVES∗ 281

Theorem 15.2.4 (Serret Frenet) Let R(s) be the parametrization with respect to arc lengthof a space curve and T (s) =R′ (s) is the unit tangent vector. Suppose

∣∣T ′ (s)∣∣ ̸= 0 so the

principal normal N (s) = T ′(s)|T ′(s)| is defined. The binormal is the vector B ≡ T ×N so

T,N,B forms a right handed system of unit vectors each of which is perpendicular toevery other. Then the following system of differential equations holds in R9.

B′ = τN, T ′ = κN, N ′ =−κT − τB

where κ is the curvature and is nonnegative and τ is the torsion.

Proof: κ ≥ 0 because κ =∣∣T ′ (s)∣∣. The first two equations are already established.

To get the third, note that B×T =N which follows because T,N,B is given to form aright handed system of unit vectors each perpendicular to the others. (Use your right hand.)Now take the derivative of this expression. thus

N ′ =B′×T +B×T ′ = τ N ×T+κB×N.

Now recall again that T,N,B is a right hand system. Thus

N ×T =−B, B×N =−T.

This establishes the Frenet Serret formulas. ■This is an important example of a system of differential equations in R9. It is a re-

markable result because it says that from knowledge of the two scalar functions τ and κ ,and initial values for B,T, and N when s = 0 you can obtain the binormal, unit tangent,and principal normal vectors. It is just the solution of an initial value problem althoughthis is for a vector valued rather than scalar valued function. Having done this, you canreconstruct the entire space curve starting at some point R0 because R′ (s) = T (s) and soR(s) =R0 +

∫ s0 T (r) dr.

The vectors B,T, and N are vectors which are functions of position on the space curve.Often, especially in applications, you deal with a space curve which is parameterized by afunction of t where t is time. Thus a value of t would correspond to a point on this curve andyou could let B (t) ,T (t) , and N (t) be the binormal, unit tangent, and principal normal atthis point of the curve. The following example is typical.

Example 15.2.5 Given the circular helix, R(t) = (acos t)i+(asin t)j+(bt)k, find thearc length s(t), the unit tangent vector T (t), the principal normal N (t) , the binormalB (t), the curvature κ (t), and the torsion, τ (t). Here t ∈ [0,T ].

The arc length is s(t) =∫ t

0

(√a2 +b2

)dr =

(√a2 +b2

)t. Now the tangent vector is

obtained using the chain rule as

T =dRds

=dRdt

dtds

=1√

a2 +b2R′ (t) =

1√a2 +b2

((−asin t)i+(acos t)j+bk)

The principal normal:

dTds

=dTdt

dtds

=1

a2 +b2 ((−acos t)i+(−asin t)j+0k)