350 CHAPTER 16. SPACE CURVES
perpendicular to B also. Therefore, B′ is a vector which is perpendicular to both vectorsT and B and since this is in three dimensions, B′ must be some scalar multiple of N , andthis multiple is called τ . Thus B′ = τN as claimed.
Lets go over this last claim a little more. The following situation is obtained. Thereare two vectors T and B which are perpendicular to each other and both B′ and N areperpendicular to these two vectors, hence perpendicular to the plane determined by them.Therefore, B′ must be a multiple of N. Take a piece of paper, draw two unit vectors on itwhich are perpendicular. Then you can see that any two vectors which are perpendicular tothis plane must be multiples of each other.
The scalar function τ is called the torsion. In case T ′ = 0, none of this is definedbecause in this case there is not a well defined osculating plane. The conclusion of thefollowing theorem is called the Serret Frenet formulas.
Theorem 16.7.4 (Serret Frenet) Let R(s) be the parametrization with respect toarc length of 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. There are ways to solve such a system of equations numerically
and even draw the graph of the resulting curve but this is not a topic for this book.
16.8 Exercises1. Find a parametrization for the intersection of the planes 2x+ y+ 3z = −2 and 3x−
2y+ z =−4.
2. Find a parametrization for the intersection of the plane 3x + y + z = −3 and thecircular cylinder x2 + y2 = 1.