When the torsion of a curve is always 0, the curve should stay in a single plane

Question

When the torsion of a curve is always 0, the curve should stay in a single plane. Suppose that we have a curve vector r(t)=<f(t), g(t), h(t)> which is twice differentiable and, for all t, the torsion is 0. Show 1) that vector dB/ds is always parallel to vector B. 2) That vector dB/ds is 0 and therefore vector B is constant. 3) There is a single plane through the origin such that, for all t, vector T(t) and vector N(t) belong to this plane. 4) There is a single plane such that, for all t, vector r(t) lies in this plane.

Explanation / Answer

1 and 3)

Let r(t) = <f(t), g(t), h(t)> be a space curve. If T is the unit tangent and N is the principal unit normal, the unit vector B = T × N is called the binormal. Note that the binormal is orthogonal to both T and N. Let’s see about its derivative dB/ds with respect to arclength s.

First, note that B B = 1, and then differentiate the equation B T = 0 with respect to t to get (dB/dt) T + B (dT/dt) = 0 . Since dT/dt is a multiple of N and B is perpendicular to N, thus B (dT/dt) =0 , which means that being orthogonal to B, the derivative dB/ds is in the plane of T and N, thus (dB/dt) is always orthogonal to B (not parallel) and T and N.

2)

It is easy to see that if  vector dB/dt is 0 then B is constant since anti-derivative of zero is a constant.

thus (dB/dt) is always orthogonal to B (not parallel) and T and N.

4) Since dB/dt vector is zero for r(t) since it's zero for all f(t), g(t), h(t). Also since dT/dt is a multiple of N and dN/dt = dB/dt × T(t) we have that N is also constant for the curve along with T this lead to a planar curve as both normal and tangent vector are constant at any point for the whole curve.

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