The figure shows a space curve. At a point of the curve we have drawn the unit tangent. T, the principal normal N, and the vector B = T times N, which,, being normal to both T and N, is called the binormal. At each point of the curve, the vectors T, N, B form what is called the Frenet frame a set of mutually perpendicular unit vectors that, in the order given, form a local right- handed coordinate system. Shpw that dB/ds is parallel to N, and therefore there is a scalar for which dB/ ds = N. HiNT: Since B has constant length one, dB/ds B. Show that d B / ds T by carrying out the differentiation dB/ds = d/ds (T times N). Now show that dN/ds = -kT - B. HINT: Since T, N, B form a right - handed system of mutually perpendicular unit vectors, we can show that N times B = T and B times T = N. You can assume these relations. The scalar is called the torsion of the curve. Give a geometric interpretation to | |.
Show transcribed image text The figure shows a space curve. At a point of the curve we have drawn the unit tangent. T, the principal normal N, and the vector B = T times N, which,, being normal to both T and N, is called the binormal. At each point of the curve, the vectors T, N, B form what is called the Frenet frame a set of mutually perpendicular unit vectors that, in the order given, form a local right- handed coordinate system. Shpw that dB/ds is parallel to N, and therefore there is a scalar for which dB/ ds = N. HiNT: Since B has constant length one, dB/ds B. Show that d B / ds T by carrying out the differentiation dB/ds = d/ds (T times N). Now show that dN/ds = -kT - B. HINT: Since T, N, B form a right - handed system of mutually perpendicular unit vectors, we can show that N times B = T and B times T = N. You can assume these relations. The scalar is called the torsion of the curve. Give a geometric interpretation to | |.