Mathematical Methods for Physics and Engineering - Matematica.NET

VECTOR CALCULUSzCˆnPˆtˆbr(u)OyxFigure 10.3 The unit tangent ˆt, normal ˆn and binormal ˆb to the space curve Cat a particular point P .since the first term is zero by (10.10), and the second is zero because it is the vector productof two parallel (in this case identical) vectors. Integrating, we obtain the required resultr × dr = c, (10.11)dtwhere c is a constant vector.As a further point of interest we may note that in an infinitesimal time dt the changein the position vector of the small mass is dr and the element of area swept out by theposition vector of the particle is simply dA = 1 |r×dr|. Dividing both sides of this equation2by dt, we conclude thatdAdt = 1 2 ∣ r × drdt ∣ = |c|2 ,and that the physical interpretation of the above result (10.11) is that the position vector rof the small mass sweeps out equal areas in equal times. This result is in fact valid formotion under any force that acts along the line joining the two particles. ◭10.3 Space curvesIn the previous section we mentioned that the velocity vector of a particle is atangent to the curve in space along which the particle moves. We now give a morecomplete discussion of curves in space and also a discussion of the geometricalinterpretation of the vector derivative.AcurveC in space can be described by the vector r(u) joining the origin O ofa coordinate system to a point on the curve (see figure 10.3). As the parameter uvaries, the end-point of the vector moves along the curve. In Cartesian coordinates,r(u) =x(u)i + y(u)j + z(u)k,where x = x(u), y = y(u) andz = z(u) aretheparametric equations of the curve.340