By Tony Bridgeman, P. C. Chatwin, C. Plumpton (auth.)
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Dr dt Thus = lim ((jr) 61-+0 (jt = lim (r(t cJt-+O + (jt) (jt - r(t)). 1) As :r is also, in general, a function of the scalar t, it may be differentiated . h respect t . t he second d· . 0 f r, d enoted by ddt2 r or r. WIt to t to gIve envatlve 2 Of special interest is the case when the scalar t is time. Then (jr is the displacement of P during the time interval (jt and the vector ~: is the average velocity (where average is with respect to time) during this interval. 3. Hence ifv denotes the velocity, v = :~ = t.
The velocity V has components VI' V2, V3 along the three coordinate axes and so P will move distances of VI t, V2t, V3t parallel to the X-, y-, z-axes respectively, in time t. Thus it has undergone a displacement equal to VIti + V2tj + V3tk = vt. Hence if time is measured from the instant when the point P is at A(a) (see Fig. 7), its position vector after time t is given by r, where r= a + vt. Of course, these results only apply when V is constant. We are now able to solve problems of the following type.
B = O. EX = 0, and state the geometrical meaning of this result. 7 (a) State the angle between two non-zero vectors a and b in each of the following cases. (i) 2a + 3b = 0; (ii) lal = Ibl = la + bl; (iii) a. (a + b) = 0 and Ibl = 21al. lop + Itq, where A. and It are constants. Given that p = 2i + 3j and q = 3i + 4j, find the numbers A. and It for the vector c which is of magnitude 10 units and makes the angle arc tan! with the positive Ox axis in the first quadrant. S A tetrahedron OABC with vertex 0 at the origin is such that OA = a, OB = band DC = c.
Vectors by Tony Bridgeman, P. C. Chatwin, C. Plumpton (auth.)