Calculus: Analytic Geometry and Calculus, with Vectors

(^258) Integrals
From this we conclude that there must be a constant vector cl such that
(2)
mxi = TCt2i.
But (dx/dt)i is the velocity v at time t, and putting t = 0 in (2) shows that
c1 = 0. Therefore,
(3)
KE=Cx=2m(Ct)2= -njv12.
From this we conclude that there is a constant vector C2 such that
(4) mxi = Ct2i + c2.
But x = 0 when t = 0, so c2 = 0. Therefore,
(5)
The kinetic energy KE of our particle at time t is defined to be the amount of
work done by the force F in bringing the particle from its state of rest at time
t = 0 to its state of motion at time t. Since IFI has the constant magnitude C
and has the direction of motion of the particle as the particle moves the distance
x, the amount of work done is Cx. Thus KE = Cx and, with the aid of (5) and
(3), we find that
(6)
Therefore,
(7) KE = "ffmIvl2.

m d i=Cti+c1.

MV = mdii = Cii.

The next problem requires that the same result be worked out by a different

method without the assumption that F is a constant.

13 A particle P of mass m is moved around in E3 by a continuous net force F

which operates over a time interval a 5 t < b. The Newton law F = ma then

0

Figure 4.694

tk_1 and tk. Tell why the scalars

(1) r(tk-1)] and L1tk

should, when IPI is small, both be good approximations

to the work done by F in forcing P from Pk_1 to Pk. Tell why it should be

reasonable to adopt either one of the formulas

(2) W = lim ' r(tk_l)]

IPI-o k=1

(3) W = lim L1tk

IPI- O k.1

shows that the displacement vector r (which is OP),

the velocity vector v, and the acceleration vector a

are continuous functions of t. Make a partition of

the interval a < t < b and look at Figure 4.694 which

k-1 shows, among other things, the positions of P at times