James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 16.3 The Fundamental Theorem for Line Integrals 1093
and comparison with (12) gives
t y sy, zd − e 3z
Thus tsy, zd − ye 3z 1 hszd and we rewrite (14) as
f sx, y, zd − xy 2 1 ye 3z 1 hszd
Finally, differentiating with respect to z and comparing with (13), we obtain h9szd − 0
and therefore hszd − K, a constant. The desired function is
f sx, y, zd − xy 2 1 ye 3z 1 K
It is easily verified that =f − F.
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Conservation of Energy
Let’s apply the ideas of this chapter to a continuous force field F that moves an object
along a path C given by rstd, a < t < b, where rsad − A is the initial point and rsbd − B
is the terminal point of C. According to Newton’s Second Law of Motion (see Section
13.4), the force Fsrstdd at a point on C is related to the acceleration astd − r0std by
the equation
Fsrstdd − mr0std
So the work done by the force on the object is
Therefore
W − y C
F dr − y b
− m 2 yb a
− m 2 yb a
a
Fsrstdd r9std dt − y b
mr0std r9std dt
a
d
fr9std r9stdg dt (Theorem 13.2.3, Formula 4)
dt
d
dt | r9std | 2 dt − m 2 f| r9std | 2 g a
b
− m 2 s | r9sbd| 2 2 | r9sad| 2 d
15 W − 1 2 m | vsbd | 2 2 1 2 m | vsad | 2
(Fundamental Theorem of Calculus)
where v − r9 is the velocity.
The quantity 1 2 m | vstd | 2 , that is, half the mass times the square of the speed, is called
the kinetic energy of the object. Therefore we can rewrite Equation 15 as
16
W − KsBd 2 KsAd
which says that the work done by the force field along C is equal to the change in kinetic
energy at the endpoints of C.
Now let’s further assume that F is a conservative force field; that is, we can write
F − =f . In physics, the potential energy of an object at the point sx, y, zd is defined as
Psx, y, zd − 2f sx, y, zd, so we have F − 2=P. Then by Theorem 2 we have
W − y C
F dr − 2y C
=P dr − 2fPsrsbdd 2 Psrsaddg − PsAd 2 PsBd
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