Kenneth Kuttler

516 CHAPTER 25. LINE INTEGRALS

or in other words,dWdt

= F (x(t)) ·x′ (t) .

Defining the total work done by the force at t = 0, corresponding to the first endpoint ofthe curve, to equal zero, the work would satisfy the following initial value problem.

dWdt

= F (x(t)) ·x′ (t) , W (a) = 0.

This motivates the following definition of work.

Definition 25.1.2 Let F (x) be given above. Then the work done by this forcefield on an object moving over the curve C in the direction determined by the specifiedorientation is defined as ∫

CF ·d R≡

∫ b

aF (x(t)) ·x′ (t) dt

where the function x is one of the allowed parameterizations of C in the given orientationof C. In other words, there is an interval [a,b] and as t goes from a to b, x(t) moves in thedirection determined from the given orientation of the curve.

Theorem 25.1.3 The symbol∫

C F ·dR, is well defined in the sense that every par-ametrization in the given orientation of C gives the same value for

∫C F ·dR.

Proof: Suppose g : [c,d]→ C is another allowed parametrization. Thus g−1 ◦f is anincreasing function φ . Then since φ is increasing, it follows from the change of variablesformula that ∫ d

cF (g (s)) ·g′ (s) ds =

∫ b

aF (g (φ (t))) ·g′ (φ (t))φ

′ (t) dt

=∫ b

aF (f (t)) · d

(g(g−1 ◦f (t)

))dt =

∫ b

aF (f (t)) ·f ′ (t) dt.

Regardless the physical interpretation of F, this is called the line integral. When Fis interpreted as a force, the line integral measures the extent to which the motion overthe curve in the indicated direction is aided by the force. If the net effect of the force onthe object is to impede rather than to aid the motion, this will show up as the work beingnegative.

Does the concept of work as defined here coincide with the earlier concept of workwhen the object moves over a straight line when acted on by a constant force? If it doesn’t,then the above is not a good definition because it will contradict earlier and more basicconstructions. Math is not like sectarian religions which are typically replete with incon-sistencies and blatant contradictions.

Let p and q be two points in Rn and suppose F is a constant force acting on anobject which moves from p to q along the straight line joining these points. Then thework done is F · (q−p). Is the same thing obtained from the above definition? Letx(t) ≡ p+t (q−p) , t ∈ [0,1] be a parametrization for this oriented curve, thestraight linein the direction from p to q. Then x′ (t) = q−p and F (x(t)) = F. Therefore, the abovedefinition yields

∫ 10 F · (q−p) dt = F · (q−p) . Therefore, the new definition adds to but

does not contradict the old one. Therefore, it is not unreasonable to use this as the defini-tion.

516 CHAPTER 25. LINE INTEGRALSor in other words,dw— =F (a(t))-a'(t).5 7 P(@()-#"(tDefining the total work done by the force at t = 0, corresponding to the first endpoint ofthe curve, to equal zero, the work would satisfy the following initial value problem.cM = F(@(1))-2! (1), W(a) =0.This motivates the following definition of work.iti el. et ax) be given above. en the work done this forceDefinition 25.1.2 Let F(x) be g b Then th k done by thfield on an object moving over the curve C in the direction determined by the specifiedorientation is defined as[vars [FP eo)-2oawhere the function x is one of the allowed parameterizations of C in the given orientationof C. In other words, there is an interval [a,b] and as t goes from a to b, x(t) moves in thedirection determined from the given orientation of the curve.Theorem 25.1.3 The symbol {- F -dR, is well defined in the sense that every par-ametrization in the given orientation of C gives the same value for J. F -dR.Proof: Suppose g : [c,d] — C is another allowed parametrization. Thus g ‘of isanincreasing function @. Then since @ is increasing, it follows from the change of variablesformula thatd b[ F909 as= | F@OO)-s OO) Oab d b=[ FPO) Soot) a= [ PFO) Fd. IRegardless the physical interpretation of F, this is called the line integral. When F'is interpreted as a force, the line integral measures the extent to which the motion overthe curve in the indicated direction is aided by the force. If the net effect of the force onthe object is to impede rather than to aid the motion, this will show up as the work beingnegative.Does the concept of work as defined here coincide with the earlier concept of workwhen the object moves over a straight line when acted on by a constant force? If it doesn’t,then the above is not a good definition because it will contradict earlier and more basicconstructions. Math is not like sectarian religions which are typically replete with incon-sistencies and blatant contradictions.Let p and q be two points in R” and suppose F' is a constant force acting on anobject which moves from p to q along the straight line joining these points. Then thework done is F'-(q—p). Is the same thing obtained from the above definition? Letx(t) = p+t(q—p),t € [0,1] be a parametrization for this oriented curve, thestraight linein the direction from p to q. Then a’ (t) = q—p and F(a (t)) = F. Therefore, the abovedefinition yields fo F'-(q—p) dt =F -(q—p). Therefore, the new definition adds to butdoes not contradict the old one. Therefore, it is not unreasonable to use this as the defini-tion.