Calculus of One and Several Variables

23.1 Line Integrals And Work

Let C be a smooth curve contained in ℝ^p. A curve C is an “oriented curve” if the only parameterizations considered are those which lie in exactly one of the two equivalence classes, each of which is called an “orientation”. In simple language, orientation specifies a direction over which motion along the curve is to take place. Thus, it specifies the order in which the points of C are encountered. The pair of concepts consisting of the set of points making up the curve along with a direction of motion along the curve is called an oriented curve.

Next the concept of work done by a force field F on an object as it moves along the curve C, in the direction determined by the given orientation of the curve will be defined. This is new. Earlier the work done by a force which acts on an object moving in a straight line was discussed but here the object moves over a curve. In order to define what is meant by the work, consider the following picture.

In this picture, the work done by a constant force F on an object which moves from the point x

to the point x along the straight line shown would equal F ⋅. It is reasonable to assume this would be a good approximation to the work done in moving along the curve joining x and x provided h is small enough. Also, provided h is small,

′ x(t+ h)− x(t) ≈ x (t)h

where the wriggly equal sign indicates the two quantities are close. In the notation of Leibniz, one writes dt for h and

dW = F (x (t))⋅x′(t)dt

or in other words,

dW--= F(x (t))⋅x′(t). dt

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

dW-- ′ dt = F (x (t))⋅x (t),W (a) = 0.

This motivates the following definition of work.

Proof: Suppose g :

→ C is another allowed parametrization. Thus g⁻¹ ∘f is an increasing function ϕ. Then since ϕ is increasing, it follows from the change of variables formula that

∫ d ∫ b F (g(s)) ⋅g′(s)ds = F(g (ϕ (t)))⋅g′(ϕ (t))ϕ′(t)dt c a

∫ b d-( ( − 1 )) ∫ b ′ = a F(f (t))⋅dt g g ∘f (t) dt = a F (f (t))⋅f (t)dt.■

Regardless 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 over the curve in the indicated direction is aided by the force. If the net effect of the force on the object is to impede rather than to aid the motion, this will show up as the work being negative.

Does the concept of work as defined here coincide with the earlier concept of work when 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 basic constructions. Math is not like religion which often abounds in apparent contradictions.

Let p and q be two points in ℝⁿ and suppose F is a constant force acting on an object which moves from p to q along the straight line joining these points. Then the work done is F ⋅

. Is the same thing obtained from the above definition? Let x ≡ p+t,t ∈ be a parametrization for this oriented curve, thestraight line in the direction from p to q. Then x^′ = q − p and F = F. Therefore, the above definition yields

∫ 1 F ⋅(q− p)dt = F⋅(q − p). 0

Therefore, the new definition adds to but does not contradict the old one. Therefore, it is not unreasonable to use this as the definition.

To find this line integral use the above definition and write

∫ ∫ π( ) F ⋅dR = 2t(cos(2t)),t2,1 ⋅(1,− 2sin(2t),2cos(2t))dt C 0

In evaluating this replace the x in the formula for F with t, the y in the formula for F with cos

and the z in the formula for F with sin because these are the values of these variables which correspond to the value of t. Taking the dot product, this equals the following integral.

∫ π (2tcos2t − 2 (sin2t)t2 + 2cos2t) dt = π2 0

You use the definition.

∫ ∫ 2( 3) ( 2) C F ⋅dR = 0 t,sin(t),(t+ 1)t ⋅ 1,cos(t),3t dt

∫ 2( 5) 1251 1 2 = 0 t+ sin(t)cos(t)+ 3(t+ 1)t dt = 14 − 2 cos (2).

Suppose you have a curve specified by r

= and it has the property that = 1 for all s ∈. Then the length of this curve for s between 0 and s₁ is ∫ ₀^s₁ds = ∫ ₀^s₁1ds = s₁. This parameter is therefore called arc length because the length of the curve up to s equals s. Now you can always change the parameter to be arc length.

Prove: Let ϕ

≡∫ _a^tdτ ≡ s. Then s is an increasing function of t because

ds -- = ϕ′(t) = |r′(t)| > 0. dt

Now define R

≡ r. Then

′ ′( −1 )( −1)′ r′(ϕ−1 (s)) R (s) = r ϕ (s) ϕ (s) = |r′(ϕ−1-(s))|

and so

= 1 as claimed. R = r = r = r and R = r = r and R delivers the same set of points in the same order as r because > 0. ■

The arc length parameter is just like any other parameter, in so far as considerations of line integrals are concerned, because it was shown above that line integrals are independent of parametrization. However, when things are defined in terms of the arc length parametrization, it is clear they depend only on geometric properties of the curve itself and for this reason, the arc length parametrization is important in differential geometry.