25.1. LINE INTEGRALS AND WORK 517
Example 25.1.4 Suppose for t ∈ [0,π] the position of an object is given by r (t) = ti+cos(2t)j+ sin(2t)k. Also suppose there is a force field defined on R3,F (x,y,z)≡ 2xyi+x2j + k. Find
∫C F · dR where C is the curve traced out by this object which has the
orientation determined by the direction of increasing t.
To find this line integral use the above definition and write∫CF ·dR=
∫π
0
(2t (cos(2t)) , t2,1
)· (1,−2sin(2t) ,2cos(2t)) dt
In evaluating this replace the x in the formula for F with t, the y in the formula for Fwith cos(2t) and the z in the formula for F with sin(2t) because these are the values ofthese variables which correspond to the value of t. Taking the dot product, this equals thefollowing integral. ∫
π
0
(2t cos2t −2(sin2t) t2 +2cos2t
)dt = π
2
Example 25.1.5 Let C denote the oriented curve obtained by r (t) =(t,sin t, t3
)where the
orientation is determined by increasing t for t ∈ [0,2]. Also let
F = (x,y,xz+ z)
Find∫
C F ·dR.
You use the definition.∫CF ·dR=
∫ 2
0
(t,sin(t) ,(t +1) t3) · (1,cos(t) ,3t2)dt
=∫ 2
0
(t + sin(t)cos(t)+3(t +1) t5
)dt =
125114
− 12
cos2 (2) .
Suppose you have a curve specified by r (s) = (x(s) ,y(s) ,z(s)) and it has the propertythat |r′ (s)| = 1 for all s ∈ [0,b]. Then the length of this curve for s between 0 and s1is∫ s1
0 |r′ (s)|ds =∫ s1
0 1ds = s1. This parameter is therefore called arc length because thelength of the curve up to s equals s. Now you can always change the parameter to be arclength.
Proposition 25.1.6 Suppose C is an oriented smooth curve parameterized by r (t) fort ∈ [a,b]. Then letting l denote the total length of C, there exists R(s), s ∈ [0, l] anotherparametrization for this curve which preserves the orientation and such that
∣∣R′ (s)∣∣ = 1
so that s is arc length.
Prove: Let φ (t)≡∫ t
a |r′ (τ)|dτ ≡ s. Then s is an increasing function of t because
dsdt
= φ′ (t) =
∣∣r′ (t)∣∣> 0.
Now define R(s)≡ r(φ−1 (s)
). Then
R′ (s) = r′(φ−1 (s)
)(φ−1)′ (s) = r′
(φ−1 (s)
)∣∣r′ (φ−1 (s))∣∣