Mathematical Tools for Physics

13—Vector Calculus 2 397

Thendx=f ̇(t)dtanddy=g ̇(t)dt, so

ds=

√(

f ̇(t)dt

) 2

+

(

g ̇(t)dt

) 2

=

√

f ̇(t)^2 +g ̇(t)^2 dt

and the integral for the length is ∫

ds=

∫b

a

dt

√

f ̇(t)^2 +g ̇(t)^2

whereaandbare the limits on the parametert. Think of this as

∫

ds=

∫

v dt, wherevis the speed.

Do the simplest example first. What is the circumference of a circle? Use the parametrization

x=Rcosθ, y=Rsinθ then ds=

√

(−Rsinθ)^2 + (Rcosθ)^2 dθ=Rdθ (3)

The circumference is then

∫

ds=

∫ 2 π

0 Rdθ= 2πR. An ellipse is a bit more of a challenge; see problem^3.

Dk

Dqk

r

rk

If the curve is expressed in polar coordinates you may find another formulation prefer-
able, though in essence it is the same. The Pythagorean Theorem is still applicable, but you
have to see what it says in these coordinates.

∆sk=

√

(∆rk)^2 + (rk∆θk)^2

If this picture doesn’t seem to show much of a right triangle, remember there’s a limit
involved, as∆rkand∆θkapproach zero this becomes more of a triangle. The integral for the length of a curve
is then ∫

ds=

∫ √

dr^2 +r^2 dθ^2

To actually do this integral you will pick a parameter to represent the curve, and that parameter may even beθ
itself. For an example, examine one loop of a logarithmic spiral:r=r 0 ekθ.

ds=

√

dr^2 +r^2 dθ^2 =

√(

dr/dθ

) 2

+r^2 dθ

The length of the arc fromθ= 0toθ= 2πis
∫ √
(
r 0 k ekθ

) 2

+

(

r 0 ekθ

) 2

dθ=

∫ 2 π

0

dθ r 0 ekθ

√

k^2 + 1 =r 0

√

k^2 + 1

1

k

[

e^2 kπ− 1

]