13—Vector Calculus 2 326Do the simplest example first. What is the circumference of a circle? Use the parametrizationx=Rcosφ, y=Rsinφ then d`=
√
(−Rsinφ)^2 + (Rcosφ)^2 dφ=Rdφ (13.4)
The circumference is then
∫
d` =
∫ 2 π0 Rdφ= 2πR. An ellipse is a bit more of a challenge; see
problem13.3.
If the curve is expressed in polar coordinates you may find another formulation preferable, though
in essence it is the same. The Pythagorean Theorem is still applicable, but you have to see what it says
in these coordinates.
∆`k=
√
(∆rk)^2 + (rk∆φk)^2 rk∆φk
∆rk
If this picture doesn’t seem to show much of a right triangle, remember there’s a limit involved, as∆rk
and∆φkapproach zero this becomes more of a triangle. The integral for the length of a curve is then
∫
d`=
∫ √
dr^2 +r^2 dφ^2
To actually do this integral you will pick a parameter to represent the curve, and that parameter may
even beφitself. For an example, examine one loop of a logarithmic spiral:r=r 0 ekφ.
d`=
√
dr^2 +r^2 dφ^2 =
√
(dr/dφ)^2 +r^2 dφ
The length of the arc fromφ= 0toφ= 2πis
∫ √(
r 0 kekφ
) 2
+
(
r 0 ekφ
) 2
dφ=
∫ 2 π0dφr 0 ekφ
√
k^2 + 1 =r 0
√
k^2 + 1
1
k
[
e^2 kπ− 1
]
Ifk→ 0 you can easily check that this give the correct answer. In the other extreme, for largek, you
can also check that it is a plausible result, but it’s a little harder to see.
Weighted Integrals
The time for a particle to travel along a short segment of a path isdt=d`/vwherevis the speed.
The total time along a path is of course the integral ofdt.
T=
∫
dt=
∫
d`
v
(13.5)
How much time does it take a particle to slide down a curve under the influence of gravity? If the speed
is determined by gravity without friction, you can use conservation of energy to compute the speed. I’ll