Mathematical Methods for Physics and Engineering : A Comprehensive Guide

10.3 SPACE CURVES

This parametric representation can be very useful, particularly in mechanics when

the parameter may be the timet. We can, however, also represent a space curve

byy=f(x),z=g(x), which can be easily converted into the above parametric

form by settingu=x,sothat

r(u)=ui+f(u)j+g(u)k.

Alternatively, a space curve can be represented in the formF(x, y, z)=0,

G(x, y, z) = 0, where each equation represents a surface and the curve is the

intersection of the two surfaces.

A curve may sometimes be described in parametric form by the vectorr(s),

where the parametersis thearc lengthalong the curve measured from a fixed

point. Even when the curve is expressed in terms of some other parameter, it is

straightforward to find the arc length between any two points on the curve. For

the curve described byr(u), let us consider an infinitesimal vector displacement

dr=dxi+dyj+dzk

along the curve. The square of the infinitesimal distance moved is then given by

(ds)^2 =dr·dr=(dx)^2 +(dy)^2 +(dz)^2 ,

fromwhichitcanbeshownthat
(
ds
du

) 2

=

dr

du

·

dr

du

.

Therefore, the arc length between two points on the curver(u), given byu=u 1

andu=u 2 ,is

s=

∫u 2

u 1

√

dr

du

·

dr

du

du. (10.12)

Acurvelyinginthexy-plane is given byy=y(x),z=0. Using (10.12), show that the

arc length along the curve betweenx=aandx=bis given bys=

∫b

a

√

1+y′^2 dx,where

y′=dy/dx.

Let us first represent the curve in parametric form by settingu=x,sothat

r(u)=ui+y(u)j.

Differentiating with respect tou, we find

dr

du

=i+

dy

du

j,

from which we obtain

dr

du

·

dr

du

=1+

(

dy

du

) 2

.