Part 3: Distance and Arclength 

If t₀, t₁,¼, t_n is a partition of a closed interval [a,b] , then the length L of the curve parameterized by r(t) over a interval [a,b] can be approximated by

L » n å j = 1  Dsj

where Ds_j is the distance between r(t_j-1) and r(t_j).  We can rewrite the summation above in the form

L » n å j = 1  Dsj Dtj   Dtj

and application of a limit results in a definite integral. Thus, the length L of the curve parameterized by r(t) over an interval [a,b] is given by

L = b a  || v( t) ||   dt

That is, total distance traveled is an integral of the speed.      

EXAMPLE 5    Find the distance traveled along the helix r( t) = á cos( t), sin( t), t ñ for t in [ 0,4p] .       

Solution: To do so, we first compute the velocity

v( t) = á -sin( t) ,cos(t) ,1 ñ

The speed is then the magnitude of the velocity:

|| v( t) || =  sin2( t)+cos2( t) +1  = Ö2

Finally, the length of the curve between r( 0) and r( 4p) is

L = 4p 0  || v( t) ||  dt = 4p 0  2 dt = 4p Ö 2

The value of L in example 5 is the length of the helix for t in [0,4p] because each point on the curve corresponds to only one value of t in [0,4p]. However, if r(t), t in [a,b], covers a curve more than once, then the total distance L is more than the length of the curve (for example, r( t) = á cos(t), sin(t) ñ for t in [ 0,6p], wraps around a circle 3 times, and correspondingly, L is 3 times the circumference of the unit circle).

However, if we are careful to use a parameterization r(t), t in [a,b], in which there is a 1-1 correspondence between t and points on the curve, then we are correct in interpreting L as the length of the curve between the point r(a) and r(b).        

EXAMPLE 6    Find the length of the curve r(t) = á cos(t), sin(t), cosh( t) ñ for t in [ 0, ln2 ] .

Solution: To do so, we first compute the velocity:

v( t) = á -sin( t), cos(t),  sinh(t) ñ

The speed is then the magnitude of the velocity:

|| v( t) || =  sin2( t) + cos2( t) + sinh2( t)

The identities cos²( t) +sin²( t) = 1 and cosh²( t) = sinh²( t) +1 imply that

|| v( t) || =  1 + sinh2( t)  =  cosh2( t)

Thus, ||v( t) || = cosh(t) and as a result,

L = ó õ ln( 2) 0  ||v( t)||dt = ó õ ln( 2) 0  cosh( t) dt

Evaluating the integral leads to

L = sinh( t) | 0ln( 2) = sinh( ln2) -sinh( 0)

Since sinh( 0) = 0, this simplifies to

L =    eln( 2) -e-ln( 2) 2   =   2-  1 2 2     =    3 4