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A formula for calculating arc length, an exercise exploring the possibility of infinite length, and some other related theorems, remarks, and exercises.

Suppose C is a piecewise smooth curve, parameterized by a function φ . Continuing to think like a physicist, we might guess that the length of this curve could be computed as follows.The particle is moving with velocity φ ' ( t ) . This velocity is thought of as a vector in R 2 , and as such it has a direction and a magnitude or speed. The speed is just the absolute value | φ ' ( t ) | of the velocity vector φ ' ( t ) . Now distance is speed multiplied by time, and so a good guess for the formula for thelength L of the curve C would be

L = a b | φ ' ( t ) | d t .

Two questions immediately present themselves. First, and of primary interest, is whether the function | φ ' | is improperly-integrable on ( a , b ) ? We know by [link] that φ ' itself is improperly-integrable, but we also know from [link] that a function can be improperly-integrable on an open interval and yet its absolute value is not.In fact, the answer to this first question is no (See [link] .). We know only that | φ ' | exists and is continuous on the open subintervals of a partition of [ a , b ] .

The second question is more subtle. What if we parameterize a curve in two different ways, i.e., withtwo different functions φ 1 and φ 2 ? How do we know that the two integral formulas for the length have to agree?Of course, maybe most important of all to us, we also must justify the physicist's intuition. That is, we must give a rigorous mathematical definition of the length of a smooth curve and showthat Formula ( [link] ) above does in fact give the length of the curve. First we deal with the independence of parameterization question.

Let C be a smooth curve joining (distinct) points z 1 to z 2 in C , and let φ 1 : [ a , b ] C and φ 2 : [ c , d ] C be two parameterizations of C . Suppose | φ 2 ' | is improperly-integrable on ( c , d ) . Then | φ 1 ' | is improperly-integrable on ( a , b ) , and

a b φ 1 ' ( t ) d t = c d φ 2 ' ( s ) d s .

We will use [link] . Thus, let g = φ 1 - 1 φ 2 , and recall that g is continuous on [ c , d ] and continuously differentiable on each open subinterval of a certain partition of [ c , d ] . Therefore, by part (d) of [link] , g ' is improperly-integrable on ( c , d ) .

Let { x 0 < x 1 < ... < x p } be a partition of [ a , b ] for which φ 1 ' is continuous and nonzeroon the subintervals ( x j - 1 , x j ) . To show that | φ 1 ' | is improperly-integrable on ( a , b ) , it will suffice to show this integrability on each subinterval ( x j - 1 , x j ) . Thus, fix a closed interval [ a ' , b ' ] ( x j - 1 , x j ) , and let [ c ' , d ' ] be the closed subinterval of [ c , d ] such that g maps [ c ' , d ' ] 1-1 and onto [ a ' , b ' ] . Hence, by part (e) of [link] , we have

a ' b ' | φ 1 ' ( t ) | d t = c ' d ' | φ 1 ' ( g ( s ) ) | g ' ( s ) d s = c ' d ' | φ 1 ' ( g ( s ) ) | | g ' ( ) s ) | d s = c ' d ' | φ 1 ' ( g ( s ) ) g ' ( s ) | d s = c ' d ' | ( φ 1 g ) ' ( s ) | d s = c ' d ' | φ 2 ' ( s ) | d s c d | φ 2 ' ( s ) | d s ,

which, by taking limits as a ' goes to x j - 1 and b ' goes to x j , shows that | φ 1 ' | is improperly-integrable over ( x j - 1 , x j ) for every j , and hence integrable over all of ( a , b ) . Using part (e) of [link] again, and a calculation similar to the one above, we deduce the equality

a b | φ 1 ' | = c d | φ 2 ' | ,

and the theorem is proved.

Let φ : [ 0 , 1 ] : R 2 be defined by φ ( 0 ) = ( 0 , 0 ) , and for t > 0 , φ ( t ) = ( t , t sin ( 1 / t ) ) . Let C be the smooth curve that is the range of φ .

  1. Graph this curve.
  2. Show that
    | φ ' ( t ) | = 1 + sin 2 ( 1 / t ) - sin ( 2 / t ) t + cos 2 ( 1 / t ) t 2 = 1 t t 2 + t 2 sin 2 ( 1 / t ) - t sin ( 2 / t ) + cos 2 ( 1 / t ) .
  3. Show that
    δ 1 | φ ' ( t ) | d t = 1 1 / δ 1 t 1 t 2 + sin 2 ( t ) t 2 - sin ( 2 t ) t + cos 2 ( t ) d t .
  4. Show that there exists an ϵ > 0 so that for each positive integer n we have cos 2 ( t ) - sin ( 2 t ) / t > 1 / 2 for all t such that | t - n π | < ϵ .
  5. Conclude that | φ ' | is not improperly-integrable on ( 0 , 1 ) . Deduce that, if Formula ( [link] ) is correct for the length of a curve, then this curve has infinite length.

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Source:  OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1
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