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Now, because φ ' is uniformly continuous on the closed interval [ t 1 , t n - 1 ] , there exists a δ > 0 such that | φ ' ( t ) - φ ' ( s ) | < ϵ if | t - s | < δ and t and s are in the interval [ t 1 , t n - 1 ] . We may assume, again by taking a finer partition if necessary, that the mesh size of P is less than this δ . Then, using part (f) of [link] , we may also assume that the partition P is such that

| t 1 t n - 1 | φ ' ( t ) | d t - j = 2 n - 1 | φ ' ( s j ) | ( t j - t j - 1 ) | < ϵ

no matter what points s j in the interval ( t j - 1 , t j ) are chosen. So, we have the following calculation, in the middle of which we use the Mean Value Theorem on the two functions u and v .

0 | L φ - t 1 t n - 1 | φ ' ( t ) | d t | | L φ - j = 2 n - 1 | φ ( t j ) - φ ( t j - 1 ) | + | j = 2 n - 1 | φ ( t j ) - φ ( t j - 1 ) | - t 1 t n - 1 | φ ' ( t ) | d t | 3 ϵ + | j = 2 n - 1 | φ ( t j ) - φ ( t j - 1 ) | - t 1 g n - 1 | φ ' ( t ) | d t | = 3 ϵ + | j = 2 n - 1 | u ( t j ) - u ( t j - 1 ) + i ( v ( t j ) - v ( t j - 1 ) | - t 1 t n - 1 | φ ' ( t ) | d t | = 3 ϵ + | j = 2 n - 1 ( u ( t j ) - u ( t j - 1 ) ) 2 + ( v ( t j ) - v ( t j - 1 ) ) 2 - t 1 g n - 1 | φ ' ( t ) | d t | = 3 ϵ + | j = 2 n - 1 ( u ' ( s j ) ) 2 + ( v ' ( r j ) ) 2 ( t j - t j - 1 ) - t 1 t n - 1 | φ ' ( t ) | d t | 3 ϵ + | j = 2 n - 1 ( u ' ( s j ) ) 2 + ( v ' ( s j ) ) 2 ( t j - t j - 1 ) - t 1 t n - 1 | φ ' ( t ) | d t | + j = 2 n - 1 | ( u ( s j ) ) 2 + ( v ' ( r j ) ) 2 - ( u ( s j ) ) 2 + ( v ' ( s j ) ) 2 | ( t j - t j - 1 ) = 3 ϵ + | j = 2 n - 1 | φ ' ( s j ) | ( t j - t j - 1 ) - t 1 t n - 1 | φ ' ( t ) | d t | + j = 2 n - 1 | ( u ( s j ) ) 2 + ( v ' ( r j ) ) 2 - ( u ( s j ) ) 2 + ( v ' ( s j ) ) 2 | ( t j - t j - 1 ) 4 ϵ + j = 2 n - 1 | ( v ' ( r j ) ) 2 - ( v ' ( s j ) ) 2 | ( u ' ( s j ) ) 2 + ( v ' ( r j ) ) 2 + ( u ' ( s j ) ) 2 + ( v ' ( s j ) ) 2 ( t j - t j - 1 ) 4 ϵ + j = 2 n - 1 | v ' ( r j ) - v ' ( s j ) | | v ' ( r j ) + v ' ( s j ) | | v ' ( r j ) | + | v ' ( s j ) | ( t j - t j - 1 ) 4 ϵ + j = 2 n - 1 | v ' ( r j ) - v ' ( s j ) | ( t j - t j - 1 ) 4 ϵ + j = 2 n - 1 | φ ' ( r j ) - φ ' ( s j ) | ( t j - t j - 1 ) 4 ϵ + j = 2 n - 1 ϵ ( t j - t j - 1 ) = 4 ϵ + ϵ ( t n - 1 - t 1 ) < ϵ ( 4 + b - a ) .

This implies that

L φ - ϵ ( 4 + b - a ) t 1 t n - 1 | φ ' | L φ + ϵ ( 4 + b - a ) .

If we now let t 1 approach a and t n - 1 approach b , we get

L φ - ϵ ( 4 + b - a ) a b | φ ' | L φ + ϵ ( 4 + b - a ) ,

which completes the proof, since ϵ is arbitrary.

  1. Take care of the piecewise case in the preceding theorem.
  2. Take care of the case when L φ is infinite in the preceding theorem.

We now have all the ingredients necessary to define the length of a smooth curve.

Let C be a piecewise smooth curve in the plane.The length or arc length L L ( C ) of C is defined by the formula

L ( C ) = L φ = sup P L P φ ,

where φ is any parameterization of C .

If z and w are two points on a piecewise smooth curve C , we will denote by L ( z , w ) the arc length of the portion of the curve between z and w .

REMARK According to [link] and [link] , we have the following formula for the length of a piecewise smooth curve:

L = a b | φ ' ( t ) | d t ,

where φ is any parameterization of C .

It should come as no surprise that the length of a curve C from z 1 to z 2 is the same as the length of that same curve C , but thought of as joining z 2 to z 1 . Nevertheless, let us make the calculation to verify this. If φ : [ a , b ] C is a parameterization of this curve from z 1 to z 2 , then we have seen in part (f) of exercise 6.1 that ψ : [ a , b ] C , defined by ψ ( t ) = φ ( a + b - t ) , is a parameterization of C from z 2 to z 1 . We just need to check that the two integrals giving the lengths are equal. Thus,

a b | ψ ' ( t ) | d t = a b | φ ' ( a + b - t ) ( - 1 ) | d t = a b | φ ' ( a + b - t ) | d t = a b | φ ' ( s ) | d s ,

where the last equality follows by changing variables, i.e., setting t = a + b - s .

We can now derive the formula for the circumference of a circle, which was one of our main goals. TRUMPETS?

Let C be a circle of radius r in the plane. Then the length of C is 2 π r .

Let the center of the circle be denoted by ( h , k ) . We can parameterize the top half of the circle by the function φ on the interval [ 0 , π ] by φ ( t ) = h + r cos ( t ) + i ( k + r sin ( t ) ) . So, the length of this half circle is given by

L = 0 π | φ ' ( t ) | d t = 0 π | - r sin ( t ) + i r cos ( t ) | d t = 0 π r d t = π r .

The same kind of calculation would show that the lower half of the circle has length π r , and hence the total length is 2 π r .

The integral formula for the length of a curve is frequently not much help, especially if you really want to know how long a curve is.The integrals that show up are frequently not easy to work out.

  1. Let C be the portion of the graph of the function y = x 2 between x = 0 and x = 1 . Let φ : [ 0 , 1 ] C be the parameterization of this curve given by φ ( t ) = t + t 2 i . Find the length of this curve.
  2. Define φ : [ - 0 , π ] C by φ ( t ) = a cos ( t ) + i b sin ( t ) . What curve does φ parameterize, and can you find its 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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