Arc Length of Parametrically De ned Curves

Jim Lambers MAT 169

Fall Semester 2009-10 Lecture 31 Notes

These notes correspond to Section 9.2 in the text.

Arc Length of Parametrically Defined Curves

In the last lecture we learned how to compute the arc length of a curve described by an equation

of the form = ( ), where . The arc length of such a curve is given by the definite

integral

= 1 + [ ( )]2 .

Now, suppose that this curve can also be defined by parametric equations

= ( ), = ( ),

(1)

where . It follows that and therefore, by the Chain Rule,

= ( ) = ( ( )),

= ( ) = ( ( )) ( ) = ( ( )) = ( ) .

In the integral defining the arc length of the curve, we make the substitution = ( ) and obtain

=

1+[

(

)]2

=

1+[

(

( ))]2

( )

=

[

( )]2 + [

(

( ))

( )]2

=

[

(

)]2 + [(

)]2

( )2 ( )2

=

+

.

1

It turns out that this formula for the arc length applies to any curve that is defined by parametric equations of the form (1), as long as and are differentiable functions of the parameter . To derive the formula in the general case, one can proceed as in the case of a curve defined by an equation of the form = ( ), and define the arc length as the limit as of the sum of the lengths of line segments whose endpoints lie on the curve.

Example Compute the length of the curve = 2 cos2 ,

= 2 cos sin ,

where 0 .

Solution This curve is plotted in Figure 1; it is a circle of radius 1 centered at the point (1, 0). It follows that its length, which we will denote by , is the circumference of the circle, which is 2 . Using the arc length formula, we can obtain the same result as follows:

( )2 ( )2

=

+

0

=

(-4 cos sin )2 + (2 cos2 - 2 sin2 )2

0

=2

(2 cos sin )2 + (cos2 - sin2 )2

0

=2

4 cos2

sin2

+ cos4

- 2 cos2

sin2

+ sin4

0

=2

cos4

+ 2 cos2

sin2

+ sin4

0

=2

(cos2 + sin2 )2

0

=2

12

0

=2

0

= 2.

Note: Double-angle and half-angle formulas could have been used in this example, but little would have been gained except during the differentiation stage, so I chose not to use them.

Example Compute the length of the curve

= sin , = cos

from = 0 to = 2 .

2

Figure 1: Curve defined by = cos2 , = 2 cos sin

Solution The length is given by the integral

2

=

(sin

+

cos )2 + (cos

-

sin )2

0

2

=

sin2 + 2 sin cos + 2 cos2 + cos2 - 2 sin cos + 2 sin2

0

2

=

(sin2 + cos2 ) + 2(cos2 + sin2 )

0

2

=

1+

2

0

tan-1(2 )

=

1

+

tan2

sec2

0

tan-1(2 )

=

sec3

0

3

1

tan-1(2 )

= [sec tan + ln sec + tan ]

2

0

=

1

[ 1

+

tan2

tan

+

ln

1

+

tan2

tan-1(2 )

] + tan

2

0

=

1 [1 + (2

)2(2

) + ln 1 + (2

)2 + 2

]

2

21.2563.

The integral of sec3 is obtained using integration by parts, with = sec and curve is displayed in Figure 2.

= sec2

. The

Figure 2: Curve defined by = sin , = cos 4

Summary

If a curve is defined by parametric equations = ( ), = ( ) for , the arc length of the curve is the integral of ( / )2 + ( / )2 = [ ( )]2 + [( )]2 from to .

5

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download