Length of a Plane Curve

Length of a Plane Curve

Objective: To find the length of a plane curve.

Arc Length

• Our first objective is to define what we mean by length (or arc length) of a plane curve y = f(x) over an interval [a, b]. Once that is done we will be able to focus on the problem of computing arc lengths. We will have the requirement that f / be continuous on [a, b] and we say that y = f(x) is a smooth curve.

Arc Length Problem

• Suppose that y = f(x) is a smooth curve on the interval [a, b]. Define and find a formula for the arc length of L of the curve y = f(x) over the interval [a, b]

Arc Length

• To define the arc length of a curve we start by breaking the curve into small segments. Then we approximate the curve segments by line segments and add the lengths of the line segments to form a Riemann Sum. As we increase the number of segments, the approximation becomes better and better.

Arc Length

• To implement our idea, divide the interval [a, b] into n subintervals by inserting points between the values . Let be the points on the curve that join the line segments. These line segments form polygonal path that we can regard as an approximation to the curve y = f(x).

nxandbxa 0

121 ,...,, nxxx

nPPP ,...,, 10

Arc Length

• The length Lk of the kth line segment in the polygonal path is

21

222 )]()([)()()( kkkkkk xfxfxyxL

Arc Length

• The length Lk of the kth line segment in the polygonal path is

• If we now add the lengths of these line segments, we obtain the following approximation to the length L of the curve.

21

222 )]()([)()()( kkkkkk xfxfxyxL

21

2

11

)]()([)(

kkk

n

k

n

kk xfxfxLL

Arc Length

• To put this in the form of a Riemann Sum we will apply the Mean-Value Theorem. This Theorem implies that there is a point between and such that

or

1kx*kx kx

)()()( */

1

1k

kk

kk xfxx

xfxf

)()()( */1

kk

kk xfx

xfxf

kkkk xxfxfxf )()()( */1

Arc Length


or

1kx*kx kx

)()()( */

1

1k

kk

kk xfxx

xfxf

21

2

11

)]()([)(

kkk

n

k

n

kk xfxfxLL

)()()( */1

kk

kk xfx

xfxf

kkkk xxfxfxf )()()( */1

2*/2

11

])([)( kkk

n

k

n

kk xxfxLL

Arc Length


or

1kx*kx kx

)()()( */

1

1k

kk

kk xfxx

xfxf

kk

n

k

xxfL

2*/

1

)]([1

)()()( */1

kk

kk xfx

xfxf

2*/2

11

])([)( kkk

n

k

n

kk xxfxLL

Arc Length

• Thus, taking the limit as n increases and the widths of the subintervals approximate zero yields the following integral that defines the arc length L:

dxxfxxfLb

a

kk

n

kx

2/2*/

10max

)]([1)]([1lim

Definition

• 7.4.2 If y = f(x) is a smooth curve on the interval [a, b], then the arc length of L of this curve over [a, b] is defined as

dxxfLb

a 2/ )]([1

Definition

• 7.4.2 If x = g(y) is a smooth curve on the interval [c, d], then the arc length of L of this curve over [c, d] is defined as

dyygLd

c 2/ )]([1

Example 1

• Find the arc length of the curve from (1, 1) to (2, ) using both formulas. 22

2/3xy

Example 1


2/3xy

2/1/

2

3)( xxf

09.22

31

2

1

22/1

dxxL

Example 1


2/3xy

3/1/

3

2)( yyg

09.23

21

22

1

23/1

dyyL

3/2yx

Homework

• Page 469

• 3, 5

Average Value of a Function

• We will use the idea of average/mean and extend the concept so that we can compute not only the arithmetic average of finitely many function values but an average of all values of f(x) as x varies over a closed interval [a, b].


• We will use the Mean-Value Theorem for Integrals, which states that if f is continuous on the interval [a, b], then there is at least one point in this interval such that

*x

))(()( * abxfdxxfb

a


• We will use the Mean-Value Theorem for Integrals, which states that if f is continuous on the interval [a, b], then there is at least one point in this interval such that

• We will look at the equation in this form as our candidate for the average value of f over the interval [a, b].

*x

))(()( * abxfdxxfb

a

ave

b

a

fdxxfab

)(1


• To explain what motivates this idea, divide the interval [a, b] into n subintervals of equal length

and choosing arbitrary points in successive subintervals. Then the arithmetic average of the values is

n

abx

**2

*1 ,...,, nxxx

)(),...,(),( **2

*1 nxfxfxf

)](...)()([1 **

2*1 nxfxfxf

nave


• Using the fact that • we can say that

• and substitute to get this equation:

n

abx

n

kkn xxf

abxxfxxfxxf

abave

1

***2

*1 )(

1])(...)()([

1

nab

x 1

)](...)()([1 **

2*1 nxfxfxf

nave


• Taking the limit as yields

b

a

n

kk

ndxxf

abxxf

ab)(

1)(

1lim

1

*

n


• Definition 7.6.1 If f is continuous on [a, b], then the average value (or mean value) of f on [a, b] is defined to be

b

a

ave dxxfab

f )(1


• If we look at the Mean-Value Theorem for Integrals together with the equation for average value, we can see the relationship.

b

a

ave dxxfab

f )(1

b

a

dxxfabxf )())(( *


• The Mean-Value Theorem for Integrals guarantees the point where the rectangle has the right height. The average value is the height.

b

a

dxxfabxf )())(( *

b

a

ave dxxfab

f )(1

*x

Example 1

• Find the average value of the function over the interval [1, 4], and find all points in the interval at which the value of f is the same as the average.

xxf )(

Example 1


• The average value of the functions is:

9

14

3

2

3

1

14

1)(

14

1

2/34

1

xdxxdxxfab

fb

a

ave

xxf )(

Example 1


• The second question is the Mean-Value Theorem for Integrals.

81

1969

14)(

*

**

x

xxf

xxf )(

Average Velocity Revisited

• When we first looked at Rectilinear Motion, we defined the average velocity of the particle over a time interval to be its displacement over the time interval divided by the time elapsed. Thus, if the particle has position s(t), then its average velocity over a time interval [t0 , t1 ] is

01

01 )()(

tt

tstsvave

avev

Average Velocity Revisited

• However, the displacement is the integral of velocity over the given time interval. We can now look at average velocity as:

01

01 )()(

tt

tstsvave

1

0

)(1

01

t

t

ave dttvtt

v

Homework

• Pages 479-480

• 1-9 odd

• Section 5.8

• Pages 388-389

• 1-11 odd

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Length of a Plane Curve