Section 6.4

Arc Length

Goals

◦ Define arc length of a smooth curve given

parametrically

◦ Give an integral formula for arc length

Introduction

What do we mean by the length of a

curve?

If the curve is a polygon, then the length is

simply the sum of the lengths of the line

segments that form the polygon.

Our approach to the length of a general

curve C is shown on the next slide:

Introduction (cont’d)

Introduction (cont’d)

We define the length by…

◦ first approximating it by a polygon, and then

◦ taking a limit as the number of segments of

the polygon is increased.

We will assume that C is described by the

parametric equations

x = f(t), y = g(t), a ≤ t ≤ b

Introduction (cont’d)

We also assume that C is smooth, that is, that the derivatives f (t) and g (t) are

◦ continuous, and

◦ not simultaneously zero, a < t < b.

This ensures that C has no sudden change

in direction.

Inscribed Polygon

We divide the parameter interval [a, b] into n

subintervals of equal width ∆t.

The preceding figure shows the points

P0, P1, …, Pn corresponding to the endpoints of

the above subintervals.

The polygon with these vertices approximates

C, and so the length of this polygon approaches that of C as n ∞

Inscribed Polygon (cont’d)

Thus we define the length of C to be the

limit of the lengths of these inscribed

polygons:

Now we work toward a more convenient

expression for L :

11

limn

i in

i

L P P

Riemann Sum

If we let ∆xi = xi – xi-1 and ∆yi = yi – yi-1,

then the length of the ith line segment of

the polygon is

But from the definition of a derivative we know that f (ti) ≈ ∆xi/∆t if ∆t is small.

Therefore ∆xi ≈ f (ti)∆t and ∆yi ≈ g (ti)∆t.

22

1i i i iP P x y

Riemann Sum (cont’d)

Therefore

so that

Riemann Sum (cont’d)

This is a Riemann sum for the function

suggests that

With the restriction that no portion of the

curve is traced out more than once, this is the

correct arc length formula:

2 2

, and so our argumentf t g t

2 2b

aL f t g t dt

Arc Length Formula

As an example we find the length of the arc of

the curve x = t2, y = t3 that lies between the

points (1, 1) and (4, 8), shown on the next

slide:

Example (cont’d)

Solution

We note that the given portion of the

curve corresponds to the parameter

interval 1 ≤ t ≤ 2, so the formula gives

Solution (cont’d)

Substituting u = 4 + 9t2 leads to

This number is just slightly larger than the

distance joining the endpoints (1, 1) and (4,

8)—as the figure would suggest.

A Special Case

If we are given a curve y = f(x), a ≤ t ≤ b,

then we can regard x as a parameter.

With the parametric equations x = x,

y = f(x), our formula becomes

Example

Find the length of one arch of the cycloid

x = r(θ – sinθ) , y = r(1 – cosθ).

Solution We know from earlier work that

one arch is described by the parameter

interval 0 ≤ θ ≤ 2π. Since

Solution (cont’d)

we have

A computer algebra system gives the value of

8r for this integral.

◦ Thus the length of one arch of a cycloid is eight

times the radius of the generating circle.

Review

Use of Riemann sums to define arc length

using polygonal approximation

Integral formula for arc length

Section 6.5

Average Value of a Function

Goals

◦ Define the average value of a function on an

interval

◦ Discuss the Mean Value Theorem for Integrals

Introduction

It is easy to calculate the average value of

finitely many numbers y1, y2, …, yn :

But how do we compute (for example) the

average temperature in a day if infinitely many

temperature readings are possible?

Average Value

Let’s try to compute the average value of a

function y = f(x), a ≤ x ≤ b :

◦ We divide the interval [a, b] into n equal

subintervals, each with length ∆x = (b – a)/n.

◦ Then we choose points x1*, … xn* in successive

subintervals and calculate the average of the

numbers f(x1*), …, f(xn*) :

Average Value (cont’d)

We can write n = (b – a)/∆x and the

average value becomes

Average Value (cont’d)

The limiting value is

by the definition of definite integral.

So we define the average value of f on the

interval [a, b] as ave

1.

b

af f x dx

b a

Example

Find the average value of the function

f(x) = 1 + x2 on the interval [–1, 2].

Solution With a = –1 and b = 2 we have

Mean Value Theorem for Integrals

We would expect there to be a time of day at

which the temperature is the same as the

average temperature for the day.

In general, is there a number c at which the

value of a function f is exactly equal to the

average value of the function?

The following theorem says that for

continuous functions, the answer is yes:

Theorem (cont’d)

The geometric interpretation of this theorem

is that for positive functions f, there is a

number c such that the rectangle has the same

area as the region under the blue curve:

Theorem (cont’d)

Example

Since f(x) = 1 + x2 is continuous on the

interval [–1, 2], the Mean Value Theorem

for Integrals says that there is a number c

in [–1, 2] such that

In this case we can find c explicitly:

2 2

11 2 1x dx f c

Example (cont’d)

Earlier we found that fave = 2, so the value

of c satisfies f(c) = fave = 2.

Therefore 1 + c2 = 2, so c = ±1.

◦ So in this case there happen to be two

numbers c = ±1 that work in the Mean Value

Theorem for Integrals.

Our two examples are illustrated on the

next slide:

Example (cont’d)

Review

Use of Riemann sums to define average

value of a function

Integral formula

Mean Value Theorem for Integrals