Section 6.4 Arc Length -...
Transcript of Section 6.4 Arc Length -...
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