5128_Ch07_pp378-433

412 Chapter 7 Applications of Definite Integrals
7.4
What you’ll learn about
• A Sine Wave
• Length of a Smooth Curve
• Vertical Tangents, Corners, and
Cusps
. . . and why
The length of a smooth curve can
be found using a definite integral.
[0, 2] by [–2, 2]
Figure 7.32 One wave of a sine curve
has to be longer than 2p.
y
x y √
2
k + 2
k
Q
y k
y = sin x
Lengths of Curves
A Sine Wave
How long is a sine wave (Figure 7.32)?
The usual meaning of wavelength refers to the fundamental period, which for y sin x
is 2p. But how long is the curve itself? If you straightened it out like a piece of string
along the positive x-axis with one end at 0, where would the other end be?
EXAMPLE 1
The Length of a Sine Wave
What is the length of the curve y sin x from x 0 to x 2p?
SOLUTION
We answer this question with integration, following our usual plan of breaking the whole
into measurable parts. We partition 0, 2p into intervals so short that the pieces of curve
(call them “arcs”) lying directly above the intervals are nearly straight. That way, each
arc is nearly the same as the line segment joining its two ends and we can take the length
of the segment as an approximation to the length of the arc.
Figure 7.33 shows the segment approximating the arc above the subinterval x k1 , x k .
The length of the segment is Δx 2 k . Δy 2 k The sum
Δx k 2 Δy 2 k
over the entire partition approximates the length of the curve. All we need now is to find
the limit of this sum as the norms of the partitions go to zero. That’s the usual plan, but
this time there is a problem. Do you see it?
The problem is that the sums as written are not Riemann sums. They do not have the
form f c k Δx. We can rewrite them as Riemann sums if we multiply and divide each
square root by Δx k .
O
P
x k–1
x k
Figure 7.33 The line segment approximating
the arc PQ of the sine curve above
the subinterval x k1 , x k . (Example 1)
Group Exploration
Later in this section we will use an integral
to find the length of the sine wave
with great precision. But there are ways
to get good approximations without
integrating. Take five minutes to come
up with a written estimate of the curve’s
length. No fair looking ahead.
x k
x
x k
2 y 2 k x k 2 y k
2 x
x
k
y
x
1 (
k
)
2
k
k
This is better, but we still need to write the last square root as a function evaluated at
some c k in the kth subinterval. For this, we call on the Mean Value Theorem for differentiable
functions (Section 4.2), which says that since sin x is continuous on x k1 , x k
and is differentiable on x k1 , x k there is a point c k in x k1 , x k at which Δy k Δx k
sin c k (Figure 7.34). That gives us
1 si nc k
2 Δx k ,
which is a Riemann sum.
Now we take the limit as the norms of the subdivisions go to zero and find that the
length of one wave of the sine function is
2p
1 si nx 2 dx 2p
1 cos 2 x dx 7.64. Using NINT
0
0
How close was your estimate? Now try Exercise 9.
x k