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