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414 Chapter 7 Applications of Definite Integrals<br />
EXAMPLE 2<br />
Applying the Definition<br />
Find the exact length of the curve<br />
y 4 2<br />
x<br />
3<br />
1 for 0 x 1.<br />
SOLUTION<br />
d y<br />
4 2<br />
• 3 dx<br />
3 2 x12 22 x 12 ,<br />
which is continuous on 0, 1. Therefore,<br />
)<br />
L 1<br />
2<br />
( 1 d y<br />
dx<br />
dx<br />
0<br />
1<br />
( 22x12)<br />
2<br />
1 <br />
0<br />
1<br />
1 8x dx<br />
0<br />
2 1<br />
3 • 1 1 <br />
8<br />
8x32]<br />
0<br />
1 3<br />
.<br />
6<br />
dx<br />
y<br />
(8, 2)<br />
Now try Exercise 11.<br />
We asked for an exact length in Example 2 to take advantage of the rare opportunity it<br />
afforded of taking the antiderivative of an arc length integrand. When you add 1 to the<br />
square of the derivative of an arbitrary smooth function and then take the square root of that<br />
sum, the result is rarely antidifferentiable by reasonable methods. We know a few more<br />
functions that give “nice” integrands, but we are saving those for the exercises.<br />
(–8, –2)<br />
Figure 7.36 The graph of y x 13 has<br />
a vertical tangent line at the origin where<br />
dydx does not exist. (Example 3)<br />
(–2, –8)<br />
(2, 8)<br />
Figure 7.37 The curve in Figure 7.36<br />
plotted with x as a function of y. The<br />
tangent at the origin is now horizontal.<br />
(Example 3)<br />
x<br />
y<br />
x<br />
Vertical Tangents, Corners, and Cusps<br />
Sometimes a curve has a vertical tangent, corner, or cusp where the derivative we need to<br />
work with is undefined. We can sometimes get around such a difficulty in ways illustrated by<br />
the following examples.<br />
EXAMPLE 3<br />
A Vertical Tangent<br />
Find the length of the curve y x 13 between 8, 2 and 8, 2.<br />
SOLUTION<br />
The derivative<br />
d y<br />
1 1<br />
dx<br />
3 x23 <br />
3x<br />
23<br />
is not defined at x 0. Graphically, there is a vertical tangent at x 0 where the derivative<br />
becomes infinite (Figure 7.36). If we change to x as a function of y, the tangent<br />
at the origin will be horizontal (Figure 7.37) and the derivative will be zero instead of<br />
undefined. Solving y x 13 for x gives x y 3 , and we have<br />
L 2<br />
2<br />
x<br />
<br />
y ) dy 2<br />
1 3y 2 <br />
2 dy 17.26. Using NINT<br />
2 1 ( d d<br />
2<br />
Now try Exercise 25.