5128_Ch07_pp378-433

More documents

Recommendations

Info

Section 7.4 Lengths of Curves 413 P Slope sin' (c k ) x k x k–1 c k x k y k Figure 7.34 The portion of the sine curve above x k1 , x k . At some c k in the interval, sin c k y k x k , the slope of segment PQ. (Example 1) d c 0 y (a, c) a √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ (x k ) 2 (y k ) 2 y f(x) x k – 1 x k Q x k y k (b, d) Figure 7.35 The graph of f, approximated by line segments. P Q b x Length of a Smooth Curve We are almost ready to define the length of a curve as a definite integral, using the procedure of Example 1. We first call attention to two properties of the sine function that came into play along the way. We obviously used differentiability when we invoked the Mean Value Theorem to replace Δy k Δx k by sinc k for some c k in the interval x k1 , x k . Less obviously, we used the continuity of the derivative of sine in passing from 1 si nc k 2 Δx k to the Riemann integral. The requirement for finding the length of a curve by this method, then, is that the function have a continuous first derivative. We call this property smoothness. A function with a continuous first derivative is smooth and its graph is a smooth curve. Let us review the process, this time with a general smooth function f x. Suppose the graph of f begins at the point a, c and ends at b, d, as shown in Figure 7.35. We partition the interval a x b into subintervals so short that the arcs of the curve above them are nearly straight. The length of the segment approximating the arc above the subinterval x k1 , x k is Δx 2 k . Δy 2 k The sum Δx 2 k Δy 2 k approximates the length of the curve. We apply the Mean Value Theorem to f on each subinterval to rewrite the sum as a Riemann sum, x k 2 y k y x 2 1 ( ) 2 k k x k 1 fc k 2 x k . Passing to the limit as the norms of the subdivisions go to zero gives the length of the curve as L b a 1 fx 2 dx b a 1 ( d d We could as easily have transformed Δx 2 k Δy 2 k into a Riemann sum by dividing and multiplying by Δy k , giving a formula that involves x as a function of y say, x gy on the interval c, d: L x k 2 y 2 2 k y y k ( 1 xk y y k k ) k y ) x 2 dx. 1 gc 2 k y k . For some point c k in (x k 1 , x k ) For some c k in (y k1 , y k ) The limit of these sums, as the norms of the subdivisions go to zero, gives another reasonable way to calculate the curve’s length, L d 1 gy 2 dy d 2 ( 1 d x d ) dy. y c c Putting these two formulas together, we have the following definition for the length of a smooth curve. DEFINITION Arc Length: Length of a Smooth Curve If a smooth curve begins at a, c and ends at b, d, a b, c d, then the length (arc length) of the curve is L b L d a 1 ( d d c 1 ( d d y ) x 2 dx 2 x y ) dy if y is a smooth function of x on a, b; if x is a smooth function of y on c, d.
414 Chapter 7 Applications of Definite Integrals EXAMPLE 2 Applying the Definition Find the exact length of the curve y 4 2 x 3 1 for 0 x 1. SOLUTION d y 4 2 • 3 dx 3 2 x12 22 x 12 , which is continuous on 0, 1. Therefore, ) L 1 2 ( 1 d y dx dx 0 1 ( 22x12) 2 1 0 1 1 8x dx 0 2 1 3 • 1 1 8 8x32] 0 1 3 . 6 dx y (8, 2) Now try Exercise 11. We asked for an exact length in Example 2 to take advantage of the rare opportunity it afforded of taking the antiderivative of an arc length integrand. When you add 1 to the square of the derivative of an arbitrary smooth function and then take the square root of that sum, the result is rarely antidifferentiable by reasonable methods. We know a few more functions that give “nice” integrands, but we are saving those for the exercises. (–8, –2) Figure 7.36 The graph of y x 13 has a vertical tangent line at the origin where dydx does not exist. (Example 3) (–2, –8) (2, 8) Figure 7.37 The curve in Figure 7.36 plotted with x as a function of y. The tangent at the origin is now horizontal. (Example 3) x y x Vertical Tangents, Corners, and Cusps Sometimes a curve has a vertical tangent, corner, or cusp where the derivative we need to work with is undefined. We can sometimes get around such a difficulty in ways illustrated by the following examples. EXAMPLE 3 A Vertical Tangent Find the length of the curve y x 13 between 8, 2 and 8, 2. SOLUTION The derivative d y 1 1 dx 3 x23 3x 23 is not defined at x 0. Graphically, there is a vertical tangent at x 0 where the derivative becomes infinite (Figure 7.36). If we change to x as a function of y, the tangent at the origin will be horizontal (Figure 7.37) and the derivative will be zero instead of undefined. Solving y x 13 for x gives x y 3 , and we have L 2 2 x y ) dy 2 1 3y 2 2 dy 17.26. Using NINT 2 1 ( d d 2 Now try Exercise 25.
Page 1 and 2: Chapter7 Applications of Definite I
Page 3 and 4: 380 Chapter 7 Applications of Defin
Page 35: 412 Chapter 7 Applications of Defin

5128_Ch07_pp378-433

Create successful ePaper yourself

Delete template?

Save as template?