1. Introduction When we say a curve has length, what does this mean? If we had a tape-measure we could fit it along the curve and measure its length with pretty good precision. For more complicated curves this might be too difficult to do with any accuracy. We require a precise definition for the length of an arc of a curve. If the curve is a polygon, the length is easily computed as the sum of the lengths of each line segment in the polygon, using the distance between the endpoints of each segment. To define the length of a general curve, we will approximate using a polygon, and take the limit as the number of segments of the polygon is increased. Supposing a curve is defined by the equation y = f(x), where f is continuous and a ≤ x ≤ b. The polygonal approximation to C is produced by dividing the interval [a, b] into n subintervals with endpoints x0, x1, ..., xn and equal width ∆x. If yi = f(xi), then the point Pi(xi, yi) lies on C and the polygon with vertices P0,P1, ..., Pn is an approximation to C. The length L of C will be approximated by the length of this polygon, and this will be a better approximation if n is increased. Hence, we define the length L of the curve C with equation y = f(x), a ≤ x ≤ b as the limit of lengths of these inscribed polygons (assuming the limit exits)
n X (1) L = lim |Pi−1Pi|. n→∞ i=1 The procedure used to define arc length is very similar to the procedure used for defining area and volume: we have divided the curve into a large number, n, of simpler parts, for which we could determine their lengths and add them together to approximate the length of the original curve, while taking the limit n → ∞ gave us the exact value.
2. An Integral Formula for Arc Length Admittedly this definition for arc length is not very helpful for computational reasons, however from this expression we can derive an integral formula for L in the case where f has a continuous derivative. We say such a function is smooth since a small change in x yields a small change in f 0(x). Denoting ∆yi = yi − yi−1 then
p 2 2 p 2 2 |Pi−1Pi| = (xi − xi−1) + (yi − yi−1) = (∆x) + (∆yi) The second term under the square root can be rewritten using the Mean Value Theorem for f on the interval [xi−1, xi], since this ensures there is some point ∗ xi−1 ≤ xi ≤ xi such that 0 ∗ 0 ∗ ∆yi = yi − yi−1 = f(xi) − f(xi−1) = f (xi )(xi − xi−1) = f (xi )∆x 1 2 ARC LENGTH
We may now rewrite this as q p 2 2 2 0 ∗ 2 |Pi−1Pi| = (∆x) + (∆yi) = (∆x) + (f (xi )∆x) q q 0 ∗ 2p 2 0 ∗ 2 = 1 + (f (xi )) (∆x) = 1 + (f (xi )) ∆x From definition (??) this is now n n q X X 0 ∗ 2 L = lim |Pi−1Pi| = lim 1 + (f (x )) ∆x n→∞ n→∞ i i=1 i=1 this is simply a definite integral now Z b L = p1 + (f 0(x))2dx. a
p 0 ∗ 2 This integral exists because the function g(x) = 1 + (f (xi )) is continuous. This proves the following proof Theorem 2.1. The Arc Length Formula If f 0 is continuous on [a, b], then the length of the curve y = f(x), a ≤ x ≤ b is Z b L = p1 + [f 0(x)]2dx a Writing this in terms of Leibniz notation for derivatives this becomes
s Z b dy 2 (2) L = 1 + dx a dx Example 2.2. Determine the length of the arc of the semicubical parabola y2 = x3 between the points (1, 1) and (4, 8).
Proof. For the top half of this curve, we have
3 dy 3 1 y = x 2 , = x 2 . dx 2 Putting this into the arc length formula we have s Z 4 dy 2 Z 4 r 9x L = 1 + dx = 1 + dx. 1 dx 1 4
9x 9 Substituting u = 1 + 4 , so that du = 4 dx, and the limits of integration are now 13 u = 4 and u = 10, then 10 10 4 Z √ 4 2 3 L = udu = u 2 9 13 9 3 13 4 4 8 133/2 1 √ √ = [103/2 − ] = (80 10 − 13 13) 27 4 27
ARC LENGTH 3
If a curve has the equation x = g(y) for c ≤ y ≤ d and g0(y) is continuous, then by interchanging the roles of x and y in the previous formulas for arc length, we have s Z d Z d dx2 (3) L = p1 + (g(y))2dy = 1 + dy c c dy Example 2.3. Find the length of the arc of the parabola y2 = x from (0, 0) to (1, 1).
2 dx Proof. With x = y , we have dy = 2y and so
1 s 2 1 Z dx Z p L = 1 + dy = 1 + 4y2dy. 0 dy 0 We now make a trigonometric substitution y = 1 tan θ, with dy = 1 sec2 θ and √ 2 2 p1 + 4y2 = 1 + tan2 θ = sec θ. When y = 0, tan θ = 0, then θ = 0 and when y = 1 this implies θ = arctan 2 = α
Z α 1 1 Z α L = sec θ ∗ sec2 θdθ = sec3 θdθ 0 2 2 0 1 = [sec θ tan θ + ln | sec θ + tan θ|]α 4 0 1 = (sec α tan α + ln | sec α + tan α|) 4 √ Notince that tan α = 2, then sec2 α = 1 + tan2 α = 5 and so sec α = 5, therefore √ √ 5 ln( 5 + 2) L = + 2 4 The square root in the integral formula for the arc length of a curve can make evaluating the integral very difficult or impossible to evaluate explicitly. In those cases, one must accept computing an approximation to the length of a curve. Example 2.4. (1) Set up an integral for the length of the arc of the hyperbola xy = 1 from the point (1, 1) to the point (2, 1/2). (2) Use Simpson’s Rule with n = 10 to estimate the arc length.
−1 dy −2 Proof. (1) Since y = x then dx = −x and the arc length is given by s √ Z 2 dy 2 Z 2 r 1 Z 2 x4 + 1 L = 1 + dx = 1 + 4 dx = 2 dx 1 dx 1 x 1 x (2) Applying the formula for Simpson’s Rule, with a = 1, b = 2, n = 10, ∆x = 0.1, and f(x) = p1 + 1/x4 we have ∆x L ≈ [f(1) + 4f(1.1) + 2f(1.2) + 4f(1.3) + 2f(1.4) + 4f(1.5) + 2f(1.6) 3 +4f(1.7) + 2f(1.8) + 4f(1.9) + f(2)] ≈ 1.1321 4 ARC LENGTH
3. The Arc Length Function The function we are about to define will be incredibly useful: it will measure the arc length of a curve from a particular starting point to any other point on the curve. For any smooth curve C defined by y = f(x) on a ≤ x ≤ b, denote s(x) be the distance along C from the initial point P0(a, f(a)) to the point Q(x, f(x)). Then s(x) is called the arc length function and by formula given in Theorem 2.1: Z x (4) s(x) = p1 + [f 0(t)]2dt. a Applying the first part of the Fundamental Theorem of Calculus to differentiate equation (??) s ds dy 2 (5) = p1 + [f 0(x)]2 = 1 + dx dx Equation (??)shows that the rate of change of s with respect to x is always at least 1 and is equal to 1 when f 0(x) (the slope of the curve) is 0. The differential of arc length is s dy 2 (6) ds = 1 + dx dx and this equation may be written in the symmetric form
(7) (ds)2 = (dx)2 + (dy)2 Writing L = R ds, equation (??) can help one recall equation (??) which produces the expression in Theorem 2.1, or by solving to get s dx2 ds = 1 + dy dy this yields equation (??).
Figure 1. The geometric interpretation of Equation (??)
2 1 Example 3.1. Find the arc length function for the cruve y = x − 8 ln x taking P0(1, 1) as the starting point. ARC LENGTH 5
2 1 Proof. If f(x) = x − 8 ln x then 1 f 0(x) = 2x − 8x 1 2 1 1 1 + [f 0(x)]2 = 1 + 2x − = 1 + 4x2 − + 8x 2 64x2 1 1 1 2 = 4x2 + + = 2x + 2 64x2 8x 1 p1 + [f 0(x)]2 = 2x + . 8x Therefore the arc length function is given by Z x s(x) = p1 + [f 0(t)]2dt 1 Z x 1 2 1 x = 2t + dt = t + ln t]1 1 8t 8 1 = x2 + ln x − 1 8 Applying this, the arc length along the curve from (1, 1) to (3, f(3)) is 1 ln 3 s(3) = 32 + ln 3 − 1 = 8 + ≈ 8.1373 8 8