416 Chapter 6: Applications of Definite Integrals Lengths of Plane Curves We know what is meant by the length of a straight line segment, but without calculus, we have no precise notion of the length of a general winding curve. The idea of approximating the length of a curve running from point A to point B by subdividing the curve into many pieces and joining successive points of division by straight line segments dates back to the ancient Greeks. Archimedes used this method to approximate the circumference of a circle by inscribing a polygon of n sides and then using geometry to compute its perimeter 6.3 HISTORICAL BIOGRAPHY Archimedes (287–212 B.C.) 4100 AWL/Thomas_ch06p396-465 9/1/04 12:55 PM Page 416 6.3 Lengths of Plane Curves 417 (Figure 6.23). The extension of this idea to a more general curve is displayed in Figure 6.24, and we now describe how that method works. Length of a Parametrically Defined Curve Let C be a curve given parametrically by the equations We assume the functions ƒ and g have continuous derivatives on the interval [a, b] that are not simultaneously zero. Such functions are said to be continuously differentiable, and the curve C defined by them is called a smooth curve. It may be helpful to imagine the curve as the path of a particle moving from point at time to point in Figure 6.24. We subdivide the path (or arc) AB into n pieces at points These points correspond to a partition of the interval [a, b] by where Join successive points of this subdivision by straight line segments (Figure 6.24). A representative line segment has length (see Figure 6.25). If is small, the length is approximately the length of arc By the Mean Value Theorem there are numbers and in such that Assuming the path from A to B is traversed exactly once as t increases from to with no doubling back or retracing, an intuitive approximation to the “length” of the curve AB is the sum of all the lengths = a n k = 1 2[ƒ¿st k * d] 2 + [g¿st k ** d] 2 ¢t k . a n k = 1 L k = a n k = 1 2s¢x k d 2 + s¢y k d 2 L k : t = b, t = a ¢y k = gst k d - gst k - 1 d = g¿st k ** d ¢t k . ¢x k = ƒst k d - ƒst k - 1 d = ƒ¿st k * d ¢t k , [t k - 1 , t k ]t k ** t k * P k - 1 P k .L k ¢t k = 2[ƒst k d - ƒst k - 1 d] 2 + [gst k d - gst k - 1 d] 2 L k = 2s¢x k d 2 + s¢y k d 2 P k = sƒst k d, gst k dd.a = t 0 6 t 1 6 t 2 6 Á 6 t n = b, A = P 0 , P 1 , P 2 , Á , P n = B. B = sƒsbd, gsbdd t = aA = sƒsad, gsadd x = ƒstd and y = gstd, a … t … b. n H11005 4 n H11005 8 n H11005 16 FIGURE 6.23 Archimedes used the perimeters of inscribed polygons to approximate the circumference of a circle. For the approximation method gives as the circumference of the unit circle.p L 3.14103 n = 96 y x 0 A H11005 P 0 B H11005 P n P 1 P 2 C P k P k–1 FIGURE 6.24 The curve C defined parametrically by the equations and The length of the curve from A to B is approximated by the sum of the lengths of the polygonal path (straight line segments) starting at then to and so on, ending at B = P n . P 1 ,A = P 0 , y = gstd, a … t … b. x = ƒstd y x 0 L k ∆x k ∆y k P k–1 H11005 ( f(t k–1 ), g(t k–1 )) P k H11005 ( f(t k ), g(t k )) FIGURE 6.25 The arc is approximated by the straight line segment shown here, which has length L k = 2s¢x k d 2 + s¢y k d 2 . P k - 1 P k 4100 AWL/Thomas_ch06p396-465 9/1/04 12:55 PM Page 417 Although this last sum on the right is not exactly a Riemann sum (because and are evaluated at different points), a theorem in advanced calculus guarantees its limit, as the norm of the partition tends to zero, to be the definite integral Therefore, it is reasonable to define the length of the curve from A to B as this integral. L b a 2[ƒ¿std] 2 + [g¿std] 2 dt. g¿ƒ¿ 418 Chapter 6: Applications of Definite Integrals DEFINITION Length of a Parametric Curve If a curve C is defined parametrically by and where and are continuous and not simultaneously zero on [a, b], and C is traversed exactly once as t increases from to then the length of C is the definite integral L = L b a 2[ƒ¿std] 2 + [g¿std] 2 dt. t = b,t = a g¿ƒ¿ y = gstd, a … t … b,x = ƒstd A smooth curve C does not double back or reverse the direction of motion over the time interval [a, b] since throughout the interval. If and then using the Leibniz notation we have the following result for arc length: (1) What if there are two different parametrizations for a curve C whose length we want to find; does it matter which one we use? The answer, from advanced calculus, is no, as long as the parametrization we choose meets the conditions stated in the definition of the length of C (see Exercise 29). EXAMPLE 1 The Circumference of a Circle Find the length of the circle of radius r defined parametrically by Solution As t varies from 0 to the circle is traversed exactly once, so the circumfer- ence is We f ind and a dx dt b 2 + a dy dt b 2 = r 2 ssin 2 t + cos 2 td = r 2 . dx dt =-r sin t, dy dt = r cos t L = L 2p 0 B a dx dt b 2 + a dy dt b 2 dt. 2p, x = r cos t and y = r sin t, 0 … t … 2p. L = L b a B a dx dt b 2 + a dy dt b 2 dt. y = gstd,x = ƒstd sƒ¿d 2 + sg¿d 2 7 0 4100 AWL/Thomas_ch06p396-465 9/1/04 12:55 PM Page 418 6.3 Lengths of Plane Curves 419 So EXAMPLE 2 Applying the Parametric Formula for Length of a Curve Find the length of the astroid (Figure 6.26) Solution Because of the curve’s symmetry with respect to the coordinate axes, its length is four times the length of the first-quadrant portion. We have Therefore, The length of the astroid is four times this: Length of a Curve Given a continuously differentiable function we can assign as a parameter. The graph of the function ƒ is then the curve C defined parametri- cally by a special case of what we considered before. Then, dx dt = 1 and dy dt = ƒ¿std. x = t and y = ƒstd, a … t … b, x = ta … x … b,y = ƒsxd, y = ƒsxd 4s3>2d = 6. =- 3 4 cos 2td 0 p>2 = 3 2 . = 3 2 L p>2 0 sin 2t dt Length of first-quadrant portion = L p>2 0 3 cos t sin t dt = 3 cos t sin t. = 3 ƒ cos t sin t ƒ = 29 cos 2 t sin 2 t B a dx dt b 2 + a dy dt b 2 = 29 cos 2 t sin 2 tscos 2 t + sin 2 td ('')''* a dy dt b 2 = [3 sin 2 tscos td] 2 = 9 sin 4 t cos 2 t a dx dt b 2 = [3 cos 2 ts -sin td] 2 = 9 cos 4 t sin 2 t x = cos 3 t, y = sin 3 t x = cos 3 t, y = sin 3 t, 0 … t … 2p. L = L 2p 0 2r 2 dt = r CtD 2p 0 = 2pr. 0 … t … p>2 cos t sin t Ú 0 for s1>2d sin 2t cos t sin t = x y 0 1 1–1 –1 x H11005 cos 3 t y H11005 sin 3 t 0 H11349 t H11349 2H9266 FIGURE 6.26 The astroid in Example 2. HISTORICAL BIOGRAPHY Gregory St. Vincent (1584–1667) 1 4100 AWL/Thomas_ch06p396-465 9/1/04 12:55 PM Page 419 From our calculations in Section 3.5, we have giving Substitution into Equation (1) gives the arc length formula for the graph of y = ƒsxd. = 1 + [ƒ¿sxd] 2 . = 1 + a dy dx b 2 a dx dt b 2 + a dy dt b 2 = 1 + [ƒ¿std] 2 dy dx = dy>dt dx>dt = ƒ¿std 420 Chapter 6: Applications of Definite Integrals Formula for the Length of If ƒ is continuously differentiable on the closed interval [a, b], the length of the curve (graph) from to is (2)L = L b a B 1 + a dy dx b 2 dx = L b a 21 + [ƒ¿sxd] 2 dx. x = bx = ay = ƒsxd y = ƒsxd, a … x … b EXAMPLE 3 Applying the Arc Length Formula for a Graph Find the length of the curve Solution We use Equation (2) with and The length of the curve from to is = 2 3 # 1 8 s1 + 8xd 3>2 d 0 1 = 13 6 . L = L 1 0 B 1 + a dy dx b 2 dx = L 1 0 21 + 8x dx x = 1x = 0 a dy dx b 2 = A222x 1>2 B 2 = 8x. dy dx = 422 3 # 3 2 x 1>2 = 222x 1>2 y = 422 3 x 3>2 - 1 a = 0, b = 1, y = 422 3 x 3>2 - 1, 0 … x … 1. Eq. (2) with a = 0, b = 1 Let integrate, and replace u by 1 + 8x. u = 1 + 8x, 4100 AWL/Thomas_ch06p396-465 9/1/04 12:55 PM Page 420 6.3 Lengths of Plane Curves 421 Dealing with Discontinuities in dy/dx At a point on a curve where dy dx fails to exist, dx dy may exist and we may be able to find the curve’s length by expressing x as a function of y and applying the following ana- logue of Equation (2): >> Formula for the Length of If g is continuously differentiable on [c, d], the length of the curve from to is (3)L = L d c B 1 + a dx dy b 2 dy = L d c 21 + [g¿s yd] 2 dy. y = dy = c x = gs yd x = gsyd, c … y … d EXAMPLE 4 Length of a Graph Which Has a Discontinuity in dy/dx Find the length of the curve from to Solution The derivative is not defined at so we cannot find the curve’s length with Equation (2). We therefore rewrite the equation to express x in terms of y: Solve for x. From this we see that the curve whose length we want is also the graph of from to (Figure 6.27). The derivative is continuous on [0, 1]. We may therefore use Equation (3) to find the curve’s length: = 2 27 A10210 - 1B L 2.27. = 1 9 # 2 3 s1 + 9yd 3>2 d 0 1 L = L d c B 1 + a dx dy b 2 dy = L 1 0 21 + 9y dy dx dy = 2a 3 2 by 1>2 = 3y 1>2 y = 1y = 0 x = 2y 3>2 x = 2y 3>2 . y 3>2 = x 2 y = a x 2 b 2>3 x = 0, dy dx = 2 3 a x 2 b -1>3 a 1 2 b = 1 3 a 2 x b 1>3 x = 2.x = 0y = sx>2d 2>3 Raise both sides to the power 3/2. Eq. (3) with c = 0, d = 1. Let integrate, and substitute back. du>9 = dy, u = 1 + 9y, x y 0 1 2 (2, 1) 1 y H11005 2/3 , 0 H11349 x H11349 2 x 2 FIGURE 6.27 The graph of from to is also the graph of from to (Example 4). y = 1y = 0x = 2y 3>2 x = 2x = 0 y = sx>2d 2>3 4100 AWL/Thomas_ch06p396-465 9/1/04 12:55 PM Page 421 The Short Differential Formula Equation (1) is frequently written in terms of differentials in place of derivatives. This is done formally by writing under the radical in place of the dt outside the radical, and then writing and It is also customary to eliminate the parentheses in and write instead, so that Equation (1) is written (4) We can think of these differentials as a way to summarize and simplify the properties of integrals. Differentials are given a precise mathematical definition in a more advanced text. To do an integral computation, dx and dy must both be expressed in terms of one and the same variable, and appropriate limits must be supplied in Equation (4). A useful way to remember Equation (4) is to write (5) and treat ds as the differential of arc length, which can be integrated between appropriate limits to give the total length of a curve. Figure 6.28a gives the exact interpretation of ds corresponding to Equation (5). Figure 6.28b is not strictly accurate but is to be thought of as a simplified approximation of Figure 6.28a. With Equation (5) in mind, the quickest way to recall the formulas for arc length is to remember the equation If we write and have the graph of we can rewrite Equation (5) to get resulting in Equation (2). If we have instead we rewrite Equation (5) and obtain Equation (3). ds = 2dx 2 + dy 2 = B dy 2 + dx 2 dy 2 dy 2 = B 1 + dx 2 dy 2 dy = B 1 + a dx dy b 2 dy, x = gsyd, ds = 2dx 2 + dy 2 = B dx 2 + dy 2 dx 2 dx 2 = B 1 + dy 2 dx 2 dx = B 1 + a dy dx b 2 dx, y = ƒsxd,L = 1 ds Arc length = L ds. ds = 2dx 2 + dy 2 L = L 2dx 2 + dy 2 . dx 2 sdxd 2 a dy dt b 2 sdtd 2 = a dy dt dtb 2 = sdyd 2 . a dx dt b 2 sdtd 2 = a dx dt dtb 2 = sdxd 2 sdtd 2 422 Chapter 6: Applications of Definite Integrals HISTORICAL BIOGRAPHY James Gregory (1638–1675) y x 0 dx ds dy H9278 (a) y x 0 dx ds dy H9278 (b) FIGURE 6.28 Diagrams for remembering the equation ds = 2dx 2 + dy 2 . 4100 AWL/Thomas_ch06p396-465 9/1/04 12:55 PM Page 422 Commercial_CD 4100 AWL/Thomas_ch06p396-465