Arc Length and Surface Area in Parametric Equations

Transcription

1 Arc Length and Surface Area in Parametric Equations MATH 211, Calculus II J. Robert Buchanan Department of Mathematics Spring 2011

2 Background We have developed definite integral formulas for arc length and surface area for curves of the form y = f(x) with a x b. s = S b = 2π 1 + (f (x)) 2 dx a b a f(x) 1 + (f (x)) 2 dx Today we will develop formulas for calculating arc length and surface area for curves described parametrically.

3 Riemann Sum Approach Suppose a curve is described by the parametric equations: x y = x(t) = y(t) where a t b and x (t) and y (t) are continuous as well.

4 Partition Partition [a, b] into n equal subintervals with t = b a n t k = a + k t for k = 0, 1,...,n. and s k = = = (x(t k ) x(t k 1 )) 2 + (y(t k ) y(t k 1 )) 2 (x (v k ) t) 2 + (y (w k ) t) 2 (by the MVT) (x (v k )) 2 + (y (w k )) 2 t (x (w k )) 2 + (y (w k )) 2 t

5 Riemann Sum Arc length s n k=1 = lim = (x (w k )) 2 + (y (w k )) 2 t n n k=1 b a (x (w k )) 2 + (y (w k )) 2 t (x (t)) 2 + (y (t)) 2 dt

6 Riemann Sum Arc length s n k=1 = lim = (x (w k )) 2 + (y (w k )) 2 t n n k=1 b Remark: the expression ds = differential arc length. a (x (w k )) 2 + (y (w k )) 2 t (x (t)) 2 + (y (t)) 2 dt (x (t)) 2 + (y (t)) 2 dt is called

7 Result Theorem For the curve defined parametrically by x = x(t), y = y(t), a t b, if x (t) and y (t) are continuous on [a, b] and the curve does not intersect itself (except possibly at a finite number of points), then the arc length s of the curve is given by s = b a b (x (t)) 2 + (y (t)) 2 dt = a (dx ) 2 + dt ( ) dy 2 dt. dt

8 Example Find the arc length of the curve given by x = 2t 2 y = 2t 3 for 0 t 2. y x

9 Solution s = = (4t) 2 + (6t 2 ) 2 dt 2t 4 + 9t 2 dt Integrate by substitution, letting u = 4 + 9t 2 and 1 du = 2t dt. 9 s = u1/2 du = 2 27 u3/2 40 = 2 27 ( ) = ( ) 4

10 Example Find the arc length of the curve given by x = 3 sin t y = 3 cos t 3 for 0 t 2π. y x

11 Solution s = = = 2π 0 2π 0 2π 0 = 6π (3 cos t) 2 + ( 3 sin t) 2 dt 9 cos 2 t + 9 sin 2 t dt 3 dt

12 Brachistochrone Problem Problem: suppose an object is to slide down a path in the xy-plane from the origin to a point below the origin but not on the y-axis. Which path should the object follow to accomplish the trip in the least time? y x

13 Physical Background distance d = speed time = v t Conservation of energy: potential energy = kinetic energy ( (dx mgy = 1 ) 2 2 m + dt ( ) ) dy 2 dt To avoid confusion with time use the symbol u as the parameter of the parametric equations. To simplify the mathematics we will assume that gy 0.

14 Formula for Trip Time Suppose the path is parameterized by x = x(u), y = y(u) for 0 u 1, then ( (dx mgy = 1 ) 2 ( ) ) dy 2 2 m + 2gy = du ( ) dx 2 + du 2gy = (v(u)) 2 2gy = v(u). ( dy du ) 2 du

15 Formula for Trip Time Suppose the path is parameterized by x = x(u), y = y(u) for 0 u 1, then ( (dx mgy = 1 ) 2 ( ) ) dy 2 2 m + 2gy = du ( ) dx 2 + du 2gy = (v(u)) 2 2gy = v(u). ( dy du Therefore the trip time is ( t = d dx ) 2 ( ) 2 1 v = du + dy du du = 1 1 ( ) dx 2 ( ) 2 du + dy du du. 0 2gy 2g 0 y ) 2 du

16 Comparison of Paths Suppose the object is to move from (0, 0) to (π, 2) along a straight line parameterized by x y = πu = 2u for 0 u 1. Find the time of travel

17 Comparison of Paths Suppose the object is to move from (0, 0) to (π, 2) along a straight line parameterized by x y = πu = 2u for 0 u 1. Find the time of travel t = π g

18 Comparison of Paths Suppose the object is to move from (0, 0) to (π, 2) along a cycloid parameterized by x y = πu sin πu = 1 + cosπu for 0 u 1. Find the time of travel. y x

19 Comparison of Paths Suppose the object is to move from (0, 0) to (π, 2) along a cycloid parameterized by x y = πu sin πu = 1 + cosπu for 0 u 1. Find the time of travel. y x t = π 2 g

20 Result π 2 g < π g Even though the cycloid is a greater distance (arc length), it requires less time. y x

21 Surface Area Geometrically we may think of the definite integral for the surface area of a solid of revolution as S = b a 2π(radius)(arc length) dx. Thus for the surface generated when the parametric curve x y = x(t) = y(t) for a t b is revolved around the x-axis is b S = 2π y(t) (x (t)) 2 + (y (t)) 2 dt a

22 Example Find the surface area of the solid of revolution generated when the parametric curve x = 3t y = 7t + 1 for 0 t 1 is revolved around the x-axis

23 Solution S = 2π = 2π = 8π ( 7t + 1) (3) 2 + ( 7) 2 dt 4( 7t + 1) dt ( 7t + 1) dt ( ) 7 = 8π 2 t2 + t ( ) 7 = 8π = 4( 7 + 2)π 1 0

24 Example Find the surface area of the solid of revolution generated when the parametric curve x = 4t 2 1 y = 3 2t for 2 t 0 is revolved around the x-axis

25 Solution (1 of 2) S = 2π = 2π = 4π = 12π (3 2t) (8t) 2 + ( 2) 2 dt (3 2t) 64t dt (3 2t) 16t dt t dt 8π t 16t dt 2 The first integral can be handled via the trigonometric substitution, t = 1 4 tanθ and dt = 1 4 sec2 θ dθ. The second integral can be handled via the substitution, u = 16t and 1 du = t dt. 32

26 Solution (1 of 2) S 0 = 12π = 3π t dt 8π tan 1 8 sec 3 θ dθ π t 16t dt u 1/2 du = 3π 2 (secθ tanθ ln[secθ + tanθ]) 0 tan 1 8 ( π 6 u3/2) 1 = 12 65π + 3π 2 ln(8 + 65) π 6 ( )

27 Homework Read Section 9.3 Exercises: 1 19 odd (arc length), odd (surface area)