Lesson 23: Areas and Distances (Section 10 version)
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Lesson 23: Areas and Distances (Section 10 version)

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We trace the computation of area through the centuries. The process known known as Riemann Sums has applications to not just area but many fields of science.

We trace the computation of area through the centuries. The process known known as Riemann Sums has applications to not just area but many fields of science.

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Lesson 23: Areas and Distances (Section 10 version) Lesson 23: Areas and Distances (Section 10 version) Presentation Transcript

  • Section 5.1 Areas and Distances V63.0121, Calculus I April 13, 2009 Announcements Moving to 624 today (no OH) Quiz 5 this week on §§4.1–4.4 Back HW returned in recitation this week . . . . . .
  • Outline Archimedes Cavalieri Generalizing Cavalieri’s method Distances Other applications . . . . . .
  • Meet the mathematician: Archimedes 287 BC – 212 BC (after Euclid) Geometer Weapons engineer . . . . . .
  • Meet the mathematician: Archimedes 287 BC – 212 BC (after Euclid) Geometer Weapons engineer . . . . . .
  • Meet the mathematician: Archimedes 287 BC – 212 BC (after Euclid) Geometer Weapons engineer . . . . . .
  • . Archimedes found areas of a sequence of triangles inscribed in a parabola. A= . . . . . .
  • 1 . . Archimedes found areas of a sequence of triangles inscribed in a parabola. A=1 . . . . . .
  • 1 . .1 .1 8 8 . Archimedes found areas of a sequence of triangles inscribed in a parabola. 1 A=1+2· 8 . . . . . .
  • 1 1 .64 .64 1 . .1 .1 8 8 1 1 .64 .64 . Archimedes found areas of a sequence of triangles inscribed in a parabola. 1 1 A=1+2· +4· + ··· 8 64 . . . . . .
  • 1 1 .64 .64 1 . .1 .1 8 8 1 1 .64 .64 . Archimedes found areas of a sequence of triangles inscribed in a parabola. 1 1 A=1+2· +4· + ··· 8 64 1 1 1 + ··· + n + ··· =1+ + 4 16 4 . . . . . .
  • We would then need to know the value of the series 1 1 1 + ··· + n + ··· 1+ + 4 16 4 . . . . . .
  • We would then need to know the value of the series 1 1 1 + ··· + n + ··· 1+ + 4 16 4 But for any number r and any positive integer n, (1 − r)(1 + r + · · · + rn ) = 1 − rn+1 So 1 − rn+1 1 + r + · · · + rn = 1−r . . . . . .
  • We would then need to know the value of the series 1 1 1 + ··· + n + ··· 1+ + 4 16 4 But for any number r and any positive integer n, (1 − r)(1 + r + · · · + rn ) = 1 − rn+1 So 1 − rn+1 1 + r + · · · + rn = 1−r Therefore 1 − (1/4)n+1 1 1 1 + ··· + n = 1+ + 1 − 1/4 4 16 4 . . . . . .
  • We would then need to know the value of the series 1 1 1 + ··· + n + ··· 1+ + 4 16 4 But for any number r and any positive integer n, (1 − r)(1 + r + · · · + rn ) = 1 − rn+1 So 1 − rn+1 1 + r + · · · + rn = 1−r Therefore 1 − (1/4)n+1 1 1 1 1 4 + ··· + n = → 1+ + = 1− 4 16 4 3 1/4 3/4 as n → ∞. . . . . . .
  • Outline Archimedes Cavalieri Generalizing Cavalieri’s method Distances Other applications . . . . . .
  • Cavalieri Italian, 1598–1647 Revisited the area problem with a different perspective . . . . . .
  • Cavalieri’s method Divide up the interval into pieces and measure the area . = x2 y of the inscribed rectangles: . . . . . . .
  • Cavalieri’s method Divide up the interval into pieces and measure the area . = x2 y of the inscribed rectangles: 1 L2 = 8 . . . . . . .
  • Cavalieri’s method Divide up the interval into pieces and measure the area . = x2 y of the inscribed rectangles: 1 L2 = 8 L3 = . . . . . . .
  • Cavalieri’s method Divide up the interval into pieces and measure the area . = x2 y of the inscribed rectangles: 1 L2 = 8 1 4 5 L3 = + = 27 27 27 . . . . . . .
  • Cavalieri’s method Divide up the interval into pieces and measure the area . = x2 y of the inscribed rectangles: 1 L2 = 8 1 4 5 L3 = + = 27 27 27 L4 = . . . . . . .
  • Cavalieri’s method Divide up the interval into pieces and measure the area . = x2 y of the inscribed rectangles: 1 L2 = 8 1 4 5 L3 = + = 27 27 27 1 4 9 14 L4 = + + = 64 64 64 64 . . . . . . .
  • Cavalieri’s method Divide up the interval into pieces and measure the area . = x2 y of the inscribed rectangles: 1 L2 = 8 1 4 5 L3 = + = 27 27 27 1 4 9 14 L4 = + + = 64 64 64 64 . L5 = . . . . . .
  • Cavalieri’s method Divide up the interval into pieces and measure the area . = x2 y of the inscribed rectangles: 1 L2 = 8 1 4 5 L3 = + = 27 27 27 1 4 9 14 L4 = + + = 64 64 64 64 1 4 9 16 30 . L5 = + + + = 125 125 125 25 125 . . . . . .
  • Cavalieri’s method Divide up the interval into pieces and measure the area . = x2 y of the inscribed rectangles: 1 L2 = 8 1 4 5 L3 = + = 27 27 27 1 4 9 14 L4 = + + = 64 64 64 64 1 4 9 16 30 . L5 = + + + = 125 125 125 25 125 Ln =? . . . . . .
  • What is Ln ? 1 Divide the interval [0, 1] into n pieces. Then each has width . n . . . . . .
  • What is Ln ? 1 Divide the interval [0, 1] into n pieces. Then each has width . n The rectangle over the ith interval and under the parabola has area ( ) i − 1 2 (i − 1)2 1 · = . n3 n n . . . . . .
  • What is Ln ? 1 Divide the interval [0, 1] into n pieces. Then each has width . n The rectangle over the ith interval and under the parabola has area ( ) i − 1 2 (i − 1)2 1 · = . n3 n n So 22 1 + 22 + 32 + · · · + (n − 1)2 (n − 1)2 1 + 3 + ··· + Ln = = n3 n n3 n3 . . . . . .
  • What is Ln ? 1 Divide the interval [0, 1] into n pieces. Then each has width . n The rectangle over the ith interval and under the parabola has area ( ) i − 1 2 (i − 1)2 1 · = . n3 n n So 22 1 + 22 + 32 + · · · + (n − 1)2 (n − 1)2 1 + 3 + ··· + Ln = = n3 n n3 n3 The Arabs knew that n(n − 1)(2n − 1) 1 + 22 + 32 + · · · + (n − 1)2 = 6 So n(n − 1)(2n − 1) Ln = 6n3 . . . . . .
  • What is Ln ? 1 Divide the interval [0, 1] into n pieces. Then each has width . n The rectangle over the ith interval and under the parabola has area ( ) i − 1 2 (i − 1)2 1 · = . n3 n n So 22 1 + 22 + 32 + · · · + (n − 1)2 (n − 1)2 1 + 3 + ··· + Ln = = n3 n n3 n3 The Arabs knew that n(n − 1)(2n − 1) 1 + 22 + 32 + · · · + (n − 1)2 = 6 So n(n − 1)(2n − 1) 1 → Ln = 6n3 3 as n → ∞. . . . . . .
  • Cavalieri’s method for different functions Try the same trick with f(x) = x3 . We have () () ( ) n−1 1 1 1 2 1 Ln = · f + ·f + ··· + · f n n n n n n . . . . . .
  • Cavalieri’s method for different functions Try the same trick with f(x) = x3 . We have () () ( ) n−1 1 1 1 2 1 Ln = · f + ·f + ··· + · f n n n n n n 1 23 1 (n − 1)3 11 · 3 + · 3 + ··· + · = n3 nn nn n . . . . . .
  • Cavalieri’s method for different functions Try the same trick with f(x) = x3 . We have () () ( ) n−1 1 1 1 2 1 Ln = · f + ·f + ··· + · f n n n n n n 1 23 1 (n − 1)3 11 · 3 + · 3 + ··· + · = n3 nn nn n 3 3 3 1 + 2 + 3 + · · · + (n − 1) = n3 . . . . . .
  • Cavalieri’s method for different functions Try the same trick with f(x) = x3 . We have () () ( ) n−1 1 1 1 2 1 Ln = · f + ·f + ··· + · f n n n n n n 1 23 1 (n − 1)3 11 · 3 + · 3 + ··· + · = n3 nn nn n 3 3 3 1 + 2 + 3 + · · · + (n − 1) = n3 The formula out of the hat is [1 ]2 1 + 23 + 33 + · · · + (n − 1)3 = − 1) 2 n(n . . . . . .
  • Cavalieri’s method for different functions Try the same trick with f(x) = x3 . We have () () ( ) n−1 1 1 1 2 1 Ln = · f + ·f + ··· + · f n n n n n n 1 23 1 (n − 1)3 11 · 3 + · 3 + ··· + · = n3 nn nn n 3 3 3 1 + 2 + 3 + · · · + (n − 1) = n3 The formula out of the hat is [1 ]2 1 + 23 + 33 + · · · + (n − 1)3 = − 1) 2 n(n So n2 (n − 1)2 1 → Ln = 4n4 4 as n → ∞. . . . . . .
  • Cavalieri’s method with different heights 1 13 1 2 3 1 n3 · 3 + · 3 + ··· + · 3 Rn = nn nn nn 1 3 + 23 + 33 + · · · + n3 = n4 1 [1 ]2 = 4 2 n(n + 1) n n2 (n + 1)2 1 → = 4 4n 4 . as n → ∞. . . . . . .
  • Cavalieri’s method with different heights 1 13 1 2 3 1 n3 · 3 + · 3 + ··· + · 3 Rn = nn nn nn 1 3 + 23 + 33 + · · · + n3 = n4 1 [1 ]2 = 4 2 n(n + 1) n n2 (n + 1)2 1 → = 4 4n 4 . as n → ∞. So even though the rectangles overlap, we still get the same answer. . . . . . .
  • Outline Archimedes Cavalieri Generalizing Cavalieri’s method Distances Other applications . . . . . .
  • Cavalieri’s method in general Let f be a positive function defined on the interval [a, b]. We want to find the area between x = a, x = b, y = 0, and y = f(x). For each positive integer n, divide up the interval into n pieces. b−a . For each i between 1 and n, let xi be the nth Then ∆x = n step between a and b. So x0 = a b−a x1 = x0 + ∆x = a + n b−a x2 = x1 + ∆x = a + 2 · n ······ b−a xi = a + i · n xx x . i . n−1. n xxx .0 .1 .2 ······ . ....... b−a a . b . xn = a + n · =b . . . . . .
  • Forming Riemann sums We have many choices of how to approximate the area: Ln = f(x0 )∆x + f(x1 )∆x + · · · + f(xn−1 )∆x Rn = f(x1 )∆x + f(x2 )∆x + · · · + f(xn )∆x ( ) ( ) ( ) x0 + x 1 x 1 + x2 xn−1 + xn ∆x + · · · + f Mn = f ∆x + f ∆x 2 2 2 . . . . . .
  • Forming Riemann sums We have many choices of how to approximate the area: Ln = f(x0 )∆x + f(x1 )∆x + · · · + f(xn−1 )∆x Rn = f(x1 )∆x + f(x2 )∆x + · · · + f(xn )∆x ( ) ( ) ( ) x0 + x 1 x 1 + x2 xn−1 + xn ∆x + · · · + f Mn = f ∆x + f ∆x 2 2 2 In general, choose ci to be a point in the ith interval [xi−1 , xi ]. Form the Riemann sum Sn = f(c1 )∆x + f(c2 )∆x + · · · + f(cn )∆x n ∑ f(ci )∆x = i=1 . . . . . .
  • Theorem of the Day Theorem If f is a continuous function on [a, b] or has finitely many jump discontinuities, then lim Sn = lim {f(c1 )∆x + f(c2 )∆x + · · · + f(cn )∆x} n→∞ n→∞ exists and is the same value no matter what choice of ci we made. . . . . . .
  • Outline Archimedes Cavalieri Generalizing Cavalieri’s method Distances Other applications . . . . . .
  • Distances Just like area = length × width, we have distance = rate × time. So here is another use for Riemann sums. . . . . . .
  • Example A sailing ship is cruising back and forth along a channel (in a straight line). At noon the ship’s position and velocity are recorded, but shortly thereafter a storm blows in and position is impossible to measure. The velocity continues to be recorded at thirty-minute intervals. Time 12:00 12:30 1:00 1:30 2:00 Speed (knots) 4 8 12 6 4 Direction E E E E W Time 2:30 3:00 3:30 4:00 Speed 3 3 5 9 Direction W E E E Estimate the ship’s position at 4:00pm. . . . . . .
  • Solution We estimate that the speed of 4 knots (nautical miles per hour) is maintained from 12:00 until 12:30. So over this time interval the ship travels ( )( ) 4 nmi 1 hr = 2 nmi hr 2 We can continue for each additional half hour and get distance = 4 × 1/2 + 8 × 1/2 + 12 × 1/2 + 6 × 1/2 − 4 × 1/2 − 3 × 1/2 + 3 × 1/2 + 5 × 1/2 = 15.5 So the ship is 15.5 nmi east of its original position. . . . . . .
  • Analysis This method of measuring position by recording velocity is known as dead reckoning. If we had velocity estimates at finer intervals, we’d get better estimates. If we had velocity at every instant, a limit would tell us our exact position relative to the last time we measured it. . . . . . .
  • Outline Archimedes Cavalieri Generalizing Cavalieri’s method Distances Other applications . . . . . .
  • Other uses of Riemann sums Anything with a product! Area, volume Anything with a density: Population, mass Anything with a “speed:” distance, throughput, power Consumer surplus Expected value of a random variable . . . . . .
  • Summary Riemann sums can be used to estimate areas, distance, and other quantities The limit of Riemann sums can get the exact value Coming up: giving this limit a name and working with it. . . . . . .