Lesson 22: Areas and Distances (handout)

825 views

Published on

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.

(Handout version of slideshow from class)

Published in: Education
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total views
825
On SlideShare
0
From Embeds
0
Number of Embeds
1
Actions
Shares
0
Downloads
19
Comments
0
Likes
0
Embeds 0
No embeds

No notes for slide

Lesson 22: Areas and Distances (handout)

  1. 1. V63.0121.006/016, Calculus I Section 5.1 : Areas and Distances April 13, 2010 Notes Section 5.1 Areas and Distances V63.0121.006/016, Calculus I New York University April 13, 2010 Announcements Quiz April 16 on §§4.1–4.4 Final Exam: Monday, May 10, 12:00noon Announcements Notes Quiz April 16 on §§4.1–4.4 Final Exam: Monday, May 10, 12:00noon V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 2 / 30 Objectives Notes Compute the area of a region by approximating it with rectangles and letting the size of the rectangles tend to zero. Compute the total distance traveled by a particle by approximating it as distance = (rate)(time) and letting the time intervals over which one approximates tend to zero. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 3 / 30 1
  2. 2. V63.0121.006/016, Calculus I Section 5.1 : Areas and Distances April 13, 2010 Outline Notes Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applications V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 4 / 30 Easy Areas: Rectangle Notes Definition The area of a rectangle with dimensions and w is the product A = w . w It may seem strange that this is a definition and not a theorem but we have to start somewhere. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 5 / 30 Easy Areas: Parallelogram Notes By cutting and pasting, a parallelogram can be made into a rectangle. h b b So Fact The area of a parallelogram of base width b and height h is A = bh V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 6 / 30 2
  3. 3. V63.0121.006/016, Calculus I Section 5.1 : Areas and Distances April 13, 2010 Easy Areas: Triangle Notes By copying and pasting, a triangle can be made into a parallelogram. h b So Fact The area of a triangle of base width b and height h is 1 A = bh 2 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 7 / 30 Easy Areas: Other Polygons Notes Any polygon can be triangulated, so its area can be found by summing the areas of the triangles: V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 8 / 30 Hard Areas: Curved Regions Notes ??? V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 9 / 30 3
  4. 4. V63.0121.006/016, Calculus I Section 5.1 : Areas and Distances April 13, 2010 Meet the mathematician: Archimedes Notes Greek (Syracuse), 287 BC – 212 BC (after Euclid) Geometer Weapons engineer V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 10 / 30 Notes 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 =1+ + + ··· + n + ··· 4 16 4 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 11 / 30 We would then need to know the value of the series Notes 1 1 1 1+ + + ··· + n + ··· 4 16 4 But for any number r and any positive integer n, (1 − r )(1 + r + · · · + r n ) = 1 − r n+1 So 1 − r n+1 1 + r + · · · + rn = 1−r Therefore 1 1 1 1 − (1/4)n+1 1 4 1+ + + ··· + n = →3 = 4 16 4 1 − 1/4 /4 3 as n → ∞. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 12 / 30 4
  5. 5. V63.0121.006/016, Calculus I Section 5.1 : Areas and Distances April 13, 2010 Cavalieri Notes Italian, 1598–1647 Revisited the area problem with a different perspective V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 13 / 30 Cavalieri’s method Notes Divide up the interval into pieces y = x2 and measure the area 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 0 1 L5 = + + + = 125 125 125 125 125 Ln =? V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 14 / 30 What is Ln ? 1 Notes Divide the interval [0, 1] into n pieces. Then each has width . The n rectangle over the ith interval and under the parabola has area 2 1 i −1 (i − 1)2 · = . n n n3 So 1 22 (n − 1)2 1 + 22 + 32 + · · · + (n − 1)2 Ln = + + ··· + = n3 n3 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 → ∞. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 15 / 30 5
  6. 6. V63.0121.006/016, Calculus I Section 5.1 : Areas and Distances April 13, 2010 Cavalieri’s method for different functions Notes Try the same trick with f (x) = x 3 . We have 1 1 1 2 1 n−1 Ln = ·f + ·f + ··· + · f n n n n n n 1 1 1 23 1 (n − 1)3 = · 3 + · 3 + ··· + · n n n n n n3 1 + 23 + 33 + · · · + (n − 1)3 = n4 The formula out of the hat is 2 1 + 23 + 33 + · · · + (n − 1)3 = 1 2 n(n − 1) So n2 (n − 1)2 1 Ln = → 4n4 4 as n → ∞. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 16 / 30 Cavalieri’s method with different heights Notes 1 13 1 23 1 n3 Rn = · + · + ··· + · 3 n n3 n n3 n n 13 + 23 + 33 + · · · + n3 = n4 1 1 2 = 4 2 n(n + 1) n n2 (n + 1)2 1 = → 4n4 4 as n → ∞. So even though the rectangles overlap, we still get the same answer. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 17 / 30 Outline Notes Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applications V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 18 / 30 6
  7. 7. V63.0121.006/016, Calculus I Section 5.1 : Areas and Distances April 13, 2010 Cavalieri’s method in general Notes 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). b−a For each positive integer n, divide up the interval into n pieces. Then ∆x = . n For each i between 1 and n, let xi be the nth 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 ······ a b b−a x0 x1 x2 . . . xi xn−1xn xn = a + n · =b n V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 19 / 30 Forming Riemann sums Notes 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 + x1 x1 + x2 xn−1 + xn Mn = f ∆x + 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 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 20 / 30 Theorem of the Day Notes Theorem If f is a continuous function or has finitely many jump discontinuities on [a, b], then n lim Sn = lim f (ci )∆x n→∞ n→∞ i=1 exists and is the same value no axxxxxxxxxxxxxxxxxxx xb xx12x3xx25x3635x911x69158x16x19 x1x132445x6xx7x91165x101513167 matter what choice of ci we 11x22543748 58412xx1317 18 x11x315x76866710812124179195 223x4355591781111x1217 234251 7872x71314141010 134 46 x8x537x8x16129 4365 7 9 x9101311131 8 x9 13 10 153 10 276 9x 11 21362 24 3611141114 20 2 4 1091012166818 3 4 4 10131057142 6 12113 151113 8 14 1815 459 15 14 12 17 7 16 12 8 6 4 made. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 21 / 30 7
  8. 8. V63.0121.006/016, Calculus I Section 5.1 : Areas and Distances April 13, 2010 Analogies Notes The Area Problem (Ch. 5) The Tangent Problem (Ch. 2–4) Want the area of a curved region Want the slope of a curve Only know the area of Only know the slope of lines polygons Approximate curve with a Approximate region with line polygons Take limit over better and Take limit over better and better approximations better approximations V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 22 / 30 Outline Notes Area through the Centuries Euclid Archimedes Cavalieri Generalizing Cavalieri’s method Analogies Distances Other applications V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 23 / 30 Distances Notes Just like area = length × width, we have distance = rate × time. So here is another use for Riemann sums. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 24 / 30 8
  9. 9. V63.0121.006/016, Calculus I Section 5.1 : Areas and Distances April 13, 2010 Application: Dead Reckoning Notes V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 25 / 30 Example Notes 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. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 26 / 30 Notes 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. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 27 / 30 9
  10. 10. V63.0121.006/016, Calculus I Section 5.1 : Areas and Distances April 13, 2010 Analysis Notes This method of measuring position by recording velocity was necessary until global-positioning satellite technology became widespread 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. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 28 / 30 Other uses of Riemann sums Notes 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 V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 29 / 30 Summary Notes We can compute the area of a curved region with a limit of Riemann sums We can compute the distance traveled from the velocity with a limit of Riemann sums Many other important uses of this process. V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 30 / 30 10

×