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# Lesson 22: Areas and Distances (handout)

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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.

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### 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