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The Problem of Area: Archimedes's Quadrature of the Parabola

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The Problem of Area: Archimedes's Quadrature of the Parabola The problem of area: Archimedes's quadrature of the parabola Francesco Cellarosi Math 120 - Lecture 25 - November 14, 2016 Math 120 Archimedes's quadrature of the parabola November 14, 2016 1 / 22 Area = b · h Area = b · h h b b The problem of area We know how to compute the area of simple geometrical figures h b Math 120 Archimedes's quadrature of the parabola November 14, 2016 2 / 22 Area = b · h h b b The problem of area We know how to compute the area of simple geometrical figures h Area = b · h b Math 120 Archimedes's quadrature of the parabola November 14, 2016 2 / 22 Area = b · h b The problem of area We know how to compute the area of simple geometrical figures h Area = b · h h b b Math 120 Archimedes's quadrature of the parabola November 14, 2016 2 / 22 b The problem of area We know how to compute the area of simple geometrical figures h Area = b · h Area = b · h h b b Math 120 Archimedes's quadrature of the parabola November 14, 2016 2 / 22 The problem of area We know how to compute the area of simple geometrical figures h Area = b · h Area = b · h h b b b Math 120 Archimedes's quadrature of the parabola November 14, 2016 2 / 22 The problem of area We know how to compute the area of simple geometrical figures h Area = b · h Area = b · h h b b 1 h Area = 2 b · h b Math 120 Archimedes's quadrature of the parabola November 14, 2016 2 / 22 The problem of area We know how to compute the area of simple geometrical figures h Area = b · h Area = b · h h b b a c Area = pp(p − a)(p − b)(p − c) 1 p = 2 (a + b + c) (Heron's formula, 1st century CE) b Math 120 Archimedes's quadrature of the parabola November 14, 2016 2 / 22 Area = p(p − a)(p − b)(p − c)(p − d) 1 p = 2 (a + b + c + d) (Brahmagupta's formula, 7th century CE) The problem of area We know how to compute the area of simple geometrical figures b a c d Math 120 Archimedes's quadrature of the parabola November 14, 2016 3 / 22 (Brahmagupta's formula, 7th century CE) The problem of area We know how to compute the area of simple geometrical figures b a Area = p(p − a)(p − b)(p − c)(p − d) 1 p = 2 (a + b + c + d) c d Math 120 Archimedes's quadrature of the parabola November 14, 2016 3 / 22 The problem of area We know how to compute the area of simple geometrical figures b a Area = p(p − a)(p − b)(p − c)(p − d) 1 p = 2 (a + b + c + d) c d (Brahmagupta's formula, 7th century CE) Math 120 Archimedes's quadrature of the parabola November 14, 2016 3 / 22 Area = πr 2 π = circumference diameter (due to Archimedes, c. 260 BCE) The problem of area We know how to compute the area of simple geometrical figures r Math 120 Archimedes's quadrature of the parabola November 14, 2016 4 / 22 (due to Archimedes, c. 260 BCE) The problem of area We know how to compute the area of simple geometrical figures Area = πr 2 r π = circumference diameter Math 120 Archimedes's quadrature of the parabola November 14, 2016 4 / 22 The problem of area We know how to compute the area of simple geometrical figures Area = πr 2 r π = circumference diameter (due to Archimedes, c. 260 BCE) Math 120 Archimedes's quadrature of the parabola November 14, 2016 4 / 22 Today we will prove a beautiful formula, due to Archimedes, for the area between a parabola and a segment whose endpoints are on the parabola. He proved this in a letter (later titled \Quadrature of the parabola") to his friend Dositheus of Pelusium (who succeeded Conon of Samos {also friend of Archimedes{ as director of the mathematical school of Alexandria). The problem of area What about more general curved regions? Math 120 Archimedes's quadrature of the parabola November 14, 2016 5 / 22 The problem of area What about more general curved regions? Today we will prove a beautiful formula, due to Archimedes, for the area between a parabola and a segment whose endpoints are on the parabola. He proved this in a letter (later titled \Quadrature of the parabola") to his friend Dositheus of Pelusium (who succeeded Conon of Samos {also friend of Archimedes{ as director of the mathematical school of Alexandria). Math 120 Archimedes's quadrature of the parabola November 14, 2016 5 / 22 Math 120 Archimedes's quadrature of the parabola November 14, 2016 6 / 22 Math 120 Archimedes's quadrature of the parabola November 14, 2016 7 / 22 Math 120 Archimedes's quadrature of the parabola November 14, 2016 8 / 22 _ Theorem (Archimedes). Consider the point P on the arc AB which is the 4 farthest from the segment AB. Then the area of the region R equals 3 times the area of the triangle P0 = 4ABP. Archimedes's Theorem _ Consider the region R between the parabolic arc AB and the segment AB. B A R Math 120 Archimedes's quadrature of the parabola November 14, 2016 9 / 22 4 Then the area of the region R equals 3 times the area of the triangle P0 = 4ABP. Archimedes's Theorem _ Consider the region R between the parabolic arc AB and the segment AB. B A R P _ Theorem (Archimedes). Consider the point P on the arc AB which is the farthest from the segment AB. Math 120 Archimedes's quadrature of the parabola November 14, 2016 9 / 22 Archimedes's Theorem _ Consider the region R between the parabolic arc AB and the segment AB. B A PR0 P _ Theorem (Archimedes). Consider the point P on the arc AB which is the 4 farthest from the segment AB. Then the area of the region R equals 3 times the area of the triangle P0 = 4ABP. Math 120 Archimedes's quadrature of the parabola November 14, 2016 9 / 22 FACT 1: The tangent line to the parabola at P is parallel to AB. FACT 2: The line through P and parallel to the... ....axis of the parabola meets AB in its middle point M Preliminary facts Here are some facts that we will assume as proven (they were known to Archimedes since the had already been proved by Euclid and Aristarchus). B A P Math 120 Archimedes's quadrature of the parabola November 14, 2016 10 / 22 FACT 2: The line through P and parallel to the... ....axis of the parabola meets AB in its middle point M Preliminary facts Here are some facts that we will assume as proven (they were known to Archimedes since the had already been proved by Euclid and Aristarchus). B A P FACT 1: The tangent line to the parabola at P is parallel to AB. Math 120 Archimedes's quadrature of the parabola November 14, 2016 10 / 22 Preliminary facts Here are some facts that we will assume as proven (they were known to Archimedes since the had already been proved by Euclid and Aristarchus). M B A P FACT 1: The tangent line to the parabola at P is parallel to AB. FACT 2: The line through P and parallel to the... ....axis of the parabola meets AB in its middle point M Math 120 Archimedes's quadrature of the parabola November 14, 2016 10 / 22 FACT 3: Every chord CD parallel to AB is bisected by PM, say at N. 2 2 FACT 4: PN=PM = ND =MB . We will assume FACTS 1÷4, without proof. Preliminary facts B M A P Math 120 Archimedes's quadrature of the parabola November 14, 2016 11 / 22 , say at N. 2 2 FACT 4: PN=PM = ND =MB . We will assume FACTS 1÷4, without proof. Preliminary facts B M A D C P FACT 3: Every chord CD parallel to AB is bisected by PM Math 120 Archimedes's quadrature of the parabola November 14, 2016 11 / 22 2 2 FACT 4: PN=PM = ND =MB . We will assume FACTS 1÷4, without proof. Preliminary facts B M A N D C P FACT 3: Every chord CD parallel to AB is bisected by PM, say at N. Math 120 Archimedes's quadrature of the parabola November 14, 2016 11 / 22 We will assume FACTS 1÷4, without proof. Preliminary facts B M A N D C P FACT 3: Every chord CD parallel to AB is bisected by PM, say at N. 2 2 FACT 4: PN=PM = ND =MB . Math 120 Archimedes's quadrature of the parabola November 14, 2016 11 / 22 Preliminary facts B M A N D C P FACT 3: Every chord CD parallel to AB is bisected by PM, say at N. 2 2 FACT 4: PN=PM = ND =MB . We will assume FACTS 1÷4, without proof. Math 120 Archimedes's quadrature of the parabola November 14, 2016 11 / 22 4APP1 from AP and 4PBP2 from PB. Proof. Two new triangles B M A P Recall: 4ABP was constructed from the chord AB. Now construct two new triangles in the same way: Math 120 Archimedes's quadrature of the parabola November 14, 2016 12 / 22 4APP1 from AP and 4PBP2 from PB. Proof. Two new triangles B M A P1 P Recall: 4ABP was constructed from the chord AB.
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