The problem of area: Archimedess quadrature of the parabola - PowerPoint PPT Presentation

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

The problem of area We know how to compute the area of simple geometrical figures h b

The problem of area We know how to compute the area of simple geometrical figures h 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

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

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 Area = 1 2 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 � a c Area = p ( p − a )( p − b )( p − c ) p = 1 2 ( a + b + c ) (Heron’s formula, 1st century CE) 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 b a c d

The problem of area We know how to compute the area of simple geometrical figures b a � Area = ( p − a )( p − b )( p − c )( p − d ) p = 1 2 ( a + b + c + d ) c d

The problem of area We know how to compute the area of simple geometrical figures b a � Area = ( p − a )( p − b )( p − c )( p − d ) p = 1 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

The problem of area We know how to compute the area of simple geometrical figures r

The problem of area We know how to compute the area of simple geometrical figures Area = π r 2 r π = circumference diameter

The problem of area We know how to compute the area of simple geometrical figures Area = π r 2 r π = circumference diameter (due to Archimedes, c. 260 BCE) Math 120 Archimedes’s quadrature of the parabola November 14, 2016 4 / 22

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

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

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 P 0 R P ⌢ Theorem (Archimedes). Consider the point P on the arc AB which is the farthest from the segment AB . Then the area of the region R equals 4 3 times the area of the triangle P 0 = △ ABP . Math 120 Archimedes’s quadrature of the parabola November 14, 2016 9 / 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). B A P 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). 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). B M 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

Preliminary facts B M A P Math 120 Archimedes’s quadrature of the parabola November 14, 2016 11 / 22

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

Preliminary facts B M A D N 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

Preliminary facts B M A D N C P FACT 3 : Every chord CD parallel to AB is bisected by PM , say at N . 2 / MB 2 . FACT 4 : PN / PM = ND Math 120 Archimedes’s quadrature of the parabola November 14, 2016 11 / 22

Preliminary facts B M A D N C P FACT 3 : Every chord CD parallel to AB is bisected by PM , say at N . 2 / MB 2 . FACT 4 : PN / PM = ND We will assume FACTS 1 ÷ 4 , without proof. Math 120 Archimedes’s quadrature of the parabola November 14, 2016 11 / 22

Proof. Two new triangles B M A P Recall: △ ABP 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

Proof. Two new triangles B M A P 1 P Recall: △ ABP 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

Proof. Two new triangles B M A P 1 P 1 P Recall: △ ABP was constructed from the chord AB . Now construct two new triangles in the same way: △ APP 1 from AP Math 120 Archimedes’s quadrature of the parabola November 14, 2016 12 / 22

Proof. Two new triangles B M A P 2 P 1 P 1 P Recall: △ ABP was constructed from the chord AB . Now construct two new triangles in the same way: △ APP 1 from AP Math 120 Archimedes’s quadrature of the parabola November 14, 2016 12 / 22

Proof. Two new triangles B M A P 2 P 2 P 1 P 1 P Recall: △ ABP was constructed from the chord AB . Now construct two new triangles in the same way: △ APP 1 from AP and △ PBP 2 from PB . Math 120 Archimedes’s quadrature of the parabola November 14, 2016 12 / 22

Proof. A bigger polygon B A P 2 P 1 P Math 120 Archimedes’s quadrature of the parabola November 14, 2016 13 / 22

Proof. A bigger polygon B A P 2 P 1 P The area of the polygon AP 1 PP 2 B is bigger than the area of the triangle △ ABP , but smaller than the area of the parabolic region Math 120 Archimedes’s quadrature of the parabola November 14, 2016 13 / 22

Proof. A bigger polygon B A P 2 P 1 P The area of the polygon AP 1 PP 2 B is bigger than the area of the triangle △ ABP , but smaller than the area of the parabolic region (because the parabola is concave up!). Math 120 Archimedes’s quadrature of the parabola November 14, 2016 13 / 22

Proof. A bigger polygon B A P 2 P 1 P The area of the polygon AP 1 PP 2 B is bigger than the area of the triangle △ ABP , but smaller than the area of the parabolic region (because the parabola is concave up!). Now that we have 4 new chords AP 1 , P 1 P , PP 2 , P 2 B we can repeat the construction above and obtain 4 new triangles. Math 120 Archimedes’s quadrature of the parabola November 14, 2016 13 / 22

Proof. Bigger and bigger polygons B A P area ( APB ) Math 120 Archimedes’s quadrature of the parabola November 14, 2016 14 / 22

Proof. Bigger and bigger polygons B A P 2 P 1 P area ( APB ) < area ( AP 1 PP 2 B ) Math 120 Archimedes’s quadrature of the parabola November 14, 2016 14 / 22

Proof. Bigger and bigger polygons B A P 6 P 3 P 2 P 1 P 5 P 4 P area ( APB ) < area ( AP 1 PP 2 B ) Math 120 Archimedes’s quadrature of the parabola November 14, 2016 14 / 22

Proof. Bigger and bigger polygons B A P 6 P 3 P 2 P 1 P 5 P 4 P area ( APB ) < area ( AP 1 PP 2 B ) < area ( AP 3 P 1 P 4 PP 5 P 2 P 6 B ) Math 120 Archimedes’s quadrature of the parabola November 14, 2016 14 / 22