MATH /10/2008. The Setup. Archimedes and Quadrature

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1 MATH 60 Archimedes and uadrature The good Christian should beware of mathematicians, and all those who make empty prophecies. The danger already exists that the mathematicians have made a covenant with the Devil to darken the spirit and to confine man in the bonds of Hell. St. Augustine DeGenesi ad Litteram, Book II, xviii, 37 0-Sept-00 MATH 60 The Setup y mx + b y ax 0-Sept-00 MATH 60 3

2 The Setup is the point at which the tangent line to the curve is parallel to the secant. Where does the line intersect the parabola? ax mx + b ax mx b 0 x m+ m + b m m + b, x a a 0-Sept-00 MATH 60 4 oints of Intersection Now, we can find the points of intersection of the line and the parabola, and. m m m + b+ ab ( x, y) y ax a m m + b m m m + b+ ab, a a m + m m + b+ ab ( x, y) y ax a m + m + b m + m m + b+ ab, a a 0-Sept-00 MATH 60 5 Slope of the Tangent Line Here we will use some Calculus to help us. The slope of the tangent line is the derivative of the function at the point. d ( ax ) ax m dx m x a m m ( p, p) ( x, ax ), a 0-Sept-00 MATH 60 6

3 Area of the arabolic Sector x x 0-Sept-00 MATH 60 7 Calculus Again to the escue Again, Calculus will help us find the area, A. A x ( ax ( mx + b) ) x x a 3 x m x + bx 3 x ( m A + 4 ab) 6a 3/ dx 0-Sept-00 MATH 60 Area of Δ p r q 0-Sept-00 MATH

4 Area of Triangle It does not look like we can find a usable angle here. What are our options? () Drop a perpendicular from to and then use dot products to compute angles and areas. () Drop a perpendicular from to and follow the above prescription. (3) Drop a perpendicular from to and follow the above prescription. (4) Use Heron s Formula. 0-Sept-00 MATH 60 0 Using Heron s Formula: Area of the Triangle p d(, ) ( x x ) + ( y y ) p + +m a ( m 4 ab)( ) q d(, ) ( x p ) + ( y p ) q ( m + 4 ab)(b+ 4+ 5m + 4m m + 4 ab) 0-Sept-00 MATH 60 Area of the Triangle r d(, ) ( p x ) + ( p y ) ( m + 4 ab)(b+ 4+ 5m 4m m + 4 ab) r p + q+ r Now, the semiperimeter is: s m + b s ( 4 + m a + b m + 4m m + b + b m 4m m + b 0-Sept-00 MATH 60 4

5 Area of the Triangle Uh oh!!!! Are we in trouble? Heron s Formula states that the area is the following product: K ss ( p)( s q)( s r) This does not look promising!! 0-Sept-00 MATH 60 3 ( m + 4 ab) K + m + ab+ + m + m m + ab 4096a + b m 4m m + b 4 + m + b m + 4m m + b + b m 4m m + b m ab+ + m + m m + ab + b m 4m m + b 4 + m b 4 5m 4m m b b 4 5 ( ( m 4m m + b / 0-Sept-00 MATH 60 4 and then a miracle occurs K ( m K ( m + b) ab) a 3/ Note then that: 3/ (m + 4 ab ) A 4 K 6a 3 0-Sept-00 MATH

6 How did Archimedes do this? Claim: Δ ΔS S 0-Sept-00 MATH 60 6 How did Archimedes do this? Claim: Δ ΔS What do we mean by equals here? What did Archimedes mean by equals? 0-Sept-00 MATH 60 7 What good does this do? What is the area of the quadrilateral S? A K + K 0-Sept-00 MATH 60 6

7 A better approximation What is the area of the pentelateral ST? T S 0-Sept-00 MATH 60 9 The better approximation Note that the triangle ΔT is exactly the same as ΔS so we have that A K + K + K K + K 4 0-Sept-00 MATH 60 0 An even better approximation Z 4 T Z S Z Z 3 0-Sept-00 MATH 60 7

8 The next approximation Let s go to the next level and add the four triangles given by secant lines S, S, T, and T. area( Δ ZS) area( Δ SZ) area( ΔS) K K 64 area( Δ Z 3T ) area( Δ TZ 4) area( ΔT) K K 64 0-Sept-00 MATH 60 The next approximation What is the area of this new polygon that is a much better approximation to the area of the sector of the parabola? 4 A A + K K + K + K Sept-00 MATH 60 3 The next approximation What is the area of each triangle in terms of the original stage? K K 3 3 K K K What is the area of the new approximation? A3 A + K3 K + K + K + K Sept-00 MATH 60 4

9 The next approximation Okay, we have a pattern to follow now. How many triangles to we add at the next stage? What is the area of each triangle in terms of the previous stage? K3 K 0-Sept-00 MATH 60 5 The next approximation What is the area of the next stage? We add twice as many triangles each of which has an eighth of the area of the previous triangle. Thus we see that t in general, An K + K + K + + K n This, too, Archimedes had found without the aid of modern algebraic notation. 0-Sept-00 MATH 60 6 The Final Analysis Now, Archimedes has to convince his readers that by exhaustion this infinite series converges to the area of the sector of the parabola. Now, he had to sum up the series. He knew n Sept-00 MATH

10 The Final Analysis Therefore, Archimedes arrives at the result 4 A K 3 Note that this is what we found by Calculus. Do you think that this means that Archimedes knew the basics of calculus? 0-Sept-00 MATH 60 Let M be the midpoint of. Claim : The x- coordinate of M is the same as that of, the vertex. roof of the Claim M 0-Sept-00 MATH 60 9 roof of the Claim From our coordinate geometry we find that the x- coordinate of M is given by: x + x Mx m + m + b m m + b + a a m a 0-Sept-00 MATH M 0

11 roof of the Claim Claim : If is a vertex and M is any point on the chord, then the ratio M /M is independent of M. M m y + y m( x + x ) + b m a a Mx My + b 0-Sept-00 MATH 60 3 roof of the Claim Now the square of the length M is ( M) ( x x ) + ( y y ) ( x x ) + m ( x x ) m + b ( x x) ( + m ) ( + m ) ( m + 4 ab)( + m ) M a 0-Sept-00 MATH 60 3 roof of the Claim Now length M is m m M My y + b a M M ( 4 ) m + ab ( m + 4 ab)( m + ) ( m + ) ( m + 4 ab) a 0-Sept-00 MATH 60 33

12 roof of the Claim This ratio depends only on the slope of the line and the coefficient of the parabola, not on anything else. Thus it is independent of the point along the chord. 0-Sept-00 MATH roof of the Claim Claim 3: If is a chord of a parabola, M its midpoint and N the midpoint of M. Drop perpendiculars from M and N to the x-axis and let them intersect the parabola at and T. Then 4 M NT 3 M N W T 0-Sept-00 MATH roof of the Claim Construct TW parallel to MN. Now from the previous claim we have: M TW M W Since MNTW is a parallelogram M MN TW M 4TW M 4W 4 4 V W + WM WM 3W V WM NT 3 3 M W N T 0-Sept-00 MATH 60 36

13 Claim 3: Δ ΔS roof of the Claim We know that Y is the midpoint M of. Thus SY bisects one side of ΔM and is N parallel to M. Thus ΔYN and ΔM are similar. Also, SY Y W intersects M at its midpoint and S 0-Sept-00 MATH roof of the Claim 4 YN V SN SN 3 3 YN SY Δ N Δ YN +Δ YN Δ YS + Δ YS ΔS Δ Δ M 4Δ N ΔS.E.D. uod erat demonstrandum uit, enough done. 0-Sept-00 MATH