## Archimedes of Syracuse

#### Exercises

1. Prove Archimedes' Proposition 1 algebraically as follows:

2. (a)  Orienting the parabola so that its axis of symmetry is the y-axis and its vertex  is the origin of the xy-plane, we can treat it as the graph of the equation

for some positive constant a.  Suppose the line that is tangent to the parabola at P has equation y = mx + b.  Then since the line QQ' is supposed to be parallel to the tangent line, it must have equation of the form y = mx + c for some constant c (the slopes must be equal).

By solving simultaneously the equations for the parabola and for QQ', determine the x-coordinates of the points Q and Q'.  Use the quadratic formula.
(b)  Now solve simultaneously the equations of the parabola and the tangent line to find the x-coordinate of the point P.  How do we explain the fact that the quadratic formula gives two solutions to the problem if we know that there is only one point of intersection?  [Hint: Some term in the expression for the solutions must equal 0.]
(c)  PV is parallel to the axis of the parabola, so what is the x-coordinate of V?
(d)  Use the results of parts (a) and (c) to argue why V must be midway between Q and Q'.  This ends the proof.
1. Prove Proposition 3 algebraically:  The diagram shows the situation described in the hypotheses of the proposition.  As in the previous exercise, we choose to orient the parabola so that the y-axis is its axis of symmetry and its vertex lies on the x-axis.  Hence its equation is
The points P, Q, and Q' are chosen on the parabola to have x-coordinates p, q, q', respectively, and PV is vertical (parallel to the axis).  If the line tangent to the parabola at P has equation y = mx + b, then the lines QV and Q'V', being parallel to the tangent, have the same slopes, so they have equations of the form y = mx + c and y = mx + c' respectively.
(a)  Explain why
(b)  Let W, W' be the points where the horizontal lines through Q and Q' meet the line PV.  By considering the triangles QVW and Q'V'W', show that
(c)  Since P lies on the parabola and the line y = mx + b (and similarly for Q and Q'), we can set the equations of the parabola and line equal, substituting x = p (or, appropriately, x = q and x = q') to find that
Take the first of these equations and use the quadratic formula to solve for p, noting that since the line is tangent to the curve at this point, there can only be one solution to this equation (see #1(b) above).  From this, show that m = 2ap.
(d)  Use the equations we found in part (c) to show that
(e)  Now complete the proof of the proposition.

1. Use the result of Archimedes Proposition 24 to evaluate these integrals:

1. In a manner similar to the way in which Archimedes approximated the circumference of a circle in Measurement of the Circle, we can attempt to approximate the arclength of a portion of a parabola.  Consider the parabola with equation y = x2 and suppose that we wish to estimate the length of the curve between the points P = (-1,1) and Q = (1,1).

2. (a)  As a first crude approximation we join P and Q to the origin O = (0,0) (which also lies on the curve) and approximate the curve with the pair of line segments PO and OQ.  Draw a diagram illustrating this and compute the total length of the two line segments.
(b)  A second and better approximation results if we introduce two more intermediate points R = (-1/2,1/4) and S = (1/2,1/4).  Draw a diagram illustrating this and compute the total length of the four line segments PR + RO + OS + SQ.
(c)  Carry out this procedure one more time by inserting four more points T, U, V, W midway between the ones we have and computing the total length of the resulting eight line segments.  Draw the corresponding diagram and compare your approximation with the true value of 2.95788571... .