Archimedes of Syracuse
Prove Archimedes' Proposition 1 algebraically
(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
for some positive constant a.
Suppose the line that is tangent to the parabola at P has equation
= 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.
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
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,
respectively, and PV is vertical (parallel to the axis). If
the line tangent to the parabola at P has equation y =
+ b, then the lines QV and
Q'V', being parallel
to the tangent, have the same slopes, so they have equations of the form
= mx + c and
y = mx +
(a) Explain why
(b) Let W,
be the points where the horizontal lines through Q and
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,
= 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.
Use the result of Archimedes Proposition
24 to evaluate these integrals:
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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
P = (-1,1) and Q = (1,1).
(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
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
(c) Carry out this procedure
one more time by inserting four more points T, U, V,
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... .
last modified 9/5/02
Copyright (c) 2000. Daniel