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Topic: Historyu of the integral of secant
Replies: 13   Last Post: Mar 3, 2009 3:11 PM

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Michael Press

Posts: 2,137
Registered: 12/26/06
Re: Historyu of the integral of secant
Posted: Mar 1, 2009 2:30 PM
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In article <280220091705031971%edgar@math.ohio-state.edu.invalid>,
"G. A. Edgar" <edgar@math.ohio-state.edu.invalid> wrote:

> In article <49A8C201.20705@netscape.net>, Stephen J. Herschkorn
> <sjherschko@netscape.net> wrote:
>

> > What is the history of the trick to find the indefinite inteegral of the
> > secant function? Who first used it, and when?

>
> I seem to remember it goes back to Mercator. He needed it for a map
> projection. Of course that is long before Newton & Leibniz put
> together
> tools to make it easy.


Yes, Gerhard Kremer, called Mercator (1512-1594) worked
this out by subdivision, geometry, and adding up the bits.

Ships at sea set a constant bearing and maintain it. This
curve on the globe is called a loxodrome. It is clear
that over the open ocean a course between two points can
be made in one leg of constant bearing. It would be good
to have a chart for which

I Lines of constant latitude are parallel straight lines.
II Lines of constant longitude are parallel straight lines.
III The map from globe to chart is conformal so that lines
of latitude and longitude intersect at right angles.

Then the course bearing could be found by laying a straight
edge on the departure and destination points, a protractor
laid on the straight edge, and the bearing read off directly.

Take the radius of the globe to be 1. Use a stereographic
projection from the north pole to the equatorial plane.
The map is conformal. For a point of latitude p, the
distance of the point's image from the center is

r(p) = tan(pi/4 + p/2).

Take the image plane to be the complex plane. The image
of a point with latitude p, longitude L is

r(p).exp(iL).

Log is conformal so the map

(p, L) -> log(tan(pi/4 + p/2)) + iL

satisfies I, II, & III.

Consider a course on the globe with bearing k, k being
the angle measured clockwise from north. Take a tiny right
triangle on the globe with the hypotenuse on the course
and the two legs on lines of latitude and longitude. The
leg on the line of longitude has length dp, and the leg
on the line of latitude has length cos(p) dL = tan(k) dp.
Suppose the course starts from latitude 0, longitude 0
and continues on bearing k to (p, L) . Then

L = tan(k) int_0^p sec(t) dt.

Referring to our chart we see that

L = tan(k) log(tan(pi/4 + p/2)),

proving int_0^x sec(t) dt = log(tan(pi/4 + x/2)).

Interestingly the loxodrome course is not enough longer
than a great course route to matter in practice. The
length of the loxodrome from (p1, L1) to (p2, L2) is
sec(k) (p2 - p1), where tan(k) = (L2 - L1)/(r(p2) - r(p1))

--
Michael Press



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