Editing Isaac Barrow (section)

==Calculating tangents==
The geometrical lectures contain some new ways of determining the areas and [[tangent]]s of curves. The most celebrated of these is the method given for the determination of tangents to [[curve]]s, and this is sufficiently important to require a detailed notice, because it illustrates the way in which Barrow, [[Johann Hudde|Hudde]] and Sluze were working on the lines suggested by [[Pierre de Fermat|Fermat]] towards the methods of the [[differential calculus]].

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Fermat had observed that the tangent at a point ''P'' on a curve was determined if one other point besides ''P'' on it were known; hence, if the length of the subtangent ''MT'' could be found (thus determining the point ''T''), then the line ''TP'' would be the required tangent. Now Barrow remarked that if the abscissa and ordinate at a point ''Q'' adjacent to ''P'' were drawn, he got a small [[triangle]] ''PQR'' (which he called the
differential triangle, because its sides ''QR'' and ''RP'' were the differences of the abscissae and ordinates of ''P'' and ''Q''), so that
K
:''TM'' : ''MP'' = ''QR'' : ''RP''.

To find ''QR'' : ''RP'' he supposed that ''x'', ''y''
were the co-ordinates of ''P'', and ''x'' − ''e'', ''y''
− ''a'' those of ''Q'' (Barrow actually used ''p'' for ''x'' and ''m'' for ''y'', but this article uses the standard modern notation).  Substituting the co-ordinates of ''Q'' in the equation of the curve, and neglecting the squares and higher powers of ''e'' and ''a'' as compared with their first powers, he obtained ''e'' : ''a''. The [[ratio]] ''a''/''e'' was subsequently (in accordance with a suggestion made by Sluze) termed the angular coefficient of the tangent at the point.

Barrow applied this method to the curves
#''x''<sup>2</sup> (''x''<sup>2</sup> + ''y''<sup>2</sup>) = ''r''<sup>2</sup>''y''<sup>2</sup>, the [[kappa curve]];
#''x''<sup>3</sup> + ''y''<sup>3</sup> = ''r''<sup>3</sup>;
#''x''<sup>3</sup> + ''y''<sup>3</sup> = ''rxy'', called ''[[Folium of Descartes|la galande]]'';
#''y'' = (''r'' − ''x'') tan π''x''/2''r'', the [[quadratrix]]; and
#''y'' = ''r'' tan π''x''/2''r''.

It will be sufficient here to take as an illustration the simpler case of the parabola ''y''<sup>2</sup> = ''px''.
Using the notation given above, we have for the point ''P'', ''y''<sup>2</sup> = ''px''; and for the point ''Q'':
:(''y'' − ''a'')<sup>2</sup> = ''p''(''x'' − ''e'').

Subtracting we get
:2''ay'' − ''a''<sup>2</sup> = ''pe''.
But, if ''a'' be an infinitesimal quantity,
''a''<sup>2</sup> must be infinitely smaller and therefore may be neglected
when compared with the quantities 2''ay'' and ''pe''.  Hence
:2''ay'' = ''pe'', that is, ''e'' : ''a'' = 2''y'' : ''p''.
Therefore,
:''TM'' : ''y'' = ''e'' : ''a'' = 2''y'' : ''p''.
Hence
:TM = 2''y''<sup>2</sup>/''p'' = 2''x''.

This is exactly the procedure of the differential calculus, except that there we
have a rule by which we can get the ratio ''a''/''e'' or ''dy''/''dx'' directly without the labour of going through a calculation similar to the above for every separate case.