By Herman. H. Goldstine

The calculus of diversifications is a topic whose starting might be accurately dated. it'd be acknowledged to start in the interim that Euler coined the identify calculus of adaptations yet this can be, in fact, no longer the genuine second of inception of the topic. it will no longer were unreasonable if I had long past again to the set of isoperimetric difficulties thought of through Greek mathemati cians equivalent to Zenodorus (c. two hundred B. C. ) and preserved through Pappus (c. three hundred A. D. ). i haven't performed this in view that those difficulties have been solved by means of geometric capacity. as an alternative i've got arbitrarily selected to start with Fermat's based precept of least time. He used this precept in 1662 to teach how a gentle ray used to be refracted on the interface among optical media of other densities. This research of Fermat turns out to me in particular acceptable as a kick off point: He used the tools of the calculus to reduce the time of passage cif a gentle ray during the media, and his technique was once tailored by way of John Bernoulli to unravel the brachystochrone challenge. there were numerous different histories of the topic, yet they're now hopelessly archaic. One by way of Robert Woodhouse seemed in 1810 and one other through Isaac Todhunter in 1861.

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**Extra info for A History of the Calculus of Variations from the 17th through the 19th Century**

**Sample text**

P. 479n. 2. 21 ) where 0: = BG/4. 12'). Next, following Whiteside, we see how Newton found the representation for x using the hyperbola described above. 20), it follows that 1 SN = GS + BG X (BG 1 + FS) X AE - BG X BG 2 BE -4- X log BA ' since AH = BG + FS and the equation of the hyperbola relative to the axes BS and BE is Xy= B~2 . Thus T= (BG 2/4)logBE/ BG since BA = BG. 19), GF= = (BG + BE)(5BG 2 + BE2) 16BG 2 BG 16 (q + 1)(q2 + 5) = ~ ( q + 1)( q2 + 5) so that BG + FS = BS - GF = 0: (q2 + 1)2 q - 0: '4 (q + 1)( q2 + 5).

This is reference CR. 7. John Bernoulli's First Published Solution and Some Related Work 41 to the cycloid. fax - x 2 He then goes on to say49: However (a dx - 2x dx) : 2(ax - XX)I/2 is the differential quantity whose integral is (ax - XX)I/2 or LO; and a dx : 2(ax - XX)I/2 is the differential of the arc GL. For this reason by integrating the equation dy = dx(x/(a - X»I/2 one has y = eM = [arc]GL - LO, and therefore MO = CO - [arc]GL + LO. Since however CO - [arc]GL = [arc]LK (because CO = semicircle GLK), it follows that MO = [arc]LK + LO and by deducting the common value LO, that ML = [arc]LK, which implies that the curve KMA is a cycloid.

Bcp, and arc GL .. bw - bcp ... bt. Then sin t = sin qi = LO I b and CO = A G == b'IT - Y + M L + LO = Y + arc LK + b sin t; hence y = b'IT - arc LK - b sin t - bw - bcp - b sin t = b(t - sin t). To find x, note that cos t = -cos qi = -(b - OK) and that x = AC == GK - OK - 2b - b - b cos t == b(1 - cos t). ) ~Stiickel, in a remark appended to his translation (p. 137, note 8), mentions that this idea of Bernoulli was a pet of eighteenth century thinkers, and he refers the reader to a discussion in Mach [19331.