# The Analyst Volume 1-2

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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1874 edition. Excerpt: ...Prof. Hendrickson U.S. Naval Academy proposes the following method of obtainiug the equation of the tangent in terms of its direction ratio, the equation of the curve being in the form y =Jx). (1) Let TM=%=m (2) Then x and y being implicitly functions of m, let y--mx = pim) (3) The form of the function p(m) is reqnired. Differentiate (3), and we have dy--mdx--xdm = f'(rn)dm. Since dy = mdx and dm is not zero, (1) representing a curve--x = p'?n). Considering a; as a function of m, and integrating lm) = C--/ xpm. To find p(m) from (4) it is necessary to express x in terms of m from (2), perform an integration and determine C. In many cases it is easy to determine pm) directly from (3), which requires us to express x in terms of m as before and also to express y in terms of m by elimination from (1) and (2). when on the other hand this elimination is inconvenient, p(m) may often be determined by (4).' The equation of the tangent is then y = mx + /(m). EXAPLES. 1. Given y = log x, then m =--, whence x =--and y =--logm. Henc from (3)--log m--1 = /m); and the tangeat is y--mx--(1 + logm). 2. Given y = (a? + 1) (x + 1), then m == 3c2 + 2x + 1, and x=--l x/(\$m--%) (a) Hence from (4) /(m) = C + \$m =F 2y/(m--\$)s; therefore y = mx+ C+frn 21/(im--f)3. To determine C, let x----1 (say), then y = 0, m = 2; substituting' using the lower sign because, if m--2 in (a), the upper sign gives x = 0==--2 + C+ + orC==f; therefore the tangentsared y = mx + H im + 2/%m--%f. Prof. Johnson says that his intention was, in proposing problem 33, to require an integral equation between x and y referred to rectangular axes. The special interest, he remarks, in the problem consists in the avoidance of radicals which have not properly the double sign; and he requests us...