The Elements of the Differential And Integral Calculus, With Numerous Examples

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Compare this length with Veto 2 + cl k+1 y 2 , where d k+1 y is the differential correspond- ing to A k+1 y.
limit _Vdx 2 + d k+1 y 2 _ limit 11 — CO J 1 + cto a/ 1+ d k+l y dx v^c dx Therefore Va# 2 + A k+1 y 2 may be replaced by Veto 2 + c4 +1 ?/ 2 in any problem involving the limit of the sum of infinitesimals.
Replace each term in (1) by Veto 2 -f d k+1 y 2 , where d k+l y has the proper value for that term.
Therefore the required length of arc x=h = n=*X^^+W (2) x—a \dxj 220 DIFFERENTIAL A
...ND INTEGRAL CALCULUS by dividing and multiplying by dx, -jrv dx) The length of an arc can also be expressed as j \/l + ( — ) cfa/ by dividing and multiplying (2) by dy. In this case c and d are the values of y corresponding respectively to the values a and b of x.
196. The results of the preceding article have been estab- lished only in the case where the curve satisfies the conditions imposed in that article. When either or both of these con- ditions are violated, a special investigation is necessary.


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