Show that the shortest curve that has an area (A) below it is a circular arc, [

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Show that the shortest curve that has an area \(A\) below it is a circular arc,
\[
(\lambda x-c)^{2}+(\lambda y-d)^{2}=1,
\]
as shown in Fig. 14.9. Here \(\lambda\) is a Lagrange multiplier constant and \(c\) and \(d\) are constants of integration. For this constrained problem, \(y(0)=a, y(1)=b\), and \(A=\int_{0}^{1} y(x) d x\).

FIGURE 14.9 y a 0 1/2 Optimal curve 1/2 Area below arc is A b X

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Advanced Mathematics For Engineering Students The Essential Toolbox

ISBN: 9780128236826

1st Edition

Authors: Brent J Lewis, Nihan Onder, E Nihan Onder, Andrew Prudil

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