Eq6.36 - y 2 = Cy 2-1 Thus dy p Cy 2-1 = ± dx(9 The integral of the left hand side of this equation can be found from a table of indeﬁnite

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Shina Tan Many problems appear diﬃcult at ﬁrst sight. If you attack them directly, some of them will yield beautiful solutions, so that in hindsight they do not seem hard. In the problem of minimizing the area of a soap ﬁlm, you got a diﬀerential equation p 1 + y 0 2 = d dx ± yy 0 p 1 + y 0 2 ² . (1) Let us solve it. Using the rules of diﬀerentiation to directly expand the right hand side, we get p 1 + y 0 2 = y 0 2 + y 0 4 + yy 00 (1 + y 0 2 ) 3 / 2 . (2) Multiplying both sides by (1 + y 0 2 ) 3 / 2 , and canceling out some terms, we ﬁnd yy 00 - y 0 2 = 1 , (3) which looks simple. But we are stuck - Equation (3) does not permit an easy integration. (We wish we had an equation like yy 00 + y 0 2 = 1 instead, which would yield ( y 2 / 2) 00 = 1.) It takes a trick to move forward. Note that for any two functions a ( x ) and b ( x ) ± a b ² 0 = a 0 b - ab 0 b 2 . (4) So if we divide both sides of Eq. (3) by y 2 , we get ± y 0 y ² 0 = 1 y 2 . (5) Now multiply both sides by y 0 /y : ± y 0 y ²± y 0 y ² 0 = y 0 y 3 , (6) which may be rewritten as d dx h 1 2 ± y 0 y ² 2 i = d dx ± - 1 2 y 2 ² . (7) So y 0 2 y 2 + 1 y 2 = C > 0 , C = constant . (8) 1

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Unformatted text preview: y 2 = Cy 2-1. Thus dy p Cy 2-1 = ± dx. (9) The integral of the left hand side of this equation can be found from a table of indeﬁnite integrals. We obtain ln( p Cy 2-1 + √ C y ) √ C = ± ( x-x ) , (10) where x is a constant of integration. We may change the origin of the x axis such that x = 0. Further solving the equation for y , we get y = e √ Cx + e-√ Cx 2 √ C . (11) Introducing a length parameter L = 1 / √ C , we have y = L cosh x L , (12) a neat result! It is plotted in Figure 1 (next page). The hyperbolic cosine, cosh θ ≡ e θ + e-θ 2 , is actually the cosine of an imaginary angle : cosh θ = cos( iθ ), with i ≡ √-1. 2 M 2 M 1 1 2 x L 1 2 3 4 y L Figure 1: The soap ﬁlm is formed by any segment of the above curve (say x 1 < x < x 2 ) rotated about the x axis. 3...
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This note was uploaded on 01/22/2012 for the course PHYS 3202 taught by Professor Tan during the Spring '11 term at Georgia Institute of Technology.

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Eq6.36 - y 2 = Cy 2-1 Thus dy p Cy 2-1 = ± dx(9 The integral of the left hand side of this equation can be found from a table of indeﬁnite

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