Path: news.mathworks.com!not-for-mail From: <HIDDEN> Newsgroups: comp.soft-sys.matlab Subject: Re: Integrating f(x,y) on a certain line in XY plane Date: Mon, 9 May 2011 20:58:05 +0000 (UTC) Organization: The MathWorks, Inc. Lines: 32 Message-ID: <iq9kgt$t02$1@newscl01ah.mathworks.com> References: <iq5tkg$j6e$1@newscl01ah.mathworks.com> <iq5vo1$nch$1@newscl01ah.mathworks.com> <iq6390$17d$1@newscl01ah.mathworks.com> <iq6t1h$ocu$1@newscl01ah.mathworks.com> <iq704t$1av$1@newscl01ah.mathworks.com> <iq740k$9e4$1@newscl01ah.mathworks.com> <iq7900$jj3$1@newscl01ah.mathworks.com> <iq7sns$1j5$1@newscl01ah.mathworks.com> <iq8oeq$5tv$1@newscl01ah.mathworks.com> <iq9dhr$9bb$1@newscl01ah.mathworks.com> <iq9jd9$po9$1@newscl01ah.mathworks.com> Reply-To: <HIDDEN> NNTP-Posting-Host: www-06-blr.mathworks.com Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit X-Trace: newscl01ah.mathworks.com 1304974685 29698 172.30.248.38 (9 May 2011 20:58:05 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Mon, 9 May 2011 20:58:05 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 2733406 Xref: news.mathworks.com comp.soft-sys.matlab:725932 "Roger Stafford" wrote in message <iq9jd9$po9$1@newscl01ah.mathworks.com>... > "Roger Stafford" wrote in message <iq9dhr$9bb$1@newscl01ah.mathworks.com>... > > I didn't say you could reduce your expression to log((XM-X).^2). What I said was that it could be reduced to the form log(x^2+a^2)) which is a very different entity. > > ......... > - - - - - - - - - > Mohamed, the explanation I gave you earlier could have been made much more intuitively evident. Hopefully the following is an improvement. > > You start with log(sqrt((X-XM)^2+(Y-YM)^2)) to be integrated along some straight line extending from (X1,Y1) to (X2,Y2). The quantity within the logarithm is the distance from a point (X,Y), which is moving along this line segment, to the point (XM,YM). Now find the closest point (X0,Y0) along the extended line to the point (XM,YM). The three points (X,Y), (XM,YM), and (X0,Y0) will form a right triangle, so by the theorem of Pythagoras we have > > (X-XM)^2+(Y-YM)^2 = (X-X0)^2+(Y-Y0)^2 + (X0-XM)^2+(Y0-YM)^2 > > If we call t the distance from (X,Y) to (X0,Y0) and f the distance from (X0,Y0) to (XM,YM), this can be written expressed as t^2+f^2. > > This means we seek to integrate > > log(sqrt(t^2+f^2)) = 1/2*log(t^2+f^2) > > along the line segment from (X1,Y1) to (X2,Y2). The indefinite integral of > > log(t^2+f^2) > > with respect to t is given by the formula > > t*log(t^2+f^2) - 2*t + 2*f*atan(t/f) + C > > which would allow you to express the definite integral in terms of the t values at (X1,Y1) and (X2,Y2) and the constant f. No need to do numerical integration. > > (I overlooked the fact that the a^2+c^2 quantity in my previous discussion is necessarily equal to 1, so the log(a^2+c^2) constant would have been zero.) > > Roger Stafford Thanks my friends I managed to get the desired results in the light of your efforts...Many thanks