01:42 -!- Prax01D(n=Prax01D@149.99.86.242) has joined #maxima 06:40 -!- lisppaste4(n=lisppast@common-lisp.net) has joined #maxima 07:47 -!- Prax01D(n=Prax01D@149.99.90.231) has joined #maxima 07:52 -!- paoleela|w(i=kq3bboqn@gateway/web/cgi-irc/irc.net/x-6bd857a48f304fd4) has joined #maxima 13:42 -!- Prax01D(n=Prax01D@tor58-7-83-203.dynamic.rogerstelecom.net) has joined #maxima 14:06 -!- paoleela|w(i=e06e12os@gateway/web/cgi-irc/irc.net/x-4b7d70607f519273) has joined #maxima 15:19 -!- wally345(n=op@fgb122.internetdsl.tpnet.pl) has joined #maxima 15:20 Hi, how to solve these equation with maxima ? 65-sqrt(400-x^2)=sqrt(2500-x^2); 16:34 wally345: If solve doesn't find an answer, I'd square both sides, rearrange, square again, ask solve for the answer, and then check for extraneous ones. 16:39 Oh, don't need to square again. After the first squaring, solve produces 2 roots. 16:51 rtoym: Hi 16:51 thx 16:51 i thought maxima can do this kind of stuff easily 16:52 also having problems with integrals 17:04 -!- wally345(n=op@fgb122.internetdsl.tpnet.pl) has joined #maxima 17:17 Solve is not that smart. 17:17 What kind of integral problems? 17:18 the equation above are two intersecting circles 17:19 i have to calc the intersection area with integrals 17:19 between 0 and the solution of the above equation 17:20 maple said its 11.52.... 17:20 this is the x value of intersection point 17:21 maxima produced two solutions to the equation. I think they're both right. 17:22 so in need something like "int(f(x) dx [0..11.52] 17:22 there are 2 intersection points 17:22 and 1 area 17:22 the area i have to find out 17:23 Are the centers of the circles on the y axis? 17:23 zes 17:24 Did you draw a picture? I think once you have a picture, you'll know what to do. 17:24 the question is given as two radii (20 and 50) and a centerdistance of 65 17:24 i did 17:25 The area you want is kind of an eye shape. Find the equation for the upper part of the eye and the equation for the lower part, as a function of x. 17:27 Once you've formulated that, ask maxima to evaluate the integral. If maxima can't do that, then maybe I can help. 17:27 Is this a homework problem? 17:27 yes, the upper is sqrt(400/x^2) and the lower is -sqrt(2500-x%2)+65 17:28 no, its no homework problem 17:28 i want to test some pc/math apps 17:28 Not homework? Ok. 17:29 maple do it easily but i like opensource and so i wanted to trz maxima 17:29 Well, the integral is then integrate(upper-lower,x,-11.52,11.52); 17:29 actuallz have wxmaxima and the TeXmacs installed 17:29 octave too, but i didnt tried this yet 17:30 11,52 17:30 but now we need the area under the upper and lower function in the range from 0 to 11,52 17:30 The upper curve, sqrt(400/x^2) is not a part of a circle. 17:31 the difference is the half of the "eye"-area 17:31 400-x^2 17:31 its the radius 20^2 17:32 Yes, of course, but that's not what you wrote. :-) 17:32 sorry 17:34 Ok, maxima says the value (1/2 area) is (800*asin(3*sqrt(399)/104)+5000*asin(3*sqrt(399)/26)-75*sqrt(399))/4 = 38.99 or so. 17:35 may zou paste your int-command here please ? 17:37 sorry i have little troubles with y and z , chnaging from german to us keyboard 17:47 . 17:49 Basically integrate(upper-lower,x,0,15*sqrt(399)/26); 17:49 when i do float(%), numer; on zour above expression i got some "%i" 17:49 is this a result of the 3rd root 17:49 something imaginary _ 17:49 Where upper is sqrt(400-x^2) and lower is -sqrt(2500-x^2)+65. The 15*sqrt(399)/26 is one of the roots solve found. 17:50 Oops. Mis-copy. It's asin(3*sqrt(399)/260). 17:52 ic 17:52 Does that answer match what maple says? I hope so. 17:53 yes, but i can not reproduce it on mz maxima 17:53 which version do you use ? 17:53 5.11. Shouldn't matter much, though. 17:54 i agree 17:54 integrate(sqrt(400-x^2)-(-sqrt(2500-x^2)+65),x,0,15*sqrt(399)/26); 18:00 -!- wally345_(n=op@fgb122.internetdsl.tpnet.pl) has joined #maxima 18:01 rtoym: soory , was dc 18:02 do you see the results from maxima i pasted 18:02 ? 18:03 (%i1) integrate(sqrt(400-x^2)-(-sqrt(2500-x^2)+65),x,0,15*sqrt(399)/26); 18:03 3 sqrt(399) 3 sqrt(399) 18:03 800 asin(-----------) + 5000 asin(-----------) - 75 sqrt(399) 18:03 104 260 18:03 (%o1) ------------------------------------------------------------- 18:03 4 18:05 How to let maxima calc the resulkt from this ? something near 38.9.... 18:14 float(%o1) 18:14 Or %o1,numer 18:17 he calcs this already a few minutes, is this possible ? or it hangs 18:19 No, that's not right. I should return instantly. 18:19 so anything is wrong 18:20 Don't know. 18:20 Try this: float(asin(3*sqrt(399)/104)) 18:20 ok 18:20 mom 18:21 the same problem 18:21 no output 18:22 Don't forget the semicolon ; 18:23 yes the semicolon 18:23 Does it work now? 18:24 for your last float yes 18:24 now trz the big one 18:24 float(%o1); doesn't give an answer right away? 18:24 It works for me. 18:26 it works now :) 18:27 i have no idea whats the problem was, i usually apply the semicolon always (i hope at least 18:28 thank you very much, so i see i do not make a fundamental mistake with maxima 18:30 can you recommend a good tutorial ? 18:41 another question to the above method 18:42 you recommended integrate(upper - lower) 18:43 is this the same as integrate (upper) - integrate(lower) ? 18:57 There are some tutorials on maxima's website. Pretty good, I think. 18:57 integrate(upper-lower) is, of course, the same as integrate(upper)-integrate(lower). 18:57 And maxima does that itself, anyway, I think. 18:58 ok, i have to learn a lot 18:58 thx 18:59 :) 18:59 Good luck. If you have problems, ask here, or, even better, ask on the mailing list. 19:00 ok, i will remeber this invitation 19:01 wish you a nice evening 19:01 here is alreadz 7pm 19:28 -!- gabkdlly(n=gabriel@dslb-088-073-008-045.pools.arcor-ip.net) has joined #maxima 19:50 -!- Prax01D(n=Prax01D@tor58-10-92-223.dynamic.rogerstelecom.net) has joined #maxima 21:00 -!- wally345(n=op@fgb122.internetdsl.tpnet.pl) has joined #maxima