manjees

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10 years, 71 days

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These are answers submitted by manjees

I am not able to use colors flexibly in contourplot3d. So, I am trying another way to get my plot. But I am not exactly getting my result. This is what I am trying to do. I need help with these three steps. 1) For Latitude from -10 degrees to 10 degrees and longitude from -30 degrees to 30 degrees, I need to calculate aa:=1.4*cos(latitude)-2*sin(longitude). 2)The above value of 'aa' gives color. (eg. if aa< 0 color=blue, if 13 color= red etc) 3) Then plot this point with color from step 2 at (x,y) given by x := 2*sqrt(2)*cos(latitude)*sin((1/2)*longitude)/sqrt(1+cos(latitude)*cos((1/2)*longitude)) and y:=sqrt(2)*sin(latitude)/sqrt(1+cos(latitude)*cos((1/2)*longitude)) Any help is appreciated. Thanks
Thanks. But how can I change width of this dotted line?
Thanks for your input. I just noticed it. But I got my torus from pointplot. Appreciate it.
Hi Axel, Would you mind explaning why you have used K:='op(J)[1]' in the above post? Thanks MS
I am using Axel's method (posted above) to solve my equation. I am taking an approximation and integrating from 0.02 rather than from 0. Question: The answer after integration is coming out to be negative and when multiplied by other constants (-8* pi^2), it becomes positive. But I need to compare this answer with RHS of this equation which is negative. So, LHS after solving (using Axel's method above) is coming out to be positive while RHS is negative. I need to compare my RHS to LHS. So, any suggestions on what's happening.? Thanks MS
The source is bounded in my problem and it's gravitational potential goes to zero at infinity.
Thanks for the advice. But y(r) is a type of gravitational potential. So, thats why we take it to be zero at infinity. So, by dividing by ln[max], i just give a cutoff at very large distance. I thought the whole point of finding asymptotic solution was to find a solution that goes to zero at infinity. Any further suggestions?
Hi all, this is regarding Robert's asymptotic solution above. The asymptotic solution does not go to zero at infinity ( ln(r) becomes problematic at infinity)... I can take care of this if i write all ln(r)'s as ln(r/r[max]) where r[max] is a very large no. and the limit goes to r[max] rather than to infinity. My question is : should I use r[max] at the very start or jus replace all ln[r] by ln[r/r[max]] in final asymptotic solution...(Robert's solution in a post above). Thanks MS
i is actually a function of r i.e. i(r). But this is not taken care of in the above method. So, g actually leaves out i(r) and only integrates r.
I am sorry but I did not got your point.
Thanks for your reply. But I need a number output for 'f' as 'g' is integrated from r=0 to r= 10. I have to compare the output with some other value. I guess the above method is useful if we just need to find f at specific 'r'. Any suggestions. Thanks MS
the asymptotic solution above does not go to zero at infinity as ln(r) becomes problematic at infinity... I can take care of this if i write all ln(r)'s as ln(r/r[max]) where r[max] is a very large no. and the limit goes to r[max] rather than to infinity. My question is : should I use r[max] at the very start or jus replace all ln[r] by ln[r/r[max]] in final asymptotic solution...(Robert's solution in a post above). Thanks MS
Hi, how do we get the last step in Robert's solution above... Patrick's version is a solution for y(1/r) but we need sol. for y(r). Thanks MS
Ha....In the above post it was meant to be guys. 'Guy' is a typo...thanks to both of u guys...
Hi, my prof had given this problem over summer so that we learn problem solving using maple. It has lot more to it, but I was stuck on this DE thing and thanks to Robert, i fully understand it now. I really appreciate the length you guy go to help out. Thanks again MS P.S. I am still not able to get an expression for y(x) after applying numerical dsolve. It's in another thread. Any advice?
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