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Comments:
<0> Catfive: thanks that seems to work.
<1> jhubert: are those coordinates the top left? or what?
<2> Flannel: yes. the x,y,h,w would be -875,-420,1500,1500 before zoom
<3> Dacicus : sorry I didn't see your solution
<1> jhubert: well, you could do it a number of ways, the easiest from a programatical perspective would be to determine the center (that's the point we want to keep in the center), and then just adjust the viewport accordingly (center-(width/2), center-(height/2), width, height)
<3> Dacicus : y' = (y + 9x)^2 (y+9x = v) substitution
<1> jhubert: to find the center, it's x+(width_o/2), y+(height_o/2)
<2> Flannel: width_o? old width?
<1> jhubert: and, you could simplify that, if you want to do it in one calculation, but with the 'two step' sort of thing, you could later allow for zooming in on an arbitrary location, not just the center, fairly easily
<1> jhubert: yeah.
<3> Flannel : can you help me solve a question?
<1> doojin: no idea. You're welcome to ask though
<3> Flannel : y' = (y + 9x)^2 (y+9x = v) substitution
<4> what does v have to do with anything there?
<3> _llll_ : the book said substitution for v would help me
<4> so, what does that substitution do to the equation?
<3> _llll_ : I thought further
<5> Well, you have v = y + 9x, v' = y' + 9
<5> I think
<3> y = -9x + v
<3> y' = -9 + v'
<3> -9 + v' = v^2
<4> ok, now solve that for v
<5> That's where I was going.
<3> v' = v^2 + 9
<3> dv/dx = v^2 + 9
<5> looks like a trig one
<3> 1/ ( v^2 + 9 ) * dv = dx
<3> Integrate ( 1 / (v^2 + 9) ) dv = Integrate dx
<3> there is a problem now
<3> how do I integrate theM?
<3> Integrate 1/ (v^2 + 9) dv is out of my ability
<5> It's a trig one...
<4> it's a standard trig integral
<3> Dacicus : ?
<3> what is a trig integral?'
<4> % Integrate[1/(x^2+1),x]
<6> _llll_: ArcTan[x]
<3> I know
<4> you've got 1/(v^2+3^2), you you wont quite get arctan(v)
<5> (1/3) Arctan(x/3), I think...
<2> Flannel: I need to figure out what the new x / y points of the element are. I can't move the viewport. so.. if the element was at -875,-420 and was then resized 30%, what are the new x, y positions that make sure the center is in the same place
<3> I extracted 1/3arctan(v/3) = x + c
<3> and subtitute v with y + 9x
<3> there is a problem again
<5> Why?
<4> first multiply by 3 and apply the tangent function
<5> Yes...
<3> 1/3arctan( (y+9x) / 3 ) = x + c is what I don't understand
<3> hmm
<3> wait
<4> why didnt you just say that int he first place?
<3> Arctan( v/3 ) 3x + 3c
<3> _llll_ : what
<3> Tan(3x + 3c) = (9x + y) / 3
<3> y/3 = Tan(3x + 3c) - 3x
<3> y = 3tan(3x + 3c ) - 9x
<3> right?
<3> I didn't remember arctan was tan-1
<4> i have no interest in checking every line in detail, but i didnt see anything obviously wrong there :)
<3> anyway
<3> % DSolve[y'[x] == (9x + y[x])^2, y[x], x]
<6> doojin: {{y[x] -> -3*I - 9*x + (-I/6 + E^((6*I)*x)*C[1])^(-1)}}
<3> that's a weird solution
<3> I and E
<3> and my mathematica 5.0 provides weirder solutions
<5> That's probably a complex variant
<3> maybe i^2 = -1
<4> well to be sure, calculate y' and (9x+y)^2 for your solution
<4> (but on paper, not in the channel please)
<3> hmm?
<4> ie check that waht you ended up with really is a solution to the original problem
<3> how do I calculate?
<4> you've got y=something involving x, so differentiate that and you get y'=some other expression
<4> etc
<3> ok
<3> I'm showing the derivation
<3> y = 3tan(3x + 3c) - 9x
<3> y' = (y+9x)^2
<3> y+9x = 3tan(3x+3c)
<3> (y+9x)^2 = 9tan^2(3x+3c)
<3> y' = 9sec^2(3x+3c) - 9
<3> 9tan^2(3x+3c) = 9sec^2(3x+3c) - 9
<3> tan^2(3x+3c) = tan^2(3x+3c) - 1
<3> oops
<3> tan^2(3x+3c) = sec^2(3x+3c) - 1
<3> I think they are same
<3> Why can't mathematica and mbot solve as well as me?
<3> y' = 9sec^2(3x+3c) - 9
<3> I thought if mathematica and mbot can't solve well, my solution could be wrong
<4> looks like mbot is not yet ready for world domination then
<5> There are usually multiple ways to write trig functions
<5> lol, _llll_
<3> Dacicus : do you think my solution is right?
<5> % D[3 Tan[3 x + 3 c] - 9 x, x]
<6> Dacicus: -9 + 9*Sec[3*c + 3*x]^2
<3> what?
<5> % (3 Tan[3 x + 3 c] - 9 x + 9 x)^2
<6> Dacicus: 9*Tan[3*c + 3*x]^2
<5> bah
<5> hold on
<3> hahaha
<5> so
<5> 9 tan(3c + 3x)^2 = 9 (sec(3c + 3x)^2 - 1) = -9 + 9 sec(3c + 3x)^2
<3> DSolve[
<3> % DSolve[y[x] == 3Tan[3x + 3c] - 9x, y[x], x]
<6> doojin: {{y[x] -> -3*(3*x - Tan[3*c + 3*x])}}
<5> Yes, it's the same...
<3> wait
<3> it was Differentiation
<3> % D[y[x] == 3Tan[3x + 3c] - 9x, x]
<6> doojin: Derivative[1][y][x] == -9 + 9*Sec[3*c + 3*x]^2
<3> -9 + 9Sec[3c + 3x]^2 = 9Tan^2(3x + 3c)
<3> yes
<4> yuo cant just substitute one of the y's with what you think the answer is
<3> I am not sure tan(3x + 3c) is a right usage of constant
<4> i tihnk probably this tan(3x+3c) thing is wrong
<3> My high school teacher said always put the constant separated
<3> like tan(3x) + c
<3> _llll_ : do you have an alter?
<4> several, why?
<3> tell me
<3> I want to know the exact solution
<4> oh, *alternative*
<4> i dont know. the method of substituting v looked right, but i think there was some algebra error somewhere
<7> what was the original problem?
<3> my solution is a right one as a general solution?
<4> i tihnk you have the right method but the wrong final result
<3> y' = (y+9x)^2 and y+9x = v
<3> _llll_ : are you willing to solve it?
<7> I am
<7> what is the original problem
<3> Safrole : y' = (y+9x)^2 and y+9x = v
<3> use v to make it easier
<7> when you say y' what do you mean dy/dx?
<3> yes
<3> dy/dx
<7> okay
<4> so v'=9+v^2
<7> let y = 9x - v
<7> and substitute
<4> % Integrate[1/(9+v^2),v]
<6> _llll_: ArcTan[v/3]/3
<4> hmm, maybe you were right after all :)
<3> _llll_ : maybe
<3> _llll_ : please make sure
<4> y+9x = 3tan(3x +C)
<4> that's what you ahd before i tihnk
<3> yes
<4> ok i think it is correct
<3> actually + 3c
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