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![[Post New]](/templates/default/images/icon_minipost_new.gif) 20 Feb 2008 23:43:07 IST
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21 Feb 2008 20:42:57 IST
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21 Feb 2008 20:50:19 IST
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what is thiss..??
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21 Feb 2008 21:12:30 IST
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sorry i put some graphs about ORTHOGONAL TRAJECTORIES but then realised that the question was about solving them.
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21 Feb 2008 22:21:00 IST
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take derivative of the given....then replace dy/dx by -dx/dy and then solve the new DE...
this gives the orthogonal trajectory...
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21 Feb 2008 23:27:47 IST
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I know that much using m1m2=-1 but do this one x^2 + y^2 = ax find the equation of orthogonal trajectory of this equation
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21 Feb 2008 23:36:30 IST
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c for the curve the diff eqn comes out to be
dy/ dx = a-2x/2y
replace dy/dx by -dx/dy....and solve
u get
dx/2x-a = dy/2y
integrating both sides ull get
1/2 log( 2x - a)+ log c = 1/2 log y
= C(2x-a) = y
this is the orthogonal trajectory for the given curve ...
hope u have understood...nudge me if u have any doubt
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21 Feb 2008 23:39:04 IST
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just tell me one thing how can u intergrate it without replacing the constant..?? check once again
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21 Feb 2008 23:40:14 IST
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first find derivative of dat function and as d curve r orthogonal, the product of their slopes (derivatives) will b -1...hence substitute negative reciprocal of derivative as slope of other curve....
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22 Feb 2008 11:02:07 IST
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x^2 + y^2 = ax
differentialte...
2x + 2ydy/dx =a
replace..
2x - 2ydx/dy =a...
now find its solution...
note that orthogonal trajectories are defined for curves having only 1 constant involved...
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