Orthogonal trajectories
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Orthogonal trajectories

[From: ] [author: ] [Date: 11-09-11] [Hit: ]
So we need to set dy/dx = negative reciprocal of -3x/y,http://i53.tinypic.com/168hbvs.......
Hi i'm having some trouble with this question
find the orthogonal trajectory for the following family of curves
and sketch on the same set of axes at leaast three members of each of the two families.
3x^2 + y^2 = C

I'm really confused and have no idea of even how to start. If someone could be kind enough to work through the question while explaining how to do it, it would be much appreciated.
Thank you

-
First we find slope of tangent at points (x,y), by differentiating implicitly:

3x² + y² = C
6x + 2y dy/dx = 0
2y dy/dx = -6x
dy/dx = -3x/y

Now we want a family of orthogonal trajectories. So we need to set dy/dx = negative reciprocal of -3x/y, then solve differential equation:

dy/dx = y/(3x)
3 dy/y = dx/x

Integrate both sides:
∫ 3 dy/y = ∫ dx/x
3 ln|y| = ln|x| + ln|k|
ln|y³| = ln|kx|
y³ = kx

Family of orthogonal trajectories is y³ = kx

http://i53.tinypic.com/168hbvs.png

Equations of curves are as follows:

3x² + y² = C (ellipses)
Red: C = 100
Green: C = 36
Blue: C = 16

y³ = kx
Red: k = 100
Green: k = −30
Blue: k = 5
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keywords: Orthogonal,trajectories,Orthogonal trajectories
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