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Old December 1st, 2008, 05:43 PM
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Default Orthogonal Trajectories

"Find the orthogonal trajectories of the family of curves x^2+y^2=2cx.

I know the steps: 1) solve for c...okay...c=(x^2+y^2)/2x.
2)Find y', and plug the c back into it.
3) set dy/dx equal to the negative reciprocal of y' from pt. 2, and solve for y. This is the family of orth. traj.

My problem is finding y' for part 2. Well, I can find y', but getting it into a form so that I can solve for y in part 3.

y=±(2cx-x^2)^(1/2)

y'=1/(2sqrt((x^2+y^2)/x-x^2)*((x^2+y^2)/x-2x)

Where should I go from here? Thanks!
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Old December 1st, 2008, 10:26 PM
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Originally Posted by lakesfan210 View Post
"Find the orthogonal trajectories of the family of curves x^2+y^2=2cx.

I know the steps: 1) solve for c...okay...c=(x^2+y^2)/2x.
2)Find y', and plug the c back into it.
3) set dy/dx equal to the negative reciprocal of y' from pt. 2, and solve for y. This is the family of orth. traj.

My problem is finding y' for part 2. Well, I can find y', but getting it into a form so that I can solve for y in part 3.

y=±(2cx-x^2)^(1/2)

y'=1/(2sqrt((x^2+y^2)/x-x^2)*((x^2+y^2)/x-2x)

Where should I go from here? Thanks!
Have you been taught implicit differentiation:

2x + 2y \frac{dy}{dx} = 2c \Rightarrow \frac{dy}{dx} = \frac{c - x}{y}.
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