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# Working with orthogonal trajectories

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(a)Find the orthogonal trajectories of the family of curves defined by

2cy + x2 = c2, c>0

State the differential equation of the orthogonal family, and show your steps in obtaining a solution.
(b) On the same set of "square" axes, plot at least five members of each of the given family and your family of orthogonal solutions.
(c) Identify one point of intersection, and approximate the coordinates of the point. Determine the slope of each curve at the point of intersection, and verify that the tangents are indeed perpendicular.

NOTE: So far, I've done implicit differentiation, and gotten 2cdy/dx + 2x = 0. Then I solved for c (from the original equation) by using the quadratic formula. But when I plug c back in and try to solve the differential equation (I've been getting a homogeneous differential equation and substitute x=vy), it doesn't seem to turn out right. Either the D.E. is too hard to integrate; or I can't get the tangents to be perpendicular in the end.

https://brainmass.com/math/derivatives/working-orthogonal-trajectories-13883

## SOLUTION This solution is FREE courtesy of BrainMass!

2cy + x2 = c2, c>0

Solution

The level curves constitute a family of curves, one for each value of the constant, and this family can be described by a single differential equation. We find the DE by eliminating the constant. This is accomplished by differentiating each term in the original DE with respect to x.
2cy + x2 = c2
From this,
and
Since c> 0 we will take the second value
Differentiating... 2c dy/dx + 2x = 0
Or, dy/dx = -x/c this is the diff. equation describing the level curves

Each member of the family of orthogonal trajectories is perpendicular to each intersecting member of the original family of curves. Therefore, the DE describing the orthogonal trajectories is

dy/dx = c/x substituting for c (Product of slopes of two perpendicular lines = -1)

This is the required orthogonal curve
Plotting this equation for k = 1 to 5 and the given family of curves for c = 1 to 5 we get the curves shown in the figure. I used matlab to plot the function.

Now you can easily verify that the lines are indeed perpendicular.

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