# What are the orthogonal trajectories

## Orthogonal trajectories

Hello, I have a question about trajectories and it is about setting up an orthogonal trajectory to the family.

If I derive once I get yes and from I get then I have used. Now it is said that applies to two straight lines that are perpendicular to each other. How can I mess that up?

Is there a general procedure for dealing with trajectories?

Thank you very much!

"I want to create the solution in collaboration with others."

"

we have m1⋅m2 = -1.

How can I mess that up in yʹ =?

"

So if is and you enter this into the sausage machine m1⋅m2 = -1

then come out below? - right?

so solve the DGL y´

.. and what do you get served then?

Hello panorama, can you tell me why and and is set? Isn't that kind of arbitrary? Do you always do this with orthogonal trajectories?

The solution of the DGL is then:

"

why m1 = yʹ and and m2 = yʹ is set? "

woof!

what's the nonsense?

if you think along, it looks like this: m1 = yʹ and m2 = -1 / yʹ

Orthogonal trajectories

at each intersection of such curves, the tangents should be to each other

stand vertically, ie for the slopes? ...? ...?

your example:

given the straight line through the origin, ie the straight line bundle. For

Orthogonal trajectories circles with a center at the origin

(So do not use the same parameter name for the new family of curves

That means that is true because then two straight lines are perpendicular to each other. That should settle the question. thanks

If I derive once I get yes and from I get then I have used. Now it is said that applies to two straight lines that are perpendicular to each other. How can I mess that up?

Is there a general procedure for dealing with trajectories?

Thank you very much!

**For everyone who wants to help me**(automatically generated by OnlineMathe):"I want to create the solution in collaboration with others."

"

we have m1⋅m2 = -1.

How can I mess that up in yʹ =?

"

So if is and you enter this into the sausage machine m1⋅m2 = -1

then come out below? - right?

so solve the DGL y´

.. and what do you get served then?

Hello panorama, can you tell me why and and is set? Isn't that kind of arbitrary? Do you always do this with orthogonal trajectories?

The solution of the DGL is then:

"

why m1 = yʹ and and m2 = yʹ is set? "

woof!

what's the nonsense?

if you think along, it looks like this: m1 = yʹ and m2 = -1 / yʹ

Orthogonal trajectories

at each intersection of such curves, the tangents should be to each other

stand vertically, ie for the slopes? ...? ...?

your example:

given the straight line through the origin, ie the straight line bundle. For

Orthogonal trajectories circles with a center at the origin

(So do not use the same parameter name for the new family of curves

That means that is true because then two straight lines are perpendicular to each other. That should settle the question. thanks

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