Solving for Orthogonal Trajectories

In summary, the orthogonal trajectories are circles with centres on the x-axis and all going through the origin.
  • #1
vorse
13
0

Homework Statement



Which of the following is the set of orthogonal trajectories for the family indicated by

(x-c)^2 + y^2 = c^2

a). (x-c)^2 + y^2 = c^2

b). (x-c)^2 - y^2 = c^2

c). x^2 + (y-c)^2 = c^2

d). x^2 - (y-c)^2 = c^2

e). None of the above




Homework Equations



I think this is a homogenous equation after playing with the variables and substituting for C. I came up with:

y' = (y^2 - x^2)/xy

I then substitute y=xv after setting y' = 1/y' for the orthogonal trajectories and tried to solve the equation, but ended up with answers no where close to the choices.



The Attempt at a Solution


 
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  • #2
Hi vorse! :smile:

Let's do it the easy way :wink:

draw a few curves, and that should show you the most obvious orthogonal trajectory …

which of a) to e) does that one lie on? :smile:
 
  • #3
I was afraid you'd said I have to graph the equation; I can never make heads to tail of what I draw out.
 
  • #4
ok … (x-c)2 + y2 = c2 is obviously a circle.

So, for each value of c, what is its centre, what is its radius, and what are the endpoints of its "horizontal" diameter?
 
  • #5
Ok, I figured out the answer. I think it's answer choice C. It's just my answer has something like 2c as one of the constant, and I guess the answer choice reduce it down to just C. Btw, I didn't graph the equation; it is a homogenous equation and can be solve that way; just the algebra manipulation was confusing. Thanks for all the help PF. Thanks for trying to help me tiny tim.
 
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  • #6
vorse said:
Ok, I figured out the answer. I think it's answer choice C.

Yup! :smile:

But a quick sketch always helps …

in this case, the original curves are the circles with centres on the x-axis, all touching each other (and the y-axis) at the origin.

Quickly hovering over this with a pencil shows that the orthogonal trajectories must approach the y-axis horizontally, and must all go through the origin …

which choice C does, since it's the circles with centres on the y-axis. :wink:

hmm … are those all of the orthogonal trajectories?

(btw, the wiki page http://en.wikipedia.org/wiki/Orthogonal_trajectories" )
 
Last edited by a moderator:

1. What is the concept of orthogonal trajectories?

Orthogonal trajectories refer to a set of curves that intersect another set of curves at right angles. In other words, the tangent lines of these curves are perpendicular to each other at the points of intersection.

2. Why is solving for orthogonal trajectories important?

Solving for orthogonal trajectories is important in many fields of science and engineering, such as physics, engineering, and mathematics. These trajectories can help us understand and analyze various physical phenomena, such as electric and magnetic fields, fluid flow, and heat transfer.

3. What is the process for finding orthogonal trajectories?

The process for finding orthogonal trajectories involves using a differential equation to represent the given set of curves. Then, using the chain rule, we find the differential equation for the orthogonal trajectories and solve it to obtain the orthogonal trajectories.

4. Can orthogonal trajectories exist for all curves?

No, orthogonal trajectories only exist for a specific type of curves known as orthogonal sets. These curves must have tangent lines that are perpendicular to each other at every point of intersection.

5. How can orthogonal trajectories be applied in real-world situations?

Orthogonal trajectories have various applications in real-world situations. For example, they can be used in the design of circuit boards, analyzing the flow of fluids in pipes, and predicting the path of light rays in optics. They can also help in solving problems in mechanics and other fields of physics.

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