Solving for Orthogonal Trajectories

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The discussion centers on identifying the set of orthogonal trajectories for the family of circles defined by the equation (x-c)^2 + y^2 = c^2. Participants analyze the problem using algebraic manipulation and graphical interpretation, concluding that the orthogonal trajectories approach the y-axis horizontally and pass through the origin. After examining the answer choices, it is determined that option C, x^2 + (y-c)^2 = c^2, represents the correct set of orthogonal trajectories. The conversation highlights the importance of visualizing the curves to understand their relationships better. Ultimately, the consensus is that choice C is the correct answer based on the geometric properties of the curves involved.
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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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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:
 
I was afraid you'd said I have to graph the equation; I can never make heads to tail of what I draw out.
 
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?
 
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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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" )
 
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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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