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Find the shortest path between two points in polar coordinates

  • Thread starter mataleo
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Homework Statement


Find the shortest distance between two points using polar coordinates, ie, using them as a line element:

ds^2 = dr^2 + r^2 dθ^2


Homework Equations



For an integral
I = ∫f
Euler-Lagrange Eq must hold
df/dθ - d/dr(df/dθ') = 0

The Attempt at a Solution


f = ds = √(1 + (r * θ')^2)

df/dθ = 0

df/dθ' = r^2 * θ' / √(1 + (r * θ')^2) = C

where C is a constant

Now I want to show this is a straight line so the form should be

y = m*x + b ==> r * cos(θ) = m * r * sin(theta) + b

but I'm struggling with how to prove this. I rearranged the terms and solved the integral

θ = ∫(dr / r √(r^2 - C^2))

but I get a piecewise solution so it seems like I've gone in the wrong direction or am missing something. How do I show that this is equivalent to a straight line?

Thanks
 
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Answers and Replies

  • #2
TSny
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Note that you can write the polar equation for a straight line as ##r \cos(\theta + \alpha) = C## for constants ##\alpha## and ##C##. See if this helps.

(It appears that you might be measuring ##\theta## from the y-axis. But I believe the usual convention is to measure it from the x-axis. Also, I think you are missing an overall factor of C in your integral for θ.)
 

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  • #3
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Yead, I dropped 'C' from the numerator in the integral and thanks for the suggestion on the polar form of the equation being equivalent to rcos(θ+α)=C, but I'm still getting an integral that doesn't seem to fit this form:

θ = √(-(C/r)^2 + 1) / C^2 for C^2/R^2 <= 1

Maybe if I just show that θ'' = 0 then we know that the path is straight.
 
  • #4
TSny
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If you can't find the integral in a table, try letting u = r/C and then letting u = sec##\phi##.

In general, θ'' ≠ 0 for a straight line.
 
  • #5
It works if you make the substitution for the element of integration to be d(c/r). If you work out the differential with the chain rule you get d(c/r) = -(c/r^2) dr. This is the right form you need when you factor the r^2 term out of the square root of the starting form of the integrand, (1 / r sqrt(r^2 - c^2)). Integration in the new form does give you sin^-1(c/r), but I don't really like the negative sign in the relation between the differentials. It may just be that it is "absorbed" into constant c, but there is probably a better explanation regarding that issue. However, the form of the differential makes it a lot easier to deal with looking up the integral in a table and also if you choose to use a trig substitution. The integral after substitution should have the form Int(d(c/r)/sqrt(1-(c/r)^2)). Then, once you have a RHS = sin^-1(c/r) = LHS = theta + constant, taking the sine of both sides and multiplying by r, you should get c = rsin(theta + constant) matching the form of a straight line a la polar coordinates (or a la "I Love Lucy", Simpsons joke). Any thoughts on the negative sign?
 

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