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$$c=\int_{x_1}^{x_2}f_{gr}(x)\;dx$$

The integral is a time-like curve between $x_1$ and $x_2$ and at imagine fgf(x1) is a lower left corner of the rectangle and fgf(x2) is the upper right corner and $x_2-x_1$ is the length of the base of the rectangle. The geodesic is the shortest length curve parametrized by this on the condition that the area = c. The solution isn't always a strait line but could be some curved function. How do I find the function ? I know this is a variational problem involving a length integral and an area integral.

Do I need to find the 2x2 metric tensor for the 2-d parameterization curve and solve a pair of differential equations like here ?

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    Are you trying to solve $\inf \{ l(f) | \int f = c \}$ (where $l(f) = \int \sqrt{1+(f')^2}$)?2012-07-29
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    yea...trying to find f2012-07-29
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    Are there other constraints, because the solution to that will be $f(x) = \frac{c}{x_2-x_1}$?2012-07-29
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    this is a horizontal line which gives the areas and is a minimal length, but does not ajoin the endpoints. A simple case would be if the area parameter were (x2-x1)*f(x1)+(x2-x1)*(f(x2)-f(x1))/2 and then geodesic would be of the form y=mx+b2012-07-29

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