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IsometricPion
#2
Mar3-12, 01:15 PM
P: 177
The length of a curve is [itex]\int_{s_0}^{s_1}ds=\int_{(x_0,y_0,z_0)}^{(x_1,y_1,z_1)}\sqrt{(\frac{\pa rtial{x}}{\partial{s}})^2+( \frac{\partial y}{\partial{s}})^2+(\frac{\partial{z}}{\partial {s}})^2}ds[/itex]. Setting the variation of this integral to zero one obtains [itex]\delta\int_{(x_0,y_0,z_0)}^{(x_1,y_1,z_1)}\sqrt{( \frac{\partial{x}}{\partial{s}})^2+( \frac{\partial y}{\partial{s}})^2+(\frac{\partial{z}}{\partial {s}})^2}ds=0\Rightarrow \int_{(x_0,y_0,z_0)}^{(x_1,y_1,z_1)}\frac{\frac{ \partial{x}}{\partial{s}}\delta( \frac{\partial{x}}{\partial{s}})+\frac{\partial{y}}{\partial{s}}\delta( \frac{\partial{y}}{\partial{s}})+\frac{\partial{z}}{\partial{s}}\delta( \frac{\partial{z}}{\partial{s}})}{\sqrt{(\frac{ \partial{x}}{\partial{s}})^2+( \frac{\partial y}{\partial{s}})^2+(\frac{\partial{z}}{\partial {s}})^2}}ds=0[/itex]. Integrating by parts (requiring the boundry terms to vanish) and using the fundamental theorem of variational calculus will yield the Euler equations (which turn out to be rather simple in this case). Solving the Euler equations will yield a straight line.