correct file in .modif

parent 9359cdd7
 ... ... @@ -167,19 +167,13 @@ expression with our general formula $$\int Z dy$$, we see that $$P = \frac{p\sqrt{v_{0}^{2} + 2 g y}}{\sqrt{1 + p^{2}}}$$. Minimization requires that $$M dy = dP$$ and, since $$M=0$$ in this case, $$dP = 0$$ and, thus, $$P = \sqrt{C}$$, where $$C$$ is a constant. Hence, we have the differential equation \begin{equation} \sqrt{C} = \frac{p\sqrt{v_{0}^{2} + 2 g y}}{\sqrt{1 + p^{2}}} = \frac{dx\sqrt{v_{0}^{2} + 2 g y}}{ds} \end{equation} the differential equation: $$\sqrt{C} = \frac{p\sqrt{v_{0}^{2} + 2 g y}}{\sqrt{1 + p^{2}}} = \frac{dx\sqrt{v_{0}^{2} + 2 g y}}{ds}$$ This may be rearranged to $$C dx^{2} + C dy^{2} = dx^{2} \left( v_{0}^{2} + 2 g y \right)$$ and separated $$dx = \frac{dy \sqrt{C}}{\sqrt{v_{0}^{2} - C + 2 g y}}$$. Integration yields the trajectory solution \begin{equation} x = \frac{1}{g} \sqrt{C \left( v_{0}^{2} - C + 2 g y \right)} \end{equation} Integration yields the trajectory solution: $$x = \frac{1}{g} \sqrt{C \left( v_{0}^{2} - C + 2 g y \right)}$$ \textbf{5.} This solution is obviously a parabola, but we should consider more carefully whether it agrees with experience. The initial tangent to the trajectory is clearly horizontal, i.e., $$dy=0$$ at the ... ... @@ -211,9 +205,7 @@ trajectory that minimizes the integral $$\int dy \sqrt{v_{0}^{2} + \int 2 F_{y} \, dy} \, \sqrt{1 + p^{2}}$$. By arguments analogous to those used in paragraph 4, we obtain the trajectory equation \begin{equation} \sqrt{C} = \frac{p\sqrt{v_{0}^{2} + \int 2 F_{y} \, dy}}{\sqrt{1 + p^{2}}} \end{equation} $$\sqrt{C} = \frac{p\sqrt{v_{0}^{2} + \int 2 F_{y} \, dy}}{\sqrt{1 + p^{2}}}$$ or, equivalently, $$p = \frac{\sqrt{C}}{\sqrt{v_{0}^{2} - C + \int 2 F_{y} \, dy}}$$, which may be simplified to ... ...
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