Commit 96470d48 authored by Leonhard Euler's avatar Leonhard Euler
Browse files

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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