Subject: Applied Mathematics 4

Topic: Calculus of Variation

Difficulty: Medium

$\begin{array}{lcr} Consider F=\sqrt{1+y^2}\\ \text{F does not contain 'x' explicilty}\\ \text{therefore,We shall use the formula}\\ \left\{F -y'\frac{\partial F}{\partial y'}=c \right\}\\ \left[\text{Note:- If F does not conatin 'x' explicilty then } F-y'\frac{\partial F}{\partial y'}=c \right] \\ \frac{\partial F}{\partial y'}=\frac{1}{\sqrt{1+y^2}}\left[2y' \right]\\ \frac{\partial F}{\partial y'}=\left[ \frac{y'}{\sqrt{1+y'^2}}\right]\\ F-y'\frac{\partial F}{\partial y'}=\left\{ \sqrt{1+y'^2 } - y'\left[\frac{y'}{\sqrt{1+y'^2}} \right] \right\}\\ \left[F-y'\frac{\partial F}{\partial y'} \right]=\left[\sqrt{1+y'^2}-\frac{y'^2}{\sqrt{1+y'^2}} \right]=\frac{\left(1+y'^2 \right)-y'^2}{\sqrt{1+y'^2}}=\frac{1}{\sqrt{1+y'^2}}\\ But\\ \left[F-y'\frac{\partial F}{\partial y'} \right]=c' (say)\\ =\gt\frac{1}{\sqrt{1-y'^2}}=c'\\ \sqrt{1+y'^2}=\frac{1}{c'} \text{ put 1/c=c''}\\ \sqrt{1+y'^2}=c''\\ 1+y'^2=c''^2 \text{ put } c''^2=c\\ 1+y'^2=c\\ y'^2=c-1\\ y'=\sqrt{c-1} \text{ put } \sqrt{c-1}=b\\ y'=b\\ =\gt \frac{dy}{dx}=b\\ dy=bdx\\ \int_{}^{}dy=b\int_{}^{}dx\\ \left[y=bx+c \right] \text{ which is a staight line} \end{array}$