Q&A

# Post History

50%
+0 −0

posted 1y ago by deleted user  ·  edited 1y ago by deleted user

#2: Post edited by deleted user · 2021-09-03T10:32:58Z (about 1 year ago)
• In the post, actually $\dot{y}=\frac{dy}{dx}$. So,
• $$\int_{x_1}^{x_2}\frac{\partial f}{\partial \dot{y}}\frac{\partial \dot{y}}{\partial \alpha}\mathrm dx=\int_{x_1}^{x_2}\frac{\partial f}{\partial \dot{y}}\frac{\partial^2 y}{\partial x \partial \alpha}\mathrm dx$$
• In the equation, they just wrote $\frac{dy}{dx}$ instead of $\dot{y}$
• In the post, actually $\dot{y}=\frac{dy}{dx}$. So,
• $$\int_{x_1}^{x_2}\frac{\partial f}{\partial \dot{y}}\frac{\partial \dot{y}}{\partial \alpha}\mathrm dx=\int_{x_1}^{x_2}\frac{\partial f}{\partial \dot{y}}\frac{\partial^2 y}{\partial x \partial \alpha}\mathrm dx$$
• In the equation, they just wrote $\frac{dy}{dx}$ instead of $\dot{y}$
#1: Initial revision by deleted user · 2021-09-03T10:32:06Z (about 1 year ago)
In the post, actually $\dot{y}=\frac{dy}{dx}$. So,
$$\int_{x_1}^{x_2}\frac{\partial f}{\partial \dot{y}}\frac{\partial \dot{y}}{\partial \alpha}\mathrm dx=\int_{x_1}^{x_2}\frac{\partial f}{\partial \dot{y}}\frac{\partial^2 y}{\partial x \partial \alpha}\mathrm dx$$
In the equation, they just wrote $\frac{dy}{dx}$ instead of $\dot{y}$