I don't understand why the partial derivative of $ F $ with respect to $v'$ is like this, does $y'$ depend on $v'$?

Question

I don't understand why the partial derivative of $ F $ with respect to $v'$ is like this, does $y'$ depend on $v'$?

Image attached:

Answer 1

It is a notation from Physics.

Consider that $F$ is a function of three variables $F(u,v, \dot{v})$. (Do not interpret $\dot{v}$ as related to $v$).

Consider that $f$ is a function of three variables $f(x,y, y')$. (Do not interpret $y'$ as related to $y$).

Now, given the formula:

\begin{align*} F(u,v, \dot{v})&= f\left(x(u,v),y(u,v), \frac{y_u+y_v\dot{v}}{x_u+x_v\dot{v}} \right)(x_u+x_v\dot{v}) =\\ &=(x_u+x_v\dot{v})f\left(x(u,v),y(u,v), \frac{y_u+y_v\dot{v}}{x_u+x_v\dot{v}} \right) \end{align*}

We differentiate it, still considering that $\dot{v}$ is an independent variable. We have

\begin{align*} \frac{\partial F}{\partial \dot{v}}(u,v, \dot{v})&= x_v f\left(x(u,v),y(u,v), \frac{y_u+y_v\dot{v}}{x_u+x_v\dot{v}} \right)+ \\ &+(x_u+x_v\dot{v})\frac{\partial f}{\partial y'}\left(x(u,v),y(u,v), \frac{y_u+y_v\dot{v}}{x_u+x_v\dot{v}} \right)\frac{\partial}{\partial\dot{v}}\left ( \frac{y_u+y_v\dot{v}}{x_u+x_v\dot{v}} \right) \end{align*} Or writing in a simpler way $$\frac{\partial F}{\partial \dot{v}}= x_v f + (x_u+x_v\dot{v})\frac{\partial f}{\partial y'}\frac{\partial}{\partial\dot{v}}\left ( \frac{y_u+y_v\dot{v}}{x_u+x_v\dot{v}} \right)$$

Answer 2

You have that $$ y=y(u,v) , x=x(u,v) $$ and according to your image you are supposing too that $v$ depends on $u$. So, using chain rule

$$ dy=y_{u}+y_{v} v'(u) $$ Similarly, $$ dx=x_{u}+x_{v}v'(u) $$ Which implies that $$ y'(x)= \displaystyle \frac{dy}{dx} = \displaystyle \frac{y_{u}+y_{v} v'(u)}{x_{u}+x_{v}v'(u)} $$ Is like when you consider the change of variables in polar coordinates : $x=r(\theta)cos(\theta)$, $y=r(\theta)sin(\theta)$.

Best regards.