Finding $\frac{dr}{du}$ and $\frac{dr}{dv}$ from $(1-\frac{r}{2m})\,\exp(\frac{r}{2m})=uv$

Question

I have the following relation

$$\left( 1-\dfrac{r}{2m} \right)\, \exp\left( \dfrac{r}{2m} \right)= u v$$

that defines implicitly $r$ as a function of $u$ and $v$. I need to find $\dfrac{dr}{du}$ and $\dfrac{dr}{dv}$. This is what I have done so far:

I have derived in $u$ and got:

$$\dfrac{1}{2m} \dfrac{dr}{du} \left[ -\exp\left( \dfrac{r}{2m} \right)+\left( 1-\dfrac{r}{2m} \right)\, \exp\left( \dfrac{r}{2m} \right) \right]=v$$

The second term in the brackets is $uv$, so I have:

$$\dfrac{1}{2m}\dfrac{dr}{du}\left[ -\exp\left( \dfrac{r}{2m} \right) +uv \right]=v$$

How do I eliminate the exponent inside the brackets to get a function of only $u$ and $v$?

Answer 1

Let $R\left(r,m,u\right)=\left(1-\frac{r}{2m}\right)e^{\frac{r}{2m}}-u$. By the Implicit Function Theorem applied to $R\left(r,m,u\right)=0$ we have

$\frac{dr}{du}=-\frac{\frac{\partial R}{\partial u}}{\frac{\partial R}{\partial r}}=-\frac{4m^{2}e^{-\frac{r}{2m}}v}{r}$

Similarly: $\frac{dr}{dv}=-\frac{4m^{2}e^{-\frac{r}{2m}}u}{r}$.

If this answers your question please accept it. If not, let me know what else is required.

Answer 2

Comment to the @Avocaddo's post. The chain rule suggests:

$\frac{\partial R}{\partial u} = \frac{\partial R}{\partial r} \cdot \frac{\partial r}{\partial u} + \frac{\partial R}{\partial u} = 0 \Rightarrow \frac{\partial r}{\partial u} = - \frac{\frac{\partial R}{\partial u}}{\frac{\partial R}{\partial r}}$