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I have an equation of the form $$\frac{x^2}{y} = F(r), $$ where F is a function of $r$. This equation has a solution $r(x,y)$. Suppose we perturb this solution to $r(x + \delta x, y + \delta y)$ for small $\delta x$ and $\delta y$. Can we say anything about how $\delta x$ and $\delta y$ are related to each other?

I tried Taylor expanding this expression and keeping first order terms, but this reduces to $0=0$.

Geigercounter
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1 Answers1

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You can differentiate the given equation implicitly, so that $$ \frac{2x}{y}\mathrm{d}x - \frac{x^2}{y^2}\mathrm{d}y = F'(r)\mathrm{d}r = F'(r)\left(\frac{\partial r}{\partial x}\mathrm{d}x + \frac{\partial r}{\partial y}\mathrm{d}y\right) $$ and thus $$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{x^2}{y^2} + F'(r)\frac{\partial r}{\partial y}}{\frac{2x}{y} - F'(r)\frac{\partial r}{\partial x}} $$

Abezhiko
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  • Just to be clear, this can also be done without introducing "bare" differentials. For example on the LHS Taylor expansion gives $\frac{(x+\delta x)^2}{y+\delta y}=\frac{x^2}{y}+\frac{2x}{y} \delta x - \frac{x}{y^2} \delta y + o(\delta x) + o(\delta y)$ and you can do the same on the RHS. Or without Taylor, you can do "ordinary" implicit differentiation, anticipating which variable is independent vs. dependent beforehand. (Thus, if the goal is $\frac{dy}{dx}$, then all your $dy$'s get replaced by $\frac{dy}{dx}$ and your $dx$'s get replaced by $1$). – Ian Oct 30 '23 at 14:11