For some function $F(x,y) = 0$, $$\frac{dy}{dx} = \frac{-F_x}{F_y}$$
Can someone give me a geometric interpretatio of this? ($F_x$ and $F_y$ are the partial derivatives)
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Sign up to join this communityFor some function $F(x,y) = 0$, $$\frac{dy}{dx} = \frac{-F_x}{F_y}$$
Can someone give me a geometric interpretatio of this? ($F_x$ and $F_y$ are the partial derivatives)
Consider a function $y=y(x)$ satisfying the equation $F(x,y(x))=0.$ That is, along the graph of $y(x)$ the function $F$ is a constant function, or the graph of $y(x)$ is contained on a level set of $F$. But the gradient of $F,$ $(F_x,F_y),$ is perpendicular to level sets. So, $(1,y'(x))$ is perpendicular to $(F_x,F_y),$ or equivalently, proportional to $(F_y,-F_x).$ From this relation, you can get the equality in the question.