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Quick partial derivatives question

I've got a quick question: for a function f(x,y)f(x,y) does dydx=fyfx\frac{dy}{dx} = \frac{f_y}{f_x}?
Original post by Monaa123
I've got a quick question: for a function f(x,y)f(x,y) does dydx=fyfx\frac{dy}{dx} = \frac{f_y}{f_x}?


I have very little experience with partial derivatives but I swear I recall dydx=−fxfy\displaystyle \frac{dy}{dx}=-\frac{f_x}{f_y} in this context. I could be wrong though.
Reply 2
Avatar for Monaa123
Monaa123
OP
4
Original post by alex2100x
I have very little experience with partial derivatives but I swear I recall dydx=−fxfy\displaystyle \frac{dy}{dx}=-\frac{f_x}{f_y} in this context. I could be wrong though.


It might be? I know that the contours at f = constant have a slope dydx=−fxfy\displaystyle \frac{dy}{dx}=-\frac{f_x}{f_y} I don't really understand the relationship between contours and derivatives though.
Reply 3
Avatar for Phichi
Phichi
11
Let z=f(x,y)=0z = f(x,y) = 0

δz=∂z∂xδx+∂z∂yδy \delta z = \dfrac{\partial z}{\partial x} \delta x + \dfrac{\partial z}{\partial y} \delta y

Divide through by δx\delta x and let δx→0 \delta x \rightarrow 0

dzdx=∂z∂x+∂z∂ydydx \dfrac{dz}{dx} = \dfrac{\partial z}{\partial x} + \dfrac{\partial z}{\partial y} \dfrac{dy}{dx}

Our initial statement z=f(x,y)=0z = f(x,y) = 0 means that dzdx=0 \dfrac{dz}{dx} = 0

Thus:

0=∂z∂x+∂z∂ydydx 0 = \dfrac{\partial z}{\partial x} + \dfrac{\partial z}{\partial y} \dfrac{dy}{dx}

dydx=−fxfy \dfrac{dy}{dx} = -\dfrac{f_x}{f_y}
Reply 4
Avatar for Monaa123
Monaa123
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4
Original post by Phichi
Let z=f(x,y)=0z = f(x,y) = 0

δz=∂z∂xδx+∂z∂yδy \delta z = \dfrac{\partial z}{\partial x} \delta x + \dfrac{\partial z}{\partial y} \delta y

Divide through by δx\delta x and let δx→0 \delta x \rightarrow 0

dzdx=∂z∂x+∂z∂ydydx \dfrac{dz}{dx} = \dfrac{\partial z}{\partial x} + \dfrac{\partial z}{\partial y} \dfrac{dy}{dx}

Our initial statement z=f(x,y)=0z = f(x,y) = 0 means that dzdx=0 \dfrac{dz}{dx} = 0

Thus:

0=∂z∂x+∂z∂ydydx 0 = \dfrac{\partial z}{\partial x} + \dfrac{\partial z}{\partial y} \dfrac{dy}{dx}

dydx=−fxfy \dfrac{dy}{dx} = -\dfrac{f_x}{f_y}



Ohh, that makes sense. Thank you!

So the gradient of the plane is the same as the gradient of the contour? But then what is the gradient vector ▽f\bigtriangledown f and why is that orthogonal to the contours?
Reply 5
Avatar for Monaa123
Monaa123
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4
So I was going over it. And the line perpendicular to the contour has the slope −1fx/fy=fyfx \frac{-1}{f_x/f_y} = \frac{f_y}{f_x} so to be orthogonal it has to be parallel to ∇f=(fx,fy)\nabla f = (f_x,f_y) but I don't understand how it's parallel? It's probably something really obvious but I just can't see it :/

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