Finding d^2y/dx^2 in terms of F(x,y)

[diff]

Homework Statement

If F(x,y)=0 find d^2y/dx^2 in terms of the partial derivatives of F(x,y).

Homework Equations

dy/dx= -Fx/Fy

The Attempt at a Solution

I tried to differentiate dy/dx with respect to x again, I thought I can get d^2y/dx^2 . But I couldn't find it. Actually I don't know how to deal with Fy or Fx. When I differentiate Fx with respect to x, Fx becomes Fxx? Or when I differentiate Fy with respect to x, it equals Fyx or not?

 

[Amok]

What is Fx or Fy? Use the total derivative:

\frac{dF}{dx} = \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} \frac{dy}{dx}

 

[diff]

What is Fx or Fy? Use the total derivative:

\frac{dF}{dx} = \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} \frac{dy}{dx}

But when I do that, I get dy/dx. But I have to find d^2y/dx^2.

 

[lanedance]

if you mean F_{xy} = \frac{\partial^2 F }{\partial y \partial x}, then yes, so i would attempt to differntiate again w.r.t. x

 

[Amok]

But when I do that, I get dy/dx. But I have to find d^2y/dx^2.

You'll have to do it twice :p

What do you mean by Fx and Fy?

 

[diff]

You'll have to do it twice :p

What do you mean by Fx and Fy?

The problem is that I couldn't do it :P

Fx means partial derivative of F w.r.t x

 

[diff]

if you mean Fxy = \frac{\partial^2 F }{\partial y \partial x} then yes, so i would attejmpt to differntiate again w.r.t. x

Sorry, but I don't understand.

 

[lanedance]

differntiate this expression again w.r.t. x using the chain rule & partial differntiation
<br /> \frac{dF}{dx} = \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} \frac{dy}{dx} <br />

 

[HallsofIvy]

Since, as Amok and lanedance have said,
\frac{dF}{dx}= \frac{\partial F}{\partial x}+ \frac{\partial F}{\partial y}\frac{dy}{dx}

\frac{d^2F}{dx^2}= \frac{d}{dx}\left(\frac{\partial F}{\partial x}+ \frac{\partial F}{\partial y}\frac{dy}{dx}\right)
= \frac{\partial}{\partial x}\left(\frac{\partial F}{\partial x}+ \frac{\partial F}{\partial y}\frac{dy}{dx}\right)+ \frac{\partial}{\partial y}\left(\frac{\partial F}{\partial x}+ \frac{\partial F}{\partial y}\frac{dy}{dx}\right)\frac{dy}{dx}

 

[mege]

A chain rule problem perhaps?

<br /> \frac{\partial}{\partial x} (-\frac{Fx}{Fy}) = -\frac{Fxx}{Fy} + Fyx\frac{Fx}{(Fy)^2}<br />

If you treat Fx and Fy as just 'ordinary' functions you can manipulate them by taking another partial differential and still be independent of the total function (since the outcome is all wrt x anyhow).

 

[diff]

differntiate this expression again w.r.t. x using the chain rule & partial differntiation
<br /> \frac{dF}{dx} = \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} \frac{dy}{dx} <br />

I did it in the first step, thanks anyway.

 

[diff]

Since, as Amok and lanedance have said,
\frac{dF}{dx}= \frac{\partial F}{\partial x}+ \frac{\partial F}{\partial y}\frac{dy}{dx}

\frac{d^2F}{dx^2}= \frac{d}{dx}\left(\frac{\partial F}{\partial x}+ \frac{\partial F}{\partial y}\frac{dy}{dx}\right)
= \frac{\partial}{\partial x}\left(\frac{\partial F}{\partial x}+ \frac{\partial F}{\partial y}\frac{dy}{dx}\right)+ \frac{\partial}{\partial y}\left(\frac{\partial F}{\partial x}+ \frac{\partial F}{\partial y}\frac{dy}{dx}\right)\frac{dy}{dx}

Thanks for your help, I need to find d^2y/dx^2 not d^2F/dx^2. I think I can find it now.

 

[diff]

A chain rule problem perhaps?

<br /> \frac{\partial}{\partial x} (-\frac{Fx}{Fy}) = -\frac{Fxx}{Fy} + Fyx\frac{Fx}{(Fy)^2}<br />

If you treat Fx and Fy as just 'ordinary' functions you can manipulate them by taking another partial differential and still be independent of the total function (since the outcome is all wrt x anyhow).

Actually when you take the partial derivative of F w.r.t x , it is quite different from Fxx. But, thanks anyway.

 

[Amok]

Thanks for your help, I need to find d^2y/dx^2 not d^2F/dx^2. I think I can find it now.

Do what HOI said till the end and you'll see that \frac{d^2 y}{dx^2} is bound to appear.

 

[diff]

Man, just do till the end and you'll see that \frac{d^2 y}{dx^2} is bound to appear.

Yeah, I know it, thanks, finally I found it :D