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When we differentiate f do we 'forget' that x and y are functions of t, and treat them as independent?

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When we differentiate f do we 'forget' that x and y are functions of t, and treat them as independent?

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[tex] f(x,y) = x(t)^2 + y(t)^2 [/tex]

[tex] f(x,y)_x = 2x(t)x'(t) [/tex]

[tex] f(x,y)_y = 2y(t)y'(t) [/tex]

You keep one constant becuse you want to find the rate of change of f(x,y) as x changes, not y. Same thing for finding y.

[tex] f(x,y)_x = 2x(t)x'(t) [/tex]

[tex] f(x,y)_y = 2y(t)y'(t) [/tex]

You keep one constant becuse you want to find the rate of change of f(x,y) as x changes, not y. Same thing for finding y.

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The independant variables for F, are x and y. t has no concern here. Why would one not be able to be fixed while the others are moving, this is the exact same thing you do when you do partial derivatives anyway.When i was taught partial derivatives, i was told that we "keep all but one of the independent variables fixed.....". Now in this case, when differentiating f with respect to y, say, i dont see how this works. For, x (=cos(t) ) cannot be fixed while y (=sin(t) ) varies, can it?

Exactly.When we differentiate f do we 'forget' that x and y are functions of t, and treat them as independent?

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i dont understand the x'(t) part that you included?

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- #6

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I don't know this well, but if you write f(x,y)=x^2+y^2...then [tex] \partial_xf(x,y)=2x[/tex]....?? It's a weird question :

[tex]\frac{d}{dt}(\frac{\partial f}{\partial x}(x,y))=\frac{d2x(t)}{dt}=2x'(t)[/tex]

where as normally, f(x(t),y(t))=g(t) is a function of t only, hence [tex]\partial_xg(t)[/tex]=0....is like if Schwarz's thm were not valid here....

In fact I would say : it depends on at which time you apply the transformation x->x(t) (hence x as variable, or x as function of a variable...)

Because you could see x and y as function : x(t), y(t), and [tex] f(x,y)=x(t)^2+y(t)^2[/tex] as a functional of x and y....

Then you could apply Gateaux derivatives in the "direction" of the function n...(n:t->n(t))...with the usual definition :

[tex] D_{x,n(t)}F(x,y)=\lim_{h->0}\frac{F(x+hn,y)-F(x,y)}{h}=\lim_{h->0}\frac{x(t)^2+2hn(t)x(t)+y(t)^2-x(t)^2-y(t)^2}{h} [/tex]

[tex]=2n(t)x(t) [/tex]

so that we recover whozum result by functionally deriving along n(t)=x'(t)

[tex]\frac{d}{dt}(\frac{\partial f}{\partial x}(x,y))=\frac{d2x(t)}{dt}=2x'(t)[/tex]

where as normally, f(x(t),y(t))=g(t) is a function of t only, hence [tex]\partial_xg(t)[/tex]=0....is like if Schwarz's thm were not valid here....

In fact I would say : it depends on at which time you apply the transformation x->x(t) (hence x as variable, or x as function of a variable...)

Because you could see x and y as function : x(t), y(t), and [tex] f(x,y)=x(t)^2+y(t)^2[/tex] as a functional of x and y....

Then you could apply Gateaux derivatives in the "direction" of the function n...(n:t->n(t))...with the usual definition :

[tex] D_{x,n(t)}F(x,y)=\lim_{h->0}\frac{F(x+hn,y)-F(x,y)}{h}=\lim_{h->0}\frac{x(t)^2+2hn(t)x(t)+y(t)^2-x(t)^2-y(t)^2}{h} [/tex]

[tex]=2n(t)x(t) [/tex]

so that we recover whozum result by functionally deriving along n(t)=x'(t)

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thats what i was confused about. we treat the x and y as independent variables, even though they wont really vary independently (since both depend on t)whozum said:The independant variables for F, are x and y. t has no concern here. Why would one not be able to be fixed while the others are moving, this is the exact same thing you do when you do partial derivatives anyway.

.

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[tex] \frac{\delta f}{\delta x} \ and \ \frac{\delta f}{\delta t} [/tex] ?

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That's the notation for the functional derivative of "f" wrt "x" or "t".

Daniel.

Daniel.

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HallsofIvy

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This makes no sense at all: x(t)whozum said:[tex] f(x,y) = x(t)^2 + y(t)^2 [/tex]

[tex] f(x,y)_x = 2x(t)x'(t) [/tex]

[tex] f(x,y)_y = 2y(t)y'(t) [/tex]

You keep one constant becuse you want to find the rate of change of f(x,y) as x changes, not y. Same thing for finding y.

Even worse is [tex] f(x,y)_x = 2x(t)x'(t) [/tex] and [tex] f(x,y)_y = 2y(t)y'(t) [/tex]. If f is a function of t, then it makes no sense to write "f

What IS true is that if f(x,y)= x

IF, further, x and y are themselves functions of t, then f is really a function of t, f(t), and f '(t)= (2x)x'(t)+ (2y)y'(t).

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- #12

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So [itex] f_x(x,y) [/itex] doesnt really exist? What you described is what I did but just assuming that you could differentiate withrespect to each variable.IF, further, x and y are themselves functions of t, then f is really a function of t, f(t), and f '(t)= (2x)x'(t)+ (2y)y'(t).

- #13

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[tex]f(x,y)=x(t)^2+y(t)^2[/tex]....(which by the way is a functional but not a function)

Then you thought, ok I want to differentiate with respect to x...hence I should move x a bit..the problem is that x is a function, hence I don't vary with another function, but the parameter of this function : t

so you did :

[tex] \lim_{h->0}\frac{f(x(t+h),y(t))-f(x(t),y(t))}{h}=\lim_{h->0}\frac{x(t+h)^2-x(t)^2}{h}\approx\lim_{h->0}\frac{(x(t)+hx'(t)+...)^2-x(t)^2}{h}=2x(t)x'(t)[/tex]

But you could eventually vary x with a function n, instead of varying the variable of x.....

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arildno

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Somehow, I doubt whozum was thinking along the lines of functionals here.

whozum:

Let x,y be given as functions x=X(t), y=Y(t)

What is now correct is that we may define a function F(t)=f(X(t),Y(t))

Then, we have by the chain rule:

[tex]\frac{dF}{dt}=\frac{\partial{f}}{\partial{x}}\mid_{(x,y)=(X(t),Y(t))}}\frac{dX}{dt}+\frac{\partial{f}}{\partial{y}}\mid_{(x,y)=(X(t),Y(t))}}\frac{dY}{dt}[/tex]

As HallsofIvy has already said.

whozum:

Let x,y be given as functions x=X(t), y=Y(t)

What is now correct is that we may define a function F(t)=f(X(t),Y(t))

Then, we have by the chain rule:

[tex]\frac{dF}{dt}=\frac{\partial{f}}{\partial{x}}\mid_{(x,y)=(X(t),Y(t))}}\frac{dX}{dt}+\frac{\partial{f}}{\partial{y}}\mid_{(x,y)=(X(t),Y(t))}}\frac{dY}{dt}[/tex]

As HallsofIvy has already said.

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Thanks for doubting me ;) Its justified though.arildno said:Somehow, I doubt whozum was thinking along the lines of functionals here.

whozum:

Let x,y be given as functions x=X(t), y=Y(t)

What is now correct is that we may define a function F(t)=f(X(t),Y(t))

Then, we have by the chain rule:

[tex]\frac{dF}{dt}=\frac{\partial{f}}{\partial{x}}\mid_{(x,y)=(X(t),Y(t))}}\frac{dX}{dt}+\frac{\partial{f}}{\partial{y}}\mid_{(x,y)=(X(t),Y(t))}}\frac{dY}{dt}[/tex]

As HallsofIvy has already said.

You are correct, and I understand what HallsofIvy said and what you are saying, but the question presnted by the OP was the last question I posted two/three posts ago, this is the only question I have left.

I'm also trying to think of a similiar instance where the indep. variable is a function of another variable, I will let you know.

Thanks.

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