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Partial derivative

  1. May 14, 2005 #1
    if you are given f(x,y)=x^2+y^2 and y=cos(t) x=sin(t), then when you differentiate f with respect to t, you use the partial derivatives of f with respect to x and y in the process. 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?

    When we differentiate f do we 'forget' that x and y are functions of t, and treat them as independent?
     
  2. jcsd
  3. May 14, 2005 #2
    [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.
     
    Last edited: May 14, 2005
  4. May 14, 2005 #3
    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.

    Exactly.
     
  5. May 14, 2005 #4
    ok, thanks. is the partial derivative of f with respect to x just 2x then?
    i dont understand the x'(t) part that you included?
     
  6. May 14, 2005 #5
    Its an application of the chain rule, which come to think of it doesnt belong there unless your finding the partial wrt to t.
     
  7. May 14, 2005 #6
    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)
     
    Last edited: May 14, 2005
  8. May 14, 2005 #7
    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)
     
  9. May 14, 2005 #8
    So what is your conclusion klein? What is

    [tex] \frac{\delta f}{\delta x} \ and \ \frac{\delta f}{\delta t} [/tex] ?
     
  10. May 14, 2005 #9

    dextercioby

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

    Daniel.
     
  11. May 14, 2005 #10

    HallsofIvy

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    This makes no sense at all: x(t)2+ y(t)2 is a function of t, not x and y, and so cannot be equal to f(x,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 "fx(x,y).

    What IS true is that if f(x,y)= x2+ y2, then fx(x,y)= 2x and fy(x,y)= 2y.
    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).
     
  12. May 15, 2005 #11
    what about if you had f(x,y) = x^2+y^2 again, but with y a function of x. The for the partial derivative of f with respect to x, you keep y fixed (?) and let x vary, even though y is a function of x (?). Thanks.
     
  13. May 15, 2005 #12
    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.
     
  14. May 15, 2005 #13
    What whozum did is the following :

    [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.....
     
  15. May 15, 2005 #14

    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.
     
    Last edited: May 15, 2005
  16. May 15, 2005 #15
    Thanks for doubting me ;) Its justified though.

    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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