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Having trouble understanding Domains and Multi Variable Functions

  1. Sep 28, 2011 #1
    Hey guys, I'm doing some multivariable calculus atm, and I need some help with the Domains of some multivariable functions...
    1) f(x,y) = 3x^2 + 2y
    The problem I'm having here is I basically forget the definition of domain... would it be for all x and y even though there are two whole quadrants this function doesn't hit? (looking at the x-y plane here..)

    2) f x(x,y) = 6x
    I'm looking at the partial derivative in regards to x here. Would I just assume this is a single variable function and only write the Domain in terms of x? If not, what would I write for the Domain?

    3)f y(x,y) = 2
    This is the partial derivative in terms of y, I know on the x-y plane it would be defined for all x... is it the same here?
     
  2. jcsd
  3. Sep 28, 2011 #2

    dynamicsolo

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    The domain of a function is the set of all values for which the function is defined. Are there any real values of x and y for which you cannot compute 3x2 + 2y ?

    It is not a single variable function, or it wouldn't have been written as f(x,y) . If there were a term for f(x,y) which only depended on y alone, would it appear in fx ?


    Yes, same issue, different variable...

    ("Multivariate calculus ATM"? I have to evaluate a triple integral to get money out of the machine now!?)
     
  4. Sep 28, 2011 #3
    Hahaha atm/at the moment, same thing! (god wouldn't that be terrifying!!)

    Ok! I think I understand. I'm just trying to see where you CAN plug in numbers for...
    For 1) you can plug in any x,y and get an answer
    For 2) any x, y works, even though there's no y, it just means it works for any y
    For 3) any x, y works because it's just constant, like in single variable if we have y=2 it works for all x
    So for each function the Domain would be any x, y.

    Now if we were looking at the range for 3), would it be z is only defined at 2?
     
  5. Sep 28, 2011 #4

    dynamicsolo

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    What fx = 6x implies is that after integrating with respect to x , we would have f( x, y ) = 3x2 + g(y) ; instead of an "arbitrary constant" as in single-variable integration, we get an "arbitrary function" which has no dependence on the variable of integration.

    So as far as the domain goes, f( x , y ) would be defined for all real x , provided that y is in the domain of g(y) . In other words, there could still be whole lines or bands which all not in the domain of f( x, y ) because g(y) is undefined for those values of y (regardless of whether terms involving x are defined).

    By the same reasoning, since fy = 2 , integration in y tells us that
    f( x , y ) = 2y + g(x) . We can always calculate 2y , but if g(x) = 1/x , for instance, then the domain of f( x , y ) would not include the line x = 0 .
     
  6. Sep 28, 2011 #5
    Ok! I get it now, thanks a lot for all the help! I'm starting to dread integration of multivariable functions now though :P
     
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