Having trouble understanding Domains and Multi Variable Functions

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The discussion focuses on understanding the domains of multivariable functions in calculus. For the function f(x,y) = 3x^2 + 2y, the domain is all real numbers for x and y, despite the function not covering certain quadrants in the x-y plane. The partial derivatives, fx(x,y) = 6x and fy(x,y) = 2, also have domains that include all real numbers for their respective variables, as they are defined for any x and y. It is emphasized that while the functions can be evaluated for any values, the overall domain may still be affected by other components of the function, such as g(y) or g(x) in integration. The conversation concludes with a clearer understanding of how to approach domains and the potential complexities in multivariable calculus.
_Steve_
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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?
 
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_Steve_ said:
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..)

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 ?

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?

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 ?
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?

Yes, same issue, different variable...

("Multivariate calculus ATM"? I have to evaluate a triple integral to get money out of the machine now!?)
 
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?
 
_Steve_ said:
For 2) any x, y works, even though there's no y, it just means it works for any y

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).

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?

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 .
 
Ok! I get it now, thanks a lot for all the help! I'm starting to dread integration of multivariable functions now though :P
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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