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 2 years ago
I have a question about existence and uniqueness. It says that if f(x,y)=dy/dx and partialf/partialy are both continuous on the xy plane, then there is one and ony one solution etc... My question is, how can you tell if the function f and the partial derivative are continuous on the xy plane if they aren't even functions in the xy plane. They are functions in 3 space. You cannot even graph dy/dx= f(x,y). For instance, if dy/dx== 2y/x, that is not in the xy plane. That is a function of 2 variables in 3 space. I am confused?
 2 years ago
I have a question about existence and uniqueness. It says that if f(x,y)=dy/dx and partialf/partialy are both continuous on the xy plane, then there is one and ony one solution etc... My question is, how can you tell if the function f and the partial derivative are continuous on the xy plane if they aren't even functions in the xy plane. They are functions in 3 space. You cannot even graph dy/dx= f(x,y). For instance, if dy/dx== 2y/x, that is not in the xy plane. That is a function of 2 variables in 3 space. I am confused?

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hellow
 2 years ago
Best ResponseYou've already chosen the best response.0This statement of the theorem might be helpful to you: www.math.uiuc.edu/~tyson/existence.pdf. (For f and its partials you are considering a selection of inputs which you are interested in.) The function f which you gave is not continuous at (0, y) because it is not defined there. Since f is a rational function, it is continuous on its domain, so, for x and y any reals except x=0. For a function of 2 variables to be continuous at a point, you must have: f(a,b) exists, and \[\lim_{(x,y) \rightarrow (a,b)} f(x,y)\] exists, and \[\lim_{(x,y) \rightarrow (a,b)} f(x,y) = f(a,b)\] (This must hold as the (x,y) approaches (a,b) along any direction or path.)
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