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I'm working through multivariable functions and derivatives of multivariable functions. Since I am not very familiar yet with multivariable functions I wondered about the following:

In a function like $f(x,y)=x^2+y$, are x and y independent of each other and are we allowed to pick values for each deliberately?

Say for 1 for x and 99 for y?

As far as I have understood that topic it seems to me that indeed I can deliberately pick any value and will get an output in a third dimension. Rather than a curve I will receive a surface representing every possible combination of x and y. As long as the function is not limited like the equation for a circle, $x^2+y^2=r^2$. But I am not sure if I concluded this correctly.

I guess it's a pretty easy or maybe even silly question but I haven't fully figured that out yet.

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  • $\begingroup$ @PJTraill thank you for reviewing and correcting the spelling mistakes and improving used vocabulary, it helps me to improve :) $\endgroup$ – LurioTabasco Jan 4 '18 at 8:39
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This question is no "silly" at all! I would consider it rather "to-the-point" of multivariable calculus.

In general, as you mentioned, the set $$G(f)=\{(x,y,z)\in\mathbb{R}^3|z=f(x,y)\}$$ is indeed a subset of $\mathbb{R}^3$ and, in general, a surface. So, you can pick arbitrarily coordintates $x,y$ in $f$'s domain. Or, to take it to the multidimensional level, you can pick elements of $\mathbb{R}^2$ of the form $(x,y)$ that do belong in $f$'s domain.

Now, as for the circle, well, $x^2+y^2=r^2$ is not a function with free variables $x,y$ - you noticed they are not free. However, if you consider the function $g(x,y)=x^2+y^2$ this is a function of $x,y$ defined on $\mathbb{R}^2$.

Now, if the two coordinates have some kind of dependence from one-another, the thing gets more compicated. Imagine the situation where $$y=3x+2$$ In that case, it is clear that: $$f(x,y)=f(x,3x+2)=h(x)$$ for some other, one-variable, function $h$. In our case, if $f(x)=x^2+y$, then $h(x)=x^2+3x+2$. You can imagine this dependence between $y,x$ as a choice of a "path" on the surface that $f$ defines.

Lastly, in many cases, I see many-dimensional situations - since, well, for $n\geq4$, $\mathbb{R}^n$ is a little "scary" place - as a degree of freedom. With two free variables I have the ability to move over a surface, whereas with one free variable, I have the ability to move over a curve etc.

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In general $$f(x,y): D \subseteq \mathbb{R^2}\to\mathbb{R}$$is defined in a subset $D$ of $\mathbb{R^2}$ that is the real plane.

In your case $$f(x,y)=x^2+y$$

is defined for each value of (x,y) thus $D\equiv \mathbb{R^2}$.

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  • $\begingroup$ +1. Moreover, equations such as $x^2+y^2=r^2$ are not functions; they have to be true. As a result, one cannot pick arbitrary values for $x$ and $y$ since $x^2+y^2$ might not be equal to $r^2$. $\endgroup$ – wjm Jan 3 '18 at 10:19
  • $\begingroup$ Thank you @raptor I have changed that in the question $\endgroup$ – LurioTabasco Jan 3 '18 at 10:33

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