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A independent variable is "the input" and the dependent variable is the "output", atleast thats how it was explained to us.

But if you have some random function can't both variables be seen as "affecting" the other variable?

For example, in $ y = 1/x$, "x" could be seen as an input and "y" the output, but isn't the opposite also true. Why can't "y" be the input and "x" be the output?

You could say that the dependent variable is the variable that has a unique independent variable associated with it, but thats not necessary true with the independent variable. (Think $y = x^2$. $x$ is the independent variable because for every $x$ value there is only one $y$ value, but there isn't a a unique $x$ value for every $y$ value)

But how can this definition be extended to functions that has unique domain values for all range values and unique range values for all domain values? Think $y= 1/x$.

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  • $\begingroup$ Comment for clarity: I believe there is another way "independent variable" is used in fields like statistics. There it is used to mean "one dimension of a variable does not affect any other dimension of the variable" rather than the input/output conceptualization you are using. $\endgroup$
    – DanielV
    Commented Aug 27, 2014 at 5:50

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It's not generally written in stone. If we face a problem like $x^2+y^2=1$ then both $x$ and $y$ appear symmetrically so it's largely a matter of custom which is viewed as dependent and which is viewed as independent. In short, yes, it is a choice.

On the other hand, if velocity $v$ is a function of time $t$ then it is clear enough that $v$ is the dependent variable and $t$ is the independent variable. Unless, you think time can be a function of velocity. Well... Digression into special relativity aside. Sometimes a model comes with some idea of cause and effect which is not reversible.

Equations, some functions, it is the case that several variables can be used as the independent variable. If you're serious about the question, eventually you'll want to learn the implicit function theorem which sorts out this question in considerable generality. But, I'll leave that for you to discover.

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  • $\begingroup$ Hi James, I know this is an old post so I hope you don't mind me commenting as I would love your feedback. When mathematicians say things like, for example, "let the variable $t$ be time" when lecturing do they mean that we're assuming the variable $t$ represents time? Just making sure my usage of the word "represents" is correct here, thank you in advance. $\endgroup$
    – Taylor Rendon
    Commented Aug 11, 2022 at 17:25
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    $\begingroup$ @TaylorRendon yes, I think that is fair, when someone says "let the variable $t$ be time" they're probably thinking in Newtonian terms where we can use a universal background clock. In that sense, we can use $t$ for more than one path. In contrast, outside the Newtonian context, it is not natural to suppose more than one curve can be parametrized by the same parameter. I run up against these ideas when discussing colliding verses intersecting paths. $\endgroup$
    – James S. Cook
    Commented Aug 12, 2022 at 4:23
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Note that $y=x^2$ is an equation, to be more precisely a statement form. It states: “The second coordinate is the square of the first one.” For the coordinates of a point that statement may be true -- for $(-2,4)$, e.g.-- or false for, say $(7,50)$. The set of points for which the statement form yields a true statement is called the solution set of the statement form. The statement form $y-x^2=0$ states: “The difference of the first coordinate and the square of the second equals zero.” is a different statement form, but has the same solution set as the first one.

The second one is given in a so called implicit form, wheras the first is called an explicit form. Why bother about the to forms? Well, if you want to determine the second coordinate of a point in the solution set, given the first coordinate is $7$, let's do it using the imlicit form. We have $$y-7^2=0\iff y=49,$$ so you have to solve an equation. (That's the reason for why we call it implicit.). Using the explicit form we get immediately $y=7^2=49$ without solving any equation.

Your first example $y=1/x$: “The second coordinate is the reciprocal of the first one.” reads in implicit version $xy=1$: “The product of both coordinates equals one.” Another explicit form is $x=1/y$. So if you are able to isolate one coordinate in an implicit form, this coordinate is called dependent, the other independent. In $x=1/y$ the coordinate $x$ is dependent and $y$ independent whereas in $y=1/y$ it's vice versa.

Now in $y-x^2=0$ you can't isolate $x$ in a closed form, here we have $\sqrt{y}=|x|$. So if for example $y=25$ we have $\sqrt{25}=|x|\iff5=|x|\iff x=5\lor x=-5$, so we have to solve an equation anyway. You may write $x=\pm\sqrt{y}$ and see that for all positive $y$ you'll get two values of $x$, so that explicit form is not a function, which mathematicians prefer.

In the example $x^2+y^2=1$ given by James there doesn't exist any explicit form which is a function, for an obvious reason: the solution set is the unit circle, centered at the origin.

Facit: if a statement form allows an explicit form of one of the coordinates and that explicit form is a function, you are free to call the isolated coordinate dependent and the other independent. I never use those attributes, because they are rather useless, won't gain any knowledge and are causing a lot of trouble as we see here.

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