I am studying partial differential equations, in which the class notes includes the following statement:
$$x^2 + y^2 + (z-c)^2 = a^2$$
Here z is the dependent variable; x,y are independent variables; and a,c are arbitary constants.
Mathematically, all the variables depend on every other variable, so there shouldn't be any difference between x,y or z. Is this a subjective distinction? Aren't variables objectively all dependent on each other?
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It is subjective in the sense that you have three choices. Note $$ 2xdx + 2ydy+2(z-c)dz = 0 $$ you can solve for $dx, dy$ or $dz$. The coefficient of those differentials tells you when it is possible to write that variable in terms of the remaining pair. For example, we can write $x = g(y,z)$ provided $x \neq 0$. Similarly, $y = h(x,z)$ provided $y \neq 0$ and $z = j(x,y)$ provided $z \neq c$. This is from the implicit function theorem, it supports the calculational wisdom I share with you here.
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It's an unfortunately common abuse of notation that confuses the notion of "dependent" with one of the approaches for dealing with dependence.
The dependent triple $\{x,y,z\}$ has two degrees of freedom meaning that, for the most part, you can pick two variables $s$ and $t$ so that $\{s,t\}$ are an independent pair, and write all of $x,y,z$ as functions of $s$ and $t$.
(e.g. you could pick $s,t$ so that $s=x$ and $t=y$)
The abuse of notation, then, is to call $s,t$ the "independent variables" and $x,y,z$ the "dependent variables". (or, if you picked $s=x$ and $t=y$ to call $x,y$ the "independent variables" and $z$ the "dependent variable")
The intention is to then not work with the variables $x,y,z$, but instead the functions that express them in terms of $s,t$.
There is another unfortunate abuse of notation that further confuses the issue — the function expressing, for example, $z$ as a function of $s$ and $t$ is often named $z$ as well.
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When god created the world ${\mathbb R}^3$ the three variables $x$, $y$, $z$ used to address the points $p\in{\mathbb R}^3$ were truly independent.
Now an equation $$x^2+y^2+(z-c)^2=a^2\tag{1}$$ is given, whereby for simplicity we assume $a>0$. Such an equation does not define any function per se, but a solution set $$S:=\bigl\{(x,y,z)\,\bigm|\,x^2+y^2+(z-c)^2=a^2\bigr\}\ .$$ In the case at hand this set is a $2$-sphere of radius $a$ with center $(0,0,c)$.
Nevertheless, an equation like $(1)$ does implicitly define a suitably chosen of the three variables as functions of the other two, albeit only locally. The implicit function theorem guarantees this (under "technical assumptions") even if you are not able to solve $(1)$ explicitly for the chosen variable, say $z$. To be precise:
If $f\in C^1$, and $\nabla f(x_0,y_0,z_0)\ne(0,0,0)$ at some point $(x_0,y_0,z_0)\in S$, say $f_z(x_0,y_0,z_0)\ne0$, then there is a rectangular box (a "local window")$$W=[x_0-h,x_0+h]\times[y_0-h',y_0+h']\times[z_0-h'',z_0+h'']$$ with center $(x_0,y_0,z_0)$, and a $C^1$-function $$\phi:\>[x_0-h,x_0+h]\times[y_0-h',y_0+h']\to[z_0-h'',z_0+h'']$$ such that $$S\cap W=\bigl\{(x,y,z)\,\bigm|\,|x-x_0|\leq h,\ |y-y_0|\leq h', \ z=\phi(x,y)\bigr\}\ .$$ This means that within the box $W$ the given equation $f(x,y,z)=0$ defines $z$ "implicitly" as a function $z=\phi(x,y)$.
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You are correct, the mathematical expression $x^2+y^2+(z−c)^2-a^2$ with $x,y,z$ variables and $a,c$ constants has at least 3 characteristic parametric surfaces by defined by 1 dependent and 2 independent variables.
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When a variable is independent, it means that it could take any value, whereas the dependent value is, well, dependent on the independent value. For example given
$y = f(x)$
$x$ is the input variable and could take any arbitrary value, whereas $y$ depends on $x$: You input an $x$ and you get a $y$ out of it.
Of course you could reverse that relation and write $x$ as a function of $y$:
$x = g(y)$
The terms dependent and independent are just there to tell you which variable is the input and which is the output.
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Let's start with a simpler example $$x + y = 0$$ In the above equation, if we input the value of either variable, we'd get the value of the other, so you could say that we choose our input to be some variable and that determines the value of the other. We usually choose $x$ by convention. In your example, we have the freedom to choose any 2 input variables and the third would have its value, so 2 are independent and the third is dependent
On
Here's a different take on it. Saying $x$ and $y$ are independent variables means we're universally quantifying over them, while saying $z$ is a dependent variable means we're existentially quantifying over it. In other words, a more technical way to say the same thing is: "for all $x$ and $y$, there exists a $z$ such that $x^2 + y^2 + (z-c)^2 = a^2$". In symbols, $$\forall x, y.\exists z.x^2 + y^2 + (z-c)^2 = a^2$$
If we Skolemize or use a constructive interpretation of the above, this is saying we have a function (or really just a multivalued function, i.e. a left total relation) $f(x, y)$ such that $$\forall x, y.x^2 + y^2 + (f(x, y)-c)^2 = a^2$$ Even more symbolically, $$\exists f.\forall x, y.x^2 + y^2 + (f(x, y)-c)^2 = a^2$$
Of course, if you actually check this, it isn't clear that this is a true statement. It very much depends on the range of values we are quantifying over. For example, it is true if $x$, $y$, and $z$ range over complex numbers, but it is false if they range over arbitrary real numbers. Usually the exercise is one of finding suitable domains, and the problem varies quite a bit depending on which and how many variables we choose to be "independent" and which "dependent".
The "solution set" in Christian Blatter's answer is the set theoretic way of saying we have a relation $$S(x, y, z) \equiv [x^2 + y^2 + (z-c)^2 = a^2]$$ using which our original expression becomes $$\forall x, y.\exists z.S(x, y, z)$$ and the second $$\exists f.\forall x, y.S(x, y, f(x, y))$$
Logically, we're also universally quantifying over the constants, so at this level there is no difference between an "independent variable" and a "constant". It's just a matter of scoping; constants living in an outer scope. That said, we may only want to consider Skolem functions, like $f$ above, that are continuous or differentiable or smooth. Now, the distinction between an independent variable and a constant is that we require that we depend continuously/differentiably/smoothly on an independent variable but no such requirement is made for constants. From the logical perspective, the universal quantification of constants is happening meta-logically.
It means that $z$ is a function of $x$ and $y$, so the full writing would be
$$x^2+y^2+(z(x,y)-c)^2=a^2$$