### Multivariate functions: Basic notions

### Functions and relations

In the natural sciences mathematical models are often used to describe relations between quantities so as to understand processes or to make predictions.

Mathematically a *relation* between three variables is a subset of #{\mathbb R}^3#, but in practice it is often given by just writing down an equation.

If #f# is a function of two variables, then the *graph* of #f# is a relation between the three variables; it is the subset of #{\mathbb R}^3# consisting of all points #\rv{x,y,z}# satisfying the equation #z=f(x,y)#.

There is no limit to the number of equations that can be used to describe a relation. For instance,\[\eqs{x&=&\e^{t}\cr y&=&\e^{2t}\cr}\] is a pair of equations defining a relation between the three variables #x#, #y#, and #t#.

Relations that are graphs of functions

A relation between three variables \(x\) , \(y\), and \(z\) is the graph of a function \(z=z(x,y)\) if, for any pair of admissible values for \(x\) and \(y\), there is exactly one value of \(z\) such that #\rv{x,y,z}# belongs to the relation.

When we are given an equation defining a relation between three variables, we can try to rewrite it in a form where one of the variables, say #v#, appears on the left hand side of the equation and does not appear on the right hand side. The act of creating such an equation of this form, that is, \[v = \text{a formula without }v\tiny,\]is called **isolating the variable** \(v\).

If the relation defines a function, we call that function **implicitly defined** by the relation.

Two simple relations

- Many relations are not as explicit as a function. For example, the formula for a lens with a focal length \(f\) is \[\frac{1}{b}+\frac{1}{v}=\frac{1}{f}\tiny,\] where \(v\) is the object distance and \(b\) the image distance.
- Another example of a relation between three variables is the sphere with center \(\rv{0,0,0}\) and radius \(1\). It is determined by the equation \(x^2+y^2+z^2=1\).

Sometimes, such a relation is the graph of a function, but not always. In the first example, #f# is a function of #b# and #v#. By isolating the variable #f#, we find the functional rule \[f(b,v)=\frac{1}{\dfrac{1}{b}+\dfrac{1}{v}}\tiny,\]which can, of course, be simplified to #\frac{b\cdot v}{b+v}#.

In the example of the sphere, \(z\) cannot be written as a function of \(x\) and \(y\). The reason is that for many points #\rv{x,y}#, there are two values of #z# such that #\rv{x,y,z}# satisfies the equation. We can describe the relation as the union of the graphs of the two functions #z_1# and #z_2#, where\[z_1(x,y)= \sqrt{1-x^2-y^2}\quad\text{and}\quad z_2(x,y)= -\sqrt{1-x^2-y^2}\tiny.\]

But \(z\) is also a function of \(x\) and \(y\). What is the corresponding function rule? In other words, express \(z\) in terms of \(x\) and \(y\) and enter your answer in the form \[z=\text{ an expression in } x\text{ and }y\tiny.\]

This can be found by isolating the variable \(z\) in the given equation \(y=\frac{13x+4z}{-6x+5z} \):

\[ \begin{array}{rcl}

(-6x+5z)\cdot y &=& 13x+4z\\ &&\phantom{xyz}

\color{blue}{\text{denominator cleared}}\\

-6x\cdot y+5y\cdot z-4z&=&13x\\ &&\phantom{xyz}

\color{blue}{\text{terms with } z \text{ to the left}}\\

5y\cdot z-4z&=&13x+6x\cdot y\\ &&\phantom{xyz}

\color{blue}{\text{terms without } z \text{ to the right}}\\

(5y-4)\cdot z&=&13x+6x\cdot y\\ &&\phantom{xyz}

\color{blue}{\text{terms with } z \text{ collected}}\\

z&=&\frac{13x+6x\cdot y}{5y-4}\\ &&\phantom{xyz}

\color{blue}{\text{divided by the coefficient of } z}\\

\end {array}\] We bring #x# outside the brackets and find the following function rule of \(z\) as an answer: \[z={{\left(6\cdot y+13\right)\cdot x}\over{5\cdot y-4}}\tiny.\]

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