Multivariate functions: Basic concepts of multivariate functions
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 variabele \(v\).
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 variabeles 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 function 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 \(y=\frac{11x+3z}{-6x+5z} \):
\[ \begin{array}{rcl}
(-6x+5z)\cdot y &=& 11x+3z\\ &&\phantom{xyz}
\color{blue}{\text{denominator cleared}}\\
-6x\cdot y+5y\cdot z-3z&=&11x\\ &&\phantom{xyz}
\color{blue}{\text{terms with } z \text{ to the left}}\\
5y\cdot z-3z&=&11x+6x\cdot y\\ &&\phantom{xyz}
\color{blue}{\text{terms without } z \text{ to the right}}\\
(5y-3)\cdot z&=&11x+6x\cdot y\\ &&\phantom{xyz}
\color{blue}{\text{terms with } z \text{ collected}}\\
z&=&\frac{11x+6x\cdot y}{5y-3}\\ &&\phantom{xyz}
\color{blue}{\text{divided by the coefficient of } z}\\
\end {array}\] We bring #x# outside the brackets and find the following function rule of \(z\) as an answer: \[z={{\left(6\cdot y+11\right)\cdot x}\over{5\cdot y-3}}\tiny.\]
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