Functions of several variables: Basic concepts and visualization
Isolation of a variable
Rewriting implicit relationship between three or more variables in a form in which one of the variables, say \(v\), is present on its own on the left-hand side of an equation, i.e., creating an equation of the form \(v = \mathrm{formule\;without\;}v\), is called the isolaton of the variable \small.\) You get then a definition of a function of several variables The example below shows how it works.
In the relationship \(\displaystyle y=\frac{5x+3z}{-2x+5z}\) you immediately see that \(y\) is a function of \(x\) and \(z\).
But is \(z\) also a function of \(x\) and \(y\)? And, if so, what is the function definition?
In other words, can you express \(z\) in \(x\) and \(y\), in the form \(z=\mathrm{formule\;in\;} x\mathrm{\;and\;}y\).
You can also reach the solution by increments entering in the form of equations:
Then you see if you are still on track, but in the end you must get to the equation in the form \(z=\ldots\).
But is \(z\) also a function of \(x\) and \(y\)? And, if so, what is the function definition?
In other words, can you express \(z\) in \(x\) and \(y\), in the form \(z=\mathrm{formule\;in\;} x\mathrm{\;and\;}y\).
You can also reach the solution by increments entering in the form of equations:
Then you see if you are still on track, but in the end you must get to the equation in the form \(z=\ldots\).
You try to isolate the variable \(z\) te isoleren in the given relation \(\displaystyle y=\frac{5x+3z}{-2x+5z} \).
Multiply both sides of the equation with \(-2x+5z\) and simplify:
\[\begin{aligned}
(-2x+5z)y &= \frac{(5x+3z)(-2x+5z)}{(-2x+5z)}\\ \\
-2xy+5yz&=5x+3z
\end{aligned}\] Now you have obtained an equation without denominators and subsequently only have to manipulate polynomial equations. Move all terms with \(z\) to the left-hand side and move all terms without \(z\) to the right-hand side, and factorize: \[\begin{aligned}5yz-3z &= 2xy+5x\\ \\
z(5y-3) &=x(2y+5)
\end{aligned}\] Division of the last equation by \(5y-3\) leads to the requested form: \[z=\frac{x(2y+5)}{5y-3}\] This can be read as definiton of a function \(z\) of \(x\) and \(y.\)
Multiply both sides of the equation with \(-2x+5z\) and simplify:
\[\begin{aligned}
(-2x+5z)y &= \frac{(5x+3z)(-2x+5z)}{(-2x+5z)}\\ \\
-2xy+5yz&=5x+3z
\end{aligned}\] Now you have obtained an equation without denominators and subsequently only have to manipulate polynomial equations. Move all terms with \(z\) to the left-hand side and move all terms without \(z\) to the right-hand side, and factorize: \[\begin{aligned}5yz-3z &= 2xy+5x\\ \\
z(5y-3) &=x(2y+5)
\end{aligned}\] Division of the last equation by \(5y-3\) leads to the requested form: \[z=\frac{x(2y+5)}{5y-3}\] This can be read as definiton of a function \(z\) of \(x\) and \(y.\)
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